<50v>

## Lemma.

\1/[Editorial Note 1] [1] If one body move from a to b in ye same time in wch another moves from a to c & a 3d body move from a wth motion compoundee|d| of those two it shall (completeing ye parallelogram abcd ) move to d in ye same time. For those motion would severally {carry} it ye one from a to c ye other from c to d &c

[2] \2/ In ye description of any Mechanicall line what ever, {illeg} there may bee found s|t|wo such motions wch compound or make up ye motion of ye point describeing it, whose ∼ motion being by them found \by ye Lemma/, its determinacon shall bee in a tangent to ye mechanicall line.

[3] Example ye 1st. If abe is an helix, {illeg} wch {illeg} described by ye point b move {sic}ing uniformely in ye line abc from ye point center a about wch ye line abc circulates uniformely. {illeg} |the line ab increasing uniformely whilest it also circulates uniformely about ye center a| Let ye radius of ye circle dmbd bee ab. & let {dm $\mathrm{b}$  {sic}} measure ye quantity of the giration of ab (viz ad touching ye helix at ye center) let bf be a tangent to ye circle dmbd . yn is ye motion of ye line point b towards c to its motion towards f , as ab , to dmb {illeg}. therefore make $bc=fg:bf=cg\colon\colon ab:dmb$. & (by ye Lemma) ye diagonall bg shall touch ye helix in b. Or make $bc=fg=ab$. & $bf=cg=dmb$. the diagonall bg shall touch ye helix. (ye length of bf may be thus found viz; $ae:bc=ad=ab\colon\colon dmbd:dmb=bf$.)

[4] Example ye 2d. If ye center $\left(a\right)$ of a globe \$\left(bae\right)$/ moves uniformely in a streight line parallel to eh, whilest ye Globe uniformely girates. Each point $\left(b\right)$ in ye Globe will describe a Trochoides: to wch a ye point b I thus draw a tangent. Draw ye radi {sic} ab & bc perpendicular to it yn is ye circular motion of {illeg}|th|e point b determined in ye line bc , & its progesive {sic} in bf . If therefore I make $bc=fg$ to $bf=cg$ as ye circular motion of ye point b to its progressive ye Diagonall (by ye Lemma) bg shall {illeg}g touch ye Trochoides in b . As if ye Globe roule upon ye pl{illeg}|ain|e eh , & I make $bc=fg=ab$. & $bf=cg=ae$. yn do{illeg}|th| ye Diagonall bg touch ye Trochoides. (Or be , {illeg} passing through ye point in wch ye globe & plaine touch, is a perpendicular to ye {Tro}.

[5] Example ye 3d. If ye line $bh\perp ak$ moves uniformely ye length of ah whilest { ab girate} uniformely from ak to am about ye center a , ye point of their intersection $\left(b\right)$ will descibe {sic} ye Quadratrix kbn . Draw $bc\perp abp\perp pq$. & $bf\perp am$ yn \$an:am=ak\colon\colon$/ $ka:kpm\colon\colon$ motion of $\left(b\right)$ to f ∶ motion of $\left(p\right)$ to q , {(sup)}. And motion of $\left(p\right)$ to q ∶ motion of $\left(b\right)$ to c ∷ \ $am=$/ $ap:ab\colon\colon kpm:sbt$ . Therefore motion of $\left(b\right)$ to f ∶ motion of $\left(b\right)$ to c $\colon\colon ka:sbt\colon\colon an:ab$. Therefore makeing $ak:sbt\colon\colon$ $an:ab\colon\colon bf=cg:bc=fg$ . (Or wch is makeing $bf=an=cg$ , & $bc=ba=fg$ ) ye Diagonall bg shall touch ye Quadratrix at b .

[6] [Scholium. The tangents of Geometricall \lines/ may be found by their descriptions {after} ye same manner. As the Ellipsis (whose foci are a & f ) being described by ye thred abf ye thred ab lengthens so much as ye thred bf shortens, or the point b moves equally from a & to f. Therefore I take $bc=bf$. $cg=fg$ \& $fg\perp bf$./ & ye diagonall bg will touch the Ellipsis in b ]. (This should follow ye 3d Example's substitute) See fol 57.

[7] Although ye nature of a Mechanicall line is not knowne from its description but from some other principle yet may a tangnt be drawne to it by ye same method.

As if be is an Hyperbola. tad its asymptote & $cf\parallel tad$ . & $gp:ph\colon\colon cadf:adeb$. {illeg} to draw a tangent to ye tangent {illeg} line gh {illeg} ghm , I consider yt, $df:de\colon\colon$ increasing of acdf ∶ increasing of abde ∷ increase of gp ∶ increase of ph ∷ motion of ye point h towards k ∶ motion of h towards r , if $hk\parallel gp$. Therefore I make {$rh=sk:rs=hk\colon\colon$} $\colon\colon de:df\colon\colon hp:pw$. & ye diagonall hs or wh shall touch ye line ghm . Or if $vg=ta=ab=xy$ . $gp=ad$. $ph=vxy\parallel phr$. \$vp\parallel yh$/. yn doth xhs touch ye line ghm at h .

Tangents to mechanichanicall lines may sometimes bee found by finding such a point wch is immoveable in respect of ye line described & also doth {{illeg}vary} in distance from ye describing point. Then in ye for ye {Sicunf{illeg}} through yt point. Thus in ye Trochoides when ye point $\left(e\right)$ toucheth ye plaine eh tis immoveable, & tis ever equidistant from ye describing point b ({illeg} both of ym fin{illeg} points in ye Globe). Therefore ye line {illeg} drawne {fro=} ye describeing point to ye touch point of ye Globe & plaine $\left(eh\right)$ is perpendicular to {ye} trochoides. But in ye spirall though ye point {illeg} is {illeg} from yt {illeg}.

[8] Instead {in ye} third example {illeg} {illeg} {illeg}. <51r> Therefore $af:ab\colon\colon ak:gbl\colon\colon$ {illeg}|abs|olute & whole motion of b towards c (or acf ) ∶ whole motion of b towards d (or ed ). Soe yt makeing $bc:bd\colon\colon af:ab$. {$ab\parallel ed\perp ed$} & $hb\parallel ce\perp cb$. The point b will be moved {illeg} to ye line {illeg} ce & ed in same times wch cannot bee unlesse it move to e (their common intersection). The point b therefore move in ye line be wch doth therefore touch ye Quadratrix at b : (The same is done by makeing $bd=$ $bl=bd\perp de\parallel ab$ . & drawing \ye tangent/ be through ye common intersection of ed & aem .)

## To resolve these and such like Problems these following propositions may bee very usefull.

May 14. 1666.

[9] Prop 1. If ye body a being in a circumference of ye circle \or sphære/ adce doth move towards its center b its acceleration \motion or velocity/ to|wards| each point \ d , c , e / of ye circ{illeg}umference is yn as ye cordes ad , ac , ae , drawn from yt body to those points are. This may be Demonstrated by Te|h|eorem R pag 57.

[10] Prop 2d. If \ ∠  {sic}$adc$ {illeg} is a parallelogram \ sim $△aec$ although they bee not in yesame plane &// three bodys move \uniformely from a /. ye first from a to b d ye 2d from a to e , ye 3d from a to c their motions being \each to other/ as ye {illeg} directing lines ad , ae , ac , are in ye same time, & adce is a parallelogram then is ye motion of ye third body compounded of ye other two. Demonstration. For \makeing $df\perp ac$f & $eb\perp${adba}/ |{illeg} makeing $df\parallel eb\perp ac$| ye motion of ye first body towards d is to its motion towards f as ad = {illeg} is to af (prop 1); \{illeg}/ & ye \{illeg}/ motion of ye second body towards be is to its motion towards db as ae $=dc$ is to b ba (prop 1). Therefore But $af+ab=ac$. Therefore &c

Prop 3d. If a moveing line keepe parallel to it selfe all its pts have equall motion.

Prop 4th. If a line move in plano, so yt all its points keepe equidistant from some common point center the motions of those points are as their distances from yt center.

Prop 5t. If ye motion of a line in plano bee mixed of parallell & circular motion, ye motion of all its points are compound (see prop 2) of that motion which they would have, had ye line onely its centrall parallel motion, & of yt wch they would have, had ye line onely its circular motion.

Schol: All motion in plano is reducible to one of these three cases, & in ye 3d case any point in yt plaine may bee taken for a center to ye circular motion.

[11] Prop 6t. If ye \streight/ line ea doth rest & da doth move: soe yt ye point a fixed in ye line da moveth towards b : Then from ye moveing line da drawing $de\parallel ab$ , & ye same way wch ye point a moveth; These motions, viz of ye \fixed/ point a towards b , of ye intersection point a in ye line ad towards d , & of ye intersection point a towar in ye line ae towards e , shall bee one to another, as their correspondent lines de , ad , & ae are.

[12] Prop 7th. If ye \streight/ lines adm , ane , doe move, soe yt ye point a fixed in ye line amd moveth towards b , & ye point a fixed in ye line ae moveth in ye towards c : Then from ye line each line to ye other draw two lines de , nm parallell to the mo the line amd , draw ∼ $de\parallel ab$ & ye same way: & from ye line ae draw $nm\parallel ac$, & ye contrary way, to make up ye Trapezium denm . And if any two of these foure lines de , {illeg} mn , md , ne , bee to any ∼ correspondent two of these foure motions, viz: of ye point a (fixed in ye line dma ) towards b , of ye point a (fixed in ye line ane ) towards c , of ye intersection point a moveing in ye line dma according to ye order of ye letters {illeg} m , d & of ye intersection point a in ye line ane according to ye order of ye letters {illeg} n , e : Also all ye foure lines shall be one to another as those foure motions are.

Note yt \in ye two last propositions/ if ye moveing lines \may/ bee crooked {illeg} \so yt/ amd , ane , bee tangents to them in ye point a .

Note also yt by ye place of a body is meant its center of gravity.

## To resolve Problems by motion ye 6 following prop: are necessary & suffcient.

May 16. 1666.

[13] If ye body a in ye perimeter of ye circle or sphære adce moveth towards its center b . its velocity to each point \$d.c.e.$ / of yt circumference is as ye cordes ad , ac , ae , drawne from yt body to those points are.

Prop 2. If ye △s adc , aec are alike though in diverse planes; & 3 bodys move from ye point a uniformely & in equall times, ye first to d , ye 2d to e , ye 3d to c : yn is ye 3d's {sic} motion compounded of ye motion of ye 1st & {2d.}

Note yt by a body is meant its center of gravity.

Prop. 3. All ye points of a body keeping parallel to it selfe are in equall motion.

Prop. 4. If a body onely move circularly about some axis, ye motion of its points are as their distances from {illeg}|th|at axis.                     Call these 2 simple motions

Prop. 5. If ye motion of a body is considered as comp mixed of simple motions: ye motions of all its points are compounded of their simple motions, so as ye motion towards c (in prop 2d) is compounded of ye motion towards d & e .

Note yt all motion is reducible to one of these 3 cases: & in ye 3d case any line may bee taken for {the} axis (or if a line or superficies {move} in plano any point of yt plaine may bee taken for ye center) of motion.

[14] Prop. 6. If ye lines {illeg} ah being moved doe continually intersect; I describe ye Trapezium abcd {illeg} its diagonall ac : & say yt ye proportion & position of these five lines ab , ad , ac , cb , cd being determined by {requisite data} they shall designe ye proportion & position of these 5 motions: {illeg} of ye point a fixed in ye {illeg} moveing towards b ; of ye point a fixed in ye line ah {illeg} moveing towards d ; of ye intersection point a moveing in ye plaine abcd towards c (for those 5 lines are {illeg} in ye same plaine though {illeg} & ah may only touch ye plaine in their intersection point): of ye intersection point a moveing in ye line ae parallely to {illeg} cb & according to ye order of ye letters c , b : & of ye {inter}section point a moveing in ye line ah parallelly to cd & according to ye order of those {illeg}

[15] Note yt a streight line is said to designe ye position of curved motion in any point {illeg} {illeg} if toucheth ye line described by ye motion in yt point, or when tis (as ab , ad , ac ), or {illeg} tis parallell to such a {illeg} (as ad , {illeg}). Note also yt one line ah resting (as in Fig 3 & 4) ye points d & a are coincident & ye point c shall bee in ye line ah if {illeg} bee streight {illeg} (fig 3), otherwise in its tangent ac (fig 4) {illeg}. Haveing an equation expressing ye relation of two lines x & y described by two bodys A & B whose motions {illeg} q ; Translate {illeg} ye termes to one side & multiply ym, being ordered according to x {illeg} {illeg} progression {$\frac{3p}{x}.\frac{2p}{x}.\frac{p}{x}.0.\frac{-p}{x}.\frac{-pp}{x}.$ } &c: & being ordered by ye dimensions of y multiply those by {illeg} {illeg} &c. ye summa of those products {illeg} equation expressing ye relation {illeg} {illeg} motions p & q .

<51v>

## To draw a tangent to ye Ellipsis

[16] Suppose ye Ellipsis to be described by ye thred acb , & yt ce is its tangent. Since ye thred ac is diminished with ye same proportion velocity yt be increased|t|h, yt is, yt ye point c hath ye same motion towards a & d , ye angles dce , ace , must bee equall, by prop 1. I And, so of ye othe {sic} {conicks}.

## To draw a Tangent to ye Concha.

[17] Suppose yt gae , glc , alf are ye rulers by wch ye concha is usually described, & yt $gt\parallel af$ $af\perp cb=\parallel mn$, & $ng=cl\perp tn\parallel rl$. And (since equality is more simple yn proportionality) suppose yt $cb=nm$ is ye velocity of ye point c towards b , or of n towards m . Then is nt ye mot circular motion of ye point n about g \(prop 1)/; & lr ye circular motion of ye point l fixed in ye ruler ng , ({illeg} prop 4). And lg is ye motion of ye intersection point l (yt is, ye velocity of ye point c ) moveing in ye line glnc from g (prop 6). Now since a two fold velocity of ye point c is {illeg} known nemely cb toward b & lg towards d , make $fd\perp dc=lg$ ; & ye motion of ye point c shall bee in ye line fc ye diameter of ye circle passing through ye points bcdf (prop {illeg} 1) & therefore tangnt to ye Concha.

## To find ye point c wch distinguisheth twixt ye concave & convex portion of ye Concha.

[18] Those things in ye former prop: being supposed, make △ gfh like gnt or lbc : & $df\perp fr$$fr\parallel =hk=2gl\perp kp$, & draw kf . Now had ye line fd onely parallel motion directed by gd or rf , (since $dc=lg$) ye motion of all its points would bee fr , (prop 3): & if it had onely circular motion about g , ye motion of ye point f fixed in yt line df would bee fh (prop 4): But ye motion of ye point f is compounded of those two {illeg} simple motions, & is therefore fk (prop 5 & 2); & ye motion of ye intersection point f made by ye lines af df , & moveing in af , shall bee fp , (prop 6). Now if ye line cf touch ye concha in ye required point, tis easily conceived yt ye motion of of ye intersection point f is infinitely little; & therefore yt ye points p & f are coincident, df & fk being one streight line, & ye triãgles gdf , fkh being alike.

Which may bee thus calculated. Make $cb=c$. $ag=b$. $cb=y$. yn is $bl=\sqrt{cc-yy}$ . $2gl=\frac{2bc}{y}=hk$. $bl:bc\colon\colon ld=\frac{bc}{y}+c:fd=\frac{cb+cy}{\sqrt{cc-yy}}$. & $\sqrt{cc-yy}:y\colon\colon bl:bc\colon\colon gf:fh\colon\colon df:kh\colon\colon \frac{cb+cy}{\sqrt{cc-yy}}:\frac{2bc}{y}\colon\colon by+yy:2b\sqrt{cc-yy}$ . Therefore $2bcc-2byy=byy+{y}^{3}$ . Or ${y}^{3}+3byy\ast -2bcc=0$.

In stead of ye ordinary method de Maximis et minimis, it will |be| as convenient (& perhaps more naturall) to use ∼ This; Namely To find ye motion of yt line or quantity ∼ ∼ ∼ & suppose it equall to nothing, or infinitely small. But yn ye motion to wch tis compared must bee finite. That is, ye unknowne quantitys ought not to bee at their greates or least, both at once.

[19] Example, In ye triangle bcd , ye side bc being given & fixed. ye side dc being given & circulateing about ye center c , I would know when bd is ye shortest it may bee. I call $bd=x$. $da=y$. \$bc=a$. $dc=b$./ yn is ye {illeg} $\sqrt{xx-yy}=ab$ . & $\sqrt{bb-yy}=ac$. & $\sqrt{xx-yy}+\sqrt{bb-yy}=a$ . or $xx-aa-bb=-2a\sqrt{bb-yy}$. & $\begin{array}{llllllll}{x}^{4}& -& 2aa& xx& +& {a}^{4}+{b}^{4}& =& 0\\ & -& 2bb& & -& 2aabb\\ & & & & +& 4aayy\end{array}$. $\begin{array}{llllllll}4p{x}^{3}& -& 4aa& px& +& 8aaqy& =& 0\\ & -& 4bb\end{array}$ (prop 7). And makeing $p=0$, tis $8aaqy=0$. Or $\frac{0}{8aaq}=y=0$. (For q signifieing ye motion of d towards bc may bee finite though, p , its motion towards b doth perish). Wherefore $xx=aa\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2b+bb$. or $x=a-b$. $x=b-a$. $x=b+a$. $x=-b-a$. are ye greatest & ye least valors of ye line bd .

Should I have taken $ba=y$, instead of $da=y$. {illeg}|Th|e effect would not have followed because both ye motions p & q would have vanished at once in ye point {e}. But I might have taken ye tangent dm for y , or any other line wch wou{ld} {illeg} coincidere wth bc at its being greatest or least.

[20] Example 2d. If {ne} is ye Conchoid ($ga=b$. $ae=c=nl$. $nb=y$. $ab=x$.) fo{illeg} em parallell to it. Then is $\sqrt{cc-yy}:y\colon\colon bl:bn\colon\colon cg:fl\phantom{\rule{0.5em}{0ex}}\text{(supra)}\phantom{\rule{0.5em}{0ex}}=\frac{cby+cyy}{\sqrt{cc-yy}}$. Then is $\sqrt{cc-yy}=bl:c=cl\colon\colon dl=cg=c+\frac{bc}{y}\phantom{\rule{0.5em}{0ex}}\text{(vide supra)}\phantom{\rule{0.5em}{0ex}}:fl=\frac{ccy+bcc}{y\sqrt{cc-yy}}$ . & $fb=\frac{bcc+{y}^{3}}{y\sqrt{cc-yy}}$ . & $y=cb:fb\colon\colon ea=c:am=\frac{b{c}^{3}+c{y}^{3}}{yy\sqrt{cc-yy}}=z$. et $bb{c}^{6}+2b{c}^{4}{y}^{3}+cc{y}^{6}-ccy+zz-{y}^{6}zz=0$ ponatur p esse motus puncti m & q esse {illeg} motus puncti c versus b . Erit (prop {6}{)} $6b{c}^{4}yyq+6cc{y}^{5}q-4cc{y}^{3}zzq-$ {illeg}{illeg}$qzz-2ccp{y}^{4}z-2p{y}^{6}z=0$ . supposeing $q=0$ (For when am is ye least yt {illeg}{illeg} bee ye point c is yt wch distinguisheth twixt ye concave & convex porti{on} of ye Conchoid, & yn ye motion q vanisheth.) it will bee $\frac{3b{c}^{4}+3cc{y}^{3}}{2ccy-3{y}^{3}}=zz=\frac{b{c}^{3}+c{y}^{3}}{yy\sqrt{cc-yy}}\right\}2$ . $\frac{3c}{2cc-3yy}=\frac{b{c}^{3}+c{y}^{3}}{cc{y}^{3}-{y}^{5}}$ . & {illeg} $cc{y}^{3}$ {illeg}$=2b{c}^{4}-3bccyy$ . Or ${y}^{3}=-3byy+2bcc$.

<55r>

## Concerning Equations when their rootesrelation twixt\ratio of/ their rootes is considered.

[21] If to {sic} of ye rootes of an Equation are in proportion ye one to ye other as a to b Then multiplying ye termes of the{illeg} Equation by this progression
|:&c| $\frac{{a}^{4}}{{b}^{3}}+\frac{{a}^{3}}{bb}+\frac{aa}{b}.\frac{{a}^{3}}{bb}+\frac{aa}{b}.\frac{aa}{b}.0.-a.-a-b.-a-b-\frac{bb}{a}.-a-b-\frac{bb}{a}-\frac{{b}^{3}}{aa}$. &c. (And that root /or by ye same progression\ augmented or diminished by any quantity, as if it bee augmented by a it will bee $\frac{{a}^{4}}{{b}^{3}}+\frac{{a}^{3}}{bb}+\frac{aa}{b}+a.\frac{{a}^{3}}{bb}+\frac{aa}{b}+a.\frac{aa}{b}+a.a.0.-b.-b-\frac{bb}{a}.-b-\frac{bb}{a}-\frac{{b}^{3}}{aa}.-b-\frac{bb}{a}-\frac{{b}^{3}}{aa}-\frac{{b}^{4}}{{a}^{3}}$. &c. Or were it augmented by $a+b$ {illeg} it would be $\frac{{a}^{3}}{bb}+\frac{aa}{b}+a+b.\frac{aa}{b}+a+b.a+b.b.0.-\frac{bb}{a}.-\frac{bb}{a}-\frac{{b}^{3}}{aa}.-\frac{bb}{a}-\frac{{b}^{3}}{aa}-\frac{{b}^{4}}{{a}^{3}}$). Then shall ye roote wch is correspondent to $\left(b\right)$ be a roote of ye resulting equation: but inverting ye order of ye progresion {sic}, yt roote wch is correspondent to $\left(a\right)$ shall bee a roote of ye equation resulting from such multiplication.

As for example did I know yt two of ye rootes of ye Equation ${x}^{3}-8xx+9x+18=0$ were in proportion as 1 to 2 & would I have ye lesser roote (viz yt wch is correspondent to 1) I make $b=1$. $a=2$. And soe the progression will bee $28.28.12.4.0.-2.-3.-\frac{7}{2}$. &c Or $30.14.6.2.0.-1.\frac{-3}{2}.\frac{-7}{4}$. &c by {illeg} adding 2 . Or by adding one more it will bee $31.15.7.3.1.0.\frac{1}{2}.\frac{3}{4}.\frac{7}{8}$. &c. By any of wch progressions ye Equation may bee multiplyed, as by ye 1st, $\begin{array}{lllllllll}{x}^{3}& -& 8xx& +& 9x& +& 18& =& 0\\ 28& .& 12& .& 4& .& 0& .\end{array}$. Wch produceth {illeg} $7xx-24x+9=0$. Or by ye 3d $\begin{array}{lllllllll}{x}^{3}& -& 8xx& +& 9x& +& 18& =& 0\\ 7& .& 3& .& 1& .& 0& .\end{array}$. Wch produceth {illeg} $7xx-24x+9=0$. Or by the first Otherwise by destroying ye 1st terme. $\begin{array}{lllllllll}{x}^{3}& -& 8xx& +& 9x& +& 18& =& 0\\ 0& .& -2& .& -3& .& -\frac{7}{2}& .\end{array}$. Wch produceth {illeg} $16xx-27x-63=0$. &c the rootes of wch products are, viz: of ye first $x=\frac{3}{7}$, & $x=3$. Of ye last $x=-\frac{21}{16}$, & $x=3$. There I conclude 3 to be ye lesse, & consequently 6 the greater of those rootes wch are in proportion as one to of ye Equation ${x}^{3}-8xx+9x+18=0$. wch are in double proportion But was ye greater of those rootes desired yn {illeg} inverting ye progression it would bee $\begin{array}{lllllllll}{x}^{3}& -& 8xx& +& 9x& +& 18& =& 0\\ 0& & 1& & 3& & 7& \end{array}$. Or $\begin{array}{lllllllll}{x}^{3}& -& 8xx& +& 9x& +& 18& =& 0\\ -\frac{7}{2}& & -3& & -2& & 0\end{array}$. The first producing $8xx-27x-126=0$ whose rootes are $x=-\frac{21}{8}$, $x=6$. {illeg} The 2d produceth $7{x}^{2}-48x+36=0$ whose rootes are $x=\frac{6}{7}$ , $x=6$ . And consequently 6 is the greater & 3 the lesse of ye rootes in duplicate propor{tion.}

Would I draw

[22] If in ye circle adef af is ye diameter, ah a perpendicular to ye end of it from wch I would draw $he\parallel af$, wch should intersect perpendicular circle in ye points d & e soe yt $\left(de\right)$ bee triple to $\left(hd\right)$, yt is he quadruple to hd . Then calling $af=g$. $ah=y$. $\begin{array}{c}hd\\ he\end{array}\right\}=x$. The equation expresing {sic} ye relation twixt x & y is {illeg} $xx-gx+yy=0$. ye rootes of wch equation must be quadruple ye one to ye other: Therefore would I find {illeg} $\left(hd\right)$ ye lesse roote I make $b=1$. $a=4$. And ye progression will bee, $21.5.1.0.-\frac{1}{4}.-\frac{5}{16}.-\frac{21}{32}$. &c. by wch ye Equation being multiplyed ye product is $\begin{array}{ccccccc}xx& -& gx& +& yy& =& 0\\ 5& & 1& & 0\end{array}$. Or $x=\frac{1}{5}g$. Therefore drawing $ab=\frac{1}{5}g$, Or $ac=\frac{4}{5}g$ . from wch /the\ point b , or c raise ye perpendiculars db , or {illeg} ce. & soe {illeg} draw hde.

Would I have dbec to be a square yt is $db=de=y$. Then to find hd I call it x & $dh=x+y=hd+de$. Soe yt ye lesse roote is to ye greater as x to $x+y$. Making therefore $b=x$, $a=x+y$, The progression will be $\frac{xx+2xy}{x}$ $\frac{yy}{x}+3x+3y.2x+y.x.0.-\frac{xx}{x+y}.-\frac{xx}{x+y}-\frac{{x}^{3}}{xx+2xy+yy}$. Or, $\frac{yy}{x}+2x+3y.x+y.0.-x.-x-\frac{xx}{x+y}$. by wch ye Equation being multiplyed produceth $\begin{array}{ccccccc}xx& -& gx& +& yy& =& 0\\ x+y& & 0& & -x\end{array}$, Or ${x}^{3}+yxx-yyx=0$ , Or $y=\frac{x}{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{5xx}{4}}=\frac{x\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}x\sqrt{5}}{2}$ . And consequently $-xx+gx=yy=\frac{xx\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2xx\sqrt{5}+5xx}{4}$. Or $4g=10x\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2x\sqrt{5}$. Or $\frac{2g}{5\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{5}}=x$ . By wch ye Equation $xx-gx+yy=0$, must be multiplyed. & it produceth $\begin{array}{ccccccc}xx& -& gx& +& yy& =& 0\\ 2x+y& .& x& .& 0\end{array}$. Or $2{x}^{3}+yxx-gxx=0$ Or $-2x+g=y$. & congsequently $-xx+gx=yy=4xx-4gx+gg$. yt is $5xx=5gx-gg$. And $x=\frac{1}{2}g\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{1}{20}gg}$ . Or $x=\frac{g\sqrt{5}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}g}{2\sqrt{5}}=\frac{1}{2}g\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\frac{g}{2\sqrt{5}}$. And consequently $y=\frac{1}{\sqrt{5}}g$. Or it might have beene done thus. x substracted from ye precedent progression it will be, {illeg} $x+y.0.-x.-x+\frac{x}{x+y}$ . &c by wch ye Equation being multiplyed produceth $\begin{array}{ccccccc}xx& -& gx& +& yy& =& 0\\ x+y& & 0& & -x\end{array}$. Or $xx+yx=yy$. And by extracting the roote, $y=\frac{1}{2}x\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{5}{4}xx}$ . And therefore $gx-xx=yy=\frac{1}{4}xx\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}x\sqrt{\frac{5}{4}xx}+${$\frac{5}{4}xx$} Or $4g=10x\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2x\sqrt{5}$. Or $\frac{2g}{5\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{5}}=x=\frac{g\sqrt{5}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}g}{2\sqrt{5}}$ . That is $\frac{2g}{5+\sqrt{5}}=\frac{g\sqrt{5}-g}{2\sqrt{5}}=ab$ . And $\frac{2g}{5-\sqrt{5}}=\frac{g\sqrt{5}+g}{2\sqrt{5}}=ac$ . And therefore { $\frac{2g}{5-\sqrt{5}}\phantom{\rule{1em}{0ex}}\frac{-2g}{5+\sqrt{5}}=af-ab=\frac{10g+2g\sqrt{5}-10g+2g\sqrt{5}}{25+5\sqrt{5}-5\sqrt{5}-0}=\frac{g}{\sqrt{5}}=ah=de=db=bc$ }.

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## Reductions of Equations may bee very often & readily\perhaps/ performed by this method

[23] As {illeg} in that problem recited by D: Cartes pag 83, viz: The square ad & ye right line s being given , to produce ac to e , soe yt ef drawn towards ye point b may bee equall to ye given line {illeg} s . Putting $df=x$ for ye unknowne quantity. {illeg} $ef=c$, & $bd=cd=a$. The Equation will bee ${x}^{4}-2a{x}^{3}\phantom{\rule{0.5em}{0ex}}\begin{array}{l}+2aaxx\\ -cc\end{array}\phantom{\rule{0.5em}{0ex}}-2{a}^{3}+{a}^{4}=0$. wch having 4 rootes the Equation must have 4 divers resolutions; that is ye lines ac , cd , produced both ways indefinitely, there may bee 4 divers lines drawne through the point b , whose parts intercepted twixt ye crosse lines lace , mdch , are equall to the given line s : And they are bih lbk , nbm . And therefore the rootes of this equation are (two affirmative) df , dh , (& two negative) dh , dm . Because $bd=ab$, $hi=fe$, Therefore $ae=dh$, $ai=fd$, $an=dk$, $al=dm$. Soe yt $fd:bd\colon\colon ab:ae=dh=\frac{aa}{x}$. That is one roote $\left(df\right)$ of this equation is to another $\left(dh\right)$ as x to $\frac{aa}{x}$ . Therefore I may multiply this Equation by this progression $\frac{{a}^{4}}{{x}^{3}}+\frac{aa}{x},\frac{aa}{x},0,-x,-\frac{{x}^{3}}{aa},-\frac{{x}^{5}}{{a}^{3}}$. And there resulteth $\begin{array}{ccccccccccc}{x}^{4}& -& 2a{x}^{3}& \phantom{\rule{0.5em}{0ex}}& \begin{array}{l}+2aaxx\\ -cc\end{array}& -& 2{a}^{3}x& +& {a}^{4}& =& 0\\ +\frac{aa}{x}& .& 0& .& -x& .& -x-\frac{{x}^{3}}{{a}^{2}}& .& -x-\frac{{x}^{3}}{{a}^{2}}-\frac{{x}^{5}}{{a}^{4}}\end{array}$. That is $aa{x}^{3}+cc{x}^{3}-2aa{x}^{3}+2{a}^{3}xx+2a{x}^{4}-{a}^{4}x-aa{x}^{3}-{x}^{5}=0$. Or, ${x}^{4}-2a{x}^{3}\phantom{\rule{0.5em}{0ex}}\begin{array}{l}+2aaxx\\ -cc\end{array}\phantom{\rule{0.5em}{0ex}}-2{a}^{3}x+{a}^{4}=0$. Which result is ye {illeg} same wth ye first Equation the reason of wch is, yt if I make $df=x$ then is $dh=\frac{aa}{x}$. Or if $dh=x$, yn is $df=\frac{aa}{x}$. Or if $dk=x$, yn is $dm=\frac{aa}{x}$. Or if $dm=x$ then is $dk=\frac{aa}{x}$. Soe yt ye relation twixt \all/ ye rotes {sic} being{illeg} the same & reciprocally the same & {illeg} not distinguishing one roote from another, tis noe wonder if they bee all indifferently expressed in ye resulting Equation. Otherwise ye reduction must have succeeded.

Suppose 3 rootes of an Equation are in proportion to each other as a , b , c . Then if that roote wch is correspondent to a be required, multiply ye termes of ye Equation by this progression any of these Progressions
|1| $\frac{{c}^{3}+bcc+bbc+{b}^{3}}{aa}+\frac{bb+bc+cc}{a}+b+c+a.\frac{bb+bc+cc}{a}+b+c+a.+b+c+a.+a.0.0.\frac{{a}^{3}}{bc}.\frac{{a}^{3}}{bc}+\frac{{a}^{4}b+{a}^{4}c}{bbcc}.\frac{{a}^{3}}{bc}+\frac{{a}^{4}b+{a}^{4}c}{bbcc}.\frac{{a}^{3}}{bc}+\frac{{a}^{4}b+{a}^{4}c}{bbcc}+\frac{{a}^{5}bb+{a}^{5}bc+{a}^{5}cc}{{b}^{3}{c}^{3}}$. |2| $\frac{+bbc+bcc}{aa}+\frac{bb+bc+cc}{a}+b+c.\frac{bc}{a}+b+c.0.-a.0.\frac{+{a}^{3}+aab+aac}{bc}.\frac{{a}^{3}+aab+aac}{bc}+\frac{{a}^{3}bb+{a}^{3}bc+{a}^{3}bb+{a}^{4}b+{a}^{4}c}{bbcc}$. |3| {illeg} $\frac{aabb+aacc+bbcc}{{a}^{3}+aab+aab}+\frac{bc}{a}.0.\frac{-ab-ac-bc}{a+b+c}.-a.0.\frac{{a}^{3}}{bc}+\frac{+bb+bc+cc}{bac+bbc+bcc}$.                      &c.

As if \3 of/ ye rootes of this Equation ${x}^{4}\ast -35xx+90x-56=0$. ${x}^{4}\ast -35ggxx+90{g}^{3}x-56{g}^{4}=0$ were to one another as $1.2.4$. And g would find ye roote wch is correspondent to 1 . Then I make $a=1$, $b=2$, $c=4$, & soe I may have have {sic} by ye first progression this{illeg} $+155.+35.+7.+1.0.0.\frac{+1}{7}.\frac{+7}{32}.\frac{+70}{128}$. By ye 2d; $+92.+14.0.-1.0.+\frac{7}{8}.+\frac{45}{32}$. &c. By ye first of wch ye Equation being multiplyed produceth $\begin{array}{cccccccccc}{x}^{4}& \ast & -& 35ggxx& +& 90{g}^{3}x& -& 56{g}^{4}& =& 0\\ 35.& 7.& & 1.& & 0.& & 0.\end{array}$. That is $35{x}^{4}-35ggxx=0$. Or $xx=gg$. & $x=g$. Or were{illeg} it multiplyed by ye 2d progression thus $\begin{array}{cccccccccc}{x}^{4}& \ast & -& 35ggxx& +& 90{g}^{3}x& -& 56{g}^{4}& =& 0\\ 0.& -1.& & 0.& & \frac{7}{8}.& & \frac{45}{32}.\end{array}$. It would produce $\frac{630}{8}{g}^{3}x-\frac{2520}{32}{g}^{4}=0$. Or $x=g$. Soe yt g being ye least roote, ye other two rootes must be $2g$ & $4g$.

If it be desired to know the length of y \& z / when the rootes of this equation are in p ${x}^{3}-aax+yyx-aaz=0$ , when ye rootes of it are in proportion {sic} $1.2.4.$ I multiply it by one of ye precedent progressions & it is $\begin{array}{ccccccccc}{x}^{3}& -& axx& +& yyx& -& \frac{8{a}^{3}}{7}& =& 0\\ 7& & 1& & 0& & 0\end{array}$. Or $x=\frac{a}{7}$ . wch valor of x inserted into its places in ye Equation there results $\frac{{a}^{3}}{343}-\frac{{a}^{3}}{49}+\frac{a}{7}yy-\frac{8{a}^{3}}{7}=0$ . Or $y=\frac{a\sqrt{398}}{7}$. Or thus $\begin{array}{ccccccc}{x}^{3}& -& axx& +& yyx& -& \frac{8{a}^{3}}{7}\\ 1& & 0& & 0\end{array}$

If it be desired to know ye length of y & z in this equation ${x}^{3}-yxx+{a}^{2}x-aaz$ when ye rootes are in proportion as $1.2.4.$ I multiply it by ye precedent progression & ye results are $\begin{array}{ccccccccc}{x}^{3}& -& yxx& +& aax& -& aaz& =& 0\\ 7& & 1& & 0& & 0\end{array}$. Or $7x=y$ . And $\begin{array}{ccccccccc}{x}^{3}& -& yxx& +& aax& -& aaz& =& 0\\ 0& & 0& & \frac{1}{8}& & \frac{7}{32}\end{array}$. Or $4x=7z$. $\begin{array}{ccccccccc}{x}^{3}& -& yxx& +& aax& -& aaz& =& 0\\ 14& & 0& & -1& & 0\end{array}$. Or $14xx=aa$. And consequently $7=x=\frac{7a}{\sqrt{14}}=y=a\sqrt{\frac{7}{2}}$ $\frac{4x}{7}=\frac{4a}{7\sqrt{14}}=z$.

Likewise were the proportion of 4 or 5 or more rog\o/tes given I might set down progressions to find them but it will bee better to set downe ye method of finding {those}{these} progressions, And it {illeg} is this. Suppose {illeg} two of ye rootes of an Equation {illeg} That Equation will bee of this forme $xx\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-a\\ -b\end{array}x\phantom{\rule{0.5em}{0ex}}+ab=0$ , or of some forme {illeg} of it; And if a corresponds to ye \desired/ roote of ye Equation desired this equation $xx$ {illeg} will bee of this forme $aa\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-a\\ -b\end{array}a\phantom{\rule{0.5em}{0ex}}+ab=0$ . Then assuming two termes ({illeg} { $a.0$ }) {illeg} third {illeg} progression by wch I multiply this equation {illeg} { $a.0$ } by which {illeg} {illeg} multiplyed produceth $aa\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-a\\ -b\end{array}a\phantom{\rule{0.5em}{0ex}}+ab=0$ Or { $xaa-{a}^{3}-aab=0$ }. And {illeg} {illeg} termes of ye {illeg} {illeg}

<56r>

Soe yt I have thus much of ye progression $a+b+\frac{bb}{a}.a+b.a.0$ . And by ye same proceding might continue it or get termes on ye other side of ye cipher. As if I multiply ye Equation by this progression $a.0.z$ there is produced $\begin{array}{cccccc}aa& \begin{array}{c}-a\\ -b\end{array}a& +& ab& =& 0\\ ×& ×& & ×\\ a& 0& & z\end{array}$. Or ${a}^{3}+abz=0$. And $z=-\frac{aa}{b}$. Againe multiplying ye Equation by $0.x$ {illeg} $0.-\frac{aa}{b}.z$ . It is $\begin{array}{ccccccc}aa& & \begin{array}{c}-a\\ -b\end{array}a& +& ab& =& 0\\ ×& & ×& & ×\\ 0& .& -\frac{aa}{b}& .& z\end{array}$. Or $\frac{{a}^{4}+{a}^{3}b}{b}+abz=0$, And $-\frac{{a}^{3}}{bb}-\frac{aa}{b}=z$. Soe that I have thus much of ye progression viz: $a+b+\frac{bb}{a}.a+b.a.0.-\frac{aa}{b}.-\frac{aa}{b}-\frac{{a}^{3}}{bb}$.

The proceeding is same when ye proportion of 3 rootes to one another are given, but there may bee some difficulty /difference\ wn ye ciphers are {illeg} far distant, as they|r|e bee three {illeg} termes betwixt them, then ye opr|e|ration may be done thus. Let ye quanti\t/ys, wch beare such proportion to one another as ye rootes doe bee, a, b, c . let a correspond to ye roote wch wch must be knowne And yn yt Equation will bee of this forme, ${a}^{3}\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-a\\ -b\\ -c\end{array}aa\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-ab\\ -ac\\ -bc\end{array}a\phantom{\rule{0.5em}{0ex}}-abc=0$. or else compounded of it. Then assuming some quantity (as a ) for one of ye termes of ye {illeg} progression & placing it conveniently, (as {illeg} \it {illeg} {no{illeg}}/ equidistant from ye ciphers) feigne two other quantitys as z , y , for ye deficient termes and there will r progression will bee $0.z.a.y.0$. By wch I multiply the Equation $\begin{array}{ccccccccc}{a}^{3}& & \begin{array}{c}-a\\ -b\\ -c\end{array}aa& & \begin{array}{c}-ab\\ -ac\\ -bc\end{array}a& -& abc& =& 0\\ ×& & ×& & ×& & ×\\ 0& .& z& .& a& .& y\end{array}$. Or $\frac{-aaz-abz-acz}{bc}=y$. Soe have I ye progression $0.z.a.\frac{-aaz-abz-acz}{bc}.0.$ by wch I againe multiply ye Equation & there is produced a correspond to ye roote wch wch must be knowne And yn yt Equation will bee of this forme, ${a}^{3}\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-a\\ -b\\ -c\end{array}aa\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}$ Or $\frac{-aaz-abz-acz+aab+aac+abc}{bc}=y$. Soe that I have the progression, $0.z.a.\frac{+aab+aac+abc-aaz-abz-acz}{bc}.0.$ by wch I againe multiply ye Equation & there results $\begin{array}{ccccccccc}{a}^{3}& & \begin{array}{c}-a\\ -b\\ -c\end{array}aa& & \begin{array}{l}-ab\\ -ac\\ -bc\end{array}a\phantom{\rule{1em}{0ex}}...............& -& abc& =& 0\\ ×& & ×& & ×& & ×\\ z& .& a& .& \frac{+aab+aac+abc-aaz-abz-acz}{bc}& .& 0\end{array}$. Or $z=\frac{aabb+aabc+aacc+abbc+abcc+bbcc}{aab+aac+bba+acc+2abc+bbc+bcc}$. wch valor of z substituted into its place in ye valor y There will bee thus much of ye progre\ssion/ $0.\frac{aabb+aabc+aacc+abbc+abcc+bbcc}{aab+aac+bba+acc+2abc+bbc+bcc}.a.\left\{\begin{array}{c}a+\frac{aab+aac}{bc}-\frac{{a}^{4}-2{a}^{3}b-2{a}^{3}c-aabb-3aabc-aacc-abcc}{aab+aac+bba+acc+2abc+bbc+bcc}\\ \frac{-{a}^{4}bb-{a}^{3}{b}^{3}-{a}^{3}bbc-{a}^{4}cc-{a}^{3}bcc-{a}^{3}{c}^{3}}{aabbc+aabcc+a{b}^{3}c+ab{c}^{3}+2abbcc+{b}^{3}cc+bb{c}^{3}}\end{array}\right\}.0$.

## The same done otherwise.

Did I know yt 2 of ye rootes of this Equation ${x}^{3}\ast -117x-324=0$, were in proportion as 3, $-4$. Then I suppose one roote to be $3a$ , ye other $-3a$ That is $x-3a=0$. $x+4a=0$. By \one of/ wch I divide ye Equation as first by $x+4a=0$ And ye operation is; $\begin{array}{ccc}x+4a)\phantom{\rule{0.5em}{0ex}}& \underset{_}{{x}^{3}\phantom{-}\ast \phantom{ax}-117x-324}& \phantom{\rule{0.5em}{0ex}}\left(xx-4ax+16aa-117=0\right\\ & \underset{_}{0-4axx-117x-324}\\ & \hfill \underset{_}{\begin{array}{c}0+16aax-324\\ \phantom{0}-117x\hfill \end{array}}\\ & \hfill \underset{_}{\begin{array}{c}0\phantom{000}-\phantom{0}64{a}^{3}\\ \hfill \phantom{0}+468{a}^{\phantom{1}}\end{array}}\\ & \hfill \underset{_}{\phantom{0000}-\phantom{0}64{a}^{3}}& \underset{_}{+468a-324=0}\\ & & 0\hfill \end{array}$. Againe I divide ye Quotient by y{e} other roote $x-3a=0$. Thus
$\begin{array}{ccc}x-3a)& \underset{_}{xx-4ax+16aa-117}=0& \left(x-a=0\right\\ & \underset{_}{0\phantom{x}-\phantom{4}ax+16aa-117}\phantom{=0}\\ & \hfill 0\phantom{00}\underset{_}{+13aa-117}=0\\ & \hfill \phantom{000+13aa}0\phantom{-117=0}\end{array}$. By ye last division I have this equation $13aa-117=0$. Or $aa-9=0$. And $a=3$. Therefore ye rootes of ye Equation ${x}^{3}-117x-324=0$ are, $3a=9=x$. $-4a=-12=x$.

If I would have y & z of such a length yt ye rootes of this equation ${x}^{3}-3yx+gz=0$ . be in proportion as $+1$, $+2$, $+3$. I suppose $x-a=0$, $x+2a=0$. $x+3a=0$. And soe first divide ye Equation by $\begin{array}{ccc}x-a)\phantom{\rule{0.5em}{0ex}}& \underset{_}{{x}^{3}+\phantom{0}\ast \phantom{0}-3yx+gz}& \phantom{\rule{0.5em}{0ex}}\left(xx+ax+aa-3y=0\right\\ & \underset{_}{0+axx-3yx+gz}\\ & \hfill 0\phantom{\rule{0.5em}{0ex}}\begin{array}{rr}+& aa\\ -& 3y\end{array}x+gz\\ & \hfill \stackrel{_}{0\phantom{\rule{1em}{0ex}}+{a}^{3}}& \stackrel{_}{+3ay+g{z}^{\phantom{1}}=0}\hfill \end{array}$. Againe I divide this product
by $\begin{array}{ccc}x-2a)\phantom{\rule{0.5em}{0ex}}& \underset{_}{xx+\phantom{3}ax+aa-3y}& \phantom{\rule{0.5em}{0ex}}\left(x+3a=0\right\\ & \hfill 0\phantom{0}+3ax+aa-3y\\ & \hfill \stackrel{_}{\phantom{{a}^{1}}0+5{a}^{3}-3y}& \stackrel{_}{=0\phantom{{a}^{1}}}\hfill \end{array}$. Lastly were it necessary I should have again divided this quote $x+3a=0$ by ye 3d supposed roote of ye Equation (viz {illeg}$=0$). By ye 2d {operation} $7{a}^{2}-3y=0$. Or $\frac{7{a}^{2}}{3}=y$. And by ye first ${a}^{3}-3ay+gz=0$ . Or {illeg} /{$18ay=7gz$ }\ Soe yt If I make {illeg} the rootes of this Equation ${x}^{3}-3yx+gz=0$ should bee {illeg}

<57r>

[24] [25] .R. An Equation being given, expressing ye Relation of two or more lines x, y, z, & described in ye same line by two or more moveing bodys A, B, C &c to find ye relation of their velocitys p, q, r &c:

## Resolution.

Sett all ye termes on one side of ye Equation yt they become equall to nothing. And first Multiply each terme by soe many times $\frac{p}{x}$ as x hath dimensions in yt terme. Seacondly multiply each terme by soe many times $\frac{q}{y}$ as y hath dimensions in {illeg} --> it. Thirdly multiply each terme by soe many times $\frac{r}{z}$ as z hath dimensions in it &c. The sume of all these products shall be equall to nothing. Which Equation gives ye relation of p, q, r &c.

[26] Or more generally thus. Order ye Equation acording {sic} to ye dimensions of x , & (putting a & b for any two numbers whither rationall or not) multiply ye termes \of it/ by any pte of this progression viz : $\text{ↄ&}.\frac{ap-3bp}{x}.\frac{ap-2bp}{x}.\frac{ap-bp}{x}.\frac{ap}{x}.\frac{ap+bp}{x}.\frac{ap+2bp}{x}.\frac{ap+3bp}{x}.\text{&c}$ : Also order ye Equation according to y & multiply ye{illeg} {illeg}termes of it by this progression: $\text{ↄ&}.\frac{aq-2bq}{y}.\frac{aq-bq}{y}.\frac{aq}{y}.\frac{aq+bq}{y}.\frac{aq+2bq}{y}.\frac{aq+3bq}{y}\phantom{.}\text{&c}$. Also order it according to ye dimentions {sic} of z & multiply its termes by this progression viz $\text{ↄ&}.\frac{ar-3br}{z}.\frac{ar-2br}{z}.\frac{ar-br}{z}.\frac{ar}{z}.\frac{ar+br}{z}.\frac{ar+2br}{z}.\frac{ar+3br}{z}\phantom{.}\text{&c}$. The sume of all these products shall bee equall to nothing. Which Equation gives ye relation of p, q,r &c.

Example 1st. If ye propounded Equation bee ${x}^{3}-2xxy+4xx+7xyy-{y}^{3}-103=0$. By ye precedent rule ye first operation will produce $3xxp-4xyp+8xp+7yyp$ . The seacond produceth $-2xxq+14xyq-3yyq$. Which two added together make $3xxp-4xyp+8xp+7yyp-2xxq+14xyq-3yyq=0$. (Now suppose a yarde to bee an unit & yt a|A| hath moved 3 yardes, yn (by ye 1st equation) B hath moved two; i,e, {sic} $x=3$, $y=2$. And \at that time/ by ye last Equation $55p+54q=0$ . Or $55:-54\colon\colon p:q\colon\colon$ velocity of A ∶ velocity of B . Onely if x increaseth yn y decreaseth, yt is, A & B move contrary ways because p & q are affected wth divers signes).

Example ye 2d. If ye Equation bee ${x}^{3}-2{a}^{2}y+{z}^{2}x-{y}^{2}x+z{y}^{2}-{x}^{3}=0$. The first operation will produce $\begin{array}{cccccc}\frac{3p}{x}& & \frac{p}{x}& 0& 0& 0\\ {x}^{3}& \ast & +zzx-yyx& -2aay& +{z}^{3}y& -{z}^{3}\end{array}$. Or $3pxx+pzz-pyy$. The second produceth $-2aaq-2yxq+2zyq$. The third $+2zxr+yyr-3zzr$. The summe of wch is $3pxx+pzz-$ $-pyy-2aaq-2yxq+2zyq+2zxr+yyr-3zzr=0$ . (Note yt in this Example there being three unknowne quantitys x , y , z , There must be two of them & their \two/ velocitys supposed thereby to find ye 3d quantity & ye third velocity. Or else there must be some other equation expressing ye relation of ye th two of these x , y , z . (as in ye first example) whereby one quantity & one velocity being supposed ye other quantity & velocity may be found & yn by this 2d Example ye 3d quantity & ye 3d velocity may bee found)

Example 3d, Of ye more generall rule. If ye Equation bee ${x}^{4}-3gyxx+yyxx-ggyy-2{y}^{4}=0$. ye first operation gives $\begin{array}{cccccc}\frac{ap}{x}& .\ast .& \frac{ap+2bp}{x}& .\ast .& \frac{ap+4bp}{x}& .\\ {x}^{4}& \ast & \begin{array}{cc}+& \hfill yy\\ -& 3gy\end{array}xx& \ast & \begin{array}{cc}-& ggyy\\ -& 2{y}^{4}\end{array}& .\end{array}$ Or $\begin{array}{lllllllll}ap{x}^{3}& +& apyyx& -& 3apgyx& -& \frac{apggyy}{x}& -& \frac{2ap{y}^{4}}{x}\\ & +& 2bpyyx& -& 6bpgyx& -& \frac{4bpggyy}{x}& -& \frac{8bp{y}^{4}}{x}\end{array}$. the 2d gives $\stackrel{_}{aq-2bq}×\stackrel{_}{-2{y}^{3}}\phantom{\rule{1em}{0ex}}\stackrel{_}{+aq}×\stackrel{_}{xxy-ggy}\phantom{\rule{1em}{0ex}}\stackrel{_}{+aq+bq}×\stackrel{_}{-3ggx}\phantom{\rule{1em}{0ex}}\stackrel{_}{+aq-2bq}×\frac{{x}^{4}}{y}$. \The summe of/ wch two products is equall to nothing. &c.

## Demonstration.

[27] Lemma. If two bodys $\begin{array}{c}A\\ B\end{array}$ move uniformely ye $\begin{array}{c}\text{one}\\ \text{other}\end{array}$ from $\begin{array}{c}a\\ b\end{array}$ to $\begin{array}{cccccc}c& ,& e& ,& g& ,\\ d& ,& f& ,& h& ,\end{array}\phantom{\rule{0.5em}{0ex}}\text{&c}$ in ye same line yn are ye lines $\begin{array}{c}ac\\ bd\end{array}$ & $\begin{array}{c}ce\\ df\end{array}$ & $\begin{array}{c}eg\\ fh\end{array}$ &c as their velocitys $\begin{array}{c}p\\ q\end{array}$ {illeg}.And though they move not uniformely yet are ye infinitely little spac lines wch each moment they describe {illeg} as their velocitys are wch they have while they describe them. As if ye body A \wth ye velocity p / describe ye infinitely little line o in one moment. Then \In ye moment/ ye body B wth ye velocity q will describe ye line $\frac{oq}{p}$ . For $p:q\colon\colon o:\frac{oq}{p}$. Soe yt if ye described \{lines}/ be x & y in one moment, they will bee $x+o$ & $y+\frac{oq}{p}$ , in ye next. [or better $p:q\colon\colon po:qo$. &c]

Now if ye Equation expressing ye relation of ye lines x & y be $rx+xx-yy=0$. I may substitute $x+o$ & $y+\frac{qo}{p}$ into ye place of x & y because (by ye lemma) they as well as x & y doe signifie ye lines described by ye bodys A & B . By doeing so there results $rx+ro+$ $+xx+2ox+oo-yy-\frac{2qoy}{p}-\frac{qqoo}{pp}=0$ . But $rx+xx-yy=0$ by supposition: there remaines therefore $ro+2ox+oo-\frac{2qoy}{p}-\frac{qqoo}{pp}=0$ . On divideing it by o tis $r+2x+o-\frac{2qy}{p}-\frac{oqq}{pp}=0$ . Also those termes in wch o is are {sic} infinitely lesse yn those in wch o is not therefore I blotting ym out there {rests} $r+2x-\frac{2qy}{p}=0$ . Or $pr+2px=2qy$.

Hence may bee observed: First, yt those {illeg} termes \ever/ vanish in wch o is not because they are ye propounded Equation. Secondly ye remaining Equation being divided by o those termes also vanish in wch o still remaines because they are infinitely little. Thirdly yt ye \still/ remaining termes consist of y {illeg} \will ever/ have yt forme wch by ye \first/ {root}{rule} they should have. [{illeg} partly appeare by Oughtreds Analyticall table].

The {rule} may bee demonstrated after ye same manner if there 3 \or more/ unknowne quantitys x, y, z {&c.}

<57v>

## Resolution

[29] Find (by ye preceding rule) in wt proportion those two lines to wch ye crooked line {cheifly {sic}} related doe increase or decrease: produce ym in yt proportion {illeg} from ye given point in ye crooked line {{illeg}|at|} those ends draw perpendiculars to ym \lines in which those ends are {enclosed} to move/ through whose intersection ye tangent shall passe.

Example 1st. If $ab=id=x$. $ai=bd=y$. & $rx-\frac{rxx}{s}=yy$. Then is $p:q\colon\colon 2y:r-\frac{2rx}{s}$. (by ye former rule) Therefore I draw $gd:ge${illeg} $de:dg\colon\colon p:q\colon\colon 2y:r-\frac{2rx}{s}$. The point g is inclined to move in a parallel to abc & ye point e in a parallel to aik (for bg \& ie / (by supposition) moves parallel to ym selves ye $\begin{array}{c}\text{one}\\ \text{other}\end{array}$ upon $\begin{array}{c}abc\\ aik\end{array}$) Therefore I draw $gf\parallel abc$ & $ef\parallel aik$. & through \ye intersection/ $\left(f\right)$ I draw hdf touching ye crooked line at d . Soe yt $gd:de\colon\colon db:bh\colon\colon q:p\colon\colon r-\frac{2rx}{s}:2y$.

Hence may bee pronounced those theorems in pag 47 Fol 47

[30] Example ye 2d. If $ac=x$. $bc=y$. (wch move about ye centers a & b as in ye Hyperbola or Ellipsis by a thred) And ye equation bee $-a+x+y=0$. yn is $p+q=0$. or $p=-q$. therefore I make $cd:cb\colon\colon p:q\colon\colon 1:-1$. (note yt I draw cd & cB ye one forward ye other {backwd} because p & q have contrary signes) ye points d & B are inclined to move ye one in a perpendicular to acd ye other to bBc (for ye circle they {illeg} move in circles whose centers are a & b ) therefore I draw $de\perp acd$ & $Be\perp bBc$ & the tangent ce through ye point e .

## {illeg} Resolution

[33] Find yt point of ye perpendicular \to ye crooked line/ wch is in least motion, let yt bee ye center of a circle wch passing through ye given point shall bee of equall crookednesse wth ye line at yt point. This point of least motion may bee found divers ways, as First. {illeg} From any two {illeg} points in ye perpendicular \to ye crooked line/ draw 2 \parallel/ lines in such proportion as ye perpendicular moves over ym: through their ends draw another line wch shall intersect ye perpendicular in ye point required.

Example. Suppose $ab=x$. $bc=y$. $x\perp y$. {illeg} $ax-\frac{axx}{b}=yy$. $be=a=\frac{a}{2}-\frac{ax}{b}$ . $p=$ motion of b from a {illeg} $y=\frac{-ap}{b}=$ motion of e from b. & $\frac{bp-ap}{b}=$ motion of {c}{e} from a $bk=cn=\frac{yy}{v}=\frac{2bx-2xx}{b-2x}$. $nm=y$. $nd=v=\frac{ab-2ax}{2b}=\frac{yy}{x}-\frac{a}{2}$. $cd=ke$

As if $ab=x$. $bc=y$. $ce\perp ck=$ ye tangent of ye crooked line. {illeg} $be=cd\parallel abef$ . & as ye motion of b from a to ye motion of e from a so kb to ef . Then, drawing dfg through ye points d {illeg} & f , cg in ye radius of a circle as crooked as line acl at c .

Example. Suppose $ax-\frac{axx}{b}-yy=0$. yn is $be=v=\frac{a}{2}-\frac{ax}{b}$. $kb=\frac{yy}{v}=\frac{2byy}{ab-2ax}$ . And $cd=ke=\frac{aabb-4aabx+4aaxx+4bbyy}{2abb-4abx}$. $bk=p=$ motion \velocity/ of b from a , $bc=y=q=$ velocity of y 's increase $r=$ velocity of v 's increase. $2bv-ba+2ax=0$ therefore $2br+2ap=0$, or (since $p=\frac{2byy}{ab-2ax}$ ) tis $r=\frac{2yy}{-b+2x}$. & $ef=\frac{2ayy+2byy}{ab-2ax}$ \{$=r+bk=$ {illeg} }/ Lastly $cd-ef:ce\colon\colon cd:cg$ (or $v-r:y\colon\colon v+p:ck$. if { $ce\perp$ {illeg} }) yt is $\frac{-4byy+4abx-abb-4axx}{-2bb+4bx}:y\colon\colon \frac{aabb-4aabx+4aaxx+4bbyy}{2abb-4abx}:ch$. $\frac{aaby+4b{y}^{3}-4a{y}^{3}}{aab}=ch=\frac{4b{y}^{3}-4a{y}^{3}+aaby}{aab}$ . & $bh=\frac{4{y}^{3}}{aa}-\frac{4{y}^{3}}{ab}$ .

Hence may bee pronounced those theorems in Fol 49.

<62v>

Addition connects affirmative qua numbers into an affirmation sume, & negative ones into a negative
one. as $\begin{array}{l}\begin{array}{c}+1352\\ +7460\end{array}\right\}\\ +8812\end{array}$ $\begin{array}{l}\begin{array}{c}-137905\\ -\phantom{0}68432\end{array}\right\}\\ -206337\end{array}$

Substractions takes ye greater \lesse/ number from ye lesse \greater/, the difference having ye same signe prefixed wch ye greater quantity \number/ {hats} as $\begin{array}{l}\begin{array}{ccccccc}+& 6& 3& 5& 7& 9& 2\\ -& 1& 4& 7& 0& 3& 1\end{array}\right\}\\ \begin{array}{ccccccc}+& 4& 8& 3& 7& 6& \phantom{0}\end{array}\end{array}$. $\begin{array}{r}+44503\\ -26794\end{array}$ $\begin{array}{l}\begin{array}{c}-26791\\ \hfill +\phantom{0.}4503\end{array}\right\}\\ -22288\end{array}$.

Multiplicacon adds one factor soe often {illeg} to it selfe as there are units in ye other, & if ye signes of ye factors bee ye same ye product is affirmative, if divers tis negative. As to multiply $+735$ by $+47$ doe thus $735in7$ / $\begin{array}{l}\begin{array}{r}\phantom{+}\phantom{0}5145=735in7\phantom{0}\\ \underset{_}{\phantom{+}29400=735in40}\end{array}\right\}\\ \begin{array}{r}+34545=735in47\end{array}\end{array}$\ Or thus $\begin{array}{rr}& 735\\ & 47\\ & \phantom{0}5145\\ & 29400\\ +& 34545\end{array}$. Or thus $\begin{array}{cc}\begin{array}{rr}& 735\\ & \phantom{0}5145\\ & 29400\\ +& 34545\end{array}& \begin{array}{l}×\phantom{0}\\ 7\phantom{0}\\ 40\\ \phantom{0}\end{array}\end{array}$. Thus to multiply $-3241$ by /$-3175$ the operation will\ bee $\begin{array}{cc}\begin{array}{rr}& 3241\\ & 16205\\ & 22687\phantom{0}\\ & 3241\phantom{00}\\ +& 567175\end{array}& \begin{array}{l}\phantom{1}\\ 5\phantom{6}\\ 7\phantom{0}\\ 1\phantom{0}\\ \phantom{0}\end{array}\end{array}$ Alsoe 465 multiplyed by $-32$ /will produce\ $\begin{array}{cc}\begin{array}{rr}& \phantom{00}465\\ & \phantom{00}930\\ & 1395\phantom{0}\\ -& 14880\end{array}& \begin{array}{l}\phantom{0}\\ 2\phantom{0}\\ 3\\ \phantom{0}\end{array}\end{array}$

Division takes ye number wch signifies how often ye divisor $\left\{\begin{array}{l}\text{is contained in &}\\ \text{may bee substracted from}\end{array}\right\}$ ye divisor, \ye sign of/ wch number or Quote {illeg} is affirmative if ye dividend & divisor have not divers signes, but negative{illeg} if they have. For if $x=\frac{a}{b}$. then $bx=a$, Or $a-bx=0$. Suppose 34545 to be divided by 47 . First get a {illeg} Table of ye Divisor drawn into ye 9 first units {illeg} as defg . {illeg} $\begin{array}{ccc}d& & f\\ & \hfill \begin{array}{|cccccccccccccccccc|}\hline 1& & 2& & 3& & 4& & 5& & 6& & 7& & 8& & 9& \\ -47& .& -94& .& -141& .& -188& .& -235& .& -282& .& -329& .& -376& .& -423& .\\ \hline\end{array}\hfill & \\ e& & g\\ & \begin{array}{cc}\begin{array}{rr}& 345\phantom{.}4\phantom{.}5\phantom{.}\\ & 016.4\phantom{.}5\phantom{.}\\ & \phantom{0}02\phantom{.}3.5\phantom{.}\\ & \phantom{00}0\phantom{.}0\phantom{.}0.\end{array}& \begin{array}{ll}+& 735\\ -& 000\\ & 735\end{array}\end{array}\hfill \end{array}$ cut at ye bottome $\left(eg\right)$ close to the figures. Then looke wch of those 9 quantitys are most like ye dividend. As in this case ye 7th 329 is therefore substract it from ye {illeg} dividend 34545 , & there will remaine $16.45$ , & then set downe its caracteristick 7 in ye quote. I make a prick twixt those figures $\left(16\right)$ wch have or might have beene altered & those $\left(45\right)$ wch could not bee altered by the subtraction, & the places of ye pricks will will {sic} skew the places of ye figures in ye quotient. Againe I substract 141 from $16.4$ &c: & set 3 in ye quote &c.

If 19489012 was to be divided by {illeg} 732 $\begin{array}{ccc}& \hfill \begin{array}{|cccccccccccccccccc|}\hline 1& & 2& & 3& & 4& & 5& & 6& & 7& & 8& & 9& \\ 732& .& 1464& .& 2196& .& 2928& .& 3660& .& 4392& .& 5124& .& 5856& .& 6588\\ \hline\end{array}\hfill & \\ \end{array}$ $\begin{array}{l}\begin{array}{cc}\begin{array}{rr}+3.& +1948\phantom{.}9\phantom{.}012\\ -2.& -147.0\phantom{.}988\\ -9.& -000\phantom{.}6.988\end{array}& \begin{array}{rr}+& 30000\phantom{,}00355\\ -& 02009,55000\\ -& 27990,45355\end{array}\end{array}\\ \begin{array}{rrr}-5.& -0\phantom{.}400& .0\hfill \\ -5.& -\phantom{.}034& \phantom{.}0.0\hfill \\ +3.& +\phantom{.}02& \phantom{.}6\phantom{.}0.0\hfill \\ +5.& +\phantom{.}0& \phantom{.}4\phantom{.}0\phantom{.}4.0\phantom{.}\hfill \\ +5.& \phantom{.}\phantom{0}+& \phantom{.}0\phantom{.}3\phantom{.}8\phantom{.}0.\hfill \\ \hfill \phantom{+3.}& \phantom{\phantom{\rule{0.8em}{0ex}}+1948\phantom{.}9\phantom{.}012}& \phantom{.}\hfill \end{array}\end{array}$

<64r>

The resolution of ye affected Equation ${x}^{3}+pxx+qx+r=0$. Or ${x}^{3}+10{x}^{2}-7x=44$ First having found two or 3 of ye first figures of ye desired roote viz $\begin{array}{cc}\begin{array}{cc}& 2\end{array}& \begin{array}{cc}& \underset{_}{2}\end{array}\end{array}$ (wch may bee done either by rationall of Logarithmicall tryalls as Me Oughtred hath thought, or \Geometrically/ by Geometricall descriptions of lines, or by an instrument consisting of 4 or 5 or more lines of numbers made to slide by one another wch may be oblong but better circular.) this knowne pte of ye root I call g, ye other unknowne pte I call y then is $g+y=x$. Then prosecute ye Reduction /Resolution\ after this manner (making $x+pinx=a$. $a+qinx=b$. $b+rinx=c$ &c.) $\begin{array}{ll}\hfill ×\hfill & 12=x+p\\ x=2,& 24=a\end{array}$. {illeg} $\begin{array}{ll}a+q=17& ×\\ \hfill b=34& 2=x\end{array}$. $r-b=10=h$. by supposing $x=2$. Againge supposing $x=\begin{array}{cc}\begin{array}{cc}& 2\end{array}& \begin{array}{cc}& \underset{_}{2}\end{array}\end{array}$ Then $\begin{array}{rr}x+p=12,2& \\ 244& 2\hfill \\ 244\phantom{0}& 2\hfill \\ 26,84& =a\hfill \end{array}$. $\begin{array}{rr}a+q=19,84& \\ 3968& 2\hfill \\ 3968\phantom{0}& 2\hfill \\ 43,648& =b\hfill \end{array}$. $r-b=0,352=k$. $h-k=9,648$ . That is ye $\left\{\begin{array}{l}\text{latter}\phantom{\rule{0.5em}{0ex}}r-b\phantom{\rule{0.5em}{0ex}}\text{substracted}\\ \text{difference twixt}\end{array}\right\$ $\begin{array}{l}\text{from the former}\phantom{\rule{0.5em}{0ex}}r-b\phantom{\rule{0.5em}{0ex}}\text{there remaines}\\ \text{this &}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{former valor of}\phantom{\rule{0.5em}{0ex}}r-b\phantom{\rule{0.5em}{0ex}}\text{is}\end{array}}9,648$. & ye difference twixt this & ye former valor of x is 0,2 . Therefore make $9,648:0,2\colon\colon 0,352:y$. Then is $y=\frac{0,0704}{9,648}=0,00728$ &c. the first figure of wch being added to ye last valor of x makes $2,207=x$ . Then wth this valor of x presecuting ye operacon as before tis $\begin{array}{rr}x+p=12,207& \\ 24414& \hfill \\ 24414\phantom{0}& 7\hfill \end{array}$ $\begin{array}{rr}x+p=12,207& \\ 65449& 7\hfill \\ 244140\phantom{0}& 02\hfill \\ 244140\phantom{000}& 2\hfill \\ 26,940849\hfill & =a\hfill \end{array}$ $\begin{array}{rr}a+q=19,94084& \\ 13958588& 7\hfill \\ 39881680\phantom{0}& 02\hfill \\ 3988168\phantom{000}& 2\hfill \\ \hfill 44,00043388& =b\hfill \end{array}$. $r-b=-0,00943388$ . wch valor of $r-b$ substracted from ye precedent valor of $r-b$ ye {diff} is $+0,36143388$ . Also ye {diff} twixt{illeg} tis & ye precedent valor of x is 0,007 . Therefore I make {$36143388:0,007\colon\colon -0,0094388:y$}. That is $y=$ {illeg}

<67v>

[34]

## Of the construction of Problems.

[35] If ye equation to be resolved bee $\mathrm{y}\mathrm{y}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\mathrm{a}\mathrm{y}-bb=0$. Or yy−ay+bb=0 in wch ye roote of ye last terme (viz b) is knowne, they may bee resolved {illeg} conveniently resolved by D. Cartes his rules. Otherwise ye rootes of yt ter{illeg}|me| must bee first extracted as in this yy−py+q=0. Where I take $ln=\frac{1}{2}\mathrm{q}.lg=\frac{\mathrm{q}-1}{2}\mathrm{q}.gs=\frac{\mathrm{q}+1}{2}\mathrm{q}$. & soe describing ye circle smf erect lm⊥ln & from m ye point of intersection draw mr∥ln. ye rootes of ye Equation shall bee mq & mr. ln being ye radius & n the center of the circle

Or

Or it may bee done thus. Let{illeg} the Equation bee $\mathrm{y}\mathrm{y}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\mathrm{p}\mathrm{y}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\mathrm{q}=0$. {illeg}Then in the indefinite line af take $ab=\frac{\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\mathrm{p}}{2}$. erect ye perpendicular {$bc=\frac{pp}{8\mathrm{c}}\frac{\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\mathrm{q}}{2\mathrm{c}}-\frac{1}{2}\mathrm{c}$. $\left(bc=\frac{pp\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}4\mathrm{q}-4cc}{8\mathrm{c}}\right)$} & wch ye Radius cd erect the perpendicular db=c. And from ye point & towards b draw $dc=\frac{pp\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}4\mathrm{q}+4cc}{8\mathrm{c}}\left(=\frac{pp}{8\mathrm{c}}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\frac{q}{2\mathrm{c}}+\frac{\mathrm{c}}{2}\right)$ wth wch radius describe ye circle edf & ae, af shall {illeg}bee ye rootes of ye Equation. When note that any quantity may be taken for c, soe yt if it may bee ye val{illeg}or of dc may bee noe fraction, & that db & dc bee as {illeg}|equall| little differing as may bee. Soe yt ye operacon {illeg}may thereby be made convenient, & {illeg} to yt purpose the difference twixt db & dc must bee as little as may bee,|(|that is twixt $pp\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}4\mathrm{q}$ & 4cc.|)| soe yt ye circle intersect not (ef) {illeg} over obliquely nor ye circle be over greate.

[36] As if I had this Equation yy+6y−9=0 Or yy=−6y+9. Then must I make $dc=\frac{9}{\mathrm{c}}+\frac{1}{2}\mathrm{c}$. Then if I make c=6 it will bee $\mathrm{d}=\frac{3}{2}+3=\frac{9}{2}=4\frac{1}{2}$. Therefore I take $ab=-\frac{\mathrm{p}}{2}=bd=$ $ab=-\frac{1}{2}\mathrm{p}=-3$. $bd=\mathrm{c}=6$. $dc=4\frac{1}{2}=\frac{3c}{4}$. And soe describing ye circle efc, I have one affirmative roote af, another negative ae. Or had I taken any other convenient valor for c as 1, or 3. or 4 the li{illeg}ne ae & af would still have bene ye same.

Had I this equation yy−8=0. or yy=8. Then is $dc=\frac{4}{\mathrm{c}}+\frac{1}{2}\mathrm{c}$. soe yt Or makeing c=2; tis: dc=3. Soe yt since p is wanting ye p{illeg} I take ab=0. ad=c=2 dc=3. & describing a circle ye rootes will bee ea, af.

Note yt if dc is negative or not greater then $\frac{1}{2}ad$ ye circle cannot intersect ye line eaf & therefore ye rootes of ye equation are imm̄agina{illeg}|{rie}|.

< insertion from f 67v >

[37] Or they may bee construed by drawing streight lines onely{.} thus. Let ye Equation be $2\mathrm{y}\mathrm{y}=2\mathrm{a}\mathrm{y}+\mathrm{b}$. or $\mathrm{y}=\mathrm{a}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{aa+\mathrm{b}}$ First I {make} {illeg} {the} {illeg} ab equall divide aa+b into square numbers (as {for} of ym as may bee) (It may ever bee divided (though not) into (ye fewest) squares by taking the greatest {square} out of aa+b & ye greatest out of ye remainder &c) as if in numbers ye Equation were yy=2y+4 Or $\mathrm{y}=1\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{5}$. I take then square 4 out of 5 & there rests 1 wch is also a square. Then I draw ab $\sqrt{4}$. & $bc=\sqrt{1}=1$. & make ab⊥bc. soe is $ac=\sqrt{5}$. to wch I add ad=1. & soe is $dc=\mathrm{y}=1+\sqrt{5}$.

Were ye Equation yy=−4y+34. Or $\mathrm{y}=-2\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{38}$. Then is 38−36=2. 2−1=1. & 38=36+1+1 wch are square numbers. Therefore I make $ad=\sqrt{36}=6$. $ad\perp de=\sqrt{1}=1$. & draw $ae\perp ec=\sqrt{1}-1$. & draw $ac=\sqrt{38}$. from wch take ab=2, & there rests $bc=-2+\sqrt{38}=\mathrm{y}$.

Were ye Equation {$6\mathrm{y}\mathrm{y}=15\mathrm{y}-\mathrm{2}$} {illeg} $\mathrm{y}=\frac{15}{6}$ $\mathrm{y}=6+\sqrt{\frac{15}{7}}$. Or $\mathrm{y}=6+\frac{\sqrt{15}}{\sqrt{7}}$. Find $\sqrt{15}$ & $\sqrt{7}$ before, &c:

< text from f 67v resumes > <68r>

[38] If ye Probleme be sollid it may bee readily resolved by {illeg} the intersection of ye Parabola & circle. \as D: Cartes hath shewed/ If it bee of 5 or 6 dimensions it may bee resolved by ye intersection of ye line b y3−byy−cdy+bcd+dxy=0. Or y3−byy+bcd+dxy=0 & ye circle \when pp={illeg} 4q. & q & v affirmative./ as D: C: hath explained. Or it might beee done by ye intersection of a circle & one of these lines, viz y3+byy−hx=0 when ye equation is reduced to such a forme yt pp=4q. Or this y3+byy+gy−hx=0. Or this y3+gy−hx={illeg}|{0}|, s being affirmative & {illeg} p=0. Or this y3+d−fyx=0 when p=0, & q & v affirmative. &c.

But since all Equa{tions}

[39] But all Equations in Generall may bee resolved by ye line a2x=y3, after this manner. First (making a=1{.}|)| describe ye line x=y3 uppon a plate. {illeg} (as cadce. Then in wch ab=x. bc=y). Then suppose ye Equation to bee resolved bee y9*+my7+ny6+py5+qy4+ry3+syy+tq+v=0. (in wch ye letters m, n, p &c: signifie ye [40]termes knowne quantitys of each terme affected wth its signe + or −). I describe another line cdce, whose nature (making ab=x, bc=y) is the exprest $\begin{array}{ccccccccc}{\mathrm{x}}^{3}& +& \mathrm{m}\mathrm{y}\mathrm{x}\mathrm{x}& +& \mathrm{p}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& +& \mathrm{s}\mathrm{y}\mathrm{y}& \phantom{=0}\\ \phantom{{\mathrm{x}}^{3}}& +& \mathrm{n}\phantom{\mathrm{y}\mathrm{x}\mathrm{x}}& +& \mathrm{q}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}& +& \mathrm{t}\mathrm{y}\phantom{\mathrm{y}}& =0\\ \phantom{{\mathrm{x}}^{3}}& \phantom{+}& \phantom{\mathrm{n}\mathrm{y}\mathrm{x}\mathrm{x}}& +& \mathrm{r}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& +& \mathrm{v}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{y}}& \phantom{=0}\end{array}$ & letting & let fall perpendicula{{illeg}|r|} from{e} every point where these two lines intersect as, df eg, they shall bee ye rootes of ye propounded eqation.

In like manner was ye Equation to bee resolved y10*+my8+my7+py6+qy5+ry4+sy3+ty2+ry/+w=0\ the nature of ye line cdce woluld bee $\begin{array}{ccccccccccccc}{\mathrm{x}}^{4}& +& \mathrm{m}\mathrm{y}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{p}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{s}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& \phantom{+}& \phantom{\mathrm{w}\mathrm{y}\mathrm{y}}& \phantom{=0}\\ \phantom{{\mathrm{x}}^{4}}& +& \mathrm{n}\phantom{\mathrm{y}}& {\mathrm{x}}^{3}& +& \mathrm{q}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{t}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}& +& \mathrm{w}\mathrm{y}\mathrm{y}& =0\\ \phantom{{\mathrm{x}}^{4}}& \phantom{+}& \phantom{\mathrm{n}\mathrm{y}}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{r}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{v}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& \phantom{+}& \phantom{\mathrm{w}\mathrm{y}\mathrm{y}}& \phantom{=0}\end{array}$. Or else it might bee $\begin{array}{cccccccccc}\mathrm{y}{\mathrm{x}}^{3}& +& \mathrm{m}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{q}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& +& \mathrm{t}\mathrm{y}\mathrm{y}& \phantom{=0}\\ \phantom{\mathrm{y}{\mathrm{x}}^{3}}& +& \mathrm{n}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{r}\mathrm{y}\phantom{\mathrm{x}}& \mathrm{x}& +& \mathrm{v}\mathrm{y}\phantom{\mathrm{y}}& =0\\ \phantom{\mathrm{y}{\mathrm{x}}^{3}}& +& \mathrm{p}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{s}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& \phantom{+}& \mathrm{w}\phantom{\mathrm{y}\mathrm{y}}& \phantom{=0}\end{array}$. Or had I this Equation y10+{illeg}|{l}|y9+my8+ny7+py6+qy5+ry3|4|+sy3{illeg}+ty2+vy\+w/=0. The nature of ye line cdce would bee, $\begin{array}{ccccccccccccc}{\mathrm{x}}^{4}& +& \mathrm{l}\mathrm{y}\mathrm{y}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{p}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{s}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& \phantom{+}& \phantom{\mathrm{w}{\mathrm{y}}^{2}}& \phantom{=0}\\ \phantom{{\mathrm{x}}^{4}}& +& \mathrm{m}\mathrm{y}\phantom{\mathrm{y}}& {\mathrm{x}}^{3}& +& \mathrm{q}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{t}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}& +& \mathrm{w}{\mathrm{y}}^{2}& =0\\ \phantom{{\mathrm{x}}^{4}}& +& \mathrm{n}\phantom{\mathrm{y}\mathrm{y}}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{r}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{v}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& \phantom{+}& \phantom{\mathrm{w}{\mathrm{y}}^{2}}& \phantom{=0}\end{array}$. Or, $\begin{array}{cccccccccccc}\phantom{+}& \mathrm{y}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{m}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{q}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& +& \mathrm{t}\mathrm{y}\mathrm{y}& \phantom{=0}\\ +& \mathrm{l}& {\mathrm{x}}^{3}& +& \mathrm{n}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{r}\mathrm{y}\phantom{\mathrm{x}}& \mathrm{x}& +& \mathrm{v}\mathrm{y}\phantom{\mathrm{y}}& =0\\ \phantom{+}& \phantom{\mathrm{l}}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{p}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{s}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& +& \mathrm{w}\phantom{\mathrm{y}\mathrm{y}}& \phantom{=0}\end{array}$. Or {illeg} it might bee, $\begin{array}{cccccccccccc}\phantom{+}& \mathrm{y}\mathrm{y}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{n}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{r}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}}& +& \mathrm{v}\mathrm{y}\mathrm{y}& \phantom{=0}\\ +& \mathrm{l}\mathrm{y}& {\mathrm{x}}^{3}& +& \mathrm{p}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{s}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}& +& \mathrm{w}\mathrm{y}\phantom{\mathrm{y}}& =0\\ +& \mathrm{m}\phantom{\mathrm{y}}& \phantom{{\mathrm{x}}^{3}}& +& \mathrm{q}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{t}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& \phantom{+}& \phantom{\mathrm{w}\mathrm{y}\mathrm{y}}& \phantom{=0}\end{array}$. If ye resolved Equation have fewer dimensions yt is if some of ye \ultimate/ termes {illeg} as, w, v, t &c: (or intermediate termes as m, n &c \be blotted out/: Or if ye Equation have more yn 10 dimensions{{illeg}|th|}e nature of ye lines \cdce/ to bee described may \be/ known by ye same manner observing ye order of ye progression

Tis evident alsoe yt there are 3 divers lines {any of w in} by wch {illeg} \any/ Probl: may bee resolved \unless some of them {chanch} to be ye same/, the easiest whereof is to bee chosen. It appeares also how Equations of 2 & 3 dimensions may be resolved by drawing streight lines; of 4, 5, & 6 by describing some conick section; of 7, 8, 9, by describing a line of 3 dimensions; of 10, 11, 12, by a line of 4 dimensi{illeg}|ons|, {illeg} &c: but yet y is never above 2 dimensions & consequently all these lines may bee described by ye rule & compasses.

{illeg}Some Those {illeg} Equations of more then 9 dimensions may bee (though seldome soe

Had I this line y4=x. described on a plate & this Equation to bee resolved viz: y13+ly12*+ny10+py9+qy8+ry7+sy6+ty5+vy4+wy3+ayy+by+c=0. It might bee resolved by describing ye line whose nature is $\begin{array}{lllllllllll}+& \mathrm{y}{\mathrm{x}}^{3}& +& \mathrm{n}\mathrm{y}\mathrm{y}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{r}{\mathrm{y}}^{3}& \phantom{\mathrm{x}}& +& \mathrm{w}{\mathrm{y}}^{3}& \phantom{=0}\\ +& \mathrm{l}\phantom{{\mathrm{x}}^{3}}& +& \mathrm{p}\mathrm{y}\phantom{\mathrm{y}}& \mathrm{x}\mathrm{x}& +& \mathrm{s}\mathrm{y}\mathrm{y}& \mathrm{x}& +& \mathrm{a}\mathrm{y}\mathrm{y}& =0\\ \phantom{+}& \phantom{\mathrm{l}{\mathrm{x}}^{3}}& +& \mathrm{q}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{t}\mathrm{y}\phantom{\mathrm{y}}& \phantom{\mathrm{x}}& +& \mathrm{b}\mathrm{y}\phantom{\mathrm{y}}& \phantom{=0}\\ \phantom{+}& \phantom{\mathrm{l}{\mathrm{x}}^{3}}& \phantom{+}& \phantom{\mathrm{q}\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}\mathrm{x}}& +& \mathrm{v}\phantom{\mathrm{y}\mathrm{y}}& \phantom{\mathrm{x}}& +& \mathrm{c}\phantom{\mathrm{y}\mathrm{y}}& \phantom{=0}\end{array}$. A line of ye 2d sort. Whereas by ye preceding rule was required yt a line of ye 3d sort should have [41] beene described. And here observe yt taking ye square number wch is next greater yn ye number of ye resolvend equation termes \dimensions/ of the resolvend equation,|.| That Equation may bee resolved by lines, ye number of whose dimensions is not greater than the roote of ye {illeg} square number. And the rectangles of th{illeg}|os|e numbers wch signifie ye dimensions how many dimensions th{illeg}|{e}| lin{illeg}|es| have, {must \even/} either equall or greater may always bee greater or equall but never lesse yn ye number of dimensions of ye resolvend Equations. For ye number of points in wch two lines may intersect can never bee [42] greater yn ye rectangle of ye numbers of theire dimensions. And they always intersect in soe many points, excepting those wch are im̄aginarie onely. Soe that all Equations $\begin{array}{ccccc}\begin{array}{l}\text{which}\phantom{\rule{0.3em}{0ex}}\text{have}\\ \text{noe more}\phantom{\rule{0.3em}{0ex}}\text{than}\end{array}& \left\{\begin{array}{c}1\\ 2\\ 4\\ 6\\ 9\\ 12\\ 16\end{array}\right\}& \begin{array}{c}\text{dimensions, may}\phantom{\rule{0.3em}{0ex}}\text{always}\phantom{\rule{0.3em}{0ex}}\text{be}\\ \text{resolved by the}\\ \text{intersection of}\end{array}& \left\{\begin{array}{l}\text{2 streight lines}\\ \text{a streight line & a c}\text{o}\text{nick section.}\\ \text{two conic}\text{k}\phantom{\rule{0.3em}{0ex}}\text{sections.}\\ \text{a conick section}\phantom{\rule{0.3em}{0ex}}\text{of a line of 3 dimensions.}\\ \text{two lines of 3 dimensions.}\\ \text{a line of 3. & another of 4 dimensions.}\\ \text{two lines of 4 dimension; or by one of 1 & another of 6 dim: &c:}\end{array}\right\}& \begin{array}{l}\text{but not by any simpler}\\ \text{lines. (From this conside}\text{r}\\ \text{may Problems be di}\text{st}\text{in-}\end{array}\end{array}$ guished into sorts.) {St} will often bee very intricate to resolve Equations \of many dimensions/ by the simplest line by wch they may be resolved & also for ye most ꝑt will regaine a des{illeg}|cr|iption of two {illeg} lines for every probleme. And then {if maybe often} {illeg} end to use two lines whereas {illeg} compound ye other more simple {illeg}|&| {illeg} As perhaps an Equation of 16 dimension may bee more speedily resolved by two lines {illeg} of 6 dimensions then by two lines {illeg} 4 dimensions.

<68v>

[43] But it will not bee {amisse} to shew \more/ particularly how these resolutions may bee performed. A{illeg}|nd| that firs by ye parabola

T Suppose therefore I had ye parabola x=yy exactly described & would resolve \{illeg} plaine probleme/ the Equation yy+ky+{illeg}l=0. I take ag={illeg}l. gf=k. fh=1=lateri recto Parab: & so draw ye line gh & from ye intersection points d, e, draw db, ec perpendicular to ye axis gc. wch shall bee ye rootes of ye Equation wch are affirmative when they fall on ye contrary side to fh, but negative if {illeg}|on| ye same, as in this case.

[44] But were I to resolve a sollid problem the Equation being of 4 dimensions, I take away ye 2d terme|,| {illeg} makeing it of this forme y4*+lyy+my+n=0. Then take $ae=\frac{1}{2}$ $ep=\frac{1}{2}\mathrm{l}$. $pq=\frac{1}{2}\mathrm{m}$. Then perpendicular to ap draw af=aq. Also draw fk∥ap, & from ye point of intersection k draw{illeg} kh={illeg}n. lastly draw kr⊥ap, & wth the radius wr upon the center q describe ye circle tsm. {illeg} (or, wch is ye same, take $ar=\frac{1-2\mathrm{l}+ll+mm-4\mathrm{n}}{4}$. & soe erecting ye perpendicular rw, wth ye Radius rw describe ye circle tsm) & from ye points where it intersects ye Parabola let fall perpendiculars to ye axis, (tv, nm) they shall bee the rootes of ye Equation ye affirmative ones falling on ye contrary side to pq. when m is affirmative.

If n=0, that is if y3*+ly{illeg}+m=0, then must the circle bee described wth ye Radius aq; for then is wr=fa=aq.

[45] If I would resolve ye cubick Equation y3+ky2+ly+m=0 (wch multiplyed by y−k=0 produceth $\begin{array}{llllllll}{\mathrm{y}}^{4}*& +& \mathrm{l}\mathrm{y}\mathrm{y}& +& \mathrm{m}\mathrm{y}& -& \mathrm{k}\mathrm{m}& =0\\ \phantom{{\mathrm{y}}^{4}*}& -& \mathrm{k}\mathrm{k}\phantom{\mathrm{y}\mathrm{y}}& -& \mathrm{k}\mathrm{l}\phantom{\mathrm{y}}& \phantom{-}& \phantom{\mathrm{k}\mathrm{m}}& \phantom{=0}\end{array}$) I make $ae=\frac{1}{2}$. $ep=\frac{{l}-kk}{2}$. $pq=\frac{{m}-kl}{2}$. fk∥ae⊥af={illeg}aq. kd={illeg}km. And wth ye radius cg upon ye center q describe ye circle wf. Or else doe thus (since k is one of ye rootes of ye Equation $\begin{array}{llllllllll}{\mathrm{y}}^{4}*& +& \mathrm{l}& \mathrm{y}\mathrm{y}& +& \mathrm{m}& \mathrm{y}& -& \mathrm{m}\mathrm{k}& =0\\ \phantom{{\mathrm{y}}^{4}*}& -& \mathrm{k}\mathrm{k}& \phantom{\mathrm{y}\mathrm{y}}& -& \mathrm{l}\mathrm{k}& \phantom{\mathrm{y}}& \phantom{-}& \phantom{\mathrm{m}\mathrm{k}}& \phantom{=0}\end{array}$) make k=ab+ar & draw bw∥ae (or make ar=kk, & wr{illeg}⊥ap) & describe a circle wth ye radius wq. Then letting fall perpendiculars from ye intersection points, they (being ye rootes of ye Equation $\begin{array}{llllllllll}{\mathrm{y}}^{4}*& +& \mathrm{l}& \mathrm{y}\mathrm{y}& +& \mathrm{m}& \mathrm{y}& -& \mathrm{m}\mathrm{k}& =0\\ \phantom{{\mathrm{y}}^{4}*}& -& \mathrm{k}\mathrm{k}& \phantom{\mathrm{y}\mathrm{y}}& -& \mathrm{l}\mathrm{k}& \phantom{\mathrm{y}}& \phantom{-}& \phantom{\mathrm{m}\mathrm{k}}& \phantom{=0}\end{array}$) shall all, except wr=k, bee ye rootes of ye Equation y3+kyy+ly+m.

This operation will bee much shortened when ye 2d terme is wanting for {yt} since k=0. it will bee $ae=\frac{1}{2}$. $ep=\frac{1}{2}\mathrm{l}$. $pq=\frac{1}{2}\mathrm{m}$ & aq ye radius of ye circle.

[46] And if ye last terme {vanish} that is if I would resolve this equation yy+ky+l=0. by ye intersection of a circle & parabola. I must take $ae=\frac{1}{2}$. $e\pi =\frac{1}{2}\mathrm{l}$. $\pi q=\frac{kk}{2}$. $\pi q=\frac{kl}{2}$ & soe wth ye radius aq upon ye center q describe a circle, & ye perpendiculars from ye intersection points to ye axis (a{{illeg}}, tv) are ye rootes excepting one wch is equall to k.

<69r>

[47] If I had ye crooked \line/ described \fig 1st/[48] whose nature is x=y3, & would resolve ye Equation y3*+lyy+m=0. (calling ad=x, dg=y; Or a=-x. ce=-y) I take ab=m. bd=l. df=1. & df⊥bd & draw bf infinitely both ways. From ye intersection points (as e) letting fall perpendiculars, they shall bee ye {illeg} rootes of ye Equation y3*+lyy+m. as ce wch in this case is negative because on that side on wch y is negative.

Would I resolve this equation yy+ky+l=0. (wch multiplyed by y-k produceth $\begin{array}{llllll}{\mathrm{y}}^{3}*& +& \mathrm{l}& \mathrm{y}& -& kl=0\\ \phantom{{\mathrm{y}}^{3}*}& -& \mathrm{kk}& \phantom{\mathrm{y}}& \phantom{-}& \phantom{kl=0}\end{array}$) I take ab=kl, (fig 2d)[49] bδ=l, δd=kk. df=1, & soe through ye points b & f draw the streight line bfλ |(|Or wch is ye same take ab=kl. k=ah⊥ab. & draw hλ∥ab untill it intersect ye crooked line in λ (i. e. untill hλ=k3 & soe through the points λ & b draw λbfe|)|. Then from ye intersection points to ye axis letting fall perpendiculars they (being ye rootes of ye Equation $\begin{array}{llllll}{\mathrm{y}}^{3}*& +& \mathrm{l}& \mathrm{y}& -& lk=0\\ \phantom{{\mathrm{y}}^{3}*}& -& \mathrm{kk}& \phantom{\mathrm{y}}& \phantom{-}& \phantom{lk=0}\end{array}$.) shall all, except βλ=k, be ye rootes{illeg} of ye Equation yy+ky+l=0.

[50] Would I resolve the Equation {y4}+ z4{illeg}+az3+bzz+cz+d=0. It may bee done by a circle thus. M{illeg}ultiply it by this Equation zz−az+aa−b=0, & it will produce $\begin{array}{rrrrrrrrrr}{\mathrm{z}}^{6}**& +& \mathrm{c}{\mathrm{z}}^{3}& +& \mathrm{d}\mathrm{z}\mathrm{z}& -& ad\mathrm{z}& +& aad& =0\\ \phantom{{\mathrm{z}}^{6}**}& -& 2ab\phantom{{\mathrm{z}}^{3}}& -& ac\phantom{\mathrm{z}\mathrm{z}}& +& aac\phantom{\mathrm{z}}& -& bd& \phantom{=0}\\ \phantom{{\mathrm{z}}^{6}**}& +& {\mathrm{a}}^{3}\phantom{{\mathrm{z}}^{3}}& +& aab\phantom{\mathrm{z}\mathrm{z}}& -& bc\phantom{\mathrm{z}}& \phantom{-}& \phantom{bd}& \phantom{=0}\\ \phantom{{\mathrm{z}}^{6}**}& \phantom{+}& \phantom{{\mathrm{a}}^{3}{\mathrm{z}}^{3}}& -& bb\phantom{\mathrm{z}\mathrm{z}}& \phantom{-}& \phantom{bc\mathrm{z}}& \phantom{-}& \phantom{bd}& \phantom{=0}\end{array}$, Of this forme z6**+mz3+nzz+pz+q=0. In wch (n) ought to be affirmative, & if it bee not, yn augment or diminish ye rootes of ye Equation z4+az3+bzz+cz+d=0. & then repeate ye operacon again untill there bee an Equation of this forme z6**+mz3+nzz+pz+q=0 in wch n is affirmmative. Then (dividing this equation by $\sqrt{4:\mathrm{n}}$ it is ${\mathrm{z}}^{6}**+\frac{\mathrm{m}}{\sqrt{\mathrm{n}\sqrt{\mathrm{n}}}}{\mathrm{z}}^{3}+\mathrm{z}\mathrm{z}+\frac{\mathrm{p}}{\mathrm{n}\sqrt{4:\mathrm{n}}}\mathrm{z}+\frac{\mathrm{q}}{\mathrm{n}\sqrt{\mathrm{n}}}=0$ therefore) take $ab=\frac{\mathrm{m}}{2\sqrt{\mathrm{n}\sqrt{\mathrm{n}}}}$. $bc=\frac{\mathrm{p}}{2\mathrm{n}\sqrt{4:\mathrm{n}}}$. {illeg} & wth ye radius $cd=\sqrt{\frac{mmn+pp-4nq}{4nn\sqrt{\mathrm{n}}}}$, describe ye circle dk & ye perpendiculars (as dh ck) multiplyed by $\sqrt{4:\mathrm{n}}$ shall bee ye rootes of ye Equation.

<71r>

## Theoremata Optica.

2do A puncto v versus A cape VG ad VA {illeg} ut est R ad I et error radij refracti DR a loco imaginis {seu} in axe AV, seu distantia punctorum B et F erit {illeg} AC, BC, BG, VD AC, FC $\frac{AC,FC,FG,{VD}^{\mathrm{q}}}{2{AV}^{\mathrm{q}},{CV}^{\mathrm{q}}}$, sive {illeg} $\frac{AC,FC,FG,VH}{{AV}^{\mathrm{q}},CV}$ quamproxime. Sed regula prior in f{illeg} radios est.

3. Ubi punctum infinite distat ita ut radius incidens parallelus sit axi, pro AC scrito AV, et pro FG scripto $\frac{\mathrm{R},VA}{\mathrm{G}}$ (nam hæ \jam/ sunt æquipollentia) error BF {illeg}|{fist}| et $\frac{\frac{\mathrm{R}}{\mathrm{G}}FC,{MD}^{\mathrm{q}}}{2{CV}^{\mathrm{q}}}$, vel {illeg} $\frac{\frac{\mathrm{R}}{\mathrm{G}}FC,VH}{CV}$, vel $\frac{RR}{II-IR}VH$.

4. Si radius non refringi{illeg}|tu|r sed reflectitur a superficie sphærica VD, eadem regula obtinet si modo ponatur S.R∷ 1.-1. et perinde capiatur VG ad contrarias partes VA {illeg} {illeg} f{illeg} ipsi VA æqualis. Erit enim{illeg}|{adh}|uc error $FB=\frac{AC,FC,FG}{{AV}^{\mathrm{q}},CV}VH$. vel AVq×CV. ACF×FG∷VH.FG.

<71v>

## Hujus autem Theorematis inventio totis est.

${AD}^{\mathrm{q}}={AV}^{\mathrm{q}}+2AC,VH$. ${DF}^{\mathrm{q}}={VF}^{\mathrm{q}}-2CF,VH$. Et extractis radicibus, $AD=AV+\frac{AC}{AV}VH-\frac{{AC}^{\mathrm{q}}}{2{AV}^{\text{cub}}}{VH}^{\mathrm{q}}$ &c $DF=VF-\frac{CF}{VF}VH-\frac{{CF}^{\mathrm{q}}}{2{VF}^{\text{cub}}}{VH}^{\mathrm{q}}$ &c fluxio \ipsius/ $AD=\frac{AC}{AV}\text{fl}:VH-\frac{{AC}^{\mathrm{q}},VH}{{AV}^{\text{cub}}}\text{f}\text{l}VH$ &c |De|fluxio ipsius $DF=-\frac{CF}{VF}\text{fl}VH-\frac{{CF}^{\mathrm{q}},VH}{{VF}^{\text{cub}}}\text{fl}VH$ &c $\frac{\mathrm{R}}{\mathrm{I}}\text{fl}:AD+\text{defl}:DF=-\frac{\mathrm{R},{AC}^{\mathrm{q}}}{\mathrm{I},{AV}^{\text{cub}}}-{CF}^{\mathrm{q}}{VF}^{\text{cub}}$ in VH, fl VH &c {illeg} Qu{illeg} si nihil esset {illeg}|ra|dij omnes accuratè refrigerentur ad focu{illeg} F. Tunc enim AD et DF fluerent in data ratione, jux{illeg} ea quæ Cartesins in Optica probavit: {A}{'} Sed qui{a} nihil n{illeg}|on| est, error{illeg} \fluxio {obliquitatis}/ superficiei \{illeg}/ erit \VD/ ut illud $\frac{\mathrm{R}}{\mathrm{I}}\text{fl}AD+\text{defl}:DF$. Et ut error ille sive defluxio a legitima figura \obliquitate/ ita \{illeg}/ error angular radij {illeg} refracti. Jam vero est $\frac{\mathrm{R},{AC}^{}}{\mathrm{I},AV}=\frac{CF}{VF}$ Ergo {fluxio} error{illeg} angularis radij refracti \est/ ut $\frac{AC}{{AV}^{\mathrm{q}}}-\frac{CF}{{VF}^{\mathrm{q}}}$ in $\frac{CF,VH}{VF}$, fl VH. Vel etiam ut $\frac{{I},CF}{\mathrm{R},AV,VF}+\frac{CF}{{VF}^{\mathrm{q}}}$ in $\frac{CF,VH}{VF}$, fl VH seu ut {illeg}, $VF+\frac{\mathrm{R}}{\mathrm{I}}AV$ in $\frac{{CF}^{\mathrm{q}},VH}{{VF}^{\mathrm{c}},AV}$ fl VH seu $\frac{FG,{CF}^{\mathrm{q}},VH}{AV,{VF}^{\text{c}\text{ub}}}$, fl {illeg} posito fl VH=1 & $VF+\frac{\mathrm{R}}{\mathrm{I}}AV=FG$. Dat{{illeg}ur} autem ratio $\frac{CF}{VF}$ ad $\frac{AC}{AV}$ ergo substituto posteriore fiet error ille ut $\frac{FG,CF,AC,VH}{{AV}^{\mathrm{q}},{VF}^{\mathrm{q}}}$. Duc in VFq et error in axe FB erit ut $\frac{FG,CF,AC,VH}{{AV}^{\mathrm{q}}}$ qua{illeg}do circuli radius determinatur. Divide per radium circuli et fiet $\frac{FG,CF,AC,VH}{CV,{AV}^{\mathrm{q}}}$ ut error BF in omni casu. Dato igitur errore illo in uno casu datur in omni. At in eo casu ubi est radius incidens axi pall|r|allelus datur error {e}odem cum quantitate {illeg} $\frac{FG,CF,AC}{CV,{AV}^{\mathrm{q}}}VH$ ergo semper idem est cum hac quantitate.

<71a(r)>

me.Ce∷d.y. mO.CO∷e.y. $\frac{\mathrm{d},Ce}{me}=\frac{\mathrm{e},CO}{mO}$. $\frac{\mathrm{d},Cq}{mq}=\frac{\mathrm{e},Cp}{mp}$. DO.DR∷Dq.Dp d, Cq, Cp−Cp|m|=e, Cp, Cq-Cm. d, Ce, CO−Cm=e, CO, mC+Ce d, Cp $\frac{\mathrm{d},qCp-\mathrm{e},qCp}{\mathrm{d},Cq-\mathrm{e},Cp}=Cm=\frac{\mathrm{d},eCO-\mathrm{e},eCO}{\mathrm{d},Ce+\mathrm{e},CO}$. d−e=f. |$\frac{{Dn}^{\mathrm{q}}}{DR}={\mathrm{D}}^{\mathrm{q}}$.| $\frac{\mathrm{f},qCp}{\mathrm{d},Cq-\mathrm{e},Cp}=\frac{\mathrm{f},eCO}{\mathrm{d},Ce+\mathrm{e},CO}$. {Se CD = 0 sit CO.CR∷Cq.Cp.} |=Cm|

$\begin{array}{ll}-df,Ce,Cp,Cq-ef,CO,Cp,Cq=0.& \mathrm{d},\mathrm{f},Ce,Cp,pO=\mathrm{e},\mathrm{f},CO,Cp,qe=\mathrm{e},\mathrm{f},{Dn}^{\mathrm{q}},qe.\\ \frac{\hfill +df,Ce,}{7\right)20⁤\frac{1}{4}\left(Dq\right)}\genfrac{}{}{0}{}{CO,Cq-ef,CO,Cp,Ce\phantom{=0.}}{\phantom{\frac{0}{0}}}& PO=\frac{\mathrm{e},,{Dn}^{\mathrm{q}},qe}{\mathrm{d},Ce,Cq}=\mathrm{R}.\sqrt{}\frac{RR+4{Dn}^{\mathrm{q}}}{4}:±\frac{1}{2}\mathrm{R}=\genfrac{}{}{0}{}{CO.}{CP.}\end{array}$ $\frac{9,20⁤\frac{1}{4},9⁤\frac{6⁤\frac{1}{4}}{7}.}{14,7,2⁤\frac{6⁤\frac{1}{4}}{7}}=\frac{89}{14}=6⁤\frac{1}{3}=PO=\mathrm{R}.\phantom{\rule{0.5em}{0ex}}\sqrt{}30⁤\frac{1}{4}±3⁤\frac{1}{6}=\genfrac{}{}{0}{}{CO}{CP}=5⁤\frac{1}{2}±3⁤\frac{1}{6}=\genfrac{}{}{0}{}{8⁤\frac{2}{3}.}{2⁤\frac{1}{3}.}\phantom{\rule{0.5em}{0ex}}CO=8⁤\frac{2}{3}.\phantom{\rule{0.5em}{0ex}}CP=2.$ $Cm=\frac{5,60⁤\frac{2}{5}}{98+78}=\frac{303⁤\frac{1}{5}}{176}\phantom{\rule{0.5em}{0ex}}\begin{array}{ll}910& 455×{12}^{\text{inch}}\\ 528& 264\end{array}=\begin{array}{cc}455& 20,6\\ {xx}_{15}& \end{array}$ |$Cm=20⁤\frac{2}{3}\text{inches}$|.

<71a(v)>

$\mathrm{Z}=\mathrm{N}+\frac{\frac{far-fz}{\mathrm{a}-\mathrm{r}+\mathrm{f}}\mathrm{x}-aNx-ffx}{\mathrm{a}-\mathrm{r}+\mathrm{f}×\mathrm{a}-\mathrm{r}-\mathrm{f.}}\phantom{\rule{5em}{0ex}}\frac{\mathrm{e}}{\mathrm{d}}\mathrm{a}=\mathrm{f}.\phantom{\rule{3em}{0ex}}AV=\mathrm{t}=\mathrm{a}-\mathrm{r}$ $\mathrm{Z}=\mathrm{N}+\frac{\frac{far-fz}{\mathrm{t}+\mathrm{f}}\mathrm{x}-aN-ff}{tt-ff}\mathrm{x}=\mathrm{N}+\frac{far-fft-2fz-faN-taN}{\overline{\mathrm{t}×\mathrm{f}}×\overline{\mathrm{t}+\mathrm{f}}×\overline{\mathrm{t}-\mathrm{f}}}.\phantom{\rule{0.5em}{0ex}}\genfrac{}{}{0}{}{\mathrm{N}=CD.}{\mathrm{N}=\frac{fr}{\mathrm{t}-\mathrm{f}}.}$ $\mathrm{t}.\mathrm{f}\colon\colon \genfrac{}{}{0}{}{BG.CF.}{VD.CD}\phantom{\rule{0.5em}{0ex}}\mathrm{t}.\mathrm{a}\colon\colon \mathrm{BG}.\frac{\mathrm{d}}{\mathrm{e}}CF$ Be.BC\BF/∷Bn.BQ.BR. B{illeg}|{e.}|BF∷Bn.{illeg}|BR| $\frac{FBn}{Be}=\frac{BS,FC}{FS}.$ {illeg} t2ft−\2/ffar−tf{z}±tfar−ttff {illeg}

[Editorial Note 2]

$zz=\frac{2zz±rr}{\frac{Id-ee}{ee}}+\frac{Idrr-2Idax}{eeaa}.$

<72r>

## Probl.

Habita Lente convex plano-convexa, invenire tum convexitatem, tum refractionem vitri.

{Sit} Lens RS, ejus superficies plana RTS, convexa RVS \axis KF/{.} Lentis superficie plana solem respiciente, observentur imaginum sola{ntum} ubi \a/ radijs {tum} trajectis tum reflexis congr{illeg} convergentibus \in charta obversa/ distinctissimè pict{æ}rum loci duo F et F|G|; F locus imaginis trajectæ {illeg}G locus reflexæ: et mensurentur quam accuratissime distantiæ VF, TG, ut et crassities vitri TV. Dein fac [VF{+}\2/GT.GT∷TF\2TV./.KT, et erit 2KV semidiameter circuli RVS. Et sinus incidentiæ ex aere in vitrum erit ad sinum refracionis ut KT ad GT, vel ut \{illeg}/ +KF ad VF. vel] ut T|V|F+2TV ad VF−2GT, {ita} sinus incidentiæ ex aere in vitrum ad sinum refractionis, {ita} KT ad GT et erit 2KV radius circuli RVS

## Probl.

Habita Lente quavis convexo-convexa, invenire ve{illeg} etiam convexa-concava cujus concavitas {illeg} convexit{illeg} major |multo minor|, invenire tum refractionem vitri, tum convexit{illeg} Lentis.

Sit Lens RS, superficies magis convexa RVS, minus convexa vel concava RTS, axis KF, vertices V ac T. Lentis hujus superficie minus convexa vel concava RTS solem directe respiciente, observentur quam accuratissim{e} sol{{illeg}|em|} imaginis in charta obversa d{illeg}|is|tinctissimè pictæ ta{tam} trajectæ locus F quam reflexæ locus G, et mensure{illeg} distantiæ VF, TG, et crassities vitri TV. Dein alter{illeg} {illeg}|Le|ntis superficie RVS solem respiciente observetur qu{æ} locus imaginis reflexæ H et mensuretur distantia V{H} qu{illeg}|æ| est imaginis illius a vitro. Biseca TV in X. Et fac $\frac{1}{HX}-\frac{2}{FX}=\mathrm{A}$. Et $\frac{1}{GX}+\mathrm{A}=\mathrm{B}$. Et $\frac{1}{\mathrm{B}}=\mathrm{gX}$. Et {L}{illeg} plano-convexa \ex consimili vitr{illeg}|{o}| confect{illeg}|{æ}|/ cujus vertices si|u|nt T, V, et convexitas \versus F {sita}/ æqu{illeg} summæ convex{|i|tatum} RTS, RVS et versus F sita{illeg} est \in fig. 1 vel differentiæ convexitatis et concavitatis in fig. 2/, pro{jectis} solis imaginem refractam ad locum \priorem/ F, reflexam \vero/ ad locum quamproximè. Unde \[51]/ si fiat (juxta Problema prius) VF−\2/g{T} gT∷VF\+2TV/.KT erit sinus incidentiæ ex aere in vitrum ad sinum refractionis ut KT ad gT, vel ut KF ad VF. {Sec} ista ratio{.} S|ec| ad R, et erit {illeg} $\frac{4\mathrm{S}-4\mathrm{R}}{\mathrm{R}×\mathrm{A}}$ radius circuli {RVS} sit ista {C}{illeg}{}{illeg}{}{illeg}{D}{illeg}{}

<72v>

et {poni} debet {illeg} $\frac{1}{2\mathrm{gV}}+\frac{1}{\mathrm{C}}=\mathrm{D}$.

Exempli gratia. In Telescopij cujusdam vitro objectivo observabam VF=13ped.11digit. {TH=} $VH={6}^{\text{ped.}}{9⁤\frac{13}{16}}^{\text{digit.}}$ $TG={2}^{\text{ped}\text{.}}{4⁤\frac{13}{16}}^{\text{dig.}}$. Et $TV={\frac{2}{9}}^{\text{dig.}}$. Ergo Seu {VF=} VF=167dig. TG= VH={illeg}1,8125dig. TG=28,8125dig. TV=0,2222 &c \dig./. Adeo $XF={167}_{,}^{\mathrm{d}}1111$. $XH={8}_{,}^{\mathrm{d}}9236$. $XG={28}_{,}^{\mathrm{d}}9236$. Unde prodit A= 0,00023{illeg}|{84}|{illeg} dig. B=0,0348{illeg}|{127}|dig. gx=28,72{illeg}dig|56dig.| KT=34,53{2}. dT={28,536 {illeg}T} VF+2TV=167,{illeg}|{444}||4|&c VF−2gT= {109,}{illeg} 109,{illeg}{60}|771|{illeg}. Ergo 167,444. 109,7{60}|71|∷I.R. vel in minoribus numeris 29.19∷I.R aut magis accuratè $\begin{array}{}61.40\\ 90.59\end{array}$I.R. $\frac{4\mathrm{I}-2\mathrm{R}}{RA}=$$\frac{}{\mathrm{R}}{1243}^{\text{dig}}=\frac{\mathrm{R}}{\mathrm{g}}{937⁤\frac{3}{4}}^{\text{ped}}=\text{radio}\phantom{\rule{.5em}{0ex}}\text{cir}\text{culi}$ {illeg}=17161{illeg}dig=1430ped. {Cir}|Un|nde alterius RVS semidiameter erat quasi 7ped 4dig. At hæc ita se habebant in vitro objectivo Telescopij Dris Babington.

In altero Telescopio quad erat \in archivis/ Academiæ, measura{ve} distantiam imaginis trajectæ a vitro objectivo VF=14ped3{gis} $+\frac{9}{10}\text{dig}$.

<73r>

## Telescopij novi delineatio

Vitrum objectivum CD parallelos radios refringat versus O. Imago O per refractionem concavæ superficiei GEH transferatur ad P, et inde per reflexionem superficiei specularis ad Q, et inde per refractionem secundam superficiei GEH ad R ubi a speculo obliquo T detorquetur per vitrum oculare perexiguum \V/ ad oculum.

Sit \imaginis/ translatio \angularis/ ab O ad P et a P ad S tanta quanta {debendis} \corrigendis/ vitri objectivi refractionibus erroneis \ab inæquali refrangibilitate ortis / sufficit et erit refract \angularis/ translatio imaginis a Q ad R tanta quanta est a P ad S, et punctum S invenietur faciendo ut sit BE.EO∷EO.ES.

Sit X centrum circuli specularis JFK et Y centrum circuli refrigentis concavi GEH. Et quoniam imaginis angulares translationes PX, XQ æquales sunt, ut et PS, QR; erunt etiam translationes SX, RX æquales: adeo si fiat {illeg} ES.SX∷EQ|R|.RQ{illeg} \RX,/ vel ES+SX.SX∷EX.Q|R|X, \ex dato puncto X/ habebitur ultimæ imaginis locus {illeg} R, e cujus regione consistet oculus.

Sit insuper Y centrum superficiei concavæ GEH, et quoniam est QP {illeg} EP.EQ∷PX.QX, et I×P{illeg} I×OE.R×OY∷EP.YP. et I×ER.R×YR∷QE.QY: inde derivabitur hæc conclusio. Hac ER.EX∷{illeg}OR.P, et $\frac{2\mathrm{I}}{\mathrm{R}}EO.OX\colon\colon \mathrm{P}.XY$ et habebitur circuli GEH centrus Y. Ubi nota quod usurpo $\frac{\mathrm{I}}{\mathrm{R}}$ pro ratione sinus incidentiæ ex aere in vitrum ad sinum refractionis. Et suppono in super vitri crassitiem EF ad instar nihili esse. ff Fac $\frac{2\mathrm{I}×EO}{\mathrm{R}×EX}=\mathrm{a}.\frac{EO}{ER}=\mathrm{b}.\mathrm{b}×RX=\mathrm{c}$. et {illeg} $\frac{OX-\mathrm{c}}{\mathrm{a}-\mathrm{b}+1}=XY$. Et habebitur circuli GEH centrum Y. Ubi nota quod usurpo $\frac{\mathrm{I}}{\mathrm{R}}$ pro ratione sinus incidentiæ ex aere in vitrum ad sinum refractionis: et suppono insuper vitri crassitiem EF ad instar nihili esse.

<77r>

Ghetaldus in his Promotus Archimededs compu{illeg}tes ye w{h}eights of ye following equall bodys to bee in ye proportions following.

$\begin{array}{cccccccccccccc}\text{Glasse}& & \text{Oyle.}& \text{Wax.}& \text{Wine.}& \text{Water.}& \text{Honey.}& \text{Tinn.}& \text{Iron.}& \text{Brasse.}& \text{Silver.}& \text{Lead.}& \text{Quicksilver.}& \text{Gold}\\ & & 1.& 1⁤\frac{5}{121}.& 1⁤\frac{4}{55}.& 1⁤\frac{1}{11}.& 1⁤\frac{32}{55}.& 8⁤\frac{4}{55}.& 8⁤\frac{8}{11}.& 9⁤\frac{9}{11}.& 11⁤\frac{3}{11}.& 12⁤\frac{6}{11}.& 14⁤\frac{62}{77}.& 20⁤\frac{8}{11}.\\ & \text{Or}& 4235.& 4410.& 4543.& 4620.& 6699.& 34188.& 36960.& 41580.& 4774.& 53130.& 62700.& 87780.\\ 1562& \text{Or}& 55.& \frac{630}{11}.& 59.& 60.& 87.& 444.& 480.& 540.& 620.& 690.& \frac{5700}{7}& 1140.\\ & \text{Or}& & & & & & 111.& 120.& 135.& 155& 172⁤\frac{1}{2}& \frac{1425}{7}.& 285.\\ & \text{Or}& & & & & & 777.& 840& 945& 1085& 1207⁤\frac{1}{2}& 1425.& 1995.\\ & \text{Or}& 4⁤\frac{47}{57}& 5⁤\frac{5}{209}& 5⁤\frac{10}{57}.& 5⁤\frac{5}{19}.& 7⁤\frac{12}{19}.& 38⁤\frac{18}{19}.& 42⁤\frac{2}{19}.& 47⁤\frac{7}{19}.& 54⁤\frac{22}{57}& 60⁤\frac{10}{19}& 71⁤\frac{3}{7}& 100.\\ 2⁤\frac{3}{5}.& \text{Or}& \frac{11}{12}.& \frac{21}{22}.& \frac{59}{60}.& 1.& \frac{29}{20}.& \frac{37}{5}.& 8.& 9.& \frac{31}{3}.& \frac{23}{2}.& \frac{95}{7}& 19\end{array}$

If Sphæres bee made of ye following metalls \each of/ whose diameters are each one foote their weights will bee as followeth. {{illeg}} |Note yt 1li|=12′. 1′=24″. 1″=24‴

|Or| $\begin{array}{cccccc}\text{Tinn.}& \text{Iron}\text{e}\text{.}& \text{Brasse.}& \text{Silver.}& \text{Lead}& \text{Gold}\\ {304}^{lib}.& {328}^{li}.7\prime .18″.19⁤\frac{17}{37}‴& {369}^{li}.8\prime .18″.3‴\frac{33}{37}& {424}^{li}.6\prime .1″.7‴\frac{5}{37}& {472}^{li}.5\prime .4″.12‴\frac{36}{37}& {780}^{li}.6\prime .11″.16‴\frac{8}{37}& \\ {304}^{lib}.& {\frac{12160}{37}}^{li}.& {\frac{13680}{37}}^{li}.& & & & \end{array}$

A Sphære of tinn whose diameter is six inches weighs Thirty and Eight pounds. The following line being {6} of Ghetaldus his inches, wch is half ye Roman foot by Villalpandus account from ye Far{nesian urn}.     $\begin{array}{c}\text{|-----------------------------------------------------------------------|}\end{array}$Soe yt ye weight of a circumscribed cilinder is {illeg} 57 \lib/ \& of such a cilinder of water $5⁤\frac{}{57}$ lb $7⁤\frac{26}{37}$ lb/. And of a circumscribed cube (viz whose side is 6 inches)|is| $\frac{836}{}$lib $\frac{627}{}$lib {Or more} exactly ${\frac{798}{11}}^{}=\stackrel{li}{72⁤\frac{6}{11}}$ \li/. Or more exactly ${\frac{25764}{355}}^{lib}=\stackrel{lib}{72,\frac{204}{355}}=\stackrel{lib}{72,57465}$ {-.} \that is $72⁤\frac{4}{7}$./ Or more exactly $\frac{{228}^{lib}}{3,14159265359}$ And such a cube of water $9⁤\frac{4}{5}$lb or more exactly 9,80752lib. & a foot cube of water $78⁤\frac{6}{13}$lb or more exactly 78,46016lb.

The {P}es Regius Gallorum, the Rhinlandick foot, the old Roman foot, & the English foot are as $12,11⁤\frac{1}{2},10⁤\frac{5}{6}&11$ {illeg} Ga{illeg} But by ye Farnesian Urn \printed by Villalpand/ the French royal foot is to {illeg} old {illeg} foot as 12 to 11{illeg} \or perhaps 11{illeg} to 10{illeg}/. The urn conteined \a longius of water weighing/ 10 Roman pounds \of 12 {illeg}/ {illeg} Gaffendus by weighing found it contein 7 french pounds of 16 ounces. \of water/ Eight such vessels make a Roman foot cube called a Quadrantal or Amphora Romana weighing 56 French pounds. A cubic French Foot of water by Mersennus trial weighed 74lb, by the vulgar estimat{illeg} {$70⁤\frac{1}{2}$} or 72, suppose 72 & two meane proportional{s} between 7{2} & {56} will be as $24⁤\frac{1}{3}$ to 23 or 12{$\frac{1}{2}$} to {11}{$\frac{1}{2}$}. Suppose 74{lb} two {meane} will be |{as}| 11 to 10 or 12 to $10⁤\frac{10}{11}$ & this is {ye} proportion of ye french foot to ye Roman. {illeg} Royal French foot {illeg} {illeg} 12. $11⁤\frac{1}{2}$. {11} {illeg} $10⁤\frac{9}{10}$. {illeg} $\frac{7}{8}$. $12⁤\frac{13}{5}$ {illeg} ({illeg}$10⁤\frac{}{4}$ The Roman {illeg} to ye {horary} foot as 8 to 9. {illeg} 9 {illeg} $12⁤\frac{17}{52}$ {illeg} $10⁤\frac{71}{81}$.

<77v>

## Some Problems of Gravity & levity &c

[52] Prob 1. To find ye proportion of ye weights of two equall bodys ye one being sollid ye other liquid.        Resp.      If ye Sollid body A bee heavier yn ye liquid B weigh it in ye aire & in ye liquid Body B; & ye difference of those two weights is ye weight of soe much water as is equall to yt sollid body |& so much {wyer} as {deppend} in ye {water}|. But if it bee lighter yn ye liquid body B, hang a heavier body C {at} to it, yt will sinke it; & weigh ym first ye body A being in ye air & C in ye water{;} 2dly liquid body, 2dly both A & C being in ye liquid body; & ye difference of these{s} weights is ye weight of so much water as is equall to A (& also to ye soe much thred wyer or hayre as was weighed both{illeg} in ye water & air.)

[53] Note yt ye weight of soe much water as is equall to yt ꝑte of ye wyer (to wch A & C are fastened) |wch| was weighed both in ye air & water, bee subducted from ye whole weight of ye water {illeg}|&| ye remaining weight shall bee ye weight of ye water requi{illeg}|re|d.

__________________________________________________________________________________________________________________________

Or Thus, Make ye scoale B as light as may bee, yet soe yt it will sinke it selfe in ye liquid body CC, & \sinke/ ye body A also, if A chance to bee lighter yn CC. Suppose yt f ye weight of scoale B in ye water, & g in ye aire; & yt e is ye weight of B+A in ye water, & d in ye aire: yn is d−g+f−e ye weight of soe much water as is equa{illeg}|ll| to A. |Note yt ye water must reach {'}at both weighings to ye same point {illeg} of ye wyer & yt ye wyer bee as small as may bee.|

{P} Cor: Hence ye proportion of all weights of all bodys sollid or liquid or both may bee gathered. & wee may hereby make hence deduce tables of ye weights of equall bodys, & of ye quantitys of bodys equally heavy.

[54] Prob 2. Two{illeg} bodys {illeg} \D & E/ given to find ye proportion of their quantity.        Weigh ym in ye scale B, let their weight \& ye weight of ye scale/ in ye air bee h & k, in ye water, m & n yn is D{illeg} E∷h−g+f−m:k−g+f−n. For their weight in air is h−g, & k−g; in the water m−f & n−f. &c.

[55] Prob 3. A compound body cd being \given/ to find ye weights \& proportions,/ of its two compounding parts c & d.      Answer.    a & c, b & d, are of ye same matter. That ye weights of ye 5 bodys a, b, cd, c, d, in ye aire are ef e, f, n, m, n−m; & in ye water g, h, q, p, q−p. Then is $\mathrm{e}:\mathrm{g}\colon\colon \mathrm{m}:\mathrm{p}=\frac{gm}{\mathrm{e}}$. & $\mathrm{f}:\mathrm{h}\colon\colon \mathrm{n}-\mathrm{m}:\mathrm{q}-\mathrm{p}=\frac{hn-hm}{\mathrm{f}}$. The{re}fore $\mathrm{q}=\frac{gm}{\mathrm{e}}+\frac{hn-hm}{\mathrm{f}}$. Or $\frac{feq-hen}{gf-eh}=\mathrm{m}=to$ ye weight of C in ye aire. Also $\frac{gfn-feq}{gf-eh}=\mathrm{n}-\mathrm{m}$. & $\frac{gfq-hgn}{gf-eh}=\mathrm{p}$. & $\frac{hgn-qeh}{gf-eh}=\mathrm{q}-\mathrm{p}$. & h|f|eq{+} \/h|f|gq−f|h|en+f|h|gn:−gf|h|n+gh|f|n+f|h|eq−h|f|eq∷c:d. (Prob 1). feq−hen+hgn−fgq:gfn−efq+ehq{illeg}|−gh|n∷c:d. And, gfn+ehq−ehn−gfq:efq+ghn−ehn−gfq∷c+d:{c}.

[56] Prob{:} 4. A body f compounded of 3 severall sorts of matter d, e, f−d−e, begin given{;|:|} wth ye proportion of ye weight of ye two bodys d & e as {1} to r. To find ye weight of ye body d.          Resp. Suppose yt ye bodys a & d, b & e, c & f−d−e are of ye same matter; & {illeg} & yt ye weight of ye bodys a, b, c, f, d, e, f−d−e, {are} in ye aire are g, h, k, p, x, rx, p−x−rx{;} & in ye water l,   m, n, v, s, t, v−s−t.    Then is g:l∷x:s. & h:m∷rx:t. & k:n∷p−x−rx:v−s−t. Therefore $\mathrm{v}-\frac{lx}{\mathrm{g}}-\frac{mrx}{\mathrm{h}}=\mathrm{v}-\mathrm{s}-\mathrm{t}=\frac{np-nx-nrx}{\mathrm{k}}$. And ghkv−ghnp=hklx+gkmrx−ghnx−ghnrx. Or $\frac{ghkv-ghnp}{hkl+ghmr-ghn-ghnr}=\mathrm{x}$.

If ye weights g=h=k, (as may bee either by experience or cal{illeg}|cu|lacon (see coroll: Prob 1) Then is $\frac{gv-np}{\mathrm{l}+mr-\mathrm{n}-nr}=\mathrm{x}$.    Now because gold is usually allayed by mixing wth it brasse & silver of each {illeg} |an| equall weight; suppose yt a & d are brasse, b & e silver, c & f−d−e gold, & yt x=rx, or r=1. Then is $\frac{gv-np}{\mathrm{l}+\mathrm{m}-2\mathrm{n}}=\mathrm{x}$ ye weight of ye brasse or siver {sic} in ye masse f, & $\frac{pl+pm-2gv}{\mathrm{l}+\mathrm{m}-2\mathrm{n}}=\mathrm{p}-2\mathrm{x}$ ye weight of ye gold in it.

<78r>

## Descriptio cujusdam {illeg}|ge|neris curvarum {terty}\secundi/ ordinis.

Concipe lineas PED datum angulum PED continentes ita moveri ut una earum EP perpetuo transeat per polum P positione datum, et altera{illeg} {illeg} ED datæ longitudinis existens perpetuò tang{illeg}|{a}|t rectam AB positione datam . Age PA {illeg}|co|nstituentem angulum PAD æqualem angulo PED sit CD æqualis AP et quodvis punctum C in recta ED. datum describet curvam secundi ordinis.

Age CB constituentem angu{illeg}|lu|m CD{illeg}|BD| æqualem angulo PED, et ad AD demitte normalem C{illeg}|F.| et dictis {illeg}. DC=c. CE=b. AB=x BC=y. et posito 1. E∷BC.BF, erit $\begin{array}{llllllllllll}{\mathrm{y}}^{3}& +& 2\mathrm{b}& & \mathrm{y}\mathrm{y}& +& \phantom{2}\mathrm{b}\mathrm{b}& & \mathrm{y}& +& \mathrm{c}\mathrm{b}\mathrm{b}& =\\ & +& \phantom{2}\mathrm{c}& & & +& 2\mathrm{b}\mathrm{c}& & & -& \mathrm{c}\mathrm{x}\mathrm{x}& \\ & -& 4\mathrm{e}\mathrm{}\mathrm{b}& & & +& 2\mathrm{b}\mathrm{x}& & & & \\ & -& 4\mathrm{e}\mathrm{e}\mathrm{c}& & & +& 4\mathrm{c}\mathrm{x}& & & & \\ & -& 2\mathrm{e}\mathrm{x}& & & +& \mathrm{x}\mathrm{x}& & & & \end{array}$

Et nota quod ubi angulus PED rectus est, et recta ED bisecatur in C, curva erit cissoides veterum.

<80r>

## A Method Whereby to Square lines Mechanichally.

[57] Lemma: |Prop 1.| Supposing ab=x⊥bc=y. If ye valor of y (in ye Equation expressing ye relation twixt x & y) consist of simple termes, Multiply {each} terme by x, & divide it by ye number of ye dimensions of x in that terme, & ye quote shall signify ye area acb.

Example. If ax=yy. Or $\sqrt{\mathrm{a}\mathrm{x}}=\mathrm{y}$. yn is $\frac{2}{3}\sqrt{\mathrm{a}{\mathrm{x}}^{3}}=\frac{2}{3}\mathrm{x}\sqrt{\mathrm{a}\mathrm{x}}=acb$.            Soe if $\frac{\mathrm{x}\mathrm{x}}{\mathrm{a}}=\mathrm{y}$. yn is $\frac{{\mathrm{x}}^{3}}{3\mathrm{a}}=acb$. Soe if $\mathrm{a}+\mathrm{x}+\frac{\mathrm{a}\mathrm{b}\mathrm{b}}{\mathrm{x}\mathrm{b}}+\frac{{\mathrm{x}}^{3}}{\mathrm{a}\mathrm{a}}=\mathrm{y}$. yn is $\mathrm{a}\mathrm{x}+\frac{\mathrm{x}\mathrm{x}}{2}+\frac{\mathrm{a}\mathrm{b}\mathrm{b}}{\mathrm{x}}+\frac{{\mathrm{x}}^{4}}{4\mathrm{a}\mathrm{a}\mathrm{x}}=acb$ Soe if $\frac{{\mathrm{a}}^{4}}{{\mathrm{x}}^{3}}+\sqrt{\frac{{\mathrm{a}}^{5}}{{\mathrm{x}}^{3}}}-\sqrt{\frac{{\mathrm{x}}^{3}}{\mathrm{a}}}+\frac{6{\mathrm{x}}^{6}}{{\mathrm{a}}^{6}}=\mathrm{y}$. yn is $\frac{{{-}^{\mathrm{a}}}^{4}}{2\mathrm{x}\mathrm{x}}-2\sqrt{\frac{{\mathrm{a}}^{5}}{\mathrm{x}}}-\frac{2}{5}\sqrt{\frac{{\mathrm{x}}^{5}}{\mathrm{a}}}+\frac{7\mathrm{b}{\mathrm{x}}^{7}}{{\mathrm{a}}^{6}}=acb$.

Lemma 2d |Prop 2.| If {illeg}any terme in ye valor of y bee a compound ter{illeg} Reduce it to simple ones by Division or Extraction of Rootes or by Vie{illeg}|{s}| Method of Resolving Affected Equations, as you would doe in Decimall Numbers; & yn find ye Area by Prop 2d 1st.

Example. If $\frac{\mathrm{a}\mathrm{a}}{\mathrm{b}+\mathrm{x}}=\mathrm{y}$. bee divided as in decimall fractions it produce{illeg} $\frac{\mathrm{a}\mathrm{a}}{\mathrm{b}+\mathrm{x}}=\frac{\mathrm{a}\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}\mathrm{x}\mathrm{x}}{{\mathrm{b}}^{3}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{{\mathrm{b}}^{4}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{{\mathrm{b}}^{5}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{{\mathrm{b}}^{6}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{6}}{{\mathrm{b}}^{7}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{{\mathrm{b}}^{8}}$ &c. & by ye {2}|1|{d|s|t} Proposition $\frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{a}\mathrm{a}\mathrm{x}\mathrm{x}}{2\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{4{\mathrm{b}}^{4}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{6}}{6{\mathrm{b}}^{6}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{7{\mathrm{b}}^{7}}-\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{8}}{8{\mathrm{b}}^{8}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{9}}{9{\mathrm{b}}^{9}}\text{&c}=abc$. ye Hyperbolas Area.

As if a=1=b\=ab=bc./ & x=0,1=be The Calculation is as followeth,

$\begin{array}{r}\begin{array}{c}0,10033,53477,31075,58063,57265,52060,03894,52633,62869,14595,91358,63\\ \end{array}\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}=0,10000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00000\phantom{,}00\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}=33,33333,33333,33333,33333,33333,33333,33333,33333,33333,33333\phantom{,}33\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}=20000,00000,00000,00000,00000,00000,00000,00000,00000,00000\phantom{,}00\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{7{\mathrm{b}}^{7}}=142,85714,28571,42857,14285,71128,57142,85714,28571,42857\phantom{,}14\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{9}}{9{\mathrm{b}}^{9}}=1,11111,11111,11111,11111,11111,11111,11111,11111,11111\phantom{,}11\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{11}}{11{\mathrm{b}}^{11}}=909,09090,90909,09090,90909,09090,90909,09090,90909\phantom{,}09\phantom{11111111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{13}}{13{\mathrm{b}}^{13}}=7,69230,76923,07692,30769,23076,92307,69230,76923\phantom{,}07\phantom{11111111111111}\\ 6666,66666,66666,66666,66666,66666,66666,66666\phantom{,}66\phantom{11111111111111}\\ 58,82352,94117,64705,88235,29411,76470,58823\phantom{,}52\phantom{11111111111111}\\ 52631,57894,73684,21052,63157,89473,68421\phantom{,}05\phantom{11111111111111}\\ 476,19047,61904,76190,47619,04761,90476\phantom{,}19\phantom{11111111111111}\\ 4,34782,60869,56521,73913,04347,82608\phantom{,}69\phantom{11111111111111}\\ 4000,00000,00000,00000,00000,00000\phantom{,}00\phantom{11111111111111}\\ 37,03703,70370,37037,03703,70370\phantom{,}37\phantom{11111111111111}\\ 34482,75862,06896,55172,41379,31034482\phantom{11111111}\\ 322,58064,51612,90322,58064\phantom{,}51\phantom{11111111111111}\\ 3,03030,30303,03030,30303\phantom{,}03\phantom{11111111111111}\\ 2857,14285,71428,57142,85\phantom{11111111111111}\\ 27,02702,70270,27027,02\phantom{11111111111111}\\ 25641,02564,10256,41\phantom{11111111111111}\\ 243,90243,90243,90\phantom{11111111111111}\\ 2,32558,13953,48837,20930,23255\\ 2222,22222,22222\phantom{11111111111}\\ 21,27659,57446\phantom{11111111111}\\ 20408,16326\phantom{11111111111}\\ 196,07843\phantom{11111111111}\\ 1,88679\phantom{11111111111}\\ \\ \phantom{1111111111111}\end{array}$

$\begin{array}{r}3\phantom{,}10\\ 000502,51679,26750,72059,17744,28779,27385,30427,57503,83731,49363,62\\ 502,51666,66666,66666,66666,66666,66666,66666,66666,66666,66666,66\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{8}}{8{\mathrm{b}}^{8}}=12,50000,00000,00000,00000,00000,00000,00000,00000,00000,00\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{10}}{10{\mathrm{b}}^{10}}=10000,00000,00000,00000,00000,00000,00000,00000,00000,00\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{12}}{12{\mathrm{b}}^{12}}=83,33333,33333,33333,33333,33333,33333,33333,33333,33\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{14}}{14{\mathrm{b}}^{14}}=71428,57142,85714,28571,42857,14285,71428,57142,85\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{16}}{16{\mathrm{b}}^{16}}=625,00000,00000,00000,00000,00000,00000,00000,00\\ 5,55555,55555,55555,55555,55555,55555,55555,55\\ 5000,00000,00000,00000,00000,00000,00000,00\\ 45,45454,54545,45454,54545,45454,54545,45\\ 41666,66666,66666,66666,66666,66666,66\\ 384,61538,46153,84615,38461,53846,15\\ 3,57142,85714,28571,42357,14285,71\\ 3333,33333,33333,33333,33333,33\\ 31,25000,00000,00000,00000,00\\ 29411,76470,58823,52941,17\\ 277,77777,77777,77777,77\\ 2,63157,89473,68421,05\\ 2500,00000,00000,00\\ 23,80952,38095,23\\ 22727,27272,72\\ 217,39130,43\\ 2,08333,33\\ 2000,00\\ 19,23\\ 18\end{array}$ |The summe of these two summes is equall to {ye area} dbfc, supposing ad={0,9}. And their difference {is equall to} ye area bche, supposing ae=1,1. & ab=1=bc∥df∥he. {that is}|

$\begin{array}{rrr}bfc& =& 0,10536,05156,57826,30122,75009,80839,31279,83001,20372,98327,40795,43\\ he& =& 0,09531,01798,04324,86004,39521,23280,76509,22206,05365,30864,41990,83\end{array}$

<80v>

In the manner If a=b=1=ab=bc. & x=0,2=be. The calculatio is as followeth

$\begin{array}{r}0,20273,25540,54082,19098,90065,57732,17456,82859,95211,73124,70987,67=\text{summe}\\ \frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}=0,20000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00\phantom{11111111}\\ \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}=266,66666,66666,66666,66666,66666,66666,66666,66666,66666,66666,66\phantom{11111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}=6,40000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00\phantom{11111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{7{\mathrm{b}}^{7}}=18285,71428,57142,85714,28571,42857,14285,71428,57142,85714,28\phantom{11111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{9}}{9{\mathrm{b}}^{9}}=568,88888,88888,88888,88888,88888,88888,88888,88888,88888,88\phantom{11111111}\\ 18,61818,18181,81818,18181,81818,18181,81818,18181,81818,18\phantom{11111111}\\ 63015,38461,53846,15384,61538,46153,84615,38461,53846,15\phantom{11111111}\\ 2184,53333,33333,33333,33333,33333,33333,33333,33333,33\phantom{11111111}\\ 77,10117,64705,88235,29411,76470,58823,52941,17647,05\phantom{11111111}\\ 2,75941,05263,15789,47368,42105,26315,78947,36842,10\phantom{11111111}\\ 9986,43809,52380,95238,09523,80052,38095,23809,52\phantom{11111111}\\ *364,72208,69565,21739,13043,47826,08695,65217,39\phantom{11111111}\\ 13,42177,28000,00000,00000,00000,00000,00000,00\phantom{11111111}\\ 49710,26962,96296,29629,62962,96296,29629,62\phantom{11111111}\\ 1851,27900,68965,51724,13793,10344,82758,6206896551\phantom{}\\ 69,27366,60645,16129,03225,80645,16129,03\phantom{11111111}\\ 2,60301,04824,24242,42424,24242,42424,24\phantom{11111111}\\ 9817,06810,51428,57142,85714,28571,42\phantom{11111111}\\ 371,45663,10054,05405,40540,54054,05\phantom{11111111}\\ 14,09630,29202,05128,20512,82051,28\phantom{11111111}\\ 53634,71355,00487,80487,80487,80\phantom{11111111}\\ 2045,60302,84204,65116,27906,976744\phantom{1111}\\ 78,18749,35307,37777,77777,77\phantom{11111111}\\ 2,99441,46458,58042,55319,1489\phantom{111111}\\ 11488,77455,96186,12244,897959\phantom{1111}\\ 441,52937,52324,01568,6274\phantom{111111}\\ 16,99471,55749,83003,77358\phantom{11111}\\ 65506,90367,08435,78\phantom{11111111}\\ 2528,33663,29097,52\phantom{11111111}\\ 97,70521,22548,17\phantom{11111111}\\ 3,78007,05069,07\phantom{11111111}\\ 14640,27307,43\phantom{11111111}\\ 567,59212,53\phantom{11111111}\\ 22,02596,302\phantom{1111111}\\ 85550,117\phantom{1111111}\\ 3325,609\phantom{1111111}\\ 129,37\phantom{11111111}\\ 5,03\phantom{11111111}\end{array}$

$\begin{array}{r}0,2041,09972,60127,56477,72885,32577,65993,50886,05873,81676,01148,45=\text{sum}\text{m}\text{e}\phantom{11}\\ 0,2041,06666,66666,66666,66666,66666,66666,66666,66666,66666,66666,66=\frac{\mathrm{a}\mathrm{a}\mathrm{x}\mathrm{x}}{2\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{4{\mathrm{b}}^{4}}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{8}}{8{\mathrm{b}}^{8}}=3200,00000,00000,00000,00000,00000,00000,00000,00000,00000,00\phantom{111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{10}}{10{\mathrm{b}}^{10}}=102,40000,00000,00000,00000,00000,00000,00000,00000,00000,00\phantom{111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{12}}{12{\mathrm{b}}^{12}}=3,41333,33333,33333,33333,33333,33333,33333,33333,33333,33\phantom{111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{14}}{14{\mathrm{b}}^{14}}=11702,85714,28571,42857,14285,71428,57142,85714,28571,42\phantom{111111111}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{16}}{16{\mathrm{b}}^{16}}=409,60000,00000,00000,00000,00000,00000,00000,00000,00\phantom{111111111}\\ 14,56355,55555,55555,55555,55555,55555,55555,55555,55\phantom{111111111}\\ 52428,80000,00000,00000,00000,00000,00000,00000,00\phantom{111111111}\\ 1906,50181,81818,18181,81818,18181,81818,18181,81\phantom{111111111}\\ 69,90506,66666,66666,66666,66666,66666,66666,66\phantom{111111111}\\ 2,58111,01538,46153,84615,38461,53846,15384,61\phantom{111111111}\\ 9586,98057,14285,71428,57142,85714,28571,42\phantom{111111111}\\ 357,91394,13333,33333,33333,33333,33333,33\phantom{111111111}\\ 13,42177,28000,00000,00000,00000,00000,00\phantom{111111111}\\ 50529\phantom{,}02701,17647,05882,35294,11764,70\phantom{111111111}\\ \phantom{111111111}\\ 1908,87435,37777,77777,77777,77777,77\phantom{111111111}\\ 72,33629,13010,52631,57894,73684,2105263\phantom{1111}\\ 2,74877,90694,40000,00000,00000,00\phantom{111111111}\\ 10471,53931,21523,80952,38095,23\phantom{111111111}\\ 399,82241,01003,63636,36363,63\phantom{111111111}\\ 15,29755,30821,00869,56521,739130434\phantom{11}\\ 58640,62014,80533,33333,33\phantom{111111111}\\ 2251,79981,36852,48000,00\phantom{111111111}\\ 86,60768,51517,40307,692\phantom{11111111}\\ 3,33599,97239,78145,18\phantom{111111111}\\ 12867,42750,67728,45\phantom{111111111}\\ 496,94892,43995,03\phantom{111111111}\\ 19,21535,84101,14\phantom{111111111}\\ 74382,03255,528\phantom{11111111}\\ 2882,30376,151\phantom{11111111}\\ 111,79844,893\phantom{11111111}\\ 4,34041,03\phantom{111111111}\\ 16865,59\phantom{111111111}\\ 655,88\phantom{111111111}\\ 25,52\phantom{111111111}\\ 99\phantom{111111111}\end{array}$ |The summe of these two summes is Equall to yeArea dbfc, supposing ad=0.8. And their Difference is equall to ye area bche, supposing ae=1,2. & 1=ab=bc∥df∥he. {that is} $\begin{array}{rrr}dbfc& =& 0,22314,35513,14209,75576,60603,07701,13885,12006,88042,06974,63440,46.\\ bche& =& 0,18232,15567,93954,62621,14832,42545,81898,10234,76294,43622,61143,46.\end{array}$|

$\begin{array}{rrr}dbfc& =& 0,22314,35513,14209,75576,62950,90309,83450,33746,01085,54800,72136,12\\ bche& =& 0,18232,15567,93954,62621,17180,25154,51463,31973,89337,91448,69839,22\end{array}$

{illeg} such respect to ye superficies bcfd{,} bche {ye numbers} {illeg} {(viz as ye lines ad} {illeg} {illeg} {illeg} {illeg}

<81r>

Soe yt since $\frac{1,2×1,2}{0,8×0,9}=2.\phantom{\rule{1em}{0ex}}\frac{1,2×1,2×1,2}{0,8×0,8×0,9}=3=\frac{1,2×2}{0,8},\phantom{\rule{1em}{0ex}}\frac{1,2×1,2×1,2×1,2}{0,8×0,8×0,8×0,9×0,9}=5=\frac{2×2}{0,8}=\frac{4}{0,8}.\phantom{\rule{1em}{0ex}}2×5=10.\phantom{\rule{1em}{0ex}}10×10=100$      10×100=1000 &c. 10×1,1=11 &c. The Superficies answere|i|ng to these lines{,} 2. 3. 9. 11. &      their products (of one of ym multiplying another) may bee found.       Viz. if ab=1=bc⊥ab{illeg}

|If ye line ak is| Then, ye superficies bcgh is

$\begin{array}{rr}& \\ 2.\phantom{0}& 0,69314,71805,59945,30941,72321,21458,17656,80755,00134,36025,52539,99\\ 3.\phantom{0}& 1,09861,22886,68109,69139,52452,36922,52570,46474,90557,82274,94515,33\\ 10.& 2,30258,50929,94045,68401,79914,54684,36420,76011,01488,62877,29756,09\\ 100.& 4,60517,01859,88091,36803,59829,09368,72841,52022,02977,25754,59512,18\\ 1000.& 6,90775,52789,82137,05205,39743,64053,09262,28033,04465,88631,89268,27\\ 10000.& 9,21034,03719,76182,73607,19658,18737,45683,04044,05954,51509,19024,36\\ \text{&c—}& \\ 11.& 2,39789,52727,98370,54406,19435,77965,12929,98217,06853,93741,71747,92\end{array}$

______________________________________________________________________________________________________________________________________________________

Having already found ye areas correspondent to ye lines 1,1. 0,9. 1,2. 0,8. tis easy by ye help of those operations to ding ye areas correspondent to ye lines 1,01. 1,001. 1,0001. &c: 0,99. 0,999. 0,9999. &c: 1,02. 1,002 &c: 0,98. 0,998. 0,9998. &c. And since $7=\sqrt{\frac{100×0,98}{2}}.\phantom{\rule{1em}{0ex}}17=\frac{100×1,02}{6}.\phantom{\rule{1em}{0ex}}13=\frac{1000×1,001}{7×11}=\frac{1001}{77}$. &c. Therefore ye areas correspond{ent} to ye lines 7. 13. 17. {illeg}|&c|: are easily found, as followeth. Viz: if x=0.02. Then

$\begin{array}{rr}0,02000,26673,06849,58071,70371,83954,64639,04807,62055,62238,59310,49& =\text{summe}\\ \frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}=0,02000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00000,00& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}=26666,66666,66666,66666,66666,66666,66666,66666,66666,66666,66& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}=6,40000,00000,00000,00000,00000,00000,00000,00000,00000,00& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{7{\mathrm{b}}^{7}}=182,85714,28571,42857,14285,71428,57142,85714,28571,42& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{9}}{9{\mathrm{b}}^{9}}=5688,88888,88888,88888,88888,88888,88888,88888,88& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{11}}{11{\mathrm{b}}^{11}}=1,86181,81818,18181,81818,18181,81818,18181,81& \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{13}}{13{\mathrm{b}}^{13}}=63,01538,46153,84615,38461,53846,15384,61& \\ 2164,53333,33333,33333,33333,33333,33\\ 77101,17647,05882,35294,11764,70& \\ 27,59410,52631,57894,73684,21& \\ 998,64380,95238,09523,80& \\ 36472,20869,56521,73& \\ 13,42177,28000,00& \\ 497,10269,62& \\ 185,12,79& \\ 6,92& \end{array}$

$\begin{array}{rr}0,00020,00400,10669,86769,10081,17069,54599,73717,71328,11118,23899,70& =\text{summe}\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{2}}{2\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{4{\mathrm{b}}^{4}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{6}}{6{\mathrm{b}}^{6}}=0,00020,00400,10666,66666,66666,66666,66666,66666,66666,66666,66666,66\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{8}}{8{\mathrm{b}}^{8}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{10}}{10{\mathrm{b}}^{10}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{12}}{12{\mathrm{b}}^{12}}=3,20102,43413,33333,33333,33333,33333,33333,33333,33\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{14}}{14{\mathrm{b}}^{14}}=1,17028,57142,85714,28571,42857,14285,71\\ 40,97456,87984,35555,55555,55555,55\\ 19,06501,81818,18181,81\\ 699,05066,66666,66\\ 25811,10153,84\\ 9,58698,05\\ 357,91\\ 13\end{array}$

The sume & difference of wch two summes give ye areas bcfd, bche as before. That is

$\begin{array}{lll}\begin{array}{c}\text{If}\phantom{\rule{1em}{0ex}}\text{fd}=0,98\\ \phantom{\text{If}}\phantom{\rule{1em}{0ex}}\text{he}=0,98\end{array}& & \begin{array}{rrr}bcfd& =& -0,02020,27073,17519,44840,80453,01024,19238,78525,33383,73356,83210,19\\ bche& =& +0,01980,26272,96179,71302,60290,66885,10039,31089,90727,51120,35410,79\end{array}\end{array}$

And since $7=\sqrt{\frac{100×0,98}{2}}=\sqrt{\frac{98}{2}}=49$. & $17=\frac{100×1,02}{6}$. |Therefore|

|{If ye} line {illeg}|{c}|k is| The superficies bcgk is $\begin{array}{rr}7.& 1,94591,01490,55313,30510,53527,43443,17972,96370,84729,58186,11881,00.\\ 17.& 2,83321,33440,56216,08024,95346,17873,12653,55882,03012,58574,47867,65.\end{array}$

___________________________________________________________________________________________________________________________________________________

|{illeg}if x=0,001. The operation is as followeth{.}| $\begin{array}{r}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}&c=0,00100,0003,33333,53333,34761,90476,19047,61904,76190,47619,04761,90\\ 111,11120,20202,02020,20202,02020,20\\ 76923,07692,30769,23\\ 6666,66666,66\\ 588,23\\ 0,00100,00003,33333,53333,34761,90587,30167,82107,55133,82180,04806,22& =\text{summe}\\ 0,00000,05000,00250,00016,66667,91666,76666,67500,00071,42863,39286,26& =\text{su}\text{mm}\text{e of}++\\ \text{for}\phantom{\rule{1em}{0ex}}befd=0,00100,05003,33583,53350,01429,82254,06834,49607,55205,25043,44092,48& \text{If}\phantom{\rule{1em}{0ex}}ad=0.97\\ \phantom{\text{for}}\phantom{\rule{1em}{0ex}}bche=0,00099,95003,33083,53316,68093,98929,53501,14607,55062,99316,65519,96& \text{If}\phantom{\rule{1em}{0ex}}ad=\end{array}$ {Where} {illeg} $\frac{1000×1,001}{77}=13$. & $\frac{1000×0,999}{27}=\frac{999}{27}=37$. |Therefore| |{illeg}ak is {illeg}| $\begin{array}{cc}13& 2,64949527881836330058457441001318604805267549760307115931\\ 37& \end{array}$ {illeg}

<81v>

Which Area may bee otherwise thus found s{illeg}|upp|osing x=db=-0,0016. Viz. $\begin{array}{rr}\frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}=0,00160,00013,65333,33333,33333,33333,33333,33333,33333,33333,33333,33& .\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{5}}{5{\mathrm{b}}^{5}}=2,09715,20000,00000,00000,00000,00000,00000,00000,00\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{7}}{7{\mathrm{b}}^{7}}=38347,92228,57142,85714,28571,42857,14285,71\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{9}}{9{\mathrm{b}}^{9}}=7635,49741,51111,11111,11111,11111,11\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{11}}{11{\mathrm{b}}^{11}}=1599,28964,04014,54545,45454,54\\ 346,43074,05669,61230,77\\ 76\phantom{,}86143,36404,56\\ 17,36164,14\\ 3\phantom{,}98\\ \text{The Su}\text{mm}\text{e of}\phantom{\rule{1em}{0ex}}\frac{\mathrm{a}\mathrm{a}\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{3}}{3{\mathrm{b}}^{3}}&c=0,00160,00013,65335,43048,91681,33197,41816,99469,20181,33677,37988,14\\ \frac{\mathrm{a}\mathrm{a}\mathrm{x}\mathrm{x}}{2\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{4{\mathrm{b}}^{4}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{6}}{6{\mathrm{b}}^{6}}=0,00000,12800,01638,40279,62026,66666,66666,66666,66666,66666,66666,66\\ \\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{8}}{8{\mathrm{b}}^{8}}=53,68709,12000,00000,00000,00000,00000,00000,00\\ \frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{10}}{10{\mathrm{b}}^{10}}=10,99511,62777,60000,00000,00000,00\\ 2,34562,48059,22133,33333,33\\ 51469,71002,70913,82\\ 11529,21504,60\\ 2623,53\\ \text{The su}\text{mm}\text{e of}\phantom{\rule{1em}{0ex}}\frac{\mathrm{a}\mathrm{a}\mathrm{x}\mathrm{x}}{2\mathrm{b}\mathrm{b}}+\frac{\mathrm{a}\mathrm{a}{\mathrm{x}}^{4}}{4{\mathrm{b}}^{4}}&c=0,00000,12800,01638,40279,62080,35386,78180,64007,26195,71331,95041,94\\ \text{The su}\text{mm}\text{e of which two summes is equall to}\phantom{\rule{0.5em}{0ex}}\text{the}\phantom{\rule{0.5em}{0ex}}\text{area bcfd, Which summe is}\\ 0,00160,12813,66973,83328,53761,68584,19997,63476,46377,05009,33030,08& .\end{array}$

As was found before excepting yt their differenc is e in ye two last figures is 28. Which agreement could scarce thus happen in more yn 50 figures, were not ye reas, correspond{illeg}|in|g to ye lines 2. 3. 5. 7. 13. &c, calculated {aright} {illeg}|in| so many figures.

## Octob. 1676.

Memorandum. The letters baccdæ13eff7i3lgn4049 rr458t12vx {illeg}|in| my second epistle to |M.| Leibnitz contein this sentence Data æquatione quotcun fluentes quantitates involvente, fluxiones invenire: et vice versâ.

The other letters in ye same Epistle, viz: 5accdæ10effhui4l3m 9nboqqr{illeg}|{8}|s11t9v3x: 11ab3cdd10eæg10ill4m7n603p3q6r5511{illeg}|t8|vx, 3acæ 4egh5i4l4m5n80q4r3s6t4v aaddæ5eüjmmnnooprrr5sttuv, express this sentenc{illeg}|e|. Una Methodus consistit in extraction{e} fluentis quantitatis ex æquatione simul involvente fluxionem ejus. Altera tantum {illeg} in assumptionæ {seriei} pro quantitate qualibet incognita ex qua cætera commodè derivari posunt, et in collatione terminorum homologorum æquationis resultantis ad eruendos terminos assumptæ {seriei}.

<82r>

The use of these differences is for composing rules to find the differences of {ye terms of a table wch {illeg} to be interpoled by ye contiuall addition of those differences & also following a geo {illeg}}

<82v>

## Prob.

Recta aliqua AA9 in æquales quotcun partes AA2 A2A3, A3A4, A4A5, A5A6, A6A7, A7A8, A8A9 &c divisa et ad puncta divisorum erectis parallelis AB, A2B2, A3B3 &c: invenire curvam geometricam quæ per omnium erectarum extremitates B, B{2} B, B2, B3 &c transibit.

Erectarum {illeg}|AB|, A2B2. A3B3 &c quære differentias primas b, b2, b3, &c secundas c, c2, c3 &c tertias d, d2, d3 &c & sic deincep{s} us dum veneris ad ultimam differentiam \i/. Tunc incipiendo ab ul{illeg}|ti|ma differentia excerpe medias differentias in alterna|i|s $\left\{\begin{array}{l}\text{columnis}\\ \text{seriebus}\\ \text{ordinibus}\end{array}\right\$ \differentiarum/ et arithmetica media inter duas medias inte reliquorum ordinum. < insertion from the left margin of f 82v > pergendo ad us seriem \primorum/ terminorum A, A2, A3, &c. Sint hæc k, l, - - - s & Nempe k ultim{æ} di{illeg}|ff|. l medium arithmeticum inter duas penultimas, m antepe media \trium/ antepenultimarum differentiarum &c

## Cas 1

{illeg} \{Igitur} Si numerus \prignorū./ terminorum A, A2, A3, &c sit impar ultim medius {servimus} eorum erit ultimus terminus series ejus k, l, m, &c. Et tunc sic pergendum erit. Sit numerus primerum terminorum 9 et erit k=i, $\mathrm{l}=\frac{\mathrm{h}+h2}{2}$ {&c}/ < text from f 82v resumes > [{illeg}sint {la} k, l, m, n, o, p, q, r, s, {illeg}, nempe k ultim{æ}=i differentiæ ultimæ, l= $\frac{\mathrm{h}+h2}{2}$, m=g2, $\mathrm{n}=\frac{f2+f3}{2}$, o=e3, $\mathrm{p}=\frac{\mathrm{d3}+d4}{2}$, {illeg} q=c4, $\mathrm{r}=\frac{\mathrm{b4}+b5}{2}$ s=A5; nam series terminorum A, A2, A3, hic est |{in}|star serieri differentiarum, adeo ut medius {illeg}|ej|us terminus A5 {illeg} ultimo termino seriei k, l m, &c existente vel medio termino seriei hujus A, A2, A3 si constat impare|i| numero terminorum ut in hoc casi, vel arithmetico medi{illeg}|{æ}| Erige {s}{illeg}g|or|dinatim applicatam PQ et bis{e}tp|a| AA9 in A5 die A5P=x, PQ ={y} pergendo us ad seriem primorum terminorum A|{illeg}|B||, A2|B2|, A3|B3|, &c. Sint h{illeg}|æ|c k, l, m, n, o, p, q, r, s &c quorum ultimu{m} significet ultimam differentiam, penultimum medium arithmeticum inter duas pe{illeg}|nu|ltimas differentias, antepenultimu{m} mediam trium antepenultimarum differentarum, & \sic/ deinceps us ad primum quod erit vel medius terminorum A, A2, A3, vel arithmeticum medium int{illeg}|er| {illeg}|du|os medios. Prius accidit ubi numerus terminorum A, A2, A3 & est impar, posterius ubi par.

## Cas. 1.

In casu priori sit A5|B5| iste medius terminus, {illeg} \hoc est/ A5|B5|=k, $\frac{b4+b5}{2}=\mathrm{l}$, c4=m, $\frac{d3+d4}{2}=\mathrm{n}$, e3=0, $\frac{f2+f3}{2}=\mathrm{p}$, g2=q, $\frac{\mathrm{h}+h2}{2}=\mathrm{r}$ {illeg}i=s. Et erecta ordinatim applicata PQ die A5P=x, due terminus hujus pr{illeg}|ogr|essionis $1×\frac{{\mathrm{x}}^{2}}{2}×\frac{{x}\mathrm{x}-1}{3,4}×\frac{{x}\mathrm{x}-4}{5,6}×\frac{{x}\mathrm{x}-9}{7,8}×\frac{{x}\mathrm{x}-16}{9,10}×\frac{{x}\mathrm{x}-25}{11,12}×\frac{{x}\mathrm{x}-36}{13,14}×$ &c in se continuò {illeg}|&| orientur coefficientes termini $1.\frac{\mathrm{x}\mathrm{x}}{2}.\frac{{\mathrm{x}}^{4}-\mathrm{x}\mathrm{x}}{24}.\frac{{\mathrm{x}}^{6}-5{\mathrm{x}}^{4}+4\mathrm{x}\mathrm{x}}{720}.$ $\frac{{\mathrm{x}}^{8}-14{\mathrm{x}}^{6}+49{\mathrm{x}}^{4}+36\mathrm{x}\mathrm{x}}{40320}$ &c $1×\frac{\mathrm{x}}{1}×\frac{\mathrm{x}}{2}×\frac{\mathrm{x}\mathrm{x}-1}{3\mathrm{x}}×\frac{\mathrm{x}}{4}×\frac{\mathrm{x}\mathrm{x}-4}{5\mathrm{x}}×\frac{\mathrm{x}}{6}×\frac{\mathrm{x}\mathrm{x}-9}{7\mathrm{x}}$ $×\frac{\mathrm{x}}{8}×\frac{\mathrm{x}\mathrm{x}-16}{9\mathrm{x}}×\frac{\mathrm{x}}{10}×\frac{\mathrm{x}\mathrm{x}-25}{11\mathrm{x}}×\frac{\mathrm{x}}{12}×\frac{\mathrm{x}\mathrm{x}-36}{13\mathrm{x}}$ &c. \in se continuo/ Et orientur termini 1. x. $\frac{\mathrm{x}\mathrm{x}}{2}.\frac{{\mathrm{x}}^{3}-\mathrm{x}}{6}.\frac{{\mathrm{x}}^{4}-\mathrm{x}\mathrm{x}}{24}.\frac{{\mathrm{x}}^{5}-5{\mathrm{x}}^{3}+4\mathrm{x}}{120}.\frac{{\mathrm{x}}^{6}-5{\mathrm{x}}^{4}+4\mathrm{x}\mathrm{x}}{720}.\frac{{\mathrm{x}}^{7}-14{\mathrm{x}}^{5}}{5040}+$ $\frac{+49{\mathrm{x}}^{3}-36\mathrm{x}}{5040}$ &c per quos si termini seriei k, l, m, n, o, p &c resspectivè multiplicentur, prodibunt aggregatum factorum $\mathrm{k}+\mathrm{x}\mathrm{l}+\frac{\mathrm{x}\mathrm{x}}{2}\mathrm{m}$ $+\frac{{\mathrm{x}}^{3}-\mathrm{x}}{6}\mathrm{n}+\frac{{\mathrm{x}}^{4}-\mathrm{x}\mathrm{x}}{24}\mathrm{o}+\frac{{\mathrm{x}}^{5}-5{\mathrm{x}}^{3}+4\mathrm{x}}{120}\mathrm{p}$ &c erit longitudo ordinatim applicatǽ PQ

## Cas 2.

In casu posteriori sint A{illeg}, A5 \A4B4, A5B5/ duo medij termini, hoc est $\frac{A4B4+A5B5}{2}$ =k, b4=l, $\frac{c3+c4}{2}=\mathrm{m}$, d3=o|n|, e2+e3={illeg}|{o}|, f2=p, $\frac{\mathrm{g}+g2}{2}=\mathrm{q}$ & h=r{,|.|} Et erecta ordinatim applicata PQ, biseca A4|5|A5 in O {et} d{illeg} OP=x due terminos hujus progressionis $1×\frac{\mathrm{x}}{1}×\frac{\mathrm{x}\mathrm{x}-\frac{1}{4}}{2\mathrm{x}}×\frac{\mathrm{x}}{3}×\frac{\mathrm{x}\mathrm{x}-\frac{9}{4}}{4}×\frac{\mathrm{x}}{5}$ $×\frac{\mathrm{x}\mathrm{x}-\frac{25}{4}}{6}×\frac{\mathrm{x}}{7}×\frac{\mathrm{x}\mathrm{x}-\frac{49}{4}}{8}$ &c in se continuo. Et orientur termini $1.\mathrm{x}.\frac{4\mathrm{x}\mathrm{x}-1}{8}$ $\frac{4{\mathrm{x}}^{3}-\mathrm{x}}{24}.\frac{16{\mathrm{x}}^{4}-40\mathrm{x}x+9}{384}$ &c per quos si termini seriei k, l, m, n, o, p &c respectivè multiplicentur, aggregatum factorum $\mathrm{k}+\mathrm{x}\mathrm{l}+\frac{4\mathrm{x}\mathrm{x}-1}{8}\mathrm{m}$ $+\frac{4{\mathrm{x}}^{3}-\mathrm{x}}{24}\mathrm{n}+\frac{16{\mathrm{x}}^{4}-40\mathrm{x}x+9}{384}\mathrm{o}$ &c erit longitudo ordinatim, applicat{illeg} PQ.

Sed hic notandum est j. Quod intervalla A,A2, A2A3, A3A4, &c hic supponuntur esse unitates, Et quod differentiæ colligi debent {illeg}ferendo inferiores quantitates de superioribus A2B\2/de AB, A3B\3/de A2B\2/, b2 de b &c, faciendo AB−A2B\2/=b{,} A2B\2/−A3B\3/=b2{,} b−b2=c {illeg}|&|c adeo quando differentia illa {hoc modo} {illeg} negativæ sig{m}a{-}earum {ubi mutand{illeg}em|p| sunt}.

<83r>

For taking of unknowne quantitys out of intricat Equations it may be convenient to have severall formes. Now suppose x, was to be taken out of ye Equations ax3+bxx+cx+d=0 & fx3+gxx+hx+k=0.

I feighe ye 3 valors of x in ye first Equation to bee -r, -s, & -t. [{illeg} is {illeg}|[|$\mathrm{r}+\mathrm{s}+\mathrm{t}=\frac{\mathrm{b}}{\mathrm{a}}$. $\mathrm{r}\mathrm{r}+\mathrm{s}\mathrm{s}+\mathrm{t}\mathrm{t}=\frac{\mathrm{b}\mathrm{b}}{\mathrm{a}\mathrm{a}}-\frac{2\mathrm{c}}{\mathrm{a}}$. ${\mathrm{r}}^{3}+{\mathrm{s}}^{3}+{\mathrm{t}}^{3}=\frac{{\mathrm{b}}^{3}-3\mathrm{a}\mathrm{b}\mathrm{c}+3\mathrm{a}\mathrm{d}\mathrm{d}}{{\mathrm{a}}^{3}}$. $\mathrm{r}\mathrm{s}+\mathrm{r}\mathrm{t}+\mathrm{s}\mathrm{t}=\frac{\mathrm{c}}{\mathrm{a}}$. & that is] {illeg} ye summe of ye rootes is $\frac{\mathrm{b}}{\mathrm{a}}${;} of their squares is $\frac{\mathrm{b}\mathrm{b}-2\mathrm{a}\mathrm{c}}{\mathrm{a}\mathrm{a}}$ of their cubes is {illeg} $\frac{{\mathrm{b}}^{3}-3\mathrm{a}\mathrm{b}\mathrm{c}+3\mathrm{a}\mathrm{a}\mathrm{d}}{{\mathrm{a}}^{3}}$; of their rectangles is $\frac{\mathrm{c}}{\mathrm{a}}$ &c] that is,] supposing a=1, {illeg}|ev|ery r is b every rr=bb−2c. every r3=b3−3{illeg}bc+3d{illeg}, rs=c, rrs=bc-3d, r3s=bbc−2cc−bd. rrss=cc−2bd. r3ss=bcc−2bbd−cd. r3s3=c3−3bcd+3dd. rst=d. rrst=bd. r3st=bbd−2cd. rrsst=cd. r3sst=bcd−3dd. {rrsstt}r3s3t=ccd−2bdd. rrsstt=dd. r3sstt=bdd.r3s3tt=cdd. r3s3t3=d3

— — — — — — — — — — — — — — — — —

Or thus, every r4=b. rs4=c. {illeg} rr4=bb−2c{illeg}. rst4=d. rrs4=bc−3d. r34=b3−3bc+3d. rrst4=bd. rrss=cc-2bd. r3s4=bbc-2cc-bd. r44=b4−4bbc+4bd+2cc. rrsst4=cd. r3st4=bbd−2cd. rrrss4=bcc−{illeg}dc−2b{{illeg}b}d{illeg}. rrsstt4=dd. r3sst4=bcd−3dd. *r3sstt4=bdd. r3s3t4=ccd−2bdd. r3s3tt4=cdd. r3s3t34=d3. *r3s34=c3−3bcd+3dd.

Now supposing k (or any other quantity of ye second Equation) to bee {illeg} i|a|n unknow{ne} quantity, it must have 3 severall valors by reason of ye 3 valors of x in ye first Equation, & therefore x being taken away, \h will bee of three dimensions in/ ye resulting equation.

The 3 valors of {illeg}h are −grr+hr +fr3{illeg}−grr+hr=k. & fs3−gss+hs=k & {fts} ft3−gtt+ht={k.} Which I multiply into one another that they may produce an equation expressing ye 3 fold valor of k: out of wch equation I take {illeg} out r, s, t by writing b for their summe c for ye sum̄e of their rectangles rs+rt+st. bc−3d for ye summe of all theire rectangles of this for{me {illeg}|rrs|} (viz: for rrs+rrt+{illeg}|rss|+rtt+rtt+sst+stt) &c as in ye Table. Wch substitution may bee most briefly done in ye said multiplication, thus; writing a to make up six dimensions $\begin{array}{ccc}\begin{array}{l}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\\ \mathrm{k}-\mathrm{h}\mathrm{s}+\mathrm{g}\mathrm{s}\mathrm{s}-\mathrm{f}{\mathrm{s}}^{3}\\ \mathrm{k}-\mathrm{h}\mathrm{t}+\mathrm{g}\mathrm{t}\mathrm{t}-\mathrm{f}{\mathrm{t}}^{3}\end{array}& \begin{array}{c}\phantom{\begin{array}{c}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\end{array}}\\ \\ \text{Produceth,}\\ \end{array}& \begin{array}{c}\phantom{\begin{array}{c}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\end{array}}\\ \phantom{\begin{array}{c}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\end{array}}\\ \phantom{\begin{array}{c}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\end{array}}\\ \phantom{\begin{array}{c}\mathrm{k}-\mathrm{h}\mathrm{r}+\mathrm{g}\mathrm{r}\mathrm{r}-\mathrm{f}{\mathrm{r}}^{3}\end{array}}\\ \begin{array}{l}{\mathrm{a}}^{3}{\mathrm{h}}^{3}-\mathrm{a}\mathrm{a}\mathrm{b}\mathrm{h}\mathrm{k}\mathrm{k}+\mathrm{a}\mathrm{b}\mathrm{b}\mathrm{g}\mathrm{k}\mathrm{k}-2\mathrm{a}\mathrm{a}\mathrm{c}\mathrm{g}\mathrm{k}\mathrm{k}-{\mathrm{b}}^{3}\mathrm{f}\mathrm{k}\mathrm{k}+3\mathrm{a}\mathrm{b}\mathrm{c}\mathrm{f}\mathrm{k}\mathrm{k}\\ -3\mathrm{a}\mathrm{a}\mathrm{d}\mathrm{f}\mathrm{k}\mathrm{k}+\mathrm{a}\mathrm{a}\mathrm{c}\mathrm{h}\mathrm{h}\mathrm{k}-\mathrm{a}\mathrm{b}\mathrm{c}\mathrm{g}\mathrm{h}\mathrm{k}+3\mathrm{a}\mathrm{a}\mathrm{d}\mathrm{g}\mathrm{h}\mathrm{k}+\mathrm{b}\mathrm{b}\mathrm{c}\mathrm{f}\mathrm{h}\mathrm{k}\\ -2\mathrm{a}\mathrm{c}\mathrm{c}\mathrm{f}\mathrm{h}\mathrm{k}-\mathrm{a}\mathrm{b}\mathrm{d}\mathrm{f}\mathrm{h}\mathrm{k}+\mathrm{a}\mathrm{c}\mathrm{c}\mathrm{g}\mathrm{g}\mathrm{k}-2\mathrm{a}\mathrm{b}\mathrm{d}\mathrm{g}\mathrm{g}\mathrm{k}-\mathrm{b}\mathrm{c}\mathrm{c}\mathrm{f}\mathrm{g}\mathrm{k}\\ +2\mathrm{b}\mathrm{b}\mathrm{d}\mathrm{f}\mathrm{g}\mathrm{k}+\mathrm{a}\mathrm{c}\mathrm{d}\mathrm{f}\mathrm{g}\mathrm{k}+{\mathrm{c}}^{3}\mathrm{f}\mathrm{f}\mathrm{k}-3\mathrm{b}\mathrm{c}\mathrm{d}\mathrm{f}\mathrm{f}\mathrm{k}+3\mathrm{a}\mathrm{d}\mathrm{d}\mathrm{f}\mathrm{f}\mathrm{k}\\ -\mathrm{a}\mathrm{a}\mathrm{d}{\mathrm{h}}^{3}+\mathrm{a}\mathrm{b}\mathrm{d}\mathrm{g}\mathrm{h}\mathrm{h}-\mathrm{b}\mathrm{b}\mathrm{d}\mathrm{f}\mathrm{h}\mathrm{h}+2\mathrm{a}\mathrm{c}\mathrm{d}\mathrm{f}\mathrm{h}\mathrm{h}-\mathrm{a}\mathrm{c}\mathrm{d}\mathrm{g}\mathrm{g}\mathrm{h}\\ +\mathrm{b}\mathrm{c}\mathrm{d}\mathrm{f}\mathrm{g}\mathrm{h}-3\mathrm{a}\mathrm{d}\mathrm{d}\mathrm{f}\mathrm{g}\mathrm{h}-\mathrm{c}\mathrm{c}\mathrm{d}\mathrm{f}\mathrm{f}\mathrm{h}+2\mathrm{b}\mathrm{d}\mathrm{d}\mathrm{f}\mathrm{f}\mathrm{h}+\mathrm{a}\mathrm{d}\mathrm{d}{\mathrm{g}}^{3}\end{array}}=0\end{array}\end{array}$

For solving this Problem generally Datis quotcun punctis Curvam                     describere quǽ per omnia transibit: Note these differences

$\begin{array}{cc}\begin{array}{r}\mathrm{A}\phantom{+\mathrm{x}}\\ \mathrm{A}+\mathrm{x}\end{array}& \begin{array}{r}\mathrm{a}\\ \mathrm{a}& +& \mathrm{b}\mathrm{x}& +& \mathrm{c}\mathrm{x}\mathrm{x}& +& \mathrm{d}{\mathrm{x}}^{3}+& \mathrm{e}{\mathrm{x}}^{4}& +& \mathrm{f}{\mathrm{x}}^{5}& &c\end{array}\\ \begin{array}{r}\mathrm{A}+\mathrm{p}\\ \mathrm{A}+\mathrm{q}\\ \mathrm{A}+\mathrm{r}\\ \mathrm{A}+\mathrm{s}\\ \mathrm{A}+\mathrm{t}\end{array}& \begin{array}{rrrrrrrrrrrr}\mathrm{a}& +& \mathrm{b}\mathrm{p}& +& \mathrm{c}\mathrm{p}\mathrm{p}& +& \mathrm{d}{\mathrm{p}}^{3}+& \mathrm{e}{\mathrm{p}}^{4}& +& \mathrm{f}{\mathrm{p}}^{5}& =& \mathrm{\alpha }\\ \mathrm{a}& +& \mathrm{b}\mathrm{q}& +& \mathrm{c}\mathrm{q}\mathrm{q}& +& \mathrm{d}{\mathrm{q}}^{3}+& \mathrm{e}{\mathrm{q}}^{4}& +& \mathrm{f}{\mathrm{q}}^{5}& =& \mathrm{\beta }\\ \mathrm{a}& +& \mathrm{b}\mathrm{r}& +& \mathrm{c}\mathrm{r}\mathrm{r}& +& \mathrm{d}{\mathrm{r}}^{3}+& \mathrm{e}{\mathrm{r}}^{4}& +& \mathrm{f}{\mathrm{r}}^{5}& =& \mathrm{\gamma }\\ \mathrm{a}& +& \mathrm{b}\mathrm{s}& +& \mathrm{c}\mathrm{s}\mathrm{s}& +& \mathrm{d}{\mathrm{s}}^{3}+& \mathrm{e}{\mathrm{s}}^{4}& +& \mathrm{f}{\mathrm{s}}^{5}& =& \mathrm{\delta }\\ \mathrm{a}& +& \mathrm{b}\mathrm{t}& +& \mathrm{c}\mathrm{t}\mathrm{t}& +& \mathrm{d}{\mathrm{t}}^{3}+& \mathrm{e}{\mathrm{t}}^{4}& +& \mathrm{f}{\mathrm{t}}^{5}& =& \mathrm{\epsilon }\end{array}\\ \begin{array}{rrrr}\mathrm{p}-\mathrm{q}& \right)& \mathrm{\alpha }-\mathrm{\beta }& =\\ \mathrm{q}-\mathrm{r}& \right)& \mathrm{\beta }-\mathrm{\gamma }& =\\ \mathrm{r}-\mathrm{s}& \right)& \mathrm{\gamma }-\mathrm{\delta }& =\\ \mathrm{s}-\mathrm{t}& \right)& \mathrm{\delta }-\mathrm{\epsilon }& =\end{array}& \begin{array}{rrrrrrrrrrrr}\mathrm{b}& +& \mathrm{c}×\overline{\mathrm{p}+\mathrm{q}}& +& \mathrm{d}×\overline{\mathrm{p}\mathrm{p}+\mathrm{p}\mathrm{q}+\mathrm{q}\mathrm{q}}& +& \mathrm{e}×\overline{{\mathrm{p}}^{3}+\mathrm{p}\mathrm{p}\mathrm{q}+\mathrm{p}\mathrm{q}\mathrm{q}+{\mathrm{q}}^{3}}& +& \mathrm{f}×{\mathrm{p}}^{4}& &c& =& \mathrm{\zeta }\\ \mathrm{b}& +& \mathrm{c}×\overline{\mathrm{q}+\mathrm{r}}& +& \mathrm{d}×\overline{\mathrm{q}\mathrm{q}+\mathrm{q}\mathrm{r}+\mathrm{r}\mathrm{r}}& +& \mathrm{e}×\overline{{\mathrm{q}}^{3}+\mathrm{q}\mathrm{q}\mathrm{r}+\mathrm{q}\mathrm{r}\mathrm{r}+{\mathrm{r}}^{3}}& +& \mathrm{f}×{\mathrm{q}}^{4}& &c& =& \mathrm{\eta }\\ \mathrm{b}& +& \mathrm{c}×\overline{\mathrm{r}+\mathrm{s}}& +& \mathrm{d}×\overline{\mathrm{r}\mathrm{r}+\mathrm{r}\mathrm{s}+\mathrm{s}\mathrm{s}}& +& \mathrm{e}×\overline{{\mathrm{r}}^{3}+\mathrm{r}\mathrm{r}\mathrm{s}+\mathrm{r}\mathrm{s}\mathrm{s}+{\mathrm{s}}^{3}}& +& \mathrm{f}×{\mathrm{r}}^{4}& &c& =& \mathrm{\theta }\\ \mathrm{b}& +& \mathrm{c}×\overline{\mathrm{s}+\mathrm{t}}& +& \mathrm{d}×\overline{\mathrm{s}\mathrm{s}+\mathrm{s}\mathrm{t}+\mathrm{t}\mathrm{t}}& +& \mathrm{e}×\overline{{\mathrm{s}}^{3}+\mathrm{s}\mathrm{s}\mathrm{t}+\mathrm{s}\mathrm{t}\mathrm{t}+{\mathrm{t}}^{3}}& +& \mathrm{f}×{\mathrm{s}}^{4}& &c& =& \mathrm{\iota }\end{array}\\ \begin{array}{rrrr}\mathrm{p}-\mathrm{r}& \right)& \mathrm{\zeta }-\mathrm{\eta }& =\\ \mathrm{q}-\mathrm{s}& \right)& \mathrm{\eta }-\mathrm{\theta }& =\\ \mathrm{r}-\mathrm{t}& \right)& \mathrm{\theta }-\mathrm{\iota }& =\end{array}& \begin{array}{rrrrrrrrrr}\mathrm{c}& +& \mathrm{d}×\overline{\mathrm{p}+\mathrm{q}+\mathrm{r}}& +& \mathrm{e}×\overline{\mathrm{p}\mathrm{p}+\mathrm{p}\mathrm{q}+\mathrm{q}\mathrm{q}+\mathrm{p}\mathrm{r}+\mathrm{q}\mathrm{r}+\mathrm{r}\mathrm{r}}& +& \mathrm{f}{\mathrm{p}}^{3}& &c& =& \mathrm{\lambda }\\ \mathrm{c}& +& \mathrm{d}×\overline{\mathrm{q}+\mathrm{r}+\mathrm{s}}& +& \mathrm{e}×\overline{\mathrm{q}\mathrm{q}+\mathrm{q}\mathrm{r}+\mathrm{q}\mathrm{s}+\mathrm{r}\mathrm{r}+\mathrm{r}\mathrm{s}+\mathrm{s}\mathrm{s}}& +& \mathrm{f}{\mathrm{q}}^{3}& &c& =& \mathrm{\mu }\\ \mathrm{c}& +& \mathrm{d}×\overline{\mathrm{r}+\mathrm{s}+\mathrm{t}}& +& \mathrm{e}×\overline{\mathrm{r}\mathrm{r}+\mathrm{r}\mathrm{s}+\mathrm{r}\mathrm{t}+\mathrm{s}\mathrm{s}+\mathrm{s}\mathrm{t}+\mathrm{t}\mathrm{t}}& +& \mathrm{f}{\mathrm{r}}^{3}& &c& =& \mathrm{\nu }\end{array}\\ \begin{array}{rrrr}\mathrm{p}-\mathrm{s}& \right)& \mathrm{\lambda }-\mathrm{\mu }& =\\ \mathrm{q}-\mathrm{t}& \right)& \mathrm{\mu }-\mathrm{\nu }& =\end{array}& \begin{array}{rrrrrrr}\mathrm{d}& +& \mathrm{e}×\overline{\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s}}& +& \mathrm{f}×\overline{\mathrm{p}\mathrm{p}+\mathrm{p}\mathrm{q}+\mathrm{p}\mathrm{r}+\mathrm{p}\mathrm{s}+\mathrm{q}\mathrm{q}+\mathrm{q}\mathrm{r}+\mathrm{q}\mathrm{s}+\mathrm{r}\mathrm{r}+\mathrm{r}\mathrm{s}+\mathrm{s}\mathrm{s}}& =& \mathrm{\xi }\\ \mathrm{d}& +& \mathrm{e}×\overline{\mathrm{q}+\mathrm{r}+\mathrm{s}+\mathrm{t}}& +& \mathrm{f}×\overline{\mathrm{q}\mathrm{q}+\mathrm{q}\mathrm{r}+\mathrm{q}\mathrm{s}+\mathrm{q}\mathrm{t}+\mathrm{r}\mathrm{r}+\mathrm{r}\mathrm{s}+\mathrm{r}\mathrm{t}+\mathrm{s}\mathrm{s}+\mathrm{s}\mathrm{t}+\mathrm{t}\mathrm{t}}& =& \mathrm{\pi }\end{array}\\ \begin{array}{rrrr}\mathrm{p}-\mathrm{t}& \right)& \mathrm{\xi }-\mathrm{\pi }& =\end{array}& \begin{array}{rrrrr}\mathrm{e}& +& \mathrm{f}×\overline{\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s}+\mathrm{t}}& =& \mathrm{\sigma }\end{array}\end{array}$

<83v>

## Prob Curvam Geometricam describere quæ per data quotcun puncta transibit.

Sint ista puncta B, B2, B3, B4, B5, B6 B7 {illeg}|&|c. Et ad rectam quamvis {demitis} AA7 demitte perpendicula BA, B2A2, B3A3 &c Et fac $\frac{\mathrm{A}\mathrm{B}-A2B2}{\mathrm{A}A2}=\mathrm{b}$, $\frac{\mathrm{A2}\mathrm{B2}-A3B3}{A2A3}=\mathrm{b2}$, $\frac{\mathrm{A3}\mathrm{B3}-A4C4}{A3A4}=\mathrm{b3}$, $\frac{\mathrm{A4}\mathrm{C4}-A5C5}{A4A5}=\mathrm{b4}$, $\frac{\mathrm{A5}\mathrm{C5}-A6C6}{A5A6}=\mathrm{b5}${,} $\frac{\mathrm{A6}\mathrm{C6}+A7C7}{A6A7}=\mathrm{b6}$. $\frac{-\mathrm{A7}\mathrm{C7}-A8B8}{A7B8}=\mathrm{b7}$. Deinde $\frac{\mathrm{b}-b2}{\mathrm{A}A2}=\mathrm{c}$. $\frac{b2-b3}{A2A4}=c2$. $\frac{b3-b4}{A3A5}=c3$ &c. Tunc $\frac{\mathrm{c}-c2}{\mathrm{A}A4}=\mathrm{d}$. $\frac{c2-c3}{A2A5}=d2$. $\frac{c3-c4}{A3A6}=d3$ &c. Tunc $\frac{\mathrm{d}-d2}{\mathrm{A}A5}=\mathrm{e}$. $\frac{d2-d3}{A2A6}=e2$. $\frac{d3-d4}{A3A7}=e3$ &c. Differe Sic pergendum est ad ultimam differentiam. Differentijs sic collectis & divisis per intervalla ordinati{illeg} applicatorum: in alter{eris} \earum/ columnis sive {illeg} \seriem vel/ ordinibus {illeg} excerpe{illeg} medias inc{illeg}|ip|iendo ab ultima et in reliquis columnis excerpe{illeg} du{a} media arithmetica inter duas medias, pergendo us ad seriem primorum terminorum AB, A2B2, &c. Sunto hæc k, l, m, n, o, p, q, r &c non quorum ultimus \terminus/ significet ultimam diff. penultimus mediam arithmeticum inter duas penultimas, antepenultimus mediam trium antepenultima{illeg}|rum| &c. Et primus {illeg} erit medias {duarum} ordinatim applicata si numerus {illeg}p datorum punctorum est impar, vel medium a{illeg}ritthmeticum inter duas medias si numerus earum est {illeg}par.

## Cas 1

In casu priori sit A4B4 ista media ordinatim applicata, hoc est A4B4=k, $\frac{b3+b4}{2}=\mathrm{l}$. c3=m, $\frac{d2+d3}{2}=\mathrm{n}$, {illeg} e2=0. f+f2=p. g=q. Et erecta ordinatim applicata PQ et in basi AA5 sumpto quovis puncto O dic OP=x, et duc \in {se}gradatim/ terminos hujus progessionis {sic} $1×\overline{\mathrm{x}-\mathrm{O}A4}×\overline{\mathrm{x}-\frac{4\mathrm{O}A3+\mathrm{O}A5}{2}}×\frac{\overline{\mathrm{x}-\mathrm{O}A3}×\overline{\mathrm{x}-\mathrm{A}05}}{\mathrm{x}-\frac{\mathrm{O}A3+\mathrm{O}A5}{2}}×\mathrm{x}-\frac{+\mathrm{O}A2+\mathrm{O}A6}{2}×$ &c \et {illeg} ortam progressionem asserva./ Vel quod perinde est duc terminos hujus progressionis $1×\overline{\mathrm{x}-\mathrm{O}A4}×\overline{\overline{\mathrm{x}-\mathrm{O}A3}×\overline{\mathrm{x}-\mathrm{O}A5}}$ $×\overline{\overline{\mathrm{x}-\mathrm{O}A2}×\overline{\mathrm{x}-\mathrm{O}A6}}×\overline{\overline{\mathrm{x}-\mathrm{O}\mathrm{A}}×\overline{\mathrm{x}-\mathrm{O}A7}}×$ &c in se gradatim et terminos ex ortos duc \respective/ in terminos hujus progressionis $1.\mathrm{x}-\frac{+\mathrm{O}A3+\mathrm{O}A5}{2}.\mathrm{x}-\frac{+\mathrm{O}A2+\mathrm{O}A6}{2}$ $\mathrm{x}-\frac{+\mathrm{O}\mathrm{A}+\mathrm{O}A7}{2}$. &c et orientur termini intermedij: {illeg} tota progressione existente $1.\mathrm{x}-\mathrm{O}A4.\mathrm{x}\mathrm{x}-\frac{+\mathrm{O}A3+2\mathrm{O}A4+\mathrm{O}A5}{2}\mathrm{x}+\mathrm{O}A3×\mathrm{O}A4+\mathrm{O}A5×\mathrm{O}A4$ &c Vel dic OA=α, OA2=β, OA3=γ{,} OA4=δ, OA5=e, OA6=ζ, OA7=η $\frac{\mathrm{O}A3+\mathrm{O}A5}{2}=\mathrm{\theta }$, $\frac{\mathrm{O}A2+\mathrm{O}A6}{2}=\mathrm{\mu }$. $\frac{\mathrm{O}\mathrm{A}+\mathrm{O}A7}{2}=\mathrm{\lambda }$. Et ex progressione $1×\overline{\mathrm{x}-\mathrm{\delta }}×$ $\overline{\overline{\mathrm{x}-\mathrm{\gamma }}×\overline{\mathrm{x}-\mathrm{\epsilon }}}×\overline{\mathrm{x}-\mathrm{\beta }}×\overline{\mathrm{x}-\mathrm{\zeta }}×\overline{\mathrm{x}-\mathrm{\alpha }}×\overline{\mathrm{x}-\mathrm{\eta }}$ collige ter{i|m|in}os \e/ quibus multiplicatis per 1×. x−θ.× x−μ. x−λ {illeg}|&|c collige alios terminos intermedios tota serie prode{ants} 1. x−δ. xx\δ//θ\x+δθ. x3\δ//−2θ\xx+\+γε//+δθ\x−{γ.} Per cujus terminos multiplica terminos seriei k. l. m, n. o &c et aggregatum productorum erit k+x−δ×l+xx\δ//θ\×m &c erit longitudo ordinatim applicata PQ.

## Cas. 2.

In casu posteriori sint A4B4 et A5B5 dua media ordinatim applicata hoc est $\frac{A4\mathrm{B}+A5B5}{2}=\mathrm{k}${.} b4=l. $\frac{c3+c4}{2}=\mathrm{m}$. d3=n. $\frac{e2+e3}{2}=\mathrm{o}$. f2=p{illeg} Et {illeg} \{illeg} k, m, {illeg}, o, {illeg} &/ coefficientes orientur \en/ multiplicatione terminorum hujus progressiones {illeg} $1×\overline{\overline{\mathrm{x}-\mathrm{O}A4}×\overline{\mathrm{x}-\mathrm{O}A5}}×\overline{\overline{\mathrm{x}-\mathrm{O}A3}×\overline{\mathrm{x}-\mathrm{O}A6}}×\overline{\overline{\mathrm{x}-\mathrm{O}A2}×\overline{\mathrm{x}-\mathrm{O}A7}}$ $×\overline{\overline{\mathrm{x}-\mathrm{O}\mathrm{A}}×\overline{\mathrm{x}-\mathrm{O}A8}}$ &c et ubiquorum coefficientes en multiplicatione {illeg} {illeg} {c}ujus progressionis $\mathrm{x}-\frac{\overline{+\mathrm{O}A4+\mathrm{O}A5}}{2}$. $\mathrm{x}-\frac{\overline{+\mathrm{O}A3+\mathrm{O}A6}}{2}$

<84r>

## Of the nature of Equations

Every Equation as x8+px7+qx6+rx5+sx4+tx3+vxx+yx+z=0. hath so many roots as dimensions, of wch ye summ is −p, the summ of the rectangles if each two +q, of each three −r, of each foure +s: &c: & of all together z.

Also the summe of theire squares, cubes, &c{illeg} is as followeth.

$\begin{array}{lrll}\begin{array}{l}\text{summe of}\\ \text{squares}\end{array}& \mathrm{p}\mathrm{p}& -& 2\mathrm{q}\\ \text{cubes}& -{\mathrm{p}}^{3}& +& 3\mathrm{p}\mathrm{q}& -& 3\mathrm{r}\\ \text{sq: square}& {\mathrm{p}}^{4}& -& 4\mathrm{p}\mathrm{p}\mathrm{q}& +& 4\mathrm{p}\mathrm{r}& +& 2\mathrm{q}\mathrm{q}& -& 4\mathrm{s}\\ \text{sq: cubes}& -{\mathrm{p}}^{5}& +& 5{\mathrm{p}}^{3}\mathrm{q}& -& 5\mathrm{p}\mathrm{q}\mathrm{q}& -& 5\mathrm{p}\mathrm{p}\mathrm{r}& +& 5\mathrm{p}\mathrm{s}& +& 5\mathrm{q}\mathrm{r}& -& 5\mathrm{t}\\ \text{cube}\text{.}\text{cubes}& {\mathrm{p}}^{6}& -& 6{\mathrm{p}}^{4}\mathrm{q}& +& 6{\mathrm{p}}^{3}\mathrm{r}& +& 9\mathrm{p}\mathrm{p}\mathrm{q}\mathrm{q}& -& 6\mathrm{p}\mathrm{p}\mathrm{s}& -& 12\mathrm{p}\mathrm{q}\mathrm{r}\\ \text{sq: sq: cubes}\\ \text{sq: cu: cubes}\\ \end{array}$

Also ye summe of their square cubes &c is as followeth $\begin{array}{llll}\text{squares}& +\mathrm{p}\mathrm{p}& -& 2\mathrm{q}\\ \text{cubes}& -{\mathrm{p}}^{3}& +& 3\mathrm{p}\mathrm{q}& -& 3\mathrm{r}\\ \text{sq: squares}& +{\mathrm{p}}^{4}& -& 4\mathrm{p}\mathrm{p}\mathrm{q}& +& 4\mathrm{p}\mathrm{r}& -& 4\mathrm{s}& +& 2\mathrm{q}\mathrm{q}\\ & -{\mathrm{p}}^{5}& +& 5{\mathrm{p}}^{3}\mathrm{q}& -& 5\mathrm{p}\mathrm{p}\mathrm{r}& +& 5\mathrm{p}\mathrm{s}& -& 5\mathrm{t}& -& 5\mathrm{p}\mathrm{q}\mathrm{q}& +& 5\mathrm{q}\mathrm{r}\\ & +{\mathrm{p}}^{6}& -& 6{\mathrm{p}}^{4}\mathrm{q}& +& 6{\mathrm{p}}^{3}\mathrm{r}& -& 6\mathrm{p}\mathrm{p}\mathrm{s}& +& 6⁤\mathrm{p}⁤\mathrm{t}& -& 6⁤\mathrm{v}& +& 9\mathrm{p}\mathrm{p}\mathrm{q}\mathrm{q}& -& 12\mathrm{p}\mathrm{q}\mathrm{r}& +& 6⁤\mathrm{q}⁤\mathrm{s}& -& 2⁤{\mathrm{q}}^{3}\\ & -{\mathrm{p}}^{7}& +& 7{\mathrm{p}}^{5}\mathrm{q}& -& 7{\mathrm{p}}^{4}\mathrm{r}& +& 7{\mathrm{p}}^{3}\mathrm{s}& -& 7\mathrm{p}\mathrm{p}\mathrm{t}& +& 7\mathrm{p}\mathrm{v}& -& 7\mathrm{q}\\ & +{\mathrm{p}}^{8}& -& 8{\mathrm{p}}^{6}\mathrm{q}& +& 8{\mathrm{p}}^{5}\mathrm{r}& -& 8{\mathrm{p}}^{4}\mathrm{s}& +& 8{\mathrm{p}}^{3}\mathrm{t}& -& 8\mathrm{p}\mathrm{p}\mathrm{v}& +& 8\mathrm{p}\mathrm{q}& -& 8\mathrm{z}\\ \end{array}$

Or thus If their sume is {illeg}||p=\/a. Then is ye sume of their squares ap−2q=\b/ of their cubes −pb+qa−3r=c. square squares pc−qb+ra−4s=d. sq: cubes −pd+qc−rb+sa−5t =−e. cube cubes +pe+/\qd+rc+\/st|b|+ta{illeg}−6v &c.

Non of these rootes some are true {illeg} other some false & some imaginary

$\mathrm{x}-\frac{+\mathrm{O}A2+\mathrm{O}A7}{2}$. $\mathrm{x}-\frac{+\mathrm{O}\mathrm{A}+\mathrm{O}A8}{2}$ &c Hoc est eritk+x−OA4+ $\mathrm{k}+\overline{\mathrm{x}-\frac{+\mathrm{O}A4+\mathrm{O}A5}{2}}×\mathrm{l}+\overline{\mathrm{x}\mathrm{x}-\overline{\mathrm{O}A4-\mathrm{O}A5}×\mathrm{x}+\mathrm{O}A4×\mathrm{O}A5}×\mathrm{m}$ &c=Ordinatim applicatæ $PQ/=\begin{array}{llll}\mathrm{k}& \begin{array}{cc}\begin{array}{l}+\mathrm{x}\\ -\frac{1}{2}\mathrm{O}A4\\ -\frac{1}{2}\mathrm{O}A5\end{array}×& \mathrm{l}\end{array}& \begin{array}{ccc}\begin{array}{l}+\mathrm{x}\\ -\mathrm{O}A4\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\mathrm{O}A5\end{array}×& \mathrm{m}\end{array}& \begin{array}{cccc}\begin{array}{l}+\mathrm{x}\\ -\mathrm{O}A4\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\mathrm{O}A5\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\frac{1}{2}\mathrm{O}A3\\ -\frac{1}{2}\mathrm{O}A6\end{array}×& \mathrm{n}\end{array}\end{array}$ $\begin{array}{ccccc}\begin{array}{l}+\mathrm{x}\\ -\mathrm{O}A3\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\mathrm{O}A4\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\mathrm{O}A5\end{array}×& \begin{array}{l}\phantom{+}\mathrm{x}\\ -\mathrm{O}A6\end{array}×& \mathrm{o}\end{array}$. {illeg} Sive dic $\mathrm{x}-\frac{+\mathrm{O}A4+\mathrm{O}A5}{2}=\mathrm{\pi }$. $\overline{\mathrm{x}-A4}×\overline{\mathrm{x}-A5}$ =ρ. $\mathrm{\rho }×\overline{\mathrm{x}-\frac{+\mathrm{O}A3+\mathrm{O}A6}{2}}=\mathrm{\sigma }$. $\mathrm{\rho }×\overline{\mathrm{x}-\mathrm{O}A3}×\overline{\mathrm{x}-\mathrm{O}A6}=\mathrm{\tau }$. $\mathrm{\tau }×\mathrm{x}-\frac{+\mathrm{O}A2+\mathrm{O}A7}{2}$ =ν. $\mathrm{\tau }×\overline{\mathrm{x}-\mathrm{O}A2}×\overline{\mathrm{x}-\mathrm{O}A7}=\mathrm{\phi }$. $\mathrm{\phi }×\mathrm{x}-\frac{+\mathrm{O}\mathrm{A}+\mathrm{O}A8}{2}=\mathrm{\chi }$. $\mathrm{\phi }×\overline{\mathrm{x}-\mathrm{O}\mathrm{A}}×$ $\overline{\mathrm{x}-\mathrm{O}A8}=\mathrm{\psi }$ Et erit k+πl+ρm+σn+τo+νp+φq χ{illeg}r+ψs=PQ.

<84v>

<85r>

## Of Equations,

Every equation hath soe many roots as dimensions of wch some may be true some false & some imaginary or i{illeg}|mp|ossible.

If there bee some imaginary then the true & false rootes may be knowne by ye signes of ye Equations termes: Namely there are soe many true rootes as variations of signes & soe many false ones as successions of ye same signes. When any termes a{illeg}|re| wanting supply their{e} voyd places wth ±0.

But if any \because imaginary roots are properly neither true nor false/ roots bee imaginary, this r{oote} soe far admitts of exception.{illeg} Thus the signes of{illeg} this Eq: x3−pxx+3{illeg}|pp|x−p3=0. show it to have three true roots, {illeg}wherefore if it bee multiplyed by x+\2/a=0 the resulting equation have {three} true x4+px3+{illeg}|pp|xx\+6p3//−q3\x−2{illeg}|pq|3=0 should have thre true rootes & a false one, g|b|ut the signes shew it to have three false & one true,|.| I conclude therefore that the two roots wch in ye one case appeare true, |&| in a|ye|nother false are neither, but imaginary; & that of ye other two roots one is true {ye} other false.

Hence it appeares yt to know ye particular constitution of any Equation it is {chi{illeg}fely} necessary to understand\{illeg}/ wt imaginary roots it hath. And this in some of the simplest Equations is easily discovered, thus in xx±ax+bb=0, {illeg} both roots are imaginary if 4bb<aa, otherwis both reall. And thus in {illeg}|px|3{illeg}|apx|−q\{illeg}/=0 two roots are imaginary if 4p3<27qq, otherwise all reall. But to give \accurate/ rules for determining the {illeg} \number/ of these roots in all so{illeg}|rt|s of Equations would bee a thing not onely {illeg}|ver|y difficult, but usele{illeg}|sse|:|,| because in Equations of many dimensions ye rules would bee more in{illeg}|tr|icate & laborious than ye to put in practise then to solve the Equations either by lines or numbers. Soe yt ye accurate determination{s} of those roots is {illeg} for the most part {esilyest} acquired by solving the Equations.

But yet because the discovery of {illeg}|th|ese roots is very usefull I shall lay downe rules whereby they may bee many times discovered at first sight, & almost always wthout much labour.

First then {illeg} if you see any terme betwixt

First yn if in any three termes together ye two extreame termes having ye same signes bee neither of them \as little or lesse/ greater ({illeg}tis as more \little or l{illeg}|es|se/ remote from nothing) then ye terme betwixt them, conclude there are two imaginary roots. Thus \{in}/{illeg} +aaax−b3{illeg} x32|3|xx+2x{\{illeg}/}−4=0 has two roots imaginary because neither 3 nor 4 are lesse then 2. And ye like of x3−3xx−2x\{illeg}/-4=0. And soe of x3+2x−4=0, or x3+0xx+2x−4=0, because ne{illeg}|it|her 1 nor 2 {illeg} is lesse then 0.

Secondly if uppon sight you discover three \such/ termes together \that/ the two extreames having the same signes their rectangle bee as greate or greater then ye square of the meane terme, conclude there are two imaginary roots. Thus in x3+\/pxx+3ppx−q3=0 are two imaginary roots because 1×3pp<−px−p. And soe of x3−pxx+ppx−\2/p3=0 because −p{x} 1×pp{illeg}=-px-p. or −px−\2/p3= pp×pp.

Thirdly if

First then the reall roots of an Equation are not more then the number of \its/ termes.

First yn ye number of impossible roots is always eaven. If one bee {illeg}|imp|ossible there must bee two, if three there must bee foure &c. And hence Equations of odd dim{en}|sions must have one roote reall at least.|

Secondly the number of reall roots \of any Equation/ are not more then the number of its termes. If Thus x4−2x+3=0 {illeg} have all foure roots reall & therefore must have two imaginary. Thus x5−3x4+4=0 can have but thre reall roots & then other two bee \must bee/ imaginary. Hence {illeg}|are| to bee excepted equations wch {want} all their odd termes as x6−2x4+3xx−2=0. And s|i|n \{illeg} like{illeg}/ cases write y for xx. And so many {illeg}|term|e roots as ye product y3−2yy+3y−2=0 hath \{times}/ soe many reall roots, halfe true halfe false, the other shal & four times soe imaginary ones ye other shall x6−2x4+3xx−2=0 shall have.

Thirdly, if under the termes of any Equation you set a progression of {fractions} {illeg} \each having/ ye dimension of the terme{illeg} above {illeg} it for its {illeg} & number {illeg} {denominating} yt terme first second third &c for its denominator & \{illeg} if/ {illeg} {illeg} any {illeg} together so yt the \rectangle of the/ first {illeg} have {illeg} {illeg} multiplyed {viz the fraction} {illeg} first {illeg} square of the {illeg} terme multipl{illeg} by {ye fraction} {illeg} conclude there are two imaginary roots {illeg} if {illeg} {illeg} one {illeg} <85v> in all throughout ye equation conclude \also/ there are two imaginary roots at least. If equall in all cases throughout the Equation, conclude th{illeg}|at|{illeg} all the roots of the equation are equall. If it be greater or equal to it in two {illeg} pla\If the said factus be greater or equall to ye said squ/ces of ye equation & not in all places betwixt conclude there are foure imaginary roots at least. If it bee greater or equall to it in three places of ye equation & not in all places betwixt, conclude there are six imaginary roots at least. And soe of the rest.

{Thus if} \Thus if the Equat be x3−3xx+4x−2=0./ /Th{illeg} /en\ ye progre{illeg}|ss| is $\frac{3}{1}.\frac{2}{2}.\frac{1}{3}$.\ & because $-3\mathrm{x}\mathrm{x}-2×\frac{2}{2}\left(6\right)<4×4×\frac{1}{3}\left(\frac{16}{3}\right)$, I conclude there are two roots imaginary at least.

$\begin{array}{ccccccccccc}\text{Th}\text{u}\text{s}\phantom{\rule{1em}{0ex}}\text{in}& {\mathrm{x}}^{3}& -& 6{\mathrm{x}}^{2}& +& 6\mathrm{x}& -& 2& =& 0.& \phantom{\rule{4em}{0ex}}\text{because}\phantom{\rule{1em}{0ex}}-6\mathrm{x}-2×\frac{2}{2}\left(12\right)=6×6×\frac{1}{3}\left(12\right)\phantom{\rule{1em}{0ex}}\text{but}\hfill \\ & \frac{3}{1}& .& \frac{2}{2}& .& \frac{1}{3}& .& & & & \phantom{\rule{4em}{0ex}}1×6×\frac{3}{1}\left(18\right)<-6×-6×\frac{2}{2}\left(36\phantom{\rule{1em}{0ex}}\text{I conclude}\hfill \end{array}$ there are two imaginary roots.

$\begin{array}{ccccccccccc}\text{T}\text{hus}\phantom{\rule{1em}{0ex}}\text{in}& {\mathrm{x}}^{3}& -& 6\mathrm{x}\mathrm{x}& +& 12\mathrm{x}& -& 8& =& 0& \phantom{\rule{4em}{0ex}}\text{because}\phantom{\rule{1em}{0ex}}1×12×3\left(36\right)=-6×-6×1\left(36\right)\phantom{\rule{1em}{0ex}}\text{and also}\hfill \\ & 3& .& 1& .& \frac{1}{3}& .& & & & \phantom{\rule{4em}{0ex}}-6×-8×1\left(48\right)=12×12×\frac{1}{3}\left(48\right)\phantom{\rule{1em}{0ex}}\text{I conclude}\phantom{\rule{1em}{0ex}}\text{that}\hfill \end{array}$ all the 3 roots are equall.

$\begin{array}{ccccccccccc}\text{Bu}\text{t}\phantom{\rule{1em}{0ex}}\text{in}& {\mathrm{x}}^{3}& +& 6\mathrm{x}\mathrm{x}& +& 12\mathrm{x}& -& 8& =& 0,& \phantom{\rule{4em}{0ex}}\text{because}\phantom{\rule{1em}{0ex}}1×12×3\left(36\right)=6×6×1\phantom{\rule{1em}{0ex}}\text{but}\hfill \\ & 3& .& 1& .& \frac{1}{3}& & & & & \phantom{\rule{4em}{0ex}}6×-8×1\left(-48\right)<12×12×\frac{1}{3}\left(+48\right)\phantom{\rule{1em}{0ex}}\text{I}\hfill \end{array}$ conclude two rootes are imaginary.

$\begin{array}{ccccccccccccc}\text{In}& {\mathrm{x}}^{4}& -& {\mathrm{x}}^{3}& +& 2\mathrm{x}\mathrm{x}& -& 2\mathrm{x}& +& 3& =& 0,& \phantom{\rule{4em}{0ex}}\text{because}\phantom{\rule{1em}{0ex}}1×2×\frac{4}{1}\left(8\right)<-1×-1×\frac{3}{2}\hfill \\ & \frac{4}{1}& \phantom{.}& \frac{3}{2}& \phantom{.}& \frac{2}{3}& \phantom{.}& \frac{1}{4}& & & & & \phantom{\rule{4em}{0ex}}\left(\frac{3}{2}\right),\phantom{\rule{1em}{0ex}}\text{I conclude there}\phantom{\rule{1em}{0ex}}\text{are two}\hfill \end{array}$ imaginary roots at least, also by the three last termes $\frac{2×3}{}×\frac{2}{3}\left(4\right)<-2×-2×\frac{1}{4}$ (1) therefore there {illeg} are two more imaginary roots \all 4 roots are imaginary/ unlesse the like happen in the three middle termes. I try therefore & find $-1×-2×\frac{3}{2}\left(3\right)<2×2×\frac{2}{3}\left(\frac{8}{3}$ & soe can conclude but two rootes imaginary.

$\begin{array}{ccccccccccccc}\text{In}& {\mathrm{x}}^{4}& -& {\mathrm{x}}^{3}& +& 3\mathrm{x}\mathrm{x}& -& 2\mathrm{x}& +& 3& =& 0& \phantom{\rule{4em}{0ex}}\text{because}\phantom{\rule{1em}{0ex}}1×3×\frac{4}{1}\left(12\right)<-1×-1×\frac{3}{2}\left(\frac{3}{2}\right)\phantom{\rule{1em}{0ex}}\text{and also}\hfill \\ & \frac{4}{1}& .& \frac{3}{2}& .& \frac{2}{3}& .& \frac{1}{4}& .& & & & \phantom{\rule{4em}{0ex}}3×3×\frac{2}{3}\left(6\right)<-2×-2×\frac{1}{4}\left(1\right),\phantom{\rule{1em}{0ex}}\text{but not}\phantom{\rule{1em}{0ex}}-1×-2×\frac{3}{2}\left(3\right)<\hfill \end{array}$ $<3×3×\frac{2}{3}\left(6\right).$ Therefore I conclude all four roots imaginary.

$\begin{array}{ccccccccccccccccc}\text{In}& {\mathrm{x}}^{7}*& +& {\mathrm{x}}^{5}& -& 2{\mathrm{x}}^{4}& +& 3{\mathrm{x}}^{3}& -& 3{\mathrm{x}}^{2}& -& 2\mathrm{x}& -& 1& =& 0& \phantom{\rule{4em}{0ex}}\text{because the three first termes}\hfill \\ & \frac{7}{1}& .& \frac{6}{2}& .& \frac{5}{3}& .& \frac{4}{4}& .& \frac{3}{5}& .& \frac{2}{6}& \phantom{.}& \frac{1}{7}& & & \phantom{\rule{4em}{0ex}}\text{give}\phantom{\rule{1em}{0ex}}1×1×\frac{7}{1}\left(7\right)<0×0×\frac{6}{2}\left(0\right)\phantom{\rule{1em}{0ex}}\text{there}\hfill \end{array}$ are two img|a|ginary roots{.} Also the 3d 4th & 5t terme give $1×3×\frac{5}{3}\left(5\right)$ therefore since by the 2d 3d & 4th terme tis $0*-2×\frac{6}{2}\left(0\right)<1×1×\frac{5}{3}\left(\frac{5}{3}\right)$ I conclude ther are 4 roots imaginary. Also \by ye 4th 5t & 6t termes/ I find $-2×-3×\frac{4}{4}\left(6\right)<3×3×\frac{3}{5}\left(\frac{17}{5}\right)$ but thence nothing can be concluded because those three termes are of the same condition wth ye 3d 4th & 5t termes wch immediately precede them. Lastly I find by the three last termes $-3×-1×\frac{2}{6}\left(1\right)<-2×-2×\frac{1}{7}\left(\frac{4}{7}\right)$; And by the termes pr{e}ceding them $3×-2×\frac{3}{5}\left(-\frac{18}{5}\right)<-3×-3×\frac{2}{6}\left(3\right)$. Therefore I conclude there are \two more imaginary/ roots; imaginary \yt is in all 6/ & but one reall.

Thus in litterall Equations, if $\begin{array}{ccccccccc}{\mathrm{x}}^{3}& -& \mathrm{p}\mathrm{x}\mathrm{x}& +& 3\mathrm{p}\mathrm{p}\mathrm{q}& -& {\mathrm{q}}^{3}& =& 0\\ \frac{3}{1}& \phantom{-}& \frac{2}{2}& \phantom{+}& \frac{1}{3}\end{array}$, because 1×3ppx{illeg} $\left(9\mathrm{p}\mathrm{p}\right)<-\mathrm{p}\mathrm{x}-\mathrm{p}×\frac{2}{2}\left(\mathrm{p}\mathrm{p}\right)$ \therefore/ what ever numbers are taken for p and q two roots shall bee im{illeg}|ag|inary. And soe of the rest.

You may set the

This rule may be otherwise thus exprest. Over ye termes of ye equation set a series of fractions each having ye dimensions of the terme under it for its numerator, & the number denominating ye terme first, second {third} &c for its denominator. Then in every three termes observe{illeg} whither the square of the middle terme \multiplyed by the fraction above/ be greater equall or lesse yn ye factus of the {terme} before & after it multiplyed by ye fraction over ye terme before it. If greater write ye signe + underneath; if equal or lesse write {illeg} the signe − underneath ye middle terme {illeg} & lastly set + under ye first terme of ye equation. Then observe how may {sic} changes there are from + to − & conclude that there \are/ soe many pa{ires} of imaginary roots{.} |unlesse {illeg}|all| ye roots bee equall|

Thus $\begin{array}{cccccc}\frac{3}{1}& .& \frac{2}{2}& .& \frac{1}{3}& .\\ {\mathrm{x}}^{3}& -& 3\mathrm{x}\mathrm{x}& +& 4\mathrm{x}& -& 2& =& 0\\ +& .& +& .& -& .\end{array}$ hath 2: & $\begin{array}{ccccccc}\frac{4}{1}& \phantom{.}& \frac{3}{2}& \phantom{.}& \frac{2}{3}& \phantom{.}& \frac{1}{4}\\ {\mathrm{x}}^{4}& -& {\mathrm{x}}^{3}& +& 3\mathrm{x}\mathrm{x}& -& 2\mathrm{x}& +& 3& =& 0\\ +& .& -& .& +& .& -& .\end{array}$ hath 4 {illeg} /many {roots}\

<86r>

$\begin{array}{ccccc}5& 4& 3& 2& 1\\ 1& 2& 3& 4& 5\end{array}$

If you would bee more exact set downe \after their signes/ the differences {illeg} of ye said squares & rectangles multiplyed. And then if you see three di{illeg}|ff|erences together wth the same signe soe yt ye square of the meane{illeg} diff bee less then ye rectangle of the other two change the signe of the said meane difference

$\begin{array}{ccccccccc}& \frac{4}{1}& .& \frac{3}{2}& .& \frac{2}{3}& .& \frac{1}{4}& .\\ \text{Thus if}& {\mathrm{x}}^{4}& -& {\mathrm{x}}^{3}& +& 2\mathrm{x}\mathrm{x}& -& 2\mathrm{x}& +& 3& =& 0.& \phantom{\rule{4em}{0ex}}\text{becaus}\phantom{\rule{1em}{0ex}}-\frac{23}{4}×-3\left(\frac{69}{4}\right)<-\frac{1}{3}×-\frac{1}{3}\left(\frac{1}{9}\right)\phantom{\rule{1em}{0ex}}\text{I change}\hfill \\ & +\frac{4}{1}& -& \frac{23}{4}.& -& \frac{1}{3}.& -& 3.& & & & & \phantom{\rule{4em}{0ex}}\text{the signe of}-\frac{1}{3}\phantom{\rule{1em}{0ex}}\text{& soe the signes +.−.+.−}\hfill \\ & +& .& -& .& +& .& -& .& & & & \phantom{\rule{4em}{0ex}}\text{shew all foure roots imaginary.}\hfill \end{array}$

If you would bee yet more exact, augment ye roots of the Equatio the more the better, & at least soe much as to make them a{illeg}|ll| true. then set ye afforesaid differences \wth their signes/ underneath as before. And under them t{illeg}|he| progression of fractions squares. Then if you see three differences together wth ye same signe soe yt ye square of the middle difference multiplyed by the fraction under {illeg}|it| bee not greater yn ye rectangle of the other two differences multiplyed by ye fraction under the first: change ye signe of ye middle difference.

Any Equation being propounded, set down a series of \so many/ fractions as ye Equation hath dimensions, whose {illeg}|nume|rators are a prog & denominators are a progression of units backward & forward. Divide each fraction by yt peceding it & set the quotes in order overal |ye| middle termes of the Equation. Then observe whither of every {thee term such} of every middle terme whither it square multiplyed by ye fraction over it bee greater equall or lesse yn ye rectangle of ye two termes on either hand. If greater write + underneath, if equall or lesse write {illeg} −. Lastly set + under ye first \& last/ terme & soe many {illeg}|chan|ges as there are from + to {illeg} & there shall bee soe many impossible roots as there are changes of signes. |Unlesse it happen yt all ye roots are equall, for &c:|

Thus if x36|3|xx+4|6|x−2|4|=0. The \series/ fractions will bee $\frac{3}{1}.\frac{2}{2}.\frac{1}{3}.$ & dividing $\frac{2}{2}$ by $\frac{3}{1}$, & $\frac{1}{3}$ by $\frac{2}{2}$ their quotes will be $\frac{1}{3}$, $\frac{1}{3}$, to be set over ye middle termes of the equation thus $\begin{array}{ccccc}& & \frac{1}{3}.& & \frac{1}{3}.\\ {\mathrm{x}}^{6}& -& 3\mathrm{x}\mathrm{x}& +& 6\mathrm{x}& -& 4& =& 0.\\ +& & -& & -& & +\end{array}$. Then I observe in ye 2d terme that $-3×-3×\frac{1}{3}\left(3\right)$ is lesse then {illeg}4|1×6| & therefore I write − under it. so in ye 3d terme I find $6×6×\frac{1}{3}\left(12\right)=-3×-4\left(12\right)$, therfore I write − under it. Lastly seting + under ye first & last terme I find two changes of sines & so conclude there are

Thus if x5\5/{illeg}x4+5x3  xx− x52|4|x4+x3−2xx−4|5|x−6|4|=0. The series of fractions will bee $\frac{5}{1}.\frac{4}{2}.\frac{3}{3}.\frac{2}{4}.\frac{1}{5}$ And dividing $\frac{4}{2}$ by $\frac{5}{1}$ & $\frac{3}{3}$ by $\frac{4}{2}$ &c there results $\frac{2}{5}.\frac{1}{2}.\frac{1}{2}.\frac{2}{5}$ to bee set over ye middle termes of the equation th{illeg}|{illeg}s| $\begin{array}{cccccccccccccc}& & \frac{2}{5}.& & \frac{1}{2}.& & \frac{1}{2}.& & \frac{2}{5}.& & & & & \\ {\mathrm{x}}^{5}& -& 4{\mathrm{x}}^{4}& +& 4{\mathrm{x}}^{3}& -& 2\mathrm{x}\mathrm{x}& -& 5\mathrm{x}& -& 4& =& 0.& \phantom{\rule{4em}{0ex}}\text{Then in}\phantom{\rule{0.8em}{0ex}}\text{the}\phantom{\rule{0.8em}{0ex}}\text{second time I find}\phantom{\rule{1em}{0ex}}-4×-4×\frac{2}{5}\left(\frac{32}{5}\right)\hfill \\ +& \phantom{.}& +& \phantom{.}& -& \phantom{.}& +& \phantom{.}& +& \phantom{.}& +& \phantom{.}& & \phantom{\rule{4em}{0ex}}<1×4\left(4\right):\text{therefore I set + under it. In}\hfill \end{array}$ the third $4×4⁤\frac{1}{2}\left(8\right)=-4×-2\left(8\right)$: therefore I write −. In the 4th $-2×-2×\frac{1}{2}\left(2\right)<4×-5$ (-20) therfore I write +. In ye 5t $-5×-5×\frac{2}{4}\left(10\right)<-2×-4\left(8\right)$ therefore I w{illeg}|rit|e +. lastly under ye first & last terme I w{illeg}|rit|e +. And soe finding two changes of termes I conclude two roots to bee impossible.

$\begin{array}{ccccccccccccc}& & & \frac{3}{8}& & \frac{4}{8}& & \frac{3}{8}& & & & & \\ \text{T}\text{hus in}& {\mathrm{x}}^{4}& +& 3{\mathrm{x}}^{3}& -& 6{\mathrm{x}}^{2}& -& 3\mathrm{x}& -& 2& =& 0.& \phantom{\rule{4em}{0ex}}\text{two}\phantom{\rule{0.3em}{0ex}}\text{roots are impossible}\hfill \\ & +& \phantom{+}& +.& \phantom{-}& +& \phantom{-}& -& \phantom{-}& +& & & \end{array}$ $\begin{array}{ccccccccccc}& & & \frac{1}{3}& & \frac{1}{3}& & & & & \\ \text{In}& {\mathrm{x}}^{3}& +& 0\mathrm{x}\mathrm{x}& +& \mathrm{p}\mathrm{p}\mathrm{x}& -& {\mathrm{q}}^{3}& =& 0& \phantom{\rule{4em}{0ex}}\text{two roots}\phantom{\rule{0.3em}{0ex}}\text{are}\phantom{\rule{0.3em}{0ex}}\text{impossible}\hfill \\ & +& \phantom{+}& -& \phantom{+}& +& \phantom{-}& +& & & \end{array}$ $\begin{array}{ccccccccccc}& & & \frac{1}{3}& & \frac{1}{3}& & & & & \\ & {\mathrm{x}}^{3}& -& 6\mathrm{x}\mathrm{x}& +& 12\mathrm{x}& -& 8& =& 0& \phantom{\rule{4em}{0ex}}\text{All}\phantom{\rule{0.3em}{0ex}}\text{the}\phantom{\rule{0.3em}{0ex}}\text{roots are equall.}\hfill \\ & +& \phantom{-}& -& \phantom{+}& -& \phantom{-}& +& & & \end{array}$

<86v>

Sometimes there may bee impossible not by this meanes discovered, wch if you suspect, augment or diminish ye roots of the Equation a little, not soe much as to make them all affirmative or all negative, or at most not much more. & try the rule againe. And if there bee any impossible roots twill rarely happen yt they shall not bee discovered at two or three such tryalls. Nor can there bee an Equation whose impossible roots may not bee thus discovered.

Thus if x3−3ppx−3p3=0, because i{f} \in wch noe/ impossible appeare I put x−p{illeg} x=y−p & the result is y3−3pyy−p3=0 in wch two appeare, Or if I put x=y−2p the result is y3−6pyy+9ppy−5p3=0 in wch also two appeare.

Thus |if| $\begin{array}{ccccccccccccc}{\mathrm{x}}^{5}& +& {\mathrm{x}}^{4}& -& 4{\mathrm{x}}^{3}& +& 5\mathrm{x}\mathrm{x}& -& 2\mathrm{x}& +& 1& =& 0\\ +& & +& & +& & +& & -& & +\end{array}$.|,| I set ye signes + & − under it as before and find two imaginary roots & to try if it have any more I suppose x=y+1 & ye results is x4|5|{illeg}|+|6x4+10x3+9xx+5x+0

Now by this rule false roots may bee often discovered at first sight; as if you see a terme {illeg}|wan|ting twixt two others of {illeg}the same signes, or if it bee lesse \greater/ there either of those two or its square {illeg}not greater then their rectangle; conclude there a paire of impossible roots, may soe many pa{illeg}|ir|e as these {caseses} happe{n}. Thus x7+\3/x5−5x4+\4/x3 \set the signe − under that terme/ /conclude\ conclude there is a paire of impossible roots at least & set the signe − under yt terme \also set ye signe + on either side the term {wanting}/. As in this $\begin{array}{ccccccccccccccccc}{\mathrm{x}}^{7}& +& 0{\mathrm{x}}^{6}& +& 2{\mathrm{x}}^{5}& -& 2{\mathrm{x}}^{4}& +& 3{\mathrm{x}}^{3}& -& 4\mathrm{x}\mathrm{x}& +& 6\mathrm{x}& -& 2& =& 0\\ +& & -& & +& & -& & \phantom{-}& & -& & \phantom{-}& & +\end{array}$. In wch it appears there are 4 if not 6 impossible roots

If there bee {illeg}|tw|o or more termes wanting set signes under them successively to{illeg} {the best} advantage {illeg} indicating {im{illeg}|pos|:} roots \begining wth a negative{illeg} only end wth an affirmative/ /if the terms on either hand have contrary signes\. As in {illeg} $\begin{array}{ccccccccccc}{\mathrm{x}}^{5}& -& \mathrm{a}{\mathrm{x}}^{4}& *& *& *& & +& {\mathrm{a}}^{5}& =& 0\\ & & & -& +& -& & & & & \end{array}$ so in $\begin{array}{cccccccccc}{\mathrm{x}}^{5}& +& \mathrm{a}{\mathrm{x}}^{4}& *& *& *& -& {\mathrm{a}}^{5}& =& 0.\\ & & & -& +& +& & & & \end{array}$ The first shows 4, ye last two roots imaginary: Soe{.} in x9 * x8 * $\begin{array}{cccccccccccccccc}{\mathrm{x}}^{10}& *& +& {\mathrm{x}}^{8}& *& *& -& 3{\mathrm{x}}^{5}& *& *& *& *& -& 6& =& 0.\\ +& -& & +& -& +& & +& -& +& -& +& & +& & \end{array}$ wch hath {illeg}|{8}| roots imaginary

<88r>

## To reduce sev{{illeg}}erall equations of four\by divisors of {three}/ dimensions.

Get ye divisors of 6 or 7 \or 8/ such numbers as were described before. Add & sustract them from 29. 8. 1. 0. -1. -8. -27. Make some number in the Take any three numbers out of the \{three}/ middle ranks, r, s. t. Make -s=c. $\frac{\mathrm{r}+\mathrm{t}}{2}-\mathrm{s}=\mathrm{a}$ $\frac{\mathrm{r}-\mathrm{t}}{2}=\mathrm{b}$. Then see if you can find 4a+2b+c in ye rank peceding {sic} these & 9a+3b+c in ye rank preceding that also 4a−2b+c in the rank following these & 9a−3b+c in the rank after that{;} if you can try the division by x3−axx+bx−c.

Or take \better/ multiply all ye numbers in ye middle rank by \4|5| &/ 9|1|0. Let s, {illeg}|{o}| {illeg}|{v}| signify ye products, out of ye two ranks on either side take any two eaven or two odd numbers out of ye two ranks on either hand. Let those be r & t. {illeg} Then observe if you can{illeg} find \make|i|ng {of}/ $\frac{\mathrm{r}+\mathrm{t}}{2}=\mathrm{m}$. $\frac{\mathrm{r}-\mathrm{t}}{2}=\mathrm{n}$ observe if you can find m−any s {illeg} 4m+any s {illeg} 4m+2n+any s in ye rank preceding those or 4m−2n{illeg}|+|any s in ye rank following those or|&| yn 9m+3n+any v in ye 2d rank preceding those & 9m−3n+any v in {illeg}|y|e 2d rank following them. If so try ye division by $\mathrm{x}3-\mathrm{m}\mathrm{x}\mathrm{x}+\frac{1}{5}\mathrm{s}+\mathrm{n}\mathrm{x}-\frac{1}{5}\mathrm{s}$.

Or yet better. Do not add {illeg}|&| subduct ye divisors from 9, 4, 1, 0, 1, 4, 9 but try if of those in {illeg}|y|e first & last rank ye difference of any two eaven or two odd ones be divisible by 6. Call $\frac{1}{2}$ that difference G & $\frac{1}{2}$ ye summ of ye same terms H. Then {illeg}|try| in ye middle collumn there be any terms wch subducted from H, or added to it produces a number divisible by {illeg} 9. Call that \term/ + {illeg} c if it be subducted or − {illeg} c if added. [Then try if in that column \next/ before or after ye middlemost wch has least /fewest\ divisors there be any te{rm} {illeg} wch added |to| or subducted from $\frac{1}{3}\mathrm{G}$ produces a number divisible by 9] Then making & ye summ or difference {illeg}K, & putting $\frac{1}{9}\mathrm{K}-3=\mathrm{a}$, & G+ /\=b. Try if you can find 8−4a+2b−c {illeg}|on| ye {illeg} |2d| rank next above ye middle one or 84|+|4a+2b+c in ye \2d/ rank 2d next \ra{illeg}/ below {illeg}|it|. If so {illeg}|tr|y ye division by x3{illeg}ax2+bx−c. Or thus against those divisors added & subducted from 20|7|. 8. 1. 0. 1. 2|8|. 27. set 9a+ {illeg}b+c. 4a+2b+c. a+b+c. c. a−b+c. 4a−2b+c. 9a−3b+c. then choose |ye| three ranks \of the fewest terms/ & in them those numbers \one in each/ {b}by wch get ye valor of a, b, & c. & those gotten will for ye give you numbers to be sought in the other ranks, wch if you find there try the division by x3−axx−bx−c. otherwise \c{illeg}|l|ose/ there other numbers out of the first \same 3/ rak|n|ks, & doe so till you have gone through all variety.

Note that if ye last term of ye æquation divided by {c} be p ye last but one q, Then if c & $\frac{\mathrm{q}}{\mathrm{c}}$ {must {illeg}ust} have a common divisor wch divides {illeg}|{n}|ot q, that c is to be rejected. Also if $\frac{1}{4}\mathrm{b}×\frac{\mathrm{p}}{\mathrm{c}}$ is greater then {+} {×} ye grea or $\frac{1}{4}\mathrm{a}×\frac{\mathrm{p}}{\mathrm{c}}$ be greater then {illeg} {illeg} ye greatest term of ye æquation yt b or a is to be rejected.

Note also yt if ye æquation to be reduced be of six dii|m|ensions it is not necessary both to ad & substract ye divisors fro 20|7|. 4|8|. 1. 0. {illeg}1. {illeg} 8. -27

Or thus best. L|I|f t{illeg}|he| Let ye numbers wch arose by substituting 2. 1. 0. -1. -2 for x be G. H. I. K. L. Seek If I end not in 5 or {illeg}|{0}| substitute 10 & -10 for x & let ye numbers arising be F. {illeg}|& M{illeg}| but if I end in 5 {illeg}|or| 0 increase ye root of & H or K do not, increase \or decrease/ ye root of ye æquation or little by an unit. Do so also if {H} be an eaven number and H or K an o{illeg}|dd| one wth fewer divisors & then substitute 10 {&} -10 for x{.}

<88v>

## How numeral æquations are to be reduced by divisors of 3 or 4 \or more/ dimensions

Substitute 5, {illeg}|{6}|, 3, 2, 1, 0, -1, -2, -3, -{illeg}, -5 {illeg} & also 4 & -4 if need be, for x, & find \all/ ye divisors of \suppose/ ye resulting terms, wch set to be F, G, H, I, K, L, M, N, O, P, Q. Find all their divisors set those of F & Q together \by pairs/ whose last figures are equal . Let ye sum Take any one of these pairs \or differ by 5. Gather the summs & differences of these pairs/. Let Let {sic} their t summ \of any two/ be R, \the tenth parts of/ their {illeg} difference S \if their last figures/, But for finding this {illeg} differences you must subduct ye divisor of Q from ye divisor of F not yt of F from yt of Q so yt S will be negative if ye divisor of Q be ye greater < insertion from the left margin > ✝ And if ye æquatiō be not of more yn {{illeg}}5 dimensions so yt {{illeg}} it must be divisible (if at all) by a divisor of 2 dimensions set down xx∓Sx $\mp \frac{1}{2}⁤\mathrm{R}-25$ to be tried for such a divisor. Where R & S must have ye same signes if s {illeg} ye divisor of F{illeg} was greater then ye divisor of K otherwise contrary signes. < text from f 88v resumes > Quadruple ye divisors of L. L {F} Take one & if {illeg}|th|e two last figures of any one be the same wth ye two last figures of 2R, take it from 2R. Let ye residue {be T} divided by 100, be T. Or if ye two last figures of any one added to ye two last figures of 2R make 100 add it to 2R & let ye summ \divided by 100/ be {illeg} T & let ye number \whose quadruple is/ added to or subducted from 2R be a And [if ye æequation be not of more then 5 dimensions so that it must be divisible (if at all) by a divisor of \but/ two dimensions, set down xx+{25}+Sx. a to be tried for a divisor. But] if ye æquation be of 6 or 7 dimensions & no more set down x3+Txx+{S-25}+a to be tried Where note that S & T must be negative if they were found so above {illeg}|&| a must be negative if it was added to 2R to make T, or els {illeg}|ff|irmative if it was subducted. |& the same is to be observed of the signes in ye following operations.|

But if ye æquation be of more then 7 dimensions then look in ye columns{illeg} among ye divisors of K for a number wch added to or subducted from {illeg}S+T+a gives a number divisible by 24. (all This number divided by 24, call v & the divisor wch gave it call β And set down $\begin{array}{lllllllll}{\mathrm{x}}^{4}& +& \mathrm{v}{\mathrm{x}}^{3}& +& \mathrm{T}\mathrm{x}\mathrm{x}& +& \mathrm{S}{\mathrm{x}}^{}& +& \mathrm{a}\\ & -& 1& -& 25& -& 25\mathrm{v}& & & \\ & & & & & -& 25\end{array}$ to be tried for a divisor \if the æquation be not of more then 9 dimensions/. Where {illeg}|{v}| must be negative if ye number β was greater then S+T+a & subducted from it. so S+T+a {illeg} in {illeg}sed or {illeg} if it was so above.

But if ye æquation be of more then 9 dimension then Cook in ye column among the divisors of M for a number wch added to or subducted from −S{illeg}\+/T+a gives a number divisible by 24. This number divided by 24 call W & ye divisor wch gave it {γ}. And set down $\begin{array}{lllllllllllllll}{\mathrm{x}}^{5}& +& \frac{1}{2}\mathrm{w}& {\mathrm{x}}^{4}& +& \frac{1}{2}\mathrm{v}& {\mathrm{x}}^{3}& +& \mathrm{T}& \mathrm{x}\mathrm{x}& +& \mathrm{S}& \mathrm{x}& +& \mathrm{a}\\ & +& \frac{1}{2}\mathrm{v}& & -& \frac{1}{2}\mathrm{w}& & -& \frac{25}{2}\mathrm{v}& & -& \frac{25}{2}\mathrm{v}& & & \\ & & & & -& 26& & -& \frac{25}{2}\mathrm{w}& & +& \frac{25}{2}\mathrm{w}& & & \\ & & & & & & & & & & +& 25& & & \end{array}$ to be tried for a divisor if ye æquation be not{illeg} of more then 11 dimensions or supposed divisible by {illeg}|{}| Divi{illeg}|so|r of not more {yn} 5 dimensions At ita in infinitum pergitur

Now the trial of these divisors is this suppose ye divisor be a+bx+cxx+dx3+ex4+fx5 &c And observe if {illeg} be among ye divisors of L{,} if {illeg}|th|is Divisor ascend but to {two} dimensions{;} & a+b{illeg}{+c+{illeg}+1}+c+1 among ye divisors of K if {illeg}ascend but to two or 3 dimensions & a−b+c{−d−1} {illeg} divisors of {illeg} if it ascend but to 2, 3 or 4

<89r> dimensions & a+2b+4c|+d+e+1| among ye divisors of I if it ascend not to more then 5 dimensions & a−2b+4c{illeg} \−d+e−f+1/ among ye divisors of N if it ascend to no more then six dimensions & a{illeg}+3b+9c+27d+81e+243f|+|&c among ye divisor{illeg}s of H if it ascend not {illeg}|to| more then 7 dimensions, & so in infinitum In all wch put c=1 & d, e, f = 0 if ye divisor be but of 2 dimensions or d=1 & e, f, = 0 if but of 3 dimensions, & so on.[58] And when you have tried all ye divisors wch may be found by this {rale}, {If}|{&}| rejected those wch will not hold this trial: if there remain more or if ye æquation be not reducible divisible by \any of/ those wch remaine, you may conclude ye {illeg}|æ|quation irreducible by {illeg} any rational divisor.

<89v>

## Veterum Loca solida componere restituta.

Cartesius de hujus Problematis confectione se jactitat quasi aliquid præstitistet a Veteribus tantopere quæsitum̄, cujus gratia putat Apollonium {tractatum} \libros suos/ de Conicis sectionibus scripsisse. Sed {illeg}|cu|m tanti viri pace r{e}m Veteres: neutiquam latuisse cred{ide}rius{.} Tradit enim Docet enim Pappus modum ducendi Ellipsim per quin data puncta et eadem est ratio in cæteris Con. sect. Et si Veteres norint ducere Conic{illeg}m|{i}s| sectionem per quin data puncta, quis {nori} videt eos cognovisse compositionem loci solidi{.} Imo vero eorum methodus longe elegantior est Cartisiana. Ille {illeg} rem peregit per calculum Algebraicum qui in verba (pro more Veterum scriptorum) resolutus Adeo prolixum et perplexum evaderet ut nauscum crearet ne{illeg} posset intelligi. At illi rem peregerunt per simplices quasdam Analogias, nihil judicantes lectu dignum quod aliter scribereter, & proinde celantes Analysin per quam Constructiones invenerunt. Ut pateat hanc rem eos non latuisse, conabor inventum restituere insistendo vestigijs Problematis Pappiani. In {illeg}|cu|m finem propo{illeg}|no| hæc Problemata.

[59] 1 Conicam sectionem per quin \tria/ data puncta \ABC/ describere quæ datum centrum \O/ habebit. AO dat A duobus punctis \AB/ ad centrum O age rectas AO BO et {eos} produc \AO/ ad \AO/ Q|O|P ut sit AOP=AO|.| , et OQ {illeg}BO, et puncta P et Q erunt ad curvam. A tertio puncto C age CS parallelam AO et occurrentem OB in S et cape ST ad AO{illeg} $\frac{SB×SQ}{SC}\colon\colon {AO}^{\mathrm{q}}.{B0}^{\mathrm{q}}$, et erit etiam punctum T ad curvam. Biseca TC in V, et ipsi OV parallelam age CR occurrentem AO in R erit CR ordinatim applicata ad diametrum AP. Et latus rect{um} erit ad AP ut RCq ad AR×R|A|P{,|.|} Figura existente Ellipsi si punctum R |cadit intra A et P, aliter Hyperbola.|

[60]

2 Per data quinque puncta \A, B, C, D, E/ Conicam Sectionem describere {illeg}ge Junge quatuor puncta A, B, C D duo puncta AC et alia duo BE sit jungentium intersectio K. Ipsis AC BE age parallelas Dg DG occurrentis AC{,} BE, AC in punctis H et F, et facto $BK×KE.AK×KC\colon\colon \frac{BH×HE}{DH}.HI\colon\colon FG.\frac{AF×FC}{FD}$, puncta G et I erunt ad curvam. Biseca ergo ID et AC in m et n ut et BE ac GD in p et q et actarum mn, pq intersectio O erit centrum Conicæ sectionis|.| nisi ubi Habito autem centro, describe figuram per puncta A, B, C ut supra, \Quæ quidem ellipsis erit si punctum R cadit inter A et P. Aliter Hyperbola/ Quod si mn et pq parallelæ sint Figura erit Parabola, et rectæ AO Cujus determi{illeg} neutiquam difficilior. Quo casu produc PQ ad V ut sit BPq.GQq ∷PV.QV. et erit PV diameter V vertex et $\frac{{BP}^{\mathrm{q}}}{PV}$ latus rectum figuræ {illeg}.

Et hæc videtur methodus naturalissim{e} {illeg} solvend problem{a} {tam quod} non {factum} quod ad{illeg}dum simpl{illeg} {sit} {illeg} pars Problemat{illeg} {illeg} {{illeg}} <90v> ab ipso Cartesio proponitur) est invenire punctum aliquod (data{m} habens conditionem et secunda pars \deinde cum infinita sint ejusmodi puncta/, determinare omnia locum in quo ejusmodi puncta \ista/ omnia reperiuntur. Quid autem magis naturale quam reducere difficultates \hujus/ posteri oris partis ad difficultates prima prioris determinando locum ex paucis punctis inventis. Proinde cum veteribus constiterit rati ducendi conicam sectionem per quin data puncta, nullus dubi{illeg}|ta|verim eos hoc medio loca Solida composuisse.

< text from f 89v resumes >
<90r>

## Problema

Data quavis refringente superficie CD quæ radios a{illeg} puncto A divergentes quomodocun refringat: invenire aliam superficiem EF quæ refractos omnes DF convergere faciet ad aliud punctum B.

Haud secus Problema resolvitur ubi tres sunt vel plures refringentes superficies.

<91v>

## De resolutione Quæstionum circa numeros.

Primo numeri quæsili redigendi sunt ad æquationem secundum conditiones quæstionis, Deinde exponendi sunt per basem et ordinatam curvæ lineæ quam æquatio illa designabit. Sit curva ista DC et numeri AB, BC, curva existente tali ut numerus BC ordinatim applicatus ad basem A numerum AB in dato angulo ABC \semper/ terminetur ad eam. {illeg} Deinde inquirenda erunt puncta curvæ quæ efficient numeros AB, BC rationales. C{illeg}|as|us autem in quibus hoc fieri deprehendo sunt sequentes.

1. Si numeri in æquatione non ultra gradum qquadraticū ascendant ita ut curva sit Conica {illeg}|{s}|actio: et detur aliquod {illeg} punctum F \F/ in \curv{illeg}|{æ}| quo/ efficitur ut numeri \AH. HF./ sint rationales {illeg} ex hoc unico exemplo regula generalis sic elicietur. In AH cape HE cujusvis rationalis longitudinis, age EF, secet hæc curvam in G et demissa GH pararallela {sic} CB numeri AK, KG erunt rationales. Si Si punctum F reperitur in linea AH, FH existente nullâ, tunc cape HN cujusvis rationalis longitudinis, Erige NE parallelam K|B|C, & cujusvis etiam rationalis longitudinis. Age HE occurrentem curvæ in G et erunt AK KG rationales.

Quoniam {illeg}|{ita}| hic unicum s{illeg}|{a}|l|{illeg}|tem exemplum requiritur, primo inquarendum est ejusmodi exemplum, dein regula generalis inde elicietur ut supra. {illeg} Exempl{illeg}|{i}| autem inveniend{i} notand{illeg} est \hec erit methodus{.}/ ni melior occurrat. Sit æquatio axx+bxy+cyy+dx+ey+y=0 ubi x et y designant numeros. Et si terminus f desit, punctt|u|m istud erit A. Si x in æquatione{illeg} axx+b|d|x+f=0 sit rationale punctam erit duo erunt in AB. Si cyy+ey+f sit rationale puncta duo erunt in recta quæ ducitur ab A parallela BC. Si bb−4ac sit quadratus numerus affirmativus tunc curva erit Hyperbola Et recta ducta a puncto A {p} vel B parallela Asymptoto {illeg} secans curvam in G exhibebit numeros AK GK rationales. Sunt alij innumeri casus quibus enumerandis non immoror

2 Si æquatio ascendit ad tres dimensiones, et tria habentur exemplo rationalia non in arithmetica progressione possunt inde innumera alia reperiri {illeg} sint enim P, Q, R puncta Curva ad ista exempla Junge PR, \RQ, PQ/ et punctum S, \T, Y/ ubi PR, \RQ, PQ/ secant curvam dab{illeg}|un|t alios duas \tres/ numeros. Dein junge QS et punctum x dabit alui quo QS secat curvam dabit aluim numerum. et Sic in infinitum.

< insertion from the bottom of the page >

[62] Punctum B circumferentiam contingit a quo si duæ agantur rectæ una ad datum punctum A altera {illeg}|in| dato angulo c ad rectam positione dat{u}m {illeg}|&| ad dat{u}m punctū terminatam DC, quadratum prioris ABq æquales sit rectangulo sub posteriore abscissa ac data DC×E

Vel sic. Si ABq=FC×E punctum B contingit circ.

< text from f 91v resumes >
<92r>

## Loca plana.

1. Si datur linea AB, et angulus ACB, punctū est in circulo per A et B transeunte.

2. Si dantur puncta A, B et ratio AC ad BC punctum C est in circulo Et si |C| est in circulo recta BC vergit ad |locum B.|

3 Si a dato puncto A ad rectam positione datam BD ducatur r{{illeg}|ec|}ta quævis AB et in ea {illeg}atur punctum C \{si}/ ea lege ut detur rectangulum BAC punct{u}m C est in circulo, transeunte per A

4 Si detur punctum A et rectangulum BAC vel \BA×DC/ et punctum B \ac D/ est ad circulus, etiam punctum C erit ad circulum.

5 Si detur punctum A et proportio AB ad AC |vel AD ad DC.| et punctum B est ad circulum etiam C erit ad circulum. Sin punctus B est ad rectam erit C a rectam

6 Si dentur puncta duo A, B et differentia quadratorum ACq−BCq punctum C est ad rectā

7 Si dentur puncta duo A, B et summa quadratorum ACq+ BCq et punct C erit ad Circulum

8 Si dentur puncta plura A, B, D et quadratorum ex lineis AC, DC, BC vel quadratorum quæ ad ips{illeg}|a| sunt in datis, rationibus summa vel quod subducendo aliqua ab alijs restat vel proportio aggregati unius ad agr|g|regatus alterum aliud, punctus C erit in circulo.

10 Si da\n/tur puncta B, D, E et productam BD secat quævis EC in A, et sit rectangulum EAC æquale rectangulo BAD erit punctū C ad circulum.

11. Si dantur puncta E, D et concurrant rectæ CE BD ad rectas positione datam AF ex|t| {detur} rectangul{illeg}|or|us. EAC, DAB differentia vel nulla est vel æqualis rectangulo sub AF et recta data: sit autem punctum B ad circulus erit punctus C ad circulum.

12 Si detur quadrilaterum ADE quod cujus anguli oppo in ci{illeg}|rc|ulo inscribi potest, et a puncto C ad latera ejus ducantur in datis angulis rectæ quatuor CF, CF|G|, CH CI ita ut rectangulus sub duabus CG×CI datam habeat rationem ad rectangulus sub alijs duabus CF×C{illeg}H, et in uno aliquo casu pu{illeg}|nc|tus C est in ci{illeg}|rc|umferentia circuli transeuntis \per/ ABDE, semper erit in circulo illo

A{ff}ini{a} sunt 1, 2, 6, 7, 8. Item 3, 4, 5. Item 10. 11

13 Si dentur puncta A, B. Et centro circulo|u|s centro A radio dato AD descriptus utcun secetu{illeg}r {illeg}|{an}| D et E a ducta AC ac demittatur normalis CF, c|s|it c|a|utem rectangulum DCE=2ABF{illeg} punctum C erit in circulo.

Si trapezij ABDE anguli A ac D recti sint et ad AB et AE{.} demittantur perpendicula CF CI a puncto quovis C secantia BD ac DE in {{illeg}} H et G erit rectan et fuerit rectangulum FCH=GCI erit punctum C in circulo transeunte per puncta ABDE. Et viceversa. {I}dem eveniet si ad latera singula \{illeg} a puncto C/ demittantur perpendicula

<92v>

[63] Si {illeg}|{i}|n circulo quovis {illeg}|AB|CD inscrib{illeg}|{a}|tur trapezium A B C, D, et a circumferentiæ puncto quovis E ad latera trapezij ducantur lineæ EF, EG, EH, EI constituentes cum lateribus conterminis AB, BC parallelogrammum EFBG et cum alijs duobus lateribus conterminis AD, DC parallelogrammum EHDC. quod sub ductis ad opposita duo latera continetur rectangulum GE×EH æquale est rectangulo EF ×EC sub ductis ad reliqua duo latera contento.

|NB| Idem eveniet si {illeg} puncto E ad latera trapezij demi{illeg}|tt|antur pependicula. Ut et si du{illeg}|ct|{illeg}|{illeg}| ad duo latera \contermina/ [64] EH, EC ad AD, CD æquales angulos cum EHD, ECD vel EHA, ECD cum ipsis constituant conficiant et ductæ ad altera duo latera EF EG æquales angul{illeg}|os| cum {illeg}|ipsi|s.

Et hinc si ductæ quosvis angulos con{illeg}|fician|t cum lateribus trap{illeg}|ez|ij, et rectangula GEH, FEC|I| sunt in data ratione facile est cognoscere utrum punctum E {illeg}|si|t in circus ferentia circuli. Nam ad latera duo opposita AB, CD, duc EK, EL in angulis EKB ELD æqualibus illis quos ducta ad altera latera opposita {faciunt cu} in quibus ad altera duo latera duc{illeg}|t{a}| sunt lineæ æqualibus|.| {illeg} \Verbi grati{a}/ EK in angulo AKE=ang CGE et EL in angulo DELE =ang DHE. Et {illeg}|se| ratio rectanguli FEI ad rectangulum GEH componitur ex ratione FE ad KE et EI ad LEL ita ut rectangula KEL GEH æqualia sint, erit punctum E in circulo {illeg}|se|cus non erit in circulo.

[65] Si ABE sit circulus et detur |A &| rectangulum BAC erit punctum C in recta. |Et si punctus C in recta sit converget recta CA ad datus punctus A. Et in fig 2 si datur rectang BAC erit C in circulo.|

[66] Si punctarum A linearum AD, BD, CE poli ABC in linea recta sunt, et puncta {illeg}|in|tersectionum duo DE lineaa|s| rectas describunt tertia intersectio F lineam rectam describet. Idem eveve {sic}niet si linea DE parallela est lineæ BC. Ut et si si {sic} AB puncta |A|BC non sint in directum si modo loca punctorum D, E se secant in recta BC.

[67] Si dati anguli DBC|A|, DCA circa polos B, C volvantur et angulis ABC|D|, ACD æquales capiantur CBF BCF sit punctum D in recta transeunte per punctum F vel etiam in conica sectione transeunte per puncta tria BCF erit punctum A in recta. Et si D in recta sit non transeunte per punctus F aut in con. sectione transeunte per duo E tribus punctis B, C, F, non autem per omnia tria, punctus A erit in conica sectione. Si per unicum tantum E tribus punctu|is| BFC transit conica sectio, {illeg} punctum A erit in curva primi gradus tertij gener{is} Si per nullum erit primi generis quarti gradus.

< insertion from the bottom of the page >

[68] Si circ{illeg} dua \{illeg}E AD/ se secuerint in A et per A agatur recta ACDB et {illeg} CD ad DB {illeg} in circulo transeunte {illeg} intersecti{illeg} {illeg} priorum.

Si recta CD {illeg} {datas} {illeg} {data \{illeg}/} {illeg} et secetur {illeg} cujus {data} est ratio {illeg}

{Si secetur punctum} {illeg} AD, BD, CD {illeg} \vel AB, BC {illeg}/ {illeg} agei B circulum {illeg}

< text from f 92v resumes >
<93r>

[69] Datis positione lineis AE BE et punctis AB: Si {illeg} recta quævis CD secat alteras in C, D ea lege et rectangulum AC, BD æquætur dato rectangulo AE×BE, compl{e} parall{o}logrammum {illeg}AEBP et locus ad quem recta CP vergit erit punctum P.

[70] Si AB datur positione et longitudine & AD BC longitudine sint CE DE æquales cape AF.BF∷AD.BC et {illeg} CD verget ad datum {pun} locum puncti F.

[71] Se|i| a datis punctis A, B ductæ AC.|,| BC datam habeant summam \vel differentiam/ N: d|D|uc CD parallelam AB et in ratione ad AC qual|m| habet N ad AB et punctum D erit in recta quæ perpendicularis est ad AB. Debet vero CD ad plagam versus A duci ubi datur summæ AC+BC, ad plagam versus B ubi datur differentia. AC−BC vel BC−AC.

[72] Si datur circulus ABD et rectangulum ACA|B| {datam} punctus C erit in circulo idem habens centrum cum circulo ABC|D|

[73] Si per data puncta AB transit circulus secans \in E/ rectam ipsi AB perpendicularem et arcui BE æqualis sit arcus B{illeg} EF erit F in circulo cujus centrum est A et si F in {illeg} tali circulo sit, {illeg} {illeg} et bisee|c|etur BF in E erit E in recta.

## De Loco rectil{í}neo.

Si locus crura duo infinita opposita habet, et non plura, aut rectus est, aut tertij, quinti, septimi vel imparis alicujus generis curva linea.

Si rectæ locum tangentis plaga determinat est rectus est locus.

Si recta nulla ad plagam infiniti infiniti cruris tendens potest locum secare rectus est locus.

Si per datum loci punctum recta transiens non potest locus alibi secare rectus est locus.

Si a loci puncto quovis ad rectas duas \positione datas/ in datis angulis demittantur aliæ duæ rectæ, et progrediendo per additionem subductionem et rationes datas, alterutra demissarum ex altera assumpta vel utra ex assumpta tertia determinari potest rectus est locus.

Si in recta quavis ad datam non infiniti cruris plagam tendente determinabile est loci punctus per simplicem Geometri{illeg} rectus est locus.

Si rectæ per punctum datus extra locum transeuntis et loci intersectio determinabilis est per simplicem Geometriam, rectus est locus.

Si rectæ cujusvis \{illeg} / \cujusvis assignatæ/ et loci intersectio determinabilis est {illeg} per simplicem Geometriam rectus est locus.

<93v>

[74] 1 A datis punctis A, B \ductæ/ conveniant AC BC \in C/ et si dentur ipsorum A, B, {illeg} summa vel differentia (loc. \C/ solid.) proportio (loc. circ) differentia quadratorum (rect{)} summa quadratorum vel ad|l|iud quodvis compositum ex quadratis (circ) \rectangulum (lineare) Area {illeg}|AB|C (rect)/ angulus ACB (circ) differentia angulorum A|B|{illeg}\/B|A| ({solid}) vel \summa/ 2A|B|+C (solid \Hyperb &/ /rect\{illeg} diff A|B|−C vel summa {illeg}A|2B|+B|A| \vel 2C+A/ (vel lineare \ellipsis/) \2/A=B \vel {illeg}{et}. 3A+C/ (Hyperb.) 2A=C (Lin {illeg}) 2C=A (Lin {illeg}) A =B (rect) D|B|=C (circ)

[75] 2 Detur Ang AD positione et ang. DAC et punctus B. Et si dentur etiam ipsorum AC BC summa, differentia (Parab \erit C in/) Proportio (loc. sol.) summa quadrati{illeg}s differentia quadratorum (Parab) summa quadratorum vel aliud quodvis compositum ex quadratis (Loc solid) rectangulum (Lineare)

[76] $\left\{\begin{array}{l}\text{3 Detur positione DA et punctum P. Et si}\phantom{\rule{0.3em}{0ex}}\text{detur AC (Locus}\\ \text{Conchoid)}\phantom{\rule{0.3em}{0ex}}\text{d}\text{iff AC}\text{±}\text{AP (loc}\phantom{\rule{0.3em}{0ex}}\left\{\begin{array}{c}\text{circ.}\\ \text{linear.}\end{array}\right\)\phantom{\rule{0.3em}{0ex}}\text{proportio}\phantom{\rule{0.3em}{0ex}}\text{PA ad AC}\phantom{\rule{0.3em}{0ex}}\text{(loc. rect.) rectangulum}\\ \text{APC (circ) rectang}\phantom{\rule{0.3em}{0ex}}\text{P}\text{AC (lin) rectang PCA (lin.)}\phantom{\rule{0.3em}{0ex}}{AP}^{\mathrm{q}}±{PC}^{\mathrm{q}}.\\ {AP}^{\mathrm{q}}±{AC}^{\mathrm{q}}.\phantom{\rule{0.3em}{0ex}}{PC}^{\mathrm{q}}±{AC}^{\mathrm{q}}\phantom{\rule{0.3em}{0ex}}\text{(lin.)}\\ \\ \\ \text{4 Detur circ AD, punctum}\phantom{\rule{0.3em}{0ex}}\text{P}\phantom{\rule{0.3em}{0ex}}\text{et si detur etiam}\phantom{\rule{0.3em}{0ex}}\text{APC}\\ \text{vel}\phantom{\rule{0.3em}{0ex}}\frac{AP}{AC}\phantom{\rule{0.3em}{0ex}}\text{(erit C in loc. Pla}\text{n}\text{.)}\end{array}\right\$

<94v>

## Quæstionum solutio Geometrica.

1 Angulum datum DAB recta datæ longitudinis CB subtendere quæ ad datum punctum P converget Cape PQ=CD et Q erit in circulo cujus centrum P radius PQ. Age QR{}∥AD et {Q}PRD∥AD|B| et erit PD.DC=AD−QR∷ ∷PR.QR. Ergo Q{illeg} in conica sectione est. Pone QR infinitū et erit $PR=\frac{PD×QR}{AD-QR}$ AD−P|Q|R.PQ|QR|∷PD.QPQ|R|. seu PR =−PD. Pone PR infinitus et erit PD+PR.PR∷AD.QR ergo AD=QR. et AB Asymptotos. alteras Cap{er}|{e}| ergo PS=PD et per S parallelam AD age alteram Asymtoton & {his} Asymptotis per punctus P describe Hyperb{o}lam secantem circulum prædictus in Q.

2 Inter circulum PDF et rectam DF ponere rectam datæ longitudinis BC quæ ad punctum P in circumferentia circuli datum converget. Biseca DF in E. Age PD,PE,PF. Cape PQ=BC. Age QR ∥DC & occurrentem PE in R. Et erit PR.{illeg}PQ∷PE.PC. PR.PE∷PQ(BC).|PC∷RQ.EC.| {illeg}Et PQ.|(|BC).FC∷DC.PC Et {PR×PQ} Ergo PQ×PC=FC×DC =ECq−EFq PR.PQ Et $BC.\frac{PE×RQ}{PR}-EF\left(FC\right)\colon\colon$ $\frac{PE×RQ}{PR}+EF\left(DC\right).\frac{PE×BC}{PR}\left(PC\right).$ Seu BQ|C|,PR.PE,PR|RQ|−PR,EF∷ $\frac{PE×RQ}{}+PR,EF.PE,BC.{BC}^{\mathrm{q}},PE,PR={PE}^{\mathrm{q}},{RQ}^{\mathrm{q}}-{PR}^{\mathrm{q}},{EF}^{\mathrm{q}}$ Si PR infinitum{Q} Ergo Q locatur in Conica sectione cujus diameter PR, ordinata RP|{PR}|. Sit RQ=0, erit PR=0 et $\frac{-{BC}^{\mathrm{q}}PE}{{EF}^{\mathrm{q}}}$ In EP producta cape ergo $PS.\frac{1}{2}PE\colon\colon {BC}^{\mathrm{q}}.{EF}^{\mathrm{q}}$ et erit S centrum et P vertex figuræ. Pone PR infinitus et erit PEq,RQq=PRq,EFq, seu PE{:|,|}RQ=±EF,PR. Quare per S ipsis PD,PS age parallelas et hæ erunt Asymptoti figuræ His igitur Asymptotis per punctum P describe Hyperbolam, ut et cent{illeg}|ro| P radio PQ circulum & per eorum intersectionem Q age rectam PC.

Corol. si ang. PEC {illeg}|re|ctus est Problema planus erit. Nam circuli centrum incidit in {illeg}|ax|em figuræ.

3 A dato puncto P rectam PC ducere cujus pars BC inter circulum et recta productam diametrum æquab DF æquabitur \semi/ diametro EF. Age EF ac demitte ⊥ PG,BH. Est EH. HB∷GC(GE+2EH).GP. Ergo punctum B in Hyperbola est. Pone {illeg} EH=0 et erit HB×GE=0 adeo HB=0. Quare Hyperbola transit per punctum E. Pone EH infinitus et erit EH.HB∷2EH.GP. Ergo $\frac{1}{2}GP=HB$. Pone HB infinitus ergo et erit EH.GE∷HB.\/GP−2HB∷HB.−2HB Ergo $\frac{1}{2}GE=-EH$. Quare biseca PE in S et per S age Asymtotos parallelas EH et HB et per punctum E vel P describe Hyperbolam secantem circul{u}m in B|{D}|. Et per B age PC.

Corol. Hinc si ang PEG semirectus erit PE axis Hyper{illeg}|bola| adeo Problema in eo casu planum.

<95v>

## Quæstionum solutio Geometrica

1 Datis trianguli cujuvis angulo latere et summa vel differentia {re}liquorum laterum datur triangulum{,} Detur latus AB reliquorum laterum AC+BC summa {illeg}vel differentia AD. Si detur angulus datus dato lateri conterminus est, sit iste A. Et dabitur triangulum DAB. Angulorum vero |C|DB ADBD differentia in priori casu summa in posteriori est ang ABC.

Si{t} angulus datus dato lateri oponitur, sit iste C {illeg} {illeg}datum{illeg} ang DCB dabuntur ang CDB CBD est dabitur tr dabuntur anguli dabitur triangulum CDB specie. In triangulo autem ADB datis lateribus AB AD et ang D datur ang ABD. Unde datur Ang ABC ut ante.

2 Data differentia segmentorum basis uno angulo summa vel differentia laterum et uno angulorum datur triangulum. Nam si datur summa laterum dabitur ratio differentiæ laterum ad basin si differentia debitur ratio summæ laterum ad basin. In utra casu \Ex ratione utravis & uno angulorum {illeg}/ per problema superius datur triangulus specie. Deinde ex data differentia segmentorum basis et ex data ratione differentiæ segmentorum \basis/ ad latera dantur latera.

3 Data summa {illeg}|vel| differentia laterum uno angulorum et ratione basis ad perpendiculum: ex duobus posterioribus dabitur triangulus specie ex priori dabitur etiam magnitudine.

4 Data summa vel differentia laterum uno angulorum et area: \ex area/ rectangulum laterum datum angulum comprehendent{ur}. Si istorum summa \vel diff./ datur ad|a| quadrat{us}|o| summæ aufer duplum rectangulus vel ad quadratum differentiæ \{laterus}/ adde {qu}duplus rectangululum {sic} et habebitur priori casu quadratus {illeg},|differenti{æ}| posteriori quamdratus differentiæ s summæ laterum: Ex dat{illeg}|is| autem summa ac differentia laterum dantur latera. Si angulus datus basi conterminus est problema erit solidum.

5 Datis angulo \{illeg}/ /A\ latera \AC vel BC/ et differentia segmentorum basi|s| AD dabitur triangulum ADC |ut| et angulus B \quod est/ complementus est anguli ADC.

Si detur angulus verticalis C laterum alterutra AC vel BC et segmento basis AC: quiescant BC,AC et punctum D in {circulo} erit radio CB centro C descripto. [Et et AB.$\frac{{AC}^{\mathrm{q}}-{BC}^{\mathrm{q}}}{AB}=AD$ ergo D determinatur per Geometriam planam simplicicem. Sed et ubi punctum D incidit in B positio rectæ AD|B| determinatur. Ergo locus puncti D conica est sectio] {illeg}|Ut| et in Chonchoide {illeg}|Po|lo {B} {illeg}|asy|mptoto AC intervallo AD descrip{illeg}|ta|.

Vel sic. Dato angulo AC{D|B|} datur summa ang: A+B. seu A+CDB Aufer hoc de duobus rectis ac dabitur differentia ang ADC−A. Unde datur triang. per sequ. Prob.

7 Datis \basi &/ differentia angulorum ad basin una cum basi et \una cum/ latere al{illeg}|tera|tro vel summa differentia ratione \laterum/ aut summa vel differentia {lo} laterum aut area, perpendiculo vel segm{en}{illeg}|ta|to basis aut summa vel differentia lateris alterutrius et perpendicula vel segment{i} basis. &c Datur triang. Nam data basi et angulorum ad {illeg}|Basem| differentia, {illeg} C erit ad Hyperbolam; et ex dato tertio, punctum C erit ad recta aut circulus aut conican aliquā sectionem.

8 Ubi datar angulus verticalis et differentia segmentorum basis et tertium aliquod, habebitur aliud triangulum ADC ubi datur {illeg} differentia angulorum ad ba{illeg}|se|m, et tertium aliquod.

9 Dat{illeg} basi ratione laterum et tertio quovis ut ⊥ segment{illeg} basis{,} angulo aliquo{,} ratione ⊥ ad lat{eri} {peh} {illeg} ad segmentum d|D|{tis} {illeg}. Nam {illeg} data {illeg} lat. Dat{ur} circulos {illeg} {illeg}

<96r>

## Quæstionum solutio Geometrica. Prob 1

Circulum \ABE/ per data duo puncta \A, B/ describere quæ rectam FG    positione datam continget.

Junge A{,}B. D biseca eam in D. Erige normalem DF occurrentem FG in F. {illeg} Produc AB donec occurrat FG i{n}

|Sit| E punctum contactus. Produc AB donec occurrat FG in G et erit EG medium proprortionale inter datas AG,BG.

## Prob. 2

Circulum \ABE/ per datum punctum A describere qui recta duas FE, FH continget. {illeg}

Center F, radi{s} FA describe Recta FD biseca{illeg} angulum HFE. Ad FD demitte normalem AD et produc donec occurrat FE in G. Cape D ad B ut sit DB=AD et per puncta A, D describe circulum ut prius qui contingat rectam FE.

## Prob. 3

Circulum \ABE/ per data duo puncta \A, B/ describere qui alium circulum positione datum \EKL/ continget.

\Puta factum/Sit punctum contactus E. Linea contingens EM. et erit AM×BM=EMq=MK×ML. Divide ergo BK in M ut sit AM.MK∷ML.MB. Cape ME medium proportionale inter AM et BM et centro M radio ME describe circulum. Hic secabit circulum EKL in puncto contactus E. Recta autem BK sic secatur in M. Est AM= AB+MB,|.| MK=BK−BMB.ML=BL−{B}MB. ergo AB+MB.BK−MB∷BL−MB. MB. Et componendo AB+BK(AK).BK−MB∷BL.MB. et inverse AK. BL∷BK−MB.MB. et rersus componendo AK+BL.BL∷BK.MB.

Si AB non secat EKL {illeg} pro MK×ML scribe MGq+FGq−KF erit AM×BM{} AG−MG×BG−{illeg}G(AM×BM)=MGq+FGq−KFq {illeg} est AG×BG−FGq+KFq=AG+BG×MG. Seu 2AG.BL∷BK:MB. {illeg} \{Unde}/ cum sit BL.BO∷M{illeg}BN.BK, erit 2AG.BN∷BN.BM. Quæ solutio versalis est.

## Prob 4

Circulum \BDE/ per datum punctum \B/ describere qui datum circulus {illeg} & rectam lineam |AD| postione datam continget.

AB est 2CD−AH. NF est \2/CQ−NS posito CQ=CF=CS. Ergo HF est 2CD−{illeg} AB−NF est 2CD−AH−2CQ+NS seu. AB−NF+DQ est Adde 2DQ, erit AB+DQ−HF=NS−AH=$\frac{{AD}^{\mathrm{q}}}{AB}$ $=\frac{{NQ}^{\mathrm{q}}}{NF}-\frac{{AD}^{\mathrm{q}}}{AB}$. Dividendo|a| est ita \data/ AH in D ita ut $\frac{{DH}^{\mathrm{q}}}{NF}-\frac{{AD}^{\mathrm{q}}}{AB}$ dato{illeg} æquale sit, nempo dato AB(Hb)+DQ−HF, seu bk. {ADq}DHq×AB AD =AB, HK, bL. DH=A{D}{illeg} AH−AD.DHq=AHq−2DAH+ADq.AD×{illeg}Kq{−2A}{illeg}{AH} −2DAH+ADq{illeg}$\frac{HK×{AD}^{\mathrm{q}}}{AB}=HK,bk$. [A{B}{illeg}−HK(bK).AH∷A{illeg}|{B}|.AV.AHq−2DA{H} $\frac{{AD}^{\mathrm{q}},AH}{AV}=HK×BL$] $AH-2DA+\frac{AB-HK}{BAH}{AD}^{\mathrm{q}}=\frac{HK,bk}{AH}$. \Fact/ AB−HK.{illeg}AH∷BA.AV. & AH. $HK\colon\colon bk.Hp.AH-2DA+\frac{{AD}^{\mathrm{q}}}{AV}=Hp$. ADq−DA{illeg}V+D{illeg} AVq=AV×PV{illeg}\{P}/AD{}AV {illeg}PV={DV.}AD {illeg}AV{illeg}PV{illeg}−AD+AV{illeg}=AD{illeg}|{×}|PV =DV. Age ergo BK occurre{nt}em AH in VHK ad {H}A versus A si {HK} {illeg} versus {b} aliter {illeg} A {illeg} ad bk {illeg} {illeg}

<96v>

Nota etiam quod Problematis quatuor sunt sunt {sic} casus quorum duo sunt impossibiles ubi circulus datus et recta data se mutuò secant. Casus impossibiles sunt ubi punctum v cadit inter A et P.

Vel \in {a}ng GED/ agatur GD datam per A transiens posito AE quadrato, quære summam radicum Fd, FD Ad AD erige ⊥ DK age erit CK {illeg} F{illeg} Ergo AK summa illa, et CDq+CKq(DKq)+GDq=GKq Aufer BGq seu CKq et restabit CDq+GDq=BKq Datur ergo {illeg} summa AK. Quare cum ang ADK rectus; super diametro AK describe circulus secantem FE in D, d

Super datis rectis tribus AB, CD, EF, tria constituere triangula quorum vertices erunt ad \idem/ punctum G et anguli ad vertices AGB, CGD, EGF æquales.Super Junge AD, BC. {illeg}|Bi|seca eas in r, s. Produ AB {illeg} secent AB, CD se mutuo in t. Age tG∥sr. Idem fac in lineis CD FE.

Vel sic. super lineis AB, CD, EF describe similia segmenta quorumvis circulorum satis magnorum \ita ut se mutuo secent compl{illeg} segm. ad circulos {illeg}/. Per intersectionem circulorum AB, CD age rectam, ut et aliam rectam per intersectionem circulorum CD, FE: nam hæ rectæ se secabunt in puncto G

The Problem {illeg} \in Schooten/ de tribus baculis may be solved more easily by supposing ye Ellipsis to be a circle first & then reducing it to ye desired circle.

In triangulo DEF \dato ABC/ {illeg} aliud triangulum DEF {illeg} dato def simile inscribere cujus latus EF transibit per datum punctum G. Nemper vi|e|rticis trianguli DEF locus {illeg} est linea recta.

In data conica sectione ABCDE, trapezium ACDB inscribere cujus anguli op duo oppositi CAD CBD dantur et data puncta A et B consistunt. Vizt si locus puncti D est conica sectio locus punctis c erit linea recta.

<97r>

## [77] Ex observationibus proprijs Cometæ anni 168\0/.

A stella major et orientalior duarum in orientali \australi/ pede Persei, B stella minor earundem. AB stellarum distantia 1gr.46′6″. α, β, γ &c loca Comet{æ}

$\begin{array}{l}\text{Feb 25 hora}\phantom{\rule{0.3em}{0ex}}8⁤\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{vesp.}\text{Dist}\text{Aα}\phantom{\rule{0.3em}{0ex}}{2}^{gr}\text{.17′.42″. Ang BAα.}\\ \text{F}\end{array}$

$\begin{array}{lll}& \text{Dist.}\phantom{\rule{0.3em}{0ex}}\text{C}\text{ometæ}\phantom{\rule{0.3em}{0ex}}\text{stella A.}& \text{Angulus}\\ \text{Die}\phantom{\rule{0.3em}{0ex}}\text{Feb 25 hora}\phantom{\rule{0.3em}{0ex}}8⁤\frac{1}{2}\phantom{\rule{0.3em}{0ex}}\text{vesp}& A\alpha =104,\frac{7}{12}\phantom{\rule{0.3em}{0ex}}\text{part}={2}^{gr}.17\prime .42″& \text{BAα}\phantom{\rule{0.3em}{0ex}}{9}^{gr}.17\prime \\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Feb 25 hora}\phantom{\rule{0.3em}{0ex}}9⁤\frac{1}{2}& A\beta =103⁤\frac{3}{4}\phantom{\rule{1em}{0ex}}\phantom{\text{part}}={2}^{\phantom{gr}}.16\prime .36″& \text{BAβ}\phantom{\rule{0.3em}{0ex}}{9}^{\phantom{gr}}.27\prime \\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Feb 27 hora}\phantom{\rule{0.3em}{0ex}}8⁤\frac{1}{4}\phantom{\rule{0.3em}{0ex}}& A\gamma =77-& \text{BAγ}\\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Mart 1 hora}\phantom{\rule{0.3em}{0ex}}11& A\delta =52\phantom{\rule{3em}{0ex}}\phantom{\text{part}}={1}^{\phantom{gr}}.8\prime .30″& \text{BAδ}\phantom{\rule{0.3em}{0ex}}{31}^{\phantom{gr}}.40\prime \\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Mart 2 hora}\phantom{\rule{0.3em}{0ex}}8& A\epsilon =43\phantom{\rule{3em}{0ex}}\phantom{\text{part}}={0}^{\phantom{gr}}.56\prime .37″& \text{BAε}\phantom{\rule{0.3em}{0ex}}{43}^{\phantom{gr}}.8\prime \\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Mart 5 hora}\phantom{\rule{0.3em}{0ex}}11⁤\frac{1}{2}& A\zeta =& \text{BAζ}\\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Mart 7 hora}\phantom{\rule{0.3em}{0ex}}9⁤\frac{1}{2}& A\eta =& \text{BAη}\\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\text{Mart 9 h}\text{o}\text{ra}\phantom{\rule{0.3em}{0ex}}8⁤\frac{1}{2}& A\theta =78⁤\frac{7}{12}\phantom{\rule{1em}{0ex}}\phantom{\text{part}}={1}^{\phantom{gr}}.43\prime .30″& \text{BAθ}\phantom{\rule{0.3em}{0ex}}{143}^{\phantom{gr}}.00\prime \\ \phantom{\text{Die}}\phantom{\rule{0.3em}{0ex}}\phantom{}\text{Mart 9 hora}\phantom{\rule{0.3em}{0ex}}12& A\iota =80⁤\frac{1}{3}\phantom{\rule{1.7em}{0ex}}\phantom{\text{part}}={1}^{\phantom{gr}}.45\prime .46″& \text{BAι}\end{array}$

<98r>

## Observationes Comet{illeg}|æ| habitæ ab Academia Physicomathematica Romana anno 1680 et 1681, a Ponthæo æditæ.

$\begin{array}{cc}\begin{array}{cccccc}\text{Stylo vel.}& \text{Stylo novo}& \text{Hora. mat}& \text{Cometæ Long.}& \text{Lat. Aust}& \text{L}\text{ong.}\phantom{\rule{0.3em}{0ex}}\text{candæ}\\ \text{Novemb 17}& \text{Novem.}\phantom{\rule{0.3em}{0ex}}27& 6\phantom{⁤\frac{1}{2}}& \phantom{0}8.30\prime & 0.40& {15}^{gr}\phantom{\rule{0.3em}{0ex}}\text{et ultra}& \\ & \phantom{\text{Novem.}}\phantom{\rule{0.3em}{0ex}}28& 6⁤\frac{1}{2}& \phantom{}13.\phantom{0}0\phantom{\prime }& 1.20& & \\ & \text{Decem.}\phantom{\rule{0.3em}{0ex}}\phantom{0}1& 7⁤\frac{1}{4}& \phantom{}27.50\phantom{\prime }& 1.16& & \\ & \phantom{\text{Decem.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}4& 7\phantom{⁤\frac{1}{2}}& 12.\phantom{0}0\phantom{\prime }& 0.40& & \\ & \phantom{\text{Decem.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}5& 7⁤\frac{1}{4}& \phantom{}15.50\phantom{\prime }& 0.40& & \\ \\ & \phantom{\text{Decem.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}6& 7⁤\frac{1}{2}& 20.\phantom{0}0\phantom{\prime }& 0.35& & \\ & \phantom{\text{Decem.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}7& 7⁤\frac{1}{2}& 24.\phantom{0}0\phantom{\prime }& 0.30& & \end{array}& \begin{array}{l}\text{Caput Nov 27}\\ \\ \text{aquabat prim}\\ \text{notæ stellas mag}\\ \text{nitudine at lu}\\ \text{mine ab eis mul}\\ \text{tum deficiebat}\end{array}\end{array}$

## Observationes ejusdem cometæ habitæ a R. P Ango in Fleche

Novem 28 \hor. 5 matitin./ in medio erat inter stellas duas exiguas quarum una est minima {illeg} trium quæ sunt in manu australi Virginis altera est in extremitate alce: Adeo longitudo Cometæ jam erat ≏ 13, Latitudo australis 50′.

Decem 1. hora 5 matutina. erat in Libra 27. 45′

## Observationes Venetijs habitæ a M. Montenaro.

Novem 30 hora {illeg} post occasum solis duodecima \Cometa/ erat in ≏ 23gr cum lat Aust. 1gr 30′

Decem 1, Erat in hor 5 matutina erat in 27. 51 ≏.

Decem 2 erat in {illeg}|| 2. 33

Decem 4 erat in 12. 52

Credidit M. Montenari latitudinem {illeg} ad us finem harum observationem augæri.

## Observationes Hevelij destituti instrumentis

Anno 1680 Decem. 2 Cometa erat in ≏ 25 cum lat. Austr. 5gr

Decem 3 {illeg}|{Arc}|te ortum is hora sexta erat in 4 cum lat. austr 4gr.

Decem 4 mane hor 6 20′ erat in 10 cum lat austr. 3gr.

<98v>

## Observationes Cometæ Mense Novembri anni 1680

Canterburiæ per Artificem quendam nomine Hill, instrumento cujus radius erat 4 pedum Die Veneris Novemb 11 tempore matutino, |C|ometa inventus est in 12gr cum lat. b|B|oreali 2gr. Locus is 29gr. 53′

Romæ per Marcum Antonium Cellium observationes hæ factæ sunt.

$\begin{array}{ccccc}\text{Stylo veter.}& \text{Stylo novo}& \text{Cometæ Longit}& \text{Lat austral}& \text{Locus}\phantom{\rule{0.3em}{0ex}}{☉}^{\text{is}}\\ \text{Novemb}\phantom{\rule{0.3em}{0ex}}17& \text{Nov}\phantom{\text{em}}\phantom{\rule{0.3em}{0ex}}27\phantom{\rule{0.3em}{0ex}}\text{mane}& \text{≏}\phantom{0}{8\phantom{.}}^{gr}30\prime & {0}^{gr}\phantom{.}\phantom{+}30\prime \phantom{,\text{circiter}}& \text{♐}{5.}^{gr}55\prime \\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}18& \phantom{\text{Nov}\phantom{\text{em}}\phantom{\rule{0.3em}{0ex}}}28\phantom{\rule{0.3em}{0ex}}\text{mane}& \phantom{\text{≏}}{13.}^{\phantom{gr}}30\phantom{\prime }& {1}^{\phantom{gr}}.\phantom{+}00\phantom{\prime },\text{circiter}& \phantom{\text{♐}}{6.}^{\phantom{gr}}56\phantom{\prime }\\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}21& \text{Decem.}\phantom{\rule{0.3em}{0ex}}\phantom{0}1\phantom{\rule{0.3em}{0ex}}\text{mane}& \phantom{\text{≏}}{28.}^{\phantom{gr}}\phantom{0}0\phantom{\prime }& {1}^{\phantom{gr}}\phantom{.}+00\phantom{\prime },\text{circiter}& \phantom{\text{♐}}{9.}^{\phantom{gr}}58\phantom{\prime }\\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}24& \phantom{\text{Decem.}\phantom{\rule{0.3em}{0ex}}0}4\phantom{\rule{0.3em}{0ex}}\text{mane}& \text{♏}{11.}^{\phantom{gr}}40\phantom{\prime }& {1}^{\phantom{gr}}\phantom{.+}00\phantom{\prime },\text{circiter}& \phantom{\text{♐}}{13.}^{\phantom{gr}}\phantom{0}1\phantom{\prime }\\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}25& \phantom{\text{Decem.}\phantom{\rule{0.3em}{0ex}}0}5\phantom{\rule{0.3em}{0ex}}\text{mane}& \phantom{\text{♏}}{15.}^{\phantom{gr}}47\phantom{\prime }& {1}^{\phantom{gr}}\phantom{.+}00\phantom{\prime ,}\text{circiter}& \phantom{\text{♐}}{14.}^{\phantom{gr}}\phantom{0}2\phantom{\prime }\\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}26& \phantom{\text{Decem.}\phantom{\rule{0.3em}{0ex}}0}6\phantom{\rule{0.3em}{0ex}}\text{mane}& \phantom{\text{♏}}{19.}^{\phantom{gr}}45\phantom{\prime }& {1}^{\phantom{gr}}\phantom{.+}00\phantom{\prime ,}\text{circiter}& \phantom{\text{♐}}{15.}^{\phantom{gr}}\phantom{0}3\phantom{\prime }\\ \phantom{\text{Novemb}\phantom{\rule{0.3em}{0ex}}}27& \phantom{\text{Decem.}\phantom{\rule{0.3em}{0ex}}0}7\phantom{\rule{0.3em}{0ex}}\text{mane}& \phantom{\text{♏}}{23.}^{\phantom{gr}}35\phantom{\prime }& {1}^{\phantom{gr}}\phantom{.+}00\phantom{\prime ,}\text{circiter}& \phantom{\text{♐}}{16.}^{\phantom{gr}}\phantom{0}4\phantom{\prime }\end{array}$

Romeæ per Galletium hæ

$\begin{array}{cccc}\text{Stylo vet}& \text{St. Novo}& \text{Comet. Long}& \text{Lat. austral.}\\ \text{Novem}\phantom{\rule{0.3em}{0ex}}17& \text{Novem}\phantom{.}\phantom{\rule{0.3em}{0ex}}27\phantom{\rule{0.3em}{0ex}}\text{Hora}\phantom{\rule{0.3em}{0ex}}18\phantom{⁤\frac{1}{2}}& \phantom{0}8.\phantom{0}0& 0.00\\ \phantom{\text{Novem}}\phantom{\rule{0.3em}{0ex}}18& \phantom{\text{Novem}.}\phantom{\rule{0.3em}{0ex}}28\phantom{\rule{0.3em}{0ex}}\phantom{\text{Hora}}\phantom{\rule{0.3em}{0ex}}17⁤\frac{1}{2}& 13.\phantom{0}0& 1.00\\ \phantom{\text{Novem}}\phantom{\rule{0.3em}{0ex}}21& \text{Dec.}\phantom{\text{em}}\phantom{\rule{0.3em}{0ex}}\phantom{0}1\phantom{\rule{0.3em}{0ex}}\phantom{\text{Hora}}\phantom{\rule{0.3em}{0ex}}17⁤\frac{1}{2}& \phantom{0}0.\phantom{0}0& 4.00\\ \phantom{\text{Novem}}\phantom{\rule{0.3em}{0ex}}23& \phantom{\text{Dec.}\text{em}}\phantom{\rule{0.3em}{0ex}}\phantom{0}3\phantom{\rule{0.3em}{0ex}}\phantom{\text{Hora}}\phantom{\rule{0.3em}{0ex}}17\phantom{⁤\frac{1}{2}}& \phantom{0}9.\phantom{0}0& 3.00\\ \phantom{\text{Novem}}\phantom{\rule{0.3em}{0ex}}25& \phantom{\text{Dec.}\text{em}}\phantom{\rule{0.3em}{0ex}}\phantom{0}5\phantom{\rule{0.3em}{0ex}}\phantom{\text{Hora}}\phantom{\rule{0.3em}{0ex}}18\phantom{⁤\frac{1}{2}}& 16.15& 2.00\\ \phantom{\text{Novem}}\phantom{\rule{0.3em}{0ex}}26& \phantom{\text{Dec.}\text{em}}\phantom{\rule{0.3em}{0ex}}\phantom{0}6\phantom{\rule{0.3em}{0ex}}\phantom{\text{Hora}}\phantom{\rule{0.3em}{0ex}}18\phantom{⁤\frac{1}{2}}& 19.30& 1.00\end{array}$

Cantabrigiæ per juvenem quendam Cometa observatus est Novemb 19 juxta spicam Virginis, quasi duobus gradibus |supra| stellam illam, ad sive ad boream, circa horam quartam vel quintam matutinam. Et cauda extendebatur ad us stellam illam primæ magnitudinis quæ cauda Leonis dicitur.

## Observationes Parisijs habitæ Cometæ subsequentis 1680 & 1681.

[78]$\begin{array}{cclllll}& \text{Longitudo}& \text{Lat. bor}& \text{A}\text{s}\text{cbn. f.}& \text{Decl}& \text{Long. caud.}& \text{Decl. caudæ ab op.}\phantom{\rule{0.3em}{0ex}}\text{O}\\ \phantom{\rule{4em}{0ex}}\text{St. vet.}\phantom{\rule{0.3}{0ex}}\text{Dec.}\phantom{\rule{0.3em}{0ex}}19\phantom{.}\phantom{\rule{0.3em}{0ex}}\text{hor}\phantom{\rule{0.3em}{0ex}}6.30\text{p.m.}& {27.}^{gr}\phantom{0}4\phantom{⁤\frac{1}{2}}\prime & 18.53& & & {62}^{gr}& {2}^{gr}.48\prime .\\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}\phantom{\rule{0.3}{0ex}}\text{Dec.}}\phantom{\rule{0.3em}{0ex}}24\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}5.30\phantom{\text{p.m.}}& {18.}^{\phantom{gr}}36⁤\frac{1}{2}\phantom{\prime }& 25.26& & & & {3}^{gr}.43\prime \phantom{.}\\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}\phantom{\rule{0.3}{0ex}}\text{Dec.}}\phantom{\rule{0.3em}{0ex}}28\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.00\phantom{\text{p.m.}}& {\phantom{0}8.}^{\phantom{gr}}12⁤\frac{1}{2}\phantom{\prime }& 28.\phantom{0}3& & & {62}^{gr}& \\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}\phantom{\rule{0.3}{0ex}}\text{Dec.}}\phantom{\rule{0.3em}{0ex}}29\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.\phantom{0}6\phantom{\text{p.m.}}& \phantom{}{12.}^{\phantom{gr}}55\phantom{⁤\frac{1}{2}\prime }& 28.16& & & & \\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}}\phantom{\rule{0.3em}{0ex}}\text{Jan}\phantom{\text{.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}4.\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.\phantom{0}6\phantom{\text{p.m.}}& \phantom{}{\phantom{0}5.}^{\phantom{gr}}53\phantom{⁤\frac{1}{2}\prime }& 26.38& & & & \\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}}\phantom{\rule{0.3em}{0ex}}\phantom{\text{Jan}\text{.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}6.\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.30\phantom{\text{p.m.}}& \phantom{}{11.}^{\phantom{gr}}33\phantom{⁤\frac{1}{2}\prime }& 25.42& & & & {5}^{gr}.13\prime \phantom{.}\\ \phantom{\rule{4em}{0ex}}\phantom{\text{St. vet.}}\phantom{\rule{0.3em}{0ex}}\phantom{\text{Jan}\text{.}}\phantom{\rule{0.3em}{0ex}}\phantom{0}8.\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.50\phantom{\text{p.m.}}& \phantom{}{16.}^{\phantom{gr}}36\phantom{⁤\frac{1}{2}\prime }& 24.44⁤\frac{1}{2}& \phantom{0}4.40& 29.\phantom{0}7& & \\ \text{Observatio crassa}\phantom{\rule{0.3em}{0ex}}13\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.20\phantom{\text{p.m.}}& \phantom{}{26.}^{\phantom{gr}}\phantom{0}7⁤\frac{1}{2}\phantom{\prime }& 22.20.& & & & \\ \phantom{\text{Observatio crassa}}\phantom{\rule{0.3em}{0ex}}23\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.40\phantom{\text{p.m.}}& & & 28.27& 31.35& & \\ \phantom{\text{Observatio crassa}}\phantom{\rule{0.3em}{0ex}}24\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.20\phantom{\text{p.m.}}& & & 29.30⁤\frac{1}{2}& 31.36& & \\ \phantom{\text{Observatio crassa}}\phantom{\rule{0.3em}{0ex}}23\phantom{.}\phantom{\rule{0.3em}{0ex}}\phantom{\text{hor}}\phantom{\rule{0.3em}{0ex}}6.40\phantom{\text{p.m.}}& & & 30.30& 31.37& & \end{array}$

## Ejusdem posterioris Cometæ Observationes Grenovici habitæ

$\begin{array}{c}\begin{array}{cc}\phantom{\rule{59.5em}{0ex}}& \begin{array}{l}\text{Hæ æquan}\\ \text{tur}\phantom{\rule{0.3em}{0ex}}\text{Perism}\\ \text{addendo vel}\\ \text{auferado}\\ \end{array}\end{array}\\ \begin{array}{ccccccccc}\text{Loca}\phantom{\rule{0.3em}{0ex}}{}^{\text{is}}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}\phantom{\rule{0.3em}{0ex}}12}& \text{Tempus verus}& \text{Ascentio recta}& \text{Declinatio}& \text{Longitudo}& \text{Latitudo}& \text{Longit}& \text{in}\hfill \\ & & & & \text{borealis}& & \text{borealis}& \text{caudæ}& \text{long. lat}\hfill \\ \phantom{0}1.53\phantom{⁤\frac{1}{2}}& \text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}\phantom{\rule{0.3em}{0ex}}12& {4}^{\mathrm{h}}46\prime \text{p. m.}& \text{————}& \text{————}& \phantom{0}{6}^{gr}.33\prime \phantom{⁤\frac{1}{2}}& \phantom{0}{8}^{gr}.26\prime \phantom{⁤\frac{1}{2}}& 35+& \\ \phantom{}11.\phantom{0}8\phantom{⁤\frac{1}{2}}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}}\phantom{\rule{0.3em}{0ex}}21& {6}^{\phantom{\mathrm{h}}}31\phantom{\prime \text{p. m.}}& 302.20⁤\frac{1}{2}& \phantom{0}2\phantom{.}\phantom{0}5⁤\frac{1}{3}& \phantom{0}{5}^{\phantom{gr}}.08\phantom{\prime }⁤\frac{1}{5}& {21}^{\phantom{gr}}.42\phantom{\prime }⁤\frac{1}{6}& 70\phantom{+}& \\ \phantom{}14.11⁤\frac{1}{3}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}}\phantom{\rule{0.3em}{0ex}}24& {6}^{\phantom{\mathrm{h}}}24\phantom{\prime \text{p. m.}}& 313\phantom{.}33\phantom{⁤\frac{1}{2}}& \phantom{0}9\phantom{.}\phantom{0}0\phantom{⁤\frac{1}{3}}& \phantom{}{58}^{\phantom{gr}}.52\phantom{\prime ⁤\frac{1}{5}}& {25}^{\phantom{gr}}.26\phantom{\prime ⁤\frac{1}{6}}& 65\phantom{+}& -\phantom{0}3.\phantom{\frac{1}{2}}+2\\ \phantom{}16.10⁤\frac{2}{3}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}}\phantom{\rule{0.3em}{0ex}}26& {5}^{\phantom{\mathrm{h}}}16\phantom{\prime \text{p. m.}}& 321\phantom{.}15\phantom{⁤\frac{1}{2}}& 13\phantom{.}19⁤\frac{2}{3}& \phantom{}{28}^{\phantom{gr}}.28\phantom{\prime }⁤\frac{1}{2}& {27}^{\phantom{gr}}.\phantom{0}5\phantom{\prime }⁤\frac{1}{2}& 60\phantom{+}& \\ \phantom{}19.21\phantom{⁤\frac{2}{3}}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}}\phantom{\rule{0.3em}{0ex}}29& {8}^{\phantom{\mathrm{h}}}\phantom{0}0\phantom{\prime \text{p. m.}}& 333\phantom{.}27\phantom{⁤\frac{1}{2}}& 19\phantom{.}21\phantom{⁤\frac{2}{3}}& \phantom{}{13}^{\phantom{gr}}.12\phantom{\prime ⁤\frac{1}{2}}& {28}^{\phantom{gr}}.10\phantom{\prime ⁤\frac{1}{2}}& 50\phantom{+}& +\phantom{0}6.\phantom{\frac{1}{2}}+8\\ \phantom{}20.22⁤\frac{1}{2}& \phantom{\text{St. vet.}\phantom{\rule{0.3em}{0ex}}\text{Decem}}\phantom{\rule{0.3em}{0ex}}30& {8}^{\phantom{\mathrm{h}}}\phantom{0}4\phantom{\prime \text{p. m.}}& 337\phantom{.}14\phantom{⁤\frac{1}{2}}& 21\phantom{.}00\phantom{⁤\frac{2}{3}}& \phantom{}{17}^{\phantom{gr}}.39\phantom{\prime ⁤\frac{1}{2}}& {28}^{\phantom{gr}}.\phantom{}12\phantom{\prime ⁤\frac{1}{2}}& 25\phantom{+}& \\ \phantom{}26.23⁤\frac{1}{3}& \phantom{\text{St. vet.}}\phantom{\rule{2.5em}{0ex}}\text{Jan}\phantom{\rule{0.3em}{0ex}}\phantom{0}5& {5}^{\phantom{\mathrm{h}}}53\phantom{\prime \text{p. m.}}& 356\phantom{.}45⁤\frac{1}{2}& 27\phantom{.}25⁤\frac{1}{2}& \phantom{0}{8}^{\phantom{gr}}.49\phantom{\prime }⁤\frac{1}{6}& {26}^{\phantom{gr}}.\phantom{}15\phantom{\prime }⁤\frac{1}{2}& 15\phantom{+}& +\phantom{0}1\phantom{.}\frac{1}{2}-4\\ \phantom{0}0.30\phantom{⁤\frac{1}{3}}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Jan}}\phantom{\rule{0.3em}{0ex}}\phantom{0}9& {6}^{\phantom{\mathrm{h}}}50\phantom{\prime \text{p. m.}}& \phantom{0}6\phantom{.}57\phantom{⁤\frac{1}{2}}& 29\phantom{.}31\phantom{⁤\frac{1}{2}}& \phantom{}{18}^{\phantom{gr}}.44\phantom{\prime ⁤\frac{1}{6}}& {24}^{\phantom{gr}}.\phantom{}12\phantom{\prime }⁤\frac{1}{10}& & +\phantom{0}9.\phantom{\frac{1}{2}-}0\\ \phantom{}\phantom{0}1.28⁤\frac{1}{2}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Jan}}\phantom{\rule{0.3em}{0ex}}10& {5}^{\phantom{\mathrm{h}}}56\phantom{\prime \text{p. m.}}& \phantom{0}9\phantom{.}\phantom{0}2\phantom{⁤\frac{1}{2}}& 29\phantom{.}51\phantom{⁤\frac{1}{2}}& \phantom{}{20}^{\phantom{gr}}.41\phantom{\prime }⁤\frac{1}{2}& {23}^{\phantom{gr}}.\phantom{}44\phantom{\prime }⁤\frac{1}{2}& 10& +18.\phantom{\frac{1}{2}-0}\\ \phantom{}\phantom{0}4.34\phantom{⁤\frac{1}{2}}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Jan}}\phantom{\rule{0.3em}{0ex}}13& {6}^{\phantom{\mathrm{h}}}55\phantom{\prime \text{p. m.}}& 14\phantom{.}52\phantom{⁤\frac{1}{2}}& 30\phantom{.}37\phantom{⁤\frac{1}{2}}& \phantom{}{25}^{\phantom{gr}}.59\phantom{\prime }⁤\frac{1}{2}& {22}^{\phantom{gr}}.\phantom{}17\phantom{\prime }⁤\frac{1}{2}& \phantom{0}5& +12.\phantom{\frac{1}{2}}+3\\ \phantom{}16.45⁤\frac{2}{3}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Jan}}\phantom{\rule{0.3em}{0ex}}25& {7}^{\phantom{\mathrm{h}}}44\phantom{\prime \text{p. m.}}& 30\phantom{.}35\phantom{⁤\frac{1}{2}}& 31\phantom{.}37\phantom{⁤\frac{1}{2}}& \phantom{0}{9}^{\phantom{gr}}.36\phantom{\prime ⁤\frac{1}{2}}& {17}^{\phantom{gr}}.\phantom{}57\phantom{\prime ⁤\frac{1}{2}}& & \\ \phantom{}21.50\phantom{⁤\frac{2}{3}}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Jan}}\phantom{\rule{0.3em}{0ex}}30& {8}^{\phantom{\mathrm{h}}}\phantom{0}7\phantom{\prime \text{p. m.}}& 35\phantom{.}\phantom{0}2\phantom{⁤\frac{1}{2}}& 31\phantom{.}41\phantom{⁤\frac{1}{2}}& \phantom{}{13}^{\phantom{gr}}.20\phantom{\prime ⁤\frac{1}{2}}& {16}^{\phantom{gr}}.\phantom{}41\phantom{\prime ⁤\frac{1}{2}}& & \\ \phantom{}24.43⁤\frac{1}{5}& \phantom{\text{St. vet.}}\phantom{\rule{2.5em}{0ex}}\text{Feb}\phantom{\rule{0.3em}{0ex}}\phantom{0}2& {6}^{\phantom{\mathrm{h}}}20\phantom{\prime \text{p. m.}}& 37\phantom{.}18\phantom{⁤\frac{1}{2}}& 31\phantom{.}41\phantom{⁤\frac{1}{2}}& \phantom{}{15}^{\phantom{gr}}.14\phantom{\prime ⁤\frac{1}{2}}& {16}^{\phantom{gr}}.\phantom{}02\phantom{\prime ⁤\frac{1}{2}}& & \\ \phantom{}27.50⁤\frac{1}{2}& \phantom{\text{St. vet.}\phantom{\rule{2.5em}{0ex}}\text{Feb}}\phantom{\rule{0.3em}{0ex}}\phantom{0}5& {7}^{\phantom{\mathrm{h}}}\phantom{0}8\phantom{\prime \text{p. m.}}& 39\phantom{.}26\phantom{⁤\frac{1}{2}}& 31\phantom{.}41\phantom{⁤\frac{1}{2}}& \phantom{}{17}^{\phantom{gr}}.00\phantom{\prime }⁤\frac{1}{2}& {15}^{\phantom{gr}}.\phantom{}27\phantom{\prime ⁤\frac{1}{2}}& & \end{array}\end{array}$

<99r>

## Observationes de Cauda Cometæ prioris

Novemb 19 Cometa juxta spicam virginis existens caudam projiciebat ad us caudam Leonis, spectante juvenes quodam.

Postea caudam versu{illeg} per meride|i|em versus occidenti|e|m projici {illeg} longam satis & ad horizontem obliquam capite {illeg} vel sub horizonte vel pone ædificia delitescente vidit Humf. Bab. S. T. D.

## De cauda Cometæ posterioris

Decemb 8 stylo veteris Hallius noster tempore matutino Parisias versus iter faciens prope Bolonian ante ortum solis Caudam vidit Cometæ quasi perpendiculariter ex horizonte surgentem, ut ipse retulit in epistola quadam cit{illeg}|an|te Flamstedio. Unde Cometa inquit Flamstedius tunc borealem habeb{a}t latitudinem & cum solenondum conjunctus fuerat. |Apparebat autem cauda lat{illeg}|{æ}| divergens et {illeg} ex corpore {illeg} eg{illeg}|re|ssa aer prius {illeg} quam {illeg} {illeg} \{illeg}/ {illeg} {\{illeg}/} {illeg}|

Decemb 10. duabus horis post occasum Solis, {illeg}bat cauda per medium distantiæ inter caudam serpentis Ophi{illeg}{illeg}|cha| et stellam (Bayero δ) in ala austrina Aquilæ. Desinebat vero ad stellas tres exiguas (Bayero Awb) in dorso Aquilæ juxta caudam, eductione caudæ Aquilæ ejusdem, id est in linea jungente stellam|s| lucida{illeg} secundæ magnitudinis in eductione colli Aquilæ, et estellam tertiæ quæ penultima est in cauda ejus, ac stellæ illi penultim{æ} [duplo quidem] propior existebat qu{à}m alteri in eductione colli. {illeg}|F|lamstedius in Epistolis ad nos datis. Desinebat igitur cauda in $19\frac{1}{2}$ cum lat. bor. $34\frac{1}{4}$ circiter

Decemb 11 {Cau} post occasum Solis cauda instar jubaris apparuit ab horizonte erecti et lunâ latioris. Post crepusculi cessationem ex tendebat ad us stellas \duas/ quartæ m{æ}gnitudinis (Bayero α, β,) in capite \seu glyphidæ/ Sagittæ. (Flam{st}. ib.) adeo desinebat in 26gr 43′ cum lat bor. 38gr 34′.

Decemb. 12. Quamprimum non obscura facta est, cauda transibat per medium sagittæ, ne ultra medium longè extendebat. (Flam{st} ib) L{in}quebat igitur stellas 5 et 6 magnitudinis, δ et ζ in tribulo sagittæ, quasi 40′ ad occidentem, et ultra per 3grad circiter vel fort{æ} 4 extendens desinebat in {} 4 cum lat bor $42\frac{1}{2}$ circiter vel 42 43 /34\ 43. Desinebat uti e regione superior{is} duarum informium 4 magnitudinis quæ supra sagittam sunt {illeg} non et ultra extendebat{illeg}. Nam cauda ensiformis nobis visa {illeg} sagittam \paul{o}/ longius superare quam Flamstedio, in viam lacteam {illeg} nihil extendens & termino acuto paulatim languescens. Ca|e|ter {illeg} in A{s}trola{b}io Flamstedij, cauda hac nocte {illeg} desinit accurat ad stellas duas exiguas prædictas in tribulo sagittæ.

Decemb 15 hor $5\frac{3}{4}$ lucida Aquilæ erat in medio caudæ fere {Pergebat vero cauda} Ancon item austrinus Aquilæ erat {illeg} {illeg} \{illeg}/la parte caudæ prope terminum ejus \{medio} caudæ fere prope terminū ejus ad latus australe vergens./ (Ipse \ego & {Bainbro} et ellis/ part{im} ex observa{tione} partim ex circu{s} {stansijs}) Erat autem cauda 50 grad. longa ({illeg} {steed} epist. 1) {nec tutren} \tenuem/ extremitatem ejus propter Lunæ \novel/ splendorem oliquam apparuise probabila est.

< insertion from the left margin of f 99r >

Decemb. 16 hor 5 P.M. Cometa existente in 17 cum lat. bor. 15gr circiter(|]| cauda lucidam Aquilæ (quæ nocte superiori erat in medio ejus) latere suo boreali \{illeg}/ tangebat, \aut quasi;/ ut et lucidam in {ancone} austrino cygni tangebat eodem latere |{illeg} aut quasi| Tota Caudæ longitudo erat 60 grad: feré, latitudo 2 gradus. (Observator quidam Scotus.) Unde Cauda terminabatur in long. 10 vel 12 circiter 9 lat. bor. 53.

Decem. 19 Hor $5⁤\frac{1}{2}$ P.M. Transibat cauda per Delphini caput dein latere suo boreali stellam penultimam in austrina ala{illeg} Cygni stringebat, tendans inde versus lucidam in Cassiopeiæ cathedra et quasi 60 gradus longa existens ({illeg} observator Scotus) vel potius 63 aut 64 grad ut ex alijs colligo, si non et paullo ultra. Desinebat igitur in 6 cum lat. bor. 52 |vel {illeg} $81⁤\frac{1}{2}$|.

Decem 17 cauda inferiùs duos gradus lata {,} superius non-nihil latior, ad caput Cepher extendebat. Decem 22 cauda $67⁤\frac{1}{4}$ grad longa ad Cassiopeiam us extendit: ({minor tamem} \& {minor}/ {quam intra} ob D{illeg} splendorem apparuit. Decem 23 caud{a} tenuis et {illeg} per Cassiosu {illeg} extendit, 72gr Conica{illeg} existens circiter. Decem 28 {illeg} orta {illeg} fortior et clarior apparuit {illeg} sed 56gr. {illeg} inter Ala{illeg} {et lum{illeg}} ge{illeg} <99v> in femure Andromedæ ad us Persei caput extendit? (Observator quidam Hamburguesis, qui præ cæteris caudam longam \ad ultimam {illeg}/ descripsisse videtur.

< text from f 99r resumes >

Decem 18 Cauda linquebat stellas Delphini ad dextra