Newton's Waste Book (Part 3)

Author: Isaac Newton

Source: MS Add. 4004, ff. 50v-198v, Cambridge University Library, Cambridge, UK

Published online: June 2013

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- This document is part of MS Add. 4004
  The previous part of this document is Newton's Waste Book (Part 2) [MS Add. 4004, ff. 15v-50r]

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<50v>

How to Draw Tangents to Mechanicall Lines

Lemma.

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

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

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

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

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

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

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

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

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

^[8] Instead {in y^e} third example {illeg} {illeg} {illeg}. <51r> Therefore $af ∶ ab ∷ ak ∶ gbl ∷$ {illeg}|abs|olute & whole motion of b towards c (or acf ) ∶ whole motion of b towards d (or ed ). Soe y^t makeing $bc ∶ bd ∷ af ∶ ab$ . { $ab ∥ ed ⊥ ed$ } & $hb ∥ ce ⊥ cb$ . The point b will be moved {illeg} to y^e line {illeg} ce & ed in same times w^ch cannot bee unlesse it move to e (their common intersection). The point b therefore move in y^e line be w^ch doth therefore touch y^e Quadratrix at b : (The same is done by makeing $bd =$ $bl = bd ⊥ de ∥ ab$ . & drawing \y^e tangent/ be through y^e 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 y^e body a being in a circumference of y^e circle \or sphære/ adce doth move towards its center b its acceleration \motion or velocity/ to|wards| each point \ d , c , e / of y^e circ{illeg}umference is yⁿ as y^e cordes ad , ac , ae , drawn from y^t body to those points are. This may be Demonstrated by Te|h|eorem R pag 57.

^[10] Prop 2^d. If \ ∠ {sic} $adc$ {illeg} is a parallelogram \ sim $△ aec$ although they bee not in y^esame plane &// three bodys move \uniformely from a /. y^e first from a to b d y^e 2^d from a to e , y^e 3^d from a to c their motions being \each to other/ as y^e {illeg} directing lines ad , ae , ac , are in y^e same time, & adce is a parallelogram then is y^e motion of y^e third body compounded of y^e other two. Demonstration. For \makeing $df ⊥ ac$ f & $eb ⊥$ {adba}/ |{illeg} makeing $df ∥ eb ⊥ ac$ | y^e motion of y^e first body towards d is to its motion towards f as ad = {illeg} is to af (prop 1); \{illeg}/ & y^e \{illeg}/ motion of y^e 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 3^d. If a moveing line keepe parallel to it selfe all its pts have equall motion.

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

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

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

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

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

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

Note also y^t by y^e place of a body is meant its center of gravity.

To resolve Problems by motion y^e 6 following prop: are necessary & suffcient.

May 16. 1666.

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

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

Note y^t by a body is meant its center of gravity.

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

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

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

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

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

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

<51v>

To draw a tangent to y^e Ellipsis

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

To draw a Tangent to y^e Concha.

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

To find y^e point c w^ch distinguisheth twixt y^e concave & convex portion of y^e Concha.

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

Which may bee thus calculated. Make $cb = c$ . $ag = b$ . $cb = y$ . yⁿ is $bl = \sqrt{c c - y y}$ . $2 gl = \frac{2 b c}{y} = hk$ . $bl ∶ bc ∷ ld = \frac{b c}{y} + c ∶ fd = \frac{c b + c y}{\sqrt{c c - y y}}$ . & $\sqrt{c c - y y} ∶ y ∷ bl ∶ bc ∷ gf ∶ fh ∷ df ∶ kh ∷ \frac{c b + c y}{\sqrt{c c - y y}} ∶ \frac{2 b c}{y} ∷ b y + y y ∶ 2 b \sqrt{c c - y y}$ . Therefore $2 b c c - 2 b y y = b y y + y^{3}$ . Or $y^{3} + 3 b y y * - 2 b c c = 0$ .

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

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

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

^[20] Example 2^d. If {ne} is y^e Conchoid ( $ga = b$ . $ae = c = nl$ . $nb = y$ . $ab = x$ .) fo{illeg} em parallell to it. Then is $\sqrt{c c - y y} ∶ y ∷ bl ∶ bn ∷ cg ∶ fl (supra) = \frac{c b y + c y y}{\sqrt{c c - y y}}$ . Then is $\sqrt{c c - y y} = bl ∶ c = cl ∷ dl = cg = c + \frac{b c}{y} (vide supra) ∶ fl = \frac{c c y + b c c}{y \sqrt{c c - y y}}$ . & $fb = \frac{b c c + y^{3}}{y \sqrt{c c - y y}}$ . & $y = cb ∶ fb ∷ ea = c ∶ am = \frac{b c^{3} + c y^{3}}{y y \sqrt{c c - y y}} = z$ . et $b b c^{6} + 2 b c^{4} y^{3} + c c y^{6} - c c y + z z - y^{6} z z = 0$ ponatur p esse motus puncti m & q esse {illeg} motus puncti c versus b . Erit (prop {6}{)} $6 b c^{4} y y q + 6 c c y^{5} q - 4 c c y^{3} z z q -$ {illeg}{illeg} $q z z - 2 c c p y^{4} z - 2 p y^{6} z = 0$ . supposeing $q = 0$ (For when am is y^e least y^t {illeg}{illeg} bee y^e point c is y^t w^ch distinguisheth twixt y^e concave & convex porti{on} of y^e Conchoid, & yⁿ y^e motion q vanisheth.) it will bee $\frac{3 b c^{4} + 3 c c y^{3}}{2 c c y - 3 y^{3}} = z z = \frac{b c^{3} + c y^{3}}{y y \sqrt{c c - y y}}} 2$ . $\frac{3 c}{2 c c - 3 y y} = \frac{b c^{3} + c y^{3}}{c c y^{3} - y^{5}}$ . & {illeg} $c c y^{3}$ {illeg} $= 2 b c^{4} - 3 b c c y y$ . Or $y^{3} = - 3 b y y + 2 b c c$ .

<55r>

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

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

As for example did I know y^t two of y^e rootes of y^e Equation $x^{3} - 8 x x + 9 x + 18 = 0$ were in proportion as 1 to 2 & would I have y^e lesser roote (viz y^t w^ch 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 w^ch progressions y^e Equation may bee multiplyed, as by y^e 1^st, $\begin{array}{l} x^{3} & - & 8 x x & + & 9 x & + & 18 & = & 0 \\ 28 & . & 12 & . & 4 & . & 0 & . \end{array}$ . W^ch produceth {illeg} $7 x x - 24 x + 9 = 0$ . Or by y^e 3^d $\begin{array}{l} x^{3} & - & 8 x x & + & 9 x & + & 18 & = & 0 \\ 7 & . & 3 & . & 1 & . & 0 & . \end{array}$ . W^ch produceth {illeg} $7 x x - 24 x + 9 = 0$ . Or by the first Otherwise by destroying y^e 1^st terme. $\begin{array}{l} x^{3} & - & 8 x x & + & 9 x & + & 18 & = & 0 \\ 0 & . & - 2 & . & - 3 & . & - \frac{7}{2} & . \end{array}$ . W^ch produceth {illeg} $16 x x - 27 x - 63 = 0$ . &c the rootes of w^ch products are, viz: of y^e first $x = \frac{3}{7}$ , & $x = 3$ . Of y^e last $x = - \frac{21}{16}$ , & $x = 3$ . There I conclude 3 to be y^e lesse, & consequently 6 the greater of those rootes w^ch are in proportion as one to of y^e Equation $x^{3} - 8 x x + 9 x + 18 = 0$ . w^ch are in double proportion But was y^e greater of those rootes desired yⁿ {illeg} inverting y^e progression it would bee $\begin{array}{l} x^{3} & - & 8 x x & + & 9 x & + & 18 & = & 0 \\ 0 & 1 & 3 & 7 \end{array}$ . Or $\begin{array}{l} x^{3} & - & 8 x x & + & 9 x & + & 18 & = & 0 \\ - \frac{7}{2} & - 3 & - 2 & 0 \end{array}$ . The first producing $8 x x - 27 x - 126 = 0$ whose rootes are $x = - \frac{21}{8}$ , $x = 6$ . {illeg} The 2^d produceth $7 x^{2} - 48 x + 36 = 0$ whose rootes are $x = \frac{6}{7}$ , $x = 6$ . And consequently 6 is the greater & 3 the lesse of y^e rootes in duplicate propor{tion.}

Would I draw

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

Would I have dbec to be a square y^t is $db = de = y$ . Then to find hd I call it x & $dh = x + y = hd + de$ . Soe y^t y^e lesse roote is to y^e greater as x to $x + y$ . Making therefore $b = x$ , $a = x + y$ , The progression will be $\frac{x x + 2 x y}{x}$ $\frac{y y}{x} + 3 x + 3 y . 2 x + y . x . 0 . - \frac{x x}{x + y} . - \frac{x x}{x + y} - \frac{x^{3}}{x x + 2 x y + y y}$ . Or, $\frac{y y}{x} + 2 x + 3 y . x + y . 0 . - x . - x - \frac{x x}{x + y}$ . by w^ch y^e Equation being multiplyed produceth $\begin{matrix} x x & - & g x & + & y y & = & 0 \\ x + y & 0 & - x \end{matrix}$ , Or $x^{3} + y x x - y y x = 0$ , Or $y = \frac{x}{2} \overset{\cup}{O} \sqrt{\frac{5 x x}{4}} = \frac{x \overset{\cup}{O} x \sqrt{5}}{2}$ . And consequently $- x x + g x = y y = \frac{x x \overset{\cup}{O} 2 x x \sqrt{5} + 5 x x}{4}$ . Or $4 g = 10 x \overset{\cup}{O} 2 x \sqrt{5}$ . Or $\frac{2 g}{5 \overset{\cup}{O} \sqrt{5}} = x$ . By w^ch y^e Equation $x x - g x + y y = 0$ , must be multiplyed. & it produceth $\begin{matrix} x x & - & g x & + & y y & = & 0 \\ 2 x + y & . & x & . & 0 \end{matrix}$ . Or $2 x^{3} + y x x - g x x = 0$ Or $- 2 x + g = y$ . & congsequently $- x x + g x = y y = 4 x x - 4 g x + g g$ . y^t is $5 x x = 5 g x - g g$ . And $x = \frac{1}{2} g \overset{\cup}{O} \sqrt{\frac{1}{20} g g}$ . Or $x = \frac{g \sqrt{5} \overset{\cup}{O} g}{2 \sqrt{5}} = \frac{1}{2} g \overset{\cup}{O} \frac{g}{2 \sqrt{5}}$ . And consequently $y = \frac{1}{\sqrt{5}} g$ . Or it might have beene done thus. x substracted from y^e precedent progression it will be, {illeg} $x + y . 0 . - x . - x + \frac{x}{x + y}$ . &c by w^ch y^e Equation being multiplyed produceth $\begin{matrix} x x & - & g x & + & y y & = & 0 \\ x + y & 0 & - x \end{matrix}$ . Or $x x + y x = y y$ . And by extracting the roote, $y = \frac{1}{2} x \overset{\cup}{O} \sqrt{\frac{5}{4} x x}$ . And therefore $g x - x x = y y = \frac{1}{4} x x \overset{\cup}{O} x \sqrt{\frac{5}{4} x x} +$ { $\frac{5}{4} x x$ } Or $4 g = 10 x \overset{\cup}{O} 2 x \sqrt{5}$ . Or $\frac{2 g}{5 \overset{\cup}{O} \sqrt{5}} = x = \frac{g \sqrt{5} \overset{\cup}{O} g}{2 \sqrt{5}}$ . That is $\frac{2 g}{5 + \sqrt{5}} = \frac{g \sqrt{5} - g}{2 \sqrt{5}} = ab$ . And $\frac{2 g}{5 - \sqrt{5}} = \frac{g \sqrt{5} + g}{2 \sqrt{5}} = ac$ . And therefore { $\frac{2 g}{5 - \sqrt{5}} \frac{- 2 g}{5 + \sqrt{5}} = af - ab = \frac{10 g + 2 g \sqrt{5} - 10 g + 2 g \sqrt{5}}{25 + 5 \sqrt{5} - 5 \sqrt{5} - 0} = \frac{g}{\sqrt{5}} = ah = de = db = bc$ }.

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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 & y^e right line s being given , to produce ac to e , soe y^t ef drawn towards y^e point b may bee equall to y^e given line {illeg} s . Putting $df = x$ for y^e unknowne quantity. {illeg} $ef = c$ , & $bd = cd = a$ . The Equation will bee $x^{4} - 2 a x^{3} \begin{array}{l} + 2 a a x x \\ - c c \end{array} - 2 a^{3} + a^{4} = 0$ . w^ch having 4 rootes the Equation must have 4 divers resolutions; that is y^e lines ac , cd , produced both ways indefinitely, there may bee 4 divers lines drawne through the point b , whose parts intercepted twixt y^e 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 y^t $fd ∶ bd ∷ ab ∶ ae = dh = \frac{a a}{x}$ . That is one roote $(df)$ of this equation is to another $(dh)$ as x to $\frac{a a}{x}$ . Therefore I may multiply this Equation by this progression $\frac{a^{4}}{x^{3}} + \frac{a a}{x}, \frac{a a}{x}, 0, - x, - \frac{x^{3}}{a a}, - \frac{x^{5}}{a^{3}}$ . And there resulteth $\begin{matrix} x^{4} & - & 2 a x^{3} & \begin{array}{l} + 2 a a x x \\ - c c \end{array} & - & 2 a^{3} x & + & a^{4} & = & 0 \\ + \frac{a a}{x} & . & 0 & . & - x & . & - x - \frac{x^{3}}{a^{2}} & . & - x - \frac{x^{3}}{a^{2}} - \frac{x^{5}}{a^{4}} \end{matrix}$ . That is $a a x^{3} + c c x^{3} - 2 a a x^{3} + 2 a^{3} x x + 2 a x^{4} - a^{4} x - a a x^{3} - x^{5} = 0$ . Or, $x^{4} - 2 a x^{3} \begin{array}{l} + 2 a a x x \\ - c c \end{array} - 2 a^{3} x + a^{4} = 0$ . Which result is y^e {illeg} same w^th y^e first Equation the reason of w^ch is, y^t if I make $df = x$ then is $dh = \frac{a a}{x}$ . Or if $dh = x$ , yⁿ is $df = \frac{a a}{x}$ . Or if $dk = x$ , yⁿ is $dm = \frac{a a}{x}$ . Or if $dm = x$ then is $dk = \frac{a a}{x}$ . Soe y^t y^e relation twixt \all/ y^e 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 y^e resulting Equation. Otherwise y^e reduction must have succeeded.

Suppose 3 rootes of an Equation are in proportion to each other as a , b , c . Then if that roote w^ch is correspondent to a be required, multiply y^e termes of y^e Equation by this progression any of these Progressions
|1| $\frac{c^{3} + b c c + b b c + b^{3}}{a a} + \frac{b b + b c + c c}{a} + b + c + a . \frac{b b + b c + c c}{a} + b + c + a . + b + c + a . + a . 0.0 . \frac{a^{3}}{b c} . \frac{a^{3}}{b c} + \frac{a^{4} b + a^{4} c}{b b c c} . \frac{a^{3}}{b c} + \frac{a^{4} b + a^{4} c}{b b c c} . \frac{a^{3}}{b c} + \frac{a^{4} b + a^{4} c}{b b c c} + \frac{a^{5} b b + a^{5} b c + a^{5} c c}{b^{3} c^{3}}$ . |2| $\frac{+ b b c + b c c}{a a} + \frac{b b + b c + c c}{a} + b + c . \frac{b c}{a} + b + c . 0 . - a . 0 . \frac{+ a^{3} + a a b + a a c}{b c} . \frac{a^{3} + a a b + a a c}{b c} + \frac{a^{3} b b + a^{3} b c + a^{3} b b + a^{4} b + a^{4} c}{b b c c}$ . |3| {illeg} $\frac{a a b b + a a c c + b b c c}{a^{3} + a a b + a a b} + \frac{b c}{a} . 0 . \frac{- a b - a c - b c}{a + b + c} . - a . 0 . \frac{a^{3}}{b c} + \frac{+ b b + b c + c c}{b a c + b b c + b c c}$ . &c.

As if \3 of/ y^e rootes of this Equation $x^{4} * - 35 x x + 90 x - 56 = 0$ . $x^{4} * - 35 g g x x + 90 g^{3} x - 56 g^{4} = 0$ were to one another as $1.2.4$ . And g would find y^e roote w^ch is correspondent to 1 . Then I make $a = 1$ , $b = 2$ , $c = 4$ , & soe I may have have {sic} by y^e first progression this{illeg} $+ 155 . + 35 . + 7 . + 1 . 0.0 . \frac{+ 1}{7} . \frac{+ 7}{32} . \frac{+ 70}{128}$ . By y^e 2^d; $+ 92 . + 14 . 0 . - 1 . 0 . + \frac{7}{8} . + \frac{45}{32}$ . &c. By y^e first of w^ch y^e Equation being multiplyed produceth $\begin{matrix} x^{4} & * & - & 35 g g x x & + & 90 g^{3} x & - & 56 g^{4} & = & 0 \\ 35 . & 7 . & 1 . & 0 . & 0 . \end{matrix}$ . That is $35 x^{4} - 35 g g x x = 0$ . Or $x x = g g$ . & $x = g$ . Or were{illeg} it multiplyed by y^e 2^d progression thus $\begin{matrix} x^{4} & * & - & 35 g g x x & + & 90 g^{3} x & - & 56 g^{4} & = & 0 \\ 0 . & - 1 . & 0 . & \frac{7}{8} . & \frac{45}{32} . \end{matrix}$ . It would produce $\frac{630}{8} g^{3} x - \frac{2520}{32} g^{4} = 0$ . Or $x = g$ . Soe y^t g being y^e least roote, y^e other two rootes must be $2 g$ & $4 g$ .

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

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

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 y^e method of finding {those}{these} progressions, And it {illeg} is this. Suppose {illeg} two of y^e rootes of an Equation {illeg} That Equation will bee of this forme $x x \begin{matrix} - a \\ - b \end{matrix} x + a b = 0$ , or of some forme {illeg} of it; And if a corresponds to y^e \desired/ roote of y^e Equation desired this equation $x x$ {illeg} will bee of this forme $a a \begin{matrix} - a \\ - b \end{matrix} a + a b = 0$ . Then assuming two termes ({illeg} { $a . 0$ }) {illeg} third {illeg} progression by w^ch I multiply this equation {illeg} { $a . 0$ } by which {illeg} {illeg} multiplyed produceth $a a \begin{matrix} - a \\ - b \end{matrix} a + a b = 0$ Or { $x a a - a^{3} - a a b = 0$ }. And {illeg} {illeg} termes of y^e {illeg} {illeg}

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Soe y^t I have thus much of y^e progression $a + b + \frac{b b}{a} . a + b . a . 0$ . And by y^e same proceding might continue it or get termes on y^e other side of y^e cipher. As if I multiply y^e Equation by this progression $a . 0 . z$ there is produced $\begin{matrix} a a & \begin{matrix} - a \\ - b \end{matrix} a & + & a b & = & 0 \\ \times & \times & \times \\ a & 0 & z \end{matrix}$ . Or $a^{3} + a b z = 0$ . And $z = - \frac{a a}{b}$ . Againe multiplying y^e Equation by $0 . x$ {illeg} $0 . - \frac{a a}{b} . z$ . It is $\begin{matrix} a a & \begin{matrix} - a \\ - b \end{matrix} a & + & a b & = & 0 \\ \times & \times & \times \\ 0 & . & - \frac{a a}{b} & . & z \end{matrix}$ . Or $\frac{a^{4} + a^{3} b}{b} + a b z = 0$ , And $- \frac{a^{3}}{b b} - \frac{a a}{b} = z$ . Soe that I have thus much of y^e progression viz: $a + b + \frac{b b}{a} . a + b . a . 0 . - \frac{a a}{b} . - \frac{a a}{b} - \frac{a^{3}}{b b}$ .

The proceeding is same when y^e proportion of 3 rootes to one another are given, but there may bee some difficulty /difference\ wⁿ y^e ciphers are {illeg} far distant, as they|r|e bee three {illeg} termes betwixt them, then y^e opr|e|ration may be done thus. Let y^e quanti\t/ys, w^ch beare such proportion to one another as y^e rootes doe bee, a, b, c . let a correspond to y^e roote w^ch w^ch must be knowne And yⁿ y^t Equation will bee of this forme, $a^{3} \begin{matrix} - a \\ - b \\ - c \end{matrix} a a \begin{matrix} - a b \\ - a c \\ - b c \end{matrix} a - a b c = 0$ . or else compounded of it. Then assuming some quantity (as a ) for one of y^e termes of y^e {illeg} progression & placing it conveniently, (as {illeg} \it {illeg} {no{illeg}}/ equidistant from y^e ciphers) feigne two other quantitys as z , y , for y^e deficient termes and there will r progression will bee $0 . z . a . y . 0$ . By w^ch I multiply the Equation $\begin{matrix} a^{3} & \begin{matrix} - a \\ - b \\ - c \end{matrix} a a & \begin{matrix} - a b \\ - a c \\ - b c \end{matrix} a & - & a b c & = & 0 \\ \times & \times & \times & \times \\ 0 & . & z & . & a & . & y \end{matrix}$ . Or $\frac{- a a z - a b z - a c z}{b c} = y$ . Soe have I y^e progression $0 . z . a . \frac{- a a z - a b z - a c z}{b c} . 0 .$ by w^ch I againe multiply y^e Equation & there is produced a correspond to y^e roote w^ch w^ch must be knowne And yⁿ y^t Equation will bee of this forme, $a^{3} \begin{matrix} - a \\ - b \\ - c \end{matrix} a a$ Or $\frac{- a a z - a b z - a c z + a a b + a a c + a b c}{b c} = y$ . Soe that I have the progression, $0 . z . a . \frac{+ a a b + a a c + a b c - a a z - a b z - a c z}{b c} . 0 .$ by w^ch I againe multiply y^e Equation & there results $\begin{matrix} a^{3} & \begin{matrix} - a \\ - b \\ - c \end{matrix} a a & \begin{array}{l} - a b \\ - a c \\ - b c \end{array} a ............... & - & a b c & = & 0 \\ \times & \times & \times & \times \\ z & . & a & . & \frac{+ a a b + a a c + a b c - a a z - a b z - a c z}{b c} & . & 0 \end{matrix}$ . Or $z = \frac{a a b b + a a b c + a a c c + a b b c + a b c c + b b c c}{a a b + a a c + b b a + a c c + 2 a b c + b b c + b c c}$ . w^ch valor of z substituted into its place in y^e valor y There will bee thus much of y^e progre\ssion/ $0 . \frac{a a b b + a a b c + a a c c + a b b c + a b c c + b b c c}{a a b + a a c + b b a + a c c + 2 a b c + b b c + b c c} . a . \{\begin{matrix} a + \frac{a a b + a a c}{b c} - \frac{a^{4} - 2 a^{3} b - 2 a^{3} c - a a b b - 3 a a b c - a a c c - a b c c}{a a b + a a c + b b a + a c c + 2 a b c + b b c + b c c} \\ \frac{- a^{4} b b - a^{3} b^{3} - a^{3} b b c - a^{4} c c - a^{3} b c c - a^{3} c^{3}}{a a b b c + a a b c c + a b^{3} c + a b c^{3} + 2 a b b c c + b^{3} c c + b b c^{3}} \end{matrix}\} . 0$ .

The same done otherwise.

Did I know y^t 2 of y^e rootes of this Equation $x^{3} * - 117 x - 324 = 0$ , were in proportion as 3, $- 4$ . Then I suppose one roote to be $3 a$ , y^e other $- 3 a$ That is $x - 3 a = 0$ . $x + 4 a = 0$ . By \one of/ w^ch I divide y^e Equation as first by $x + 4 a = 0$ And y^e operation is; $\begin{matrix} x + 4 a) & \underline{x^{3} * - 117 x - 324} & (x x - 4 a x + 16 a a - 117 = 0 \\ \underline{0 - 4 a x x - 117 x - 324} \\ \underline{\begin{matrix} 0 + 16 a a x - 324 \\ - 117 x \end{matrix}} \\ \underline{\begin{matrix} 0 - 64 a^{3} \\ + 468 a^{} \end{matrix}} \\ \underline{- 64 a^{3}} & \underline{+ 468 a - 324 = 0} \\ 0 \end{matrix}$ . Againe I divide y^e Quotient by y^{e} other roote $x - 3 a = 0$ . Thus
$\begin{matrix} x - 3 a) & \underline{x x - 4 a x + 16 a a - 117} = 0 & (x - a = 0 \\ \underline{0 - a x + 16 a a - 117} \\ 0 \underline{+ 13 a a - 117} = 0 \\ 0 \end{matrix}$ . By y^e last division I have this equation $13 a a - 117 = 0$ . Or $a a - 9 = 0$ . And $a = 3$ . Therefore y^e rootes of y^e Equation $x^{3} - 117 x - 324 = 0$ are, $3 a = 9 = x$ . $- 4 a = - 12 = x$ .

If I would have y & z of such a length y^t y^e rootes of this equation $x^{3} - 3 y x + g z = 0$ . be in proportion as $+ 1$ , $+ 2$ , $+ 3$ . I suppose $x - a = 0$ , $x + 2 a = 0$ . $x + 3 a = 0$ . And soe first divide y^e Equation by $\begin{matrix} x - a) & \underline{x^{3} + * - 3 y x + g z} & (x x + a x + a a - 3 y = 0 \\ \underline{0 + a x x - 3 y x + g z} \\ 0 \begin{array}{r} + & a a \\ - & 3 y \end{array} x + g z \\ \overline{0 + a^{3}} & \overline{+ 3 a y + g z^{} = 0} \end{matrix}$ . Againe I divide this product
by $\begin{matrix} x - 2 a) & \underline{x x + a x + a a - 3 y} & (x + 3 a = 0 \\ 0 + 3 a x + a a - 3 y \\ \overline{0 + 5 a^{3} - 3 y} & \overline{= 0} \end{matrix}$ . Lastly were it necessary I should have again divided this quote $x + 3 a = 0$ by y^e 3^d supposed roote of y^e Equation (viz {illeg} $= 0$ ). By y^e 2^d {operation} $7 a^{2} - 3 y = 0$ . Or $\frac{7 a^{2}}{3} = y$ . And by y^e first $a^{3} - 3 a y + g z = 0$ . Or {illeg} /{ $18 a y = 7 g z$ }\ Soe y^t If I make {illeg} the rootes of this Equation $x^{3} - 3 y x + g z = 0$ should bee {illeg}

<57r>

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

Resolution.

Sett all y^e termes on one side of y^e Equation y^t they become equall to nothing. And first Multiply each terme by soe many times $\frac{p}{x}$ as x hath dimensions in y^t 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 y^e relation of p, q, r &c.

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

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

Example y^e 2^d. If y^e 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 $\begin{matrix} \frac{3 p}{x} & \frac{p}{x} & 0 & 0 & 0 \\ x^{3} & * & + z z x - y y x & - 2 a a y & + z^{3} y & - z^{3} \end{matrix}$ . Or $3 p x x + p z z - p y y$ . The second produceth $- 2 a a q - 2 y x q + 2 z y q$ . The third $+ 2 z x r + y y r - 3 z z r$ . The summe of w^ch is $3 p x x + p z z -$ $- p y y - 2 a a q - 2 y x q + 2 z y q + 2 z x r + y y r - 3 z z r = 0$ . (Note y^t in this Example there being three unknowne quantitys x , y , z , There must be two of them & their \two/ velocitys supposed thereby to find y^e 3^d quantity & y^e third velocity. Or else there must be some other equation expressing y^e relation of y^e th two of these x , y , z . (as in y^e first example) whereby one quantity & one velocity being supposed y^e other quantity & velocity may be found & yⁿ by this 2^d Example y^e 3^d quantity & y^e 3^d velocity may bee found)

Example 3^d, Of y^e more generall rule. If y^e Equation bee $x^{4} - 3 g y x x + y y x x - g g y y - 2 y^{4} = 0$ . y^e first operation gives $\begin{matrix} \frac{a p}{x} & . * . & \frac{a p + 2 b p}{x} & . * . & \frac{a p + 4 b p}{x} & . \\ x^{4} & * & \begin{matrix} + & y y \\ - & 3 g y \end{matrix} x x & * & \begin{matrix} - & g g y y \\ - & 2 y^{4} \end{matrix} & . \end{matrix}$ Or $\begin{array}{l} a p x^{3} & + & a p y y x & - & 3 a p g y x & - & \frac{a p g g y y}{x} & - & \frac{2 a p y^{4}}{x} \\ + & 2 b p y y x & - & 6 b p g y x & - & \frac{4 b p g g y y}{x} & - & \frac{8 b p y^{4}}{x} \end{array}$ . the 2^d gives $\overline{a q - 2 b q} \times \overline{- 2 y^{3}} \overline{+ a q} \times \overline{x x y - g g y} \overline{+ a q + b q} \times \overline{- 3 g g x} \overline{+ a q - 2 b q} \times \frac{x^{4}}{y}$ . \The summe of/ w^ch two products is equall to nothing. &c.

Demonstration.

^[27] Lemma. If two bodys $\begin{matrix} A \\ B \end{matrix}$ move uniformely y^e $\begin{matrix} one \\ other \end{matrix}$ from $\begin{matrix} a \\ b \end{matrix}$ to $\begin{matrix} c & , & e & , & g & , \\ d & , & f & , & h & , \end{matrix} &c$ in y^e same line yⁿ are y^e lines $\begin{matrix} ac \\ bd \end{matrix}$ & $\begin{matrix} ce \\ df \end{matrix}$ & $\begin{matrix} eg \\ fh \end{matrix}$ &c as their velocitys $\begin{matrix} p \\ q \end{matrix}$ {illeg}.And though they move not uniformely yet are y^e infinitely little spac lines w^ch each moment they describe {illeg} as their velocitys are w^ch they have while they describe them. As if y^e body A \w^th y^e velocity p / describe y^e infinitely little line o in one moment. Then \In y^e moment/ y^e body B w^th y^e velocity q will describe y^e line $\frac{o q}{p}$ . For $p ∶ q ∷ o ∶ \frac{o q}{p}$ . Soe y^t if y^e described \{lines}/ be x & y in one moment, they will bee $x + o$ & $y + \frac{o q}{p}$ , in y^e next. [or better $p ∶ q ∷ p o ∶ q o$ . &c]

Now if y^e Equation expressing y^e relation of y^e lines x & y be $r x + x x - y y = 0$ . I may substitute $x + o$ & $y + \frac{q o}{p}$ into y^e place of x & y because (by y^e lemma) they as well as x & y doe signifie y^e lines described by y^e bodys A & B . By doeing so there results $r x + r o +$ $+ x x + 2 o x + o o - y y - \frac{2 q o y}{p} - \frac{q q o o}{p p} = 0$ . But $r x + x x - y y = 0$ by supposition: there remaines therefore $r o + 2 o x + o o - \frac{2 q o y}{p} - \frac{q q o o}{p p} = 0$ . On divideing it by o tis $r + 2 x + o - \frac{2 q y}{p} - \frac{o q q}{p p} = 0$ . Also those termes in w^ch o is are {sic} infinitely lesse yⁿ those in w^ch o is not therefore I blotting y^m out there {rests} $r + 2 x - \frac{2 q y}{p} = 0$ . Or $p r + 2 p x = 2 q y$ .

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

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

<57v>

By helpe of y^e preceding probleme divers others may bee readily resolved.

^[28] 1. To draw tangents to crooked lines (however they bee related to sreight {sic} ones).

Resolution

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

Example 1^st. If $ab = id = x$ . $ai = bd = y$ . & $r x - \frac{r x x}{s} = y y$ . Then is $p ∶ q ∷ 2 y ∶ r - \frac{2 r x}{s}$ . (by y^e former rule) Therefore I draw $gd ∶ ge$ {illeg} $de ∶ dg ∷ p ∶ q ∷ 2 y ∶ r - \frac{2 r x}{s}$ . The point g is inclined to move in a parallel to abc & y^e point e in a parallel to aik (for bg \& ie / (by supposition) moves parallel to y^m selves y^e $\begin{matrix} one \\ other \end{matrix}$ upon $\begin{matrix} abc \\ aik \end{matrix}$ ) Therefore I draw $gf ∥ abc$ & $ef ∥ aik$ . & through \y^e intersection/ $(f)$ I draw hdf touching y^e crooked line at d . Soe y^t $gd ∶ de ∷ db ∶ bh ∷ q ∶ p ∷ r - \frac{2 r x}{s} ∶ 2 y$ .

Hence may bee pronounced those theorems in pag 47 Fol 47

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

^[31] 2. Hitherto may bee reduced y^e manner of drawing tangents in mechanicall lines. see Fol 50.

^[32] 3. To find y^e quantity of crookednes in Geometricall lines.

{illeg} Resolution

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

Example. Suppose $ab = x$ . $bc = y$ . $x ⊥ y$ . {illeg} $a x - \frac{a x x}{b} = y y$ . $be = a = \frac{a}{2} - \frac{a x}{b}$ . $p =$ motion of b from a {illeg} $y = \frac{- a p}{b} =$ motion of e from b. & $\frac{b p - a p}{b} =$ motion of {c}{e} from a $bk = cn = \frac{y y}{v} = \frac{2 b x - 2 x x}{b - 2 x}$ . $nm = y$ . $nd = v = \frac{a b - 2 a x}{2 b} = \frac{y y}{x} - \frac{a}{2}$ . $cd = ke$

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

Example. Suppose $a x - \frac{a x x}{b} - y y = 0$ . yⁿ is $be = v = \frac{a}{2} - \frac{a x}{b}$ . $kb = \frac{y y}{v} = \frac{2 b y y}{a b - 2 a x}$ . And $cd = ke = \frac{a a b b - 4 a a b x + 4 a a x x + 4 b b y y}{2 a b b - 4 a b x}$ . $bk = p =$ motion \velocity/ of b from a , $bc = y = q =$ velocity of y 's increase $r =$ velocity of v 's increase. $2 b v - b a + 2 a x = 0$ therefore $2 b r + 2 a p = 0$ , or (since $p = \frac{2 b y y}{a b - 2 a x}$ ) tis $r = \frac{2 y y}{- b + 2 x}$ . & $ef = \frac{2 a y y + 2 b y y}{a b - 2 a x}$ \{ $= r + bk =$ {illeg} }/ Lastly $cd - ef ∶ ce ∷ cd ∶ cg$ (or $v - r ∶ y ∷ v + p ∶ ck$ . if { $ce ⊥$ {illeg} }) y^t is $\frac{- 4 b y y + 4 a b x - a b b - 4 a x x}{- 2 b b + 4 b x} ∶ y ∷ \frac{a a b b - 4 a a b x + 4 a a x x + 4 b b y y}{2 a b b - 4 a b x} ∶ ch$ . $\frac{a a b y + 4 b y^{3} - 4 a y^{3}}{a a b} = ch = \frac{4 b y^{3} - 4 a y^{3} + a a b y}{a a b}$ . & $bh = \frac{4 y^{3}}{a a} - \frac{4 y^{3}}{a b}$ .

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 $\begin{array}{l} \begin{matrix} + 1352 \\ + 7460 \end{matrix}} \\ + 8812 \end{array}$ $\begin{array}{l} \begin{matrix} - 137905 \\ - 68432 \end{matrix}} \\ - 206337 \end{array}$

Substractions takes y^e greater \lesse/ number from y^e lesse \greater/, the difference having y^e same signe prefixed w^ch y^e greater quantity \number/ {hats} as $\begin{array}{l} \begin{matrix} + & 6 & 3 & 5 & 7 & 9 & 2 \\ - & 1 & 4 & 7 & 0 & 3 & 1 \end{matrix}} \\ \begin{matrix} + & 4 & 8 & 3 & 7 & 6 \end{matrix} \end{array}$ . $\begin{array}{r} + 44503 \\ - 26794 \end{array}$ $\begin{array}{l} \begin{matrix} - 26791 \\ + 4503 \end{matrix}} \\ - 22288 \end{array}$ .

Multiplicacon adds one factor soe often {illeg} to it selfe as there are units in y^e other, & if y^e signes of y^e factors bee y^e same y^e product is affirmative, if divers tis negative. As to multiply $+ 735$ by $+ 47$ doe thus $735 in 7$ / $\begin{array}{l} \begin{array}{r} 5145 = 735 in 7 \\ \underline{29400 = 735 in 40} \end{array}} \\ \begin{array}{r} + 34545 = 735 in 47 \end{array} \end{array}$ \ Or thus $\begin{array}{r} 735 \\ 47 \\ 5145 \\ 29400 \\ + & 34545 \end{array}$ . Or thus $\begin{matrix} \begin{array}{r} 735 \\ 5145 \\ 29400 \\ + & 34545 \end{array} & \begin{array}{l} \times \\ 7 \\ 40 \end{array} \end{matrix}$ . Thus to multiply $- 3241$ by / $- 3175$ the operation will\ bee $\begin{matrix} \begin{array}{r} 3241 \\ 16205 \\ 22687 \\ 3241 \\ + & 567175 \end{array} & \begin{array}{l} 5 \\ 7 \\ 1 \end{array} \end{matrix}$ Alsoe 465 multiplyed by $- 32$ /will produce\ $\begin{matrix} \begin{array}{r} 465 \\ 930 \\ 1395 \\ - & 14880 \end{array} & \begin{array}{l} 2 \\ 3 \end{array} \end{matrix}$

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

If 19489012 was to be divided by {illeg} 732 $\begin{matrix} \begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 732 & . & 1464 & . & 2196 & . & 2928 & . & 3660 & . & 4392 & . & 5124 & . & 5856 & . & 6588 \end{matrix} \end{matrix}$ $\begin{array}{l} \begin{matrix} \begin{array}{r} + 3 . & + 1948 9 012 \\ - 2 . & - 147.0 988 \\ - 9 . & - 000 6.988 \end{array} & \begin{array}{r} + & 30000 00355 \\ - & 02009, 55000 \\ - & 27990, 45355 \end{array} \end{matrix} \\ \begin{array}{r} - 5 . & - 0 400 & . 0 \\ - 5 . & - 034 & 0.0 \\ + 3 . & + 02 & 6 0.0 \\ + 5 . & + 0 & 4 0 4.0 \\ + 5 . & + & 0 3 8 0 . \end{array} \end{array}$

<64r>

The resolution of y^e affected Equation $x^{3} + p x x + q x + r = 0$ . Or $x^{3} + 10 x^{2} - 7 x = 44$ First having found two or 3 of y^e first figures of y^e desired roote viz $\begin{matrix} \begin{matrix} 2 \end{matrix} & \begin{matrix} \underline{2} \end{matrix} \end{matrix}$ (w^ch may bee done either by rationall of Logarithmicall tryalls as M^e 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 w^ch may be oblong but better circular.) this knowne pte of y^e root I call g, y^e other unknowne pte I call y then is $g + y = x$ . Then prosecute y^e Reduction /Resolution\ after this manner (making $x + p in x = a$ . $a + q in x = b$ . $b + r in x = c$ &c.) $\begin{array}{l} \times & 12 = x + p \\ x = 2, & 24 = a \end{array}$ . {illeg} $\begin{array}{l} a + q = 17 & \times \\ b = 34 & 2 = x \end{array}$ . $r - b = 10 = h$ . by supposing $x = 2$ . Againge supposing $x = \begin{matrix} \begin{matrix} 2 \end{matrix} & \begin{matrix} \underline{2} \end{matrix} \end{matrix}$ Then $\begin{array}{r} x + p = 12,2 \\ 244 & 2 \\ 244 & 2 \\ 26,84 & = a \end{array}$ . $\begin{array}{r} a + q = 19,84 \\ 3968 & 2 \\ 3968 & 2 \\ 43,648 & = b \end{array}$ . $r - b = 0,352 = k$ . $h - k = 9,648$ . That is y^e $\{\begin{cases} latter r - b substracted \\ difference twixt \end{cases}$ $\begin{array}{l} from the former r - b there remaines \\ this & y^{e} former valor of r - b is \end{array}\} 9,648$ . & y^e difference twixt this & y^e former valor of x is 0,2 . Therefore make $9,648 ∶ 0,2 ∷ 0,352 ∶ y$ . Then is $y = \frac{0,0704}{9,648} = 0,00728$ &c. the first figure of w^ch being added to y^e last valor of x makes $2,207 = x$ . Then w^th this valor of x presecuting y^e operacon as before tis $\begin{array}{r} x + p = 12,207 \\ 24414 \\ 24414 & 7 \end{array}$ $\begin{array}{r} x + p = 12,207 \\ 65449 & 7 \\ 244140 & 02 \\ 244140 & 2 \\ 26,940849 & = a \end{array}$ $\begin{array}{r} a + q = 19,94084 \\ 13958588 & 7 \\ 39881680 & 02 \\ 3988168 & 2 \\ 44,00043388 & = b \end{array}$ . $r - b = - 0,00943388$ . w^ch valor of $r - b$ substracted from y^e precedent valor of $r - b$ y^e {diff} is $+ 0,36143388$ . Also y^e {diff} twixt{illeg} tis & y^e precedent valor of x is 0,007 . Therefore I make { $36143388 ∶ 0,007 ∷ - 0,0094388 ∶ y$ }. That is $y =$ {illeg}

<67v>

^[34]

Of the construction of Problems.

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

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

^[36] As if I had this Equation yy+6y−9=0 Or yy=−6y+9. Then must I make $dc = \frac{9}{c} + \frac{1}{2} c$ . Then if I make c=6 it will bee $d = \frac{3}{2} + 3 = \frac{9}{2} = 4 \frac{1}{2}$ . Therefore I take $ab = - \frac{p}{2} = bd =$ $ab = - \frac{1}{2} p = -3$ . $bd = c = 6$ . $dc = 4 \frac{1}{2} = \frac{3 c}{4}$ . And soe describing y^e 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 y^e same.

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

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

< insertion from f 67v >

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

Were y^e Equation yy=−4y+34. Or $y = -2 \overset{\cup}{O} \sqrt{38}$ . Then is 38−36=2. 2−1=1. & 38=36+1+1 w^ch are square numbers. Therefore I make $ad = \sqrt{36} = 6$ . $ad ⊥ de = \sqrt{1} = 1$ . & draw $ae ⊥ ec = \sqrt{1} - 1$ . & draw $ac = \sqrt{38}$ . from w^ch take ab=2, & there rests $bc = -2 + \sqrt{38} = y$ .

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

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

^[38] If y^e Probleme be sollid it may bee readily resolved by {illeg} the intersection of y^e Parabola & circle. \as D: Cartes hath shewed/ If it bee of 5 or 6 dimensions it may bee resolved by y^e intersection of y^e line b y³−byy−cdy+bcd+dxy=0. Or y³−byy+bcd+dxy=0 & y^e circle \when pp={illeg} 4q. & q & v affirmative./ as D: C: hath explained. Or it might beee done by y^e intersection of a circle & one of these lines, viz y³+byy−hx=0 when y^e equation is reduced to such a forme y^t 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 y^e line a²x=y³, after this manner. First (making a=1{.}|)| describe y^e line x=y³ uppon a plate. {illeg} (as cadce. Then in w^ch ab=x. bc=y). Then suppose y^e Equation to bee resolved bee y⁹*+my⁷+ny⁶+py⁵+qy⁴+ry³+syy+tq+v=0. (in w^ch y^e letters m, n, p &c: signifie y^e ^[40]termes knowne quantitys of each terme affected w^th its signe + or −). I describe another line cdce, whose nature (making ab=x, bc=y) is the exprest $\begin{matrix} x^{3} & + & m y x x & + & p y y & + & s y y \\ + & n & + & q y & x & + & t y & = 0 \\ + & r & + & v \end{matrix}$ & letting & let fall perpendicula{{illeg}|r|} from{e} every point where these two lines intersect as, df eg, they shall bee y^e rootes of y^e propounded eqation.

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

Tis evident alsoe y^t there are 3 divers lines {any of w in} by w^ch {illeg} \any/ Probl: may bee resolved \unless some of them {chanch} to be y^e 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 y^e 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 y^e line whose nature is $\begin{array}{l} + & y x^{3} & + & n y y & + & r y^{3} & + & w y^{3} \\ + & l & + & p y & x x & + & s y y & x & + & a y y & = 0 \\ + & q & + & t y & + & b y \\ + & v & + & c \end{array}$ . A line of y^e 2^d sort. Whereas by y^e preceding rule was required y^t a line of y^e 3^d sort should have ^[41] beene described. And here observe y^t taking y^e square number w^ch is next greater yⁿ y^e number of y^e resolvend equation termes \dimensions/ of the resolvend equation,|.| That Equation may bee resolved by lines, y^e number of whose dimensions is not greater than the roote of y^e {illeg} square number. And the rectangles of th{illeg}|os|e numbers w^ch signifie y^e 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 y^e number of dimensions of y^e resolvend Equations. For y^e number of points in w^ch two lines may intersect can never bee ^[42] greater yⁿ y^e rectangle of y^e numbers of theire dimensions. And they always intersect in soe many points, excepting those w^ch are im̄aginarie onely. Soe that all Equations $\begin{matrix} \begin{array}{l} which have \\ noe more than \end{array} & \{\begin{matrix} 1 \\ 2 \\ 4 \\ 6 \\ 9 \\ 12 \\ 16 \end{matrix}\} & \begin{matrix} dimensions, may always be \\ resolved by the \\ intersection of \end{matrix} & \{\begin{array}{l} 2 streight lines \\ a streight line & a c o nick section. \\ two conic k sections. \\ a conick section of a line of 3 dimensions. \\ two lines of 3 dimensions. \\ a line of 3. & another of 4 dimensions. \\ two lines of 4 dimension; or by one of 1 & another of 6 dim: &c: \end{array}\} & \begin{array}{l} but not by any simpler \\ lines. (From this conside r \\ may Problems be di st in- \end{array} \end{matrix}$ guished into sorts.) {St} will often bee very intricate to resolve Equations \of many dimensions/ by the simplest line by w^ch they may be resolved & also for y^e 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 y^e 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 y^e parabola

T Suppose therefore I had y^e 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 y^e line gh & from y^e intersection points d, e, draw db, ec perpendicular to y^e axis gc. w^ch shall bee y^e rootes of y^e Equation w^ch are affirmative when they fall on y^e contrary side to fh, but negative if {illeg}|on| y^e same, as in this case.

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

^[45] If I would resolve y^e cubick Equation y³+ky²+ly+m=0 (w^ch multiplyed by y−k=0 produceth $\begin{array}{l} y^{4} * & + & l y y & + & m y & - & k m & = 0 \\ - & k k & - & k l \end{array}$ ) 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 w^th y^e radius cg upon y^e center q describe y^e circle wf. Or else doe thus (since k is one of y^e rootes of y^e Equation $\begin{array}{l} y^{4} * & + & l & y y & + & m & y & - & m k & = 0 \\ - & k k & - & l k \end{array}$ ) make k=ab+ar & draw bw∥ae (or make ar=kk, & wr{illeg}⊥ap) & describe a circle w^th y^e radius wq. Then letting fall perpendiculars from y^e intersection points, they (being y^e rootes of y^e Equation $\begin{array}{l} y^{4} * & + & l & y y & + & m & y & - & m k & = 0 \\ - & k k & - & l k \end{array}$ ) shall all, except wr=k, bee y^e rootes of y^e Equation y³+kyy+ly+m.

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

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

<69r>

^[47] If I had y^e crooked \line/ described \fig 1st/^[48] whose nature is x=y³, & would resolve y^e 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 y^e intersection points (as e) letting fall perpendiculars, they shall bee y^e {illeg} rootes of y^e Equation y³*+lyy+m. as ce w^ch in this case is negative because on that side on w^ch y is negative.

Would I resolve this equation yy+ky+l=0. (w^ch multiplyed by y-k produceth $\begin{array}{l} y^{3} * & + & l & y & - & kl = 0 \\ - & kk \end{array}$ ) I take ab=kl, (fig 2^d)^[49] bδ=l, δd=kk. df=1, & soe through y^e points b & f draw the streight line bfλ |(|Or w^ch is y^e same take ab=kl. k=ah⊥ab. & draw hλ∥ab untill it intersect y^e crooked line in λ (i. e. untill hλ=k³ & soe through the points λ & b draw λbfe|)|. Then from y^e intersection points to y^e axis letting fall perpendiculars they (being y^e rootes of y^e Equation $\begin{array}{l} y^{3} * & + & l & y & - & lk = 0 \\ - & kk \end{array}$ .) shall all, except βλ=k, be y^e rootes{illeg} of y^e 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 $\begin{array}{r} z^{6} * * & + & c z^{3} & + & d z z & - & ad z & + & aad & = 0 \\ - & 2 ab & - & ac & + & aac & - & bd \\ + & a^{3} & + & aab & - & bc \\ - & bb \end{array}$ , Of this forme z⁶**+mz³+nzz+pz+q=0. In w^ch (n) ought to be affirmative, & if it bee not, yⁿ augment or diminish y^e rootes of y^e Equation z⁴+az³+bzz+cz+d=0. & then repeate y^e operacon again untill there bee an Equation of this forme z⁶**+mz³+nzz+pz+q=0 in w^ch n is affirmmative. Then (dividing this equation by $\sqrt{4 : n}$ it is $z^{6} * * + \frac{m}{\sqrt{n \sqrt{n}}} z^{3} + z z + \frac{p}{n \sqrt{4 : n}} z + \frac{q}{n \sqrt{n}} = 0$ therefore) take $ab = \frac{m}{2 \sqrt{n \sqrt{n}}}$ . $bc = \frac{p}{2 n \sqrt{4 : n}}$ . {illeg} & w^th y^e radius $cd = \sqrt{\frac{mmn + pp - 4 nq}{4 nn \sqrt{n}}}$ , describe y^e circle dk & y^e perpendiculars (as dh ck) multiplyed by $\sqrt{4 : n}$ shall bee y^e rootes of y^e Equation.

<71r>

Theoremata Optica.

Si radius divergens a puncto \dato/ A vel convergens ad punctum \idem/ A incidit in Sphæram CVD ad punctum D, et refractus ejus convergit ad puncto B vel divergit ab eodem, sit sphæræ centrum C {illeg} in recta AB situm, secet {sic} AB sphæram in v tr sit sph{illeg}|æ|ræ centrum C, & secet AC producta sphæram in V et radium refractum in {illeg} DR in B: a punctis D et C ad AB, AD, BD demitte normales DH, CI, CR; sit sinus incidentiæ ad sinum refractionis seu Ci ad CR ut I ad R; et facto R,AC.I,AV∷CF.VF. {illeg} erit {illeg}F focus, seu locus imaginis puncti A radios quaquaversum emittentis.

2^do 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 $\frac{AC, FC, FG, {VD}^{q}}{2 {AV}^{q}, {CV}^{q}}$ , sive {illeg} $\frac{AC, FC, FG, VH}{{AV}^{q}, CV}$ 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 $\frac{R, VA}{G}$ (nam hæ \jam/ sunt æquipollentia) error BF {illeg}|{fist}| et $\frac{\frac{R}{G} FC, {MD}^{q}}{2 {CV}^{q}}$ , vel {illeg} $\frac{\frac{R}{G} FC, VH}{CV}$ , 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 $FB = \frac{AC, FC, FG}{{AV}^{q}, CV} VH$ . vel AV^q×CV. ACF×FG∷VH.FG.

<71v>

Hujus autem Theorematis inventio totis est.

${AD}^{q} = {AV}^{q} + 2 AC, VH$ . ${DF}^{q} = {VF}^{q} - 2 CF, VH$ . Et extractis radicibus, $AD = AV + \frac{AC}{AV} VH - \frac{{AC}^{q}}{2 {AV}^{cub}} {VH}^{q}$ &c $DF = VF - \frac{CF}{VF} VH - \frac{{CF}^{q}}{2 {VF}^{cub}} {VH}^{q}$ &c fluxio \ipsius/ $AD = \frac{AC}{AV} fl : VH - \frac{{AC}^{q}, VH}{{AV}^{cub}} f l VH$ &c |De|fluxio ipsius $DF = - \frac{CF}{VF} fl VH - \frac{{CF}^{q}, VH}{{VF}^{cub}} fl VH$ &c $\frac{R}{I} fl : AD + defl : DF =- \frac{R, {AC}^{q}}{I, {AV}^{cub}} - {CF}^{q} {VF}^{cub}$ 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 $\frac{R}{I} fl AD + defl : DF$ . Et ut error ille sive defluxio a legitima figura \obliquitate/ ita \{illeg}/ error angular radij {illeg} refracti. Jam vero est $\frac{R, {AC}^{}}{I, AV} = \frac{CF}{VF}$ Ergo {fluxio} error{illeg} angularis radij refracti \est/ ut $\frac{AC}{{AV}^{q}} - \frac{CF}{{VF}^{q}}$ in $\frac{CF, VH}{VF}$ , fl VH. Vel etiam ut $\frac{I, CF}{R, AV, VF} + \frac{CF}{{VF}^{q}}$ in $\frac{CF, VH}{VF}$ , fl VH seu ut {illeg}, $VF + \frac{R}{I} AV$ in $\frac{{CF}^{q}, VH}{{VF}^{c}, AV}$ fl VH seu $\frac{FG, {CF}^{q}, VH}{AV, {VF}^{c ub}}$ , fl {illeg} posito fl VH=1 & $VF + \frac{R}{I} AV = FG$ . Dat{{illeg}ur} autem ratio $\frac{CF}{VF}$ ad $\frac{AC}{AV}$ ergo substituto posteriore fiet error ille ut $\frac{FG, CF, AC, VH}{{AV}^{q}, {VF}^{q}}$ . Duc in VF^q et error in axe FB erit ut $\frac{FG, CF, AC, VH}{{AV}^{q}}$ qua{illeg}do circuli radius determinatur. Divide per radium circuli et fiet $\frac{FG, CF, AC, VH}{CV, {AV}^{q}}$ 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} $\frac{FG, CF, AC}{CV, {AV}^{q}} VH$ ergo semper idem est cum hac quantitate.

<71a(r)>

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

$\begin{array}{l} - df, Ce, Cp, Cq - ef, CO, Cp, Cq = 0 . & d, f, Ce, Cp, pO = e, f, CO, Cp, qe = e, f, {Dn}^{q}, qe . \\ \frac{+ df, Ce,}{7) 20 \frac{1}{4} (Dq)} \binom{CO, Cq - ef, CO, Cp, Ce}{} & PO = \frac{e,, {Dn}^{q}, qe}{d, Ce, Cq} = R . \sqrt{} \frac{RR + 4 {Dn}^{q}}{4} :\pm \frac{1}{2} R = \binom{CO.}{CP.} \end{array}$ $\frac{9, 20 \frac{1}{4}, 9 \frac{6 \frac{1}{4}}{7} .}{14, 7, 2 \frac{6 \frac{1}{4}}{7}} = \frac{89}{14} = 6 \frac{1}{3} = PO = R . \sqrt{} 30 \frac{1}{4} \pm 3 \frac{1}{6} = \binom{CO}{CP} = 5 \frac{1}{2} \pm 3 \frac{1}{6} = \binom{8 \frac{2}{3} .}{2 \frac{1}{3} .} CO = 8 \frac{2}{3} . CP = 2.$ $Cm = \frac{5, 60 \frac{2}{5}}{98 + 78} = \frac{303 \frac{1}{5}}{176} \begin{array}{l} 910 & 455 \times 12^{inch} \\ 528 & 264 \end{array} = \begin{matrix} 455 & 20, 6 \\ {xx}_{15} \end{matrix}$ | $Cm = 20 \frac{2}{3} inches$ |.

<71a(v)>

$Z = N + \frac{\frac{far - fz}{a - r + f} x - aNx - ffx}{a - r + f \times a - r - f.} \frac{e}{d} a = f . AV = t = a - r$ $Z = N + \frac{\frac{far - fz}{t + f} x - aN - ff}{tt - ff} x = N + \frac{far - fft - 2 fz - faN - taN}{\bar{t \times f} \times \bar{t + f} \times \bar{t - f}} . \binom{N = CD .}{N = \frac{fr}{t - f} .}$ $t . f ∷ \binom{BG . CF .}{VD . CD} t . a ∷ BG . \frac{d}{e} CF$ 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]}

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

<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 $\frac{1}{HX} - \frac{2}{FX} = A$ . Et $\frac{1}{GX} + A = B$ . Et $\frac{1}{B} = gX$ . 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} $\frac{4 S - 4 R}{R \times A}$ radius circuli {RVS} sit ista {C}{illeg}{−}{illeg}{−}{illeg}{D}{illeg}{−}

<72v>

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

Exempli gratia. In Telescopij cujusdam vitro objectivo observabam VF=13^ped.11^digit. {TH=} $VH = 6^{ped.} {9 \frac{13}{16}}^{digit.}$ $TG = 2^{ped .} {4 \frac{13}{16}}^{dig.}$ . Et $TV = {\frac{2}{9}}^{dig.}$ . Ergo Seu {VF=} VF=167^dig. TG= VH={illeg}1,8125^dig. TG=28,8125^dig. TV=0,2222 &c \dig./. Adeo $XF = 167_{,}^{d} 1111$ . $XH = 8_{,}^{d} 9236$ . $XG = 28_{,}^{d} 9236$ . Unde prodit A= 0,00023{illeg}|{84}|{illeg} dig. B=0,0348{illeg}|{127}|^dig. gx=28,72{illeg}^dig|56^dig.| 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 $\begin{matrix} 61 & . & 40 \\ 90 & . & 59 \end{matrix}$ ∷ I.R. $\frac{4 I - 2 R}{RA} =$ $\frac{}{R} 1243^{dig} = \frac{R}{g} {937 \frac{3}{4}}^{ped} = radio cir culi$ {illeg}=17161{illeg}^dig=1430^ped. {Cir}|Un|nde alterius RVS semidiameter erat quasi 7^ped 4^dig. At hæc ita se habebant in vitro objectivo Telescopij D^ris Babington.

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

<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 $\frac{2 I}{R} EO . OX ∷ P . XY$ et habebitur circuli GEH centrus Y. Ubi nota quod usurpo $\frac{I}{R}$ pro ratione sinus incidentiæ ex aere in vitrum ad sinum refractionis. Et suppono in super vitri crassitiem EF ad instar nihili esse. ff Fac $\frac{2 I \times EO}{R \times EX} = a . \frac{EO}{ER} = b . b \times RX = c$ . et {illeg} $\frac{OX - c}{a - b + 1} = XY$ . Et habebitur circuli GEH centrum Y. Ubi nota quod usurpo $\frac{I}{R}$ 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 y^e w{h}eights of y^e following equall bodys to bee in y^e proportions following.

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

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

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

A Sphære of tinn whose diameter is six inches weighs Thirty and Eight pounds. The following line being {6} of Ghetaldus his inches, w^ch is half y^e Roman foot by Villalpandus account from y^e Far{nesian urn}. $\begin{matrix} |-----------------------------------------------------------------------| \end{matrix}$ Soe y^t y^e 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 ${\frac{798}{11}}^{} = \overset{li}{72 \frac{6}{11}}$ \li/. Or more exactly ${\frac{25764}{355}}^{lib} = \overset{lib}{72, \frac{204}{355}} = \overset{lib}{72,57465}$ {-.} \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 9,80752^lib. & a foot cube of water $78 \frac{6}{13}$ ^lb or more exactly 78,46016^lb.

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 y^e 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 74^lb, 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 {y^e} proportion of y^e french foot to y^e 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 y^e {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 y^e proportion of y^e weights of two equall bodys y^e one being sollid y^e other liquid. Resp. If y^e Sollid body A bee heavier yⁿ y^e liquid B weigh it in y^e aire & in y^e liquid Body B; & y^e difference of those two weights is y^e weight of soe much water as is equall to y^t sollid body |& so much {wyer} as {deppend} in y^e {water}|. But if it bee lighter yⁿ y^e liquid body B, hang a heavier body C {at} to it, y^t will sinke it; & weigh y^m first y^e body A being in y^e air & C in y^e water{;} 2^dly liquid body, 2^dly both A & C being in y^e liquid body; & y^e difference of these{s} weights is y^e weight of so much water as is equall to A (& also to y^e soe much thred wyer or hayre as was weighed both{illeg} in y^e water & air.)

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

__________________________________________________________________________________________________________________________

Or Thus, Make y^e scoale B as light as may bee, yet soe y^t it will sinke it selfe in y^e liquid body CC, & \sinke/ y^e body A also, if A chance to bee lighter yⁿ CC. Suppose y^t f y^e weight of scoale B in y^e water, & g in y^e aire; & y^t e is y^e weight of B+A in y^e water, & d in y^e aire: yⁿ is d−g+f−e y^e weight of soe much water as is equa{illeg}|ll| to A. |Note y^t y^e water must reach {'}at both weighings to y^e same point {illeg} of y^e wyer & y^t y^e wyer bee as small as may bee.|

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

^[54] Prob 2. Two{illeg} bodys {illeg} \D & E/ given to find y^e proportion of their quantity. Weigh y^m in y^e scale B, let their weight \& y^e weight of y^e scale/ in y^e air bee h & k, in y^e 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 y^e weights \& proportions,/ of its two compounding parts c & d. Answer. a & c, b & d, are of y^e same matter. That y^e weights of y^e 5 bodys a, b, cd, c, d, in y^e aire are ef e, f, n, m, n−m; & in y^e water g, h, q, p, q−p. Then is $e : g ∷ m : p = \frac{gm}{e}$ . & $f : h ∷ n - m : q - p = \frac{hn - hm}{f}$ . The{re}fore $q = \frac{gm}{e} + \frac{hn - hm}{f}$ . Or $\frac{feq - hen}{gf - eh} = m = to$ y^e weight of C in y^e aire. Also $\frac{gfn - feq}{gf - eh} = n - m$ . & $\frac{gfq - hgn}{gf - eh} = p$ . & $\frac{hgn - qeh}{gf - eh} = q - p$ . & 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{;|:|} w^th y^e proportion of y^e weight of y^e two bodys d & e as {1} to r. To find y^e weight of y^e body d. Resp. Suppose y^t y^e bodys a & d, b & e, c & f−d−e are of y^e same matter; & {illeg} & y^t y^e weight of y^e bodys a, b, c, f, d, e, f−d−e, {are} in y^e aire are g, h, k, p, x, rx, p−x−rx{;} & in y^e 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 $v - \frac{lx}{g} - \frac{mrx}{h} = v - s - t = \frac{np - nx - nrx}{k}$ . And ghkv−ghnp=hklx+gkmrx−ghnx−ghnrx. Or $\frac{ghkv - ghnp}{hkl + ghmr - ghn - ghnr} = x$ .

If y^e weights g=h=k, (as may bee either by experience or cal{illeg}|cu|lacon (see coroll: Prob 1) Then is $\frac{gv - np}{l + mr - n - nr} = x$ . Now because gold is usually allayed by mixing w^th it brasse & silver of each {illeg} |an| equall weight; suppose y^t a & d are brasse, b & e silver, c & f−d−e gold, & y^t x=rx, or r=1. Then is $\frac{gv - np}{l + m - 2 n} = x$ y^e weight of y^e brasse or siver {sic} in y^e masse f, & $\frac{pl + pm - 2 gv}{l + m - 2 n} = p - 2 x$ y^e weight of y^e 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 $\begin{array}{l} y^{3} & + & 2 b & y y & + & b b & y & + & c b b & = \\ + & c & + & 2 b c & - & c x x \\ - & 4 e b & + & 2 b x \\ - & 4 e e c & + & 4 c x \\ - & 2 e x & + & x x \end{array}$

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 y^e valor of y (in y^e Equation expressing y^e relation twixt x & y) consist of simple termes, Multiply {each} terme by x, & divide it by y^e number of y^e dimensions of x in that terme, & y^e quote shall signify y^e area acb.

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

Lemma 2^d |Prop 2.| If {illeg}any terme in y^e 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 y^e Area by Prop 2^d 1^st.

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

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

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

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

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

<80v>

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

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

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

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

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

<81r>

Soe y^t since $\frac{1,2 \times 1,2}{0,8 \times 0,9} = 2 . \frac{1,2 \times 1,2 \times 1,2}{0,8 \times 0,8 \times 0,9} = 3 = \frac{1,2 \times 2}{0,8}, \frac{1,2 \times 1,2 \times 1,2 \times 1,2}{0,8 \times 0,8 \times 0,8 \times 0,9 \times 0,9} = 5 = \frac{2 \times 2}{0,8} = \frac{4}{0,8} . 2 \times 5 = 10 . 10 \times 10 = 100$ 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 y^m multiplying another) may bee found. Viz. if ab=1=bc⊥ab{illeg}

|If y^e line ak is| Then, y^e superficies bcgh is

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

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Having already found y^e areas correspondent to y^e lines 1,1. 0,9. 1,2. 0,8. tis easy by y^e help of those operations to ding y^e areas correspondent to y^e 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 $7 = \sqrt{\frac{100 \times 0,98}{2}} . 17 = \frac{100 \times 1,02}{6} . 13 = \frac{1000 \times 1,001}{7 \times 11} = \frac{1001}{77}$ . &c. Therefore y^e areas correspond{ent} to y^e lines 7. 13. 17. {illeg}|&c|: are easily found, as followeth. Viz: if x=0.02. Then

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

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

The sume & difference of w^ch two summes give y^e areas bcfd, bche as before. That is

$\begin{array}{l} \begin{matrix} If fd = 0,98 \\ he = 0,98 \end{matrix} & \begin{array}{r} bcfd & = & -0,02020,27073,17519,44840,80453,01024,19238,78525,33383,73356,83210,19 \\ bche & = & +0,01980,26272,96179,71302,60290,66885,10039,31089,90727,51120,35410,79 \end{array} \end{array}$

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

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

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

<81v>

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

As was found before excepting y^t their differenc is e in y^e two last figures is 28. Which agreement could scarce thus happen in more yⁿ 50 figures, were not y^e reas, correspond{illeg}|in|g to y^e 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 y^e 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>

$\begin{aligned} Series Arithmetica & Series correspondens. \\ A + x . & a + b x + c x x + d x^{3} + e x^{4} + f x^{5} + g x^{6} + h x^{7} + i x^{8} + k x^{9} + l x^{10} \\ \begin{array}{l} A + 6 \\ A + 5 \\ A + 4 \\ A + 3 \\ A + 2 \\ A + 1 \\ A + 0 \\ A - 1 \\ A - 2 \\ &c \end{array} & \begin{array}{r} a & + & 6 b & + & 36 c & + & 216 d & + & 60466176 l & + & 362797056 m \\ a & + & 5 b & + & 25 c & + & 125 d & + & 625 e & + & 3125 f & 15625 g & + & 78125 h & + & 390625 i & + & 1953125 k & + & 9765625 l & + & 48828125 m \\ a & + & 4 b & + & 16 c & + & 64 d & + & 256 e & + & 1024 f & + & 4096 g & + & 16384 h & + & 65536 i & + & 262144 k & + & 1048576 l & + & 4194304 m \\ a & + & 3 b & + & 9 c & + & 27 d & + & 81 e & + & 243 f & + & 729 g & + & 2187 h & + & 65611 i & + & 19683 k & + & 59049 l & + & 177147 m \\ a & + & 2 b & + & 4 c & + & 8 d & + & 16 e & + & 72 f & + & 64 g & + & 128 h & + & 256 i & + & 512 k & + & 1024 l & + & 2048 m \\ a & + & b & + & c & + & d & + & e & + & f & + & g & + & h & + & i & + & k & + & l & + & m \\ a \\ a & - & b & + & c & - & d & + & e & - & f & + & g & - & h & + & i & - & k & + & l & - & m \\ a & - & 2 b & + & 4 c & - & 8 d & + & 16 e & - & 32 f & + & 64 g & - & 128 h & + & 256 i & - & 512 k & + & 1024 l & - & 2048 m \\ &c \end{array} \\ \begin{array}{l} T h e first Dif- \\ ference of these \\ termes. \end{array} & \begin{array}{r} &c \\ b & + & 3 c & + & 7 d & + & 15 e & + & 31 f & + & 63 g & + & 127 h & + & 255 i & + & 511 k & + & 1023 l & + & 2047 m \\ b & + & c & + & d & + & e & + & f & + & g & + & h & + & i & + & k & + & l & + & m \\ b & - & c & + & d & - & e & + & f & - & g & + & h & - & i & + & k & - & l & + & m \\ b & - & 3 c & + & 7 d & - & 15 e & + & 31 f & - & 63 g & + & 127 h & - & 255 i & + & 511 k & - & 1023 l & + & 2047 m \\ &c \end{array} \\ \begin{matrix} The second \\ Difference \end{matrix} & \begin{array}{r} &c \\ 2 c & + & 6 d & + & 14 e & + & 30 f & + & 62 g & + & 126 h & + & 254 i & + & 510 k & + & 1022 l & + & 2046 m \\ 2 c & + & 0 & + & 2 e & + & 0 & + & 2 g & + & 0 & + & 2 i & + & 0 & + & 2 l & + & 0 \\ 2 c & - & 6 d & + & 14 e & - & 30 f & + & 62 g & - & 126 h & + & 25 i & - & 510 k & + & 1022 l & - & 2046 m \\ &c \end{array} \\ \begin{matrix} The 3^{d} \\ diff. \end{matrix} & \begin{array}{r} &c \\ 6 d & + & 36 e & + & 150 f & + & 540 g & + & 1806 h & + & 5796 i & + & 18650 k & + & + & 171006 m \\ 6 d & + & 12 e & + & 30 f & + & 60 g & + & 126 h & + & 252 i & + & 510 k & + & 1020 l & + & 2046 m \\ 6 d & - & 12 e & + & 30 f & - & 60 g & + & 126 h & - & 252 i & + & 510 k & - & 1020 l & + & 2046 m \\ 6 d & - & 36 e & + & 150 f & - & 540 g & + & 1806 h & - & 5796 i & + & 18650 k & + & 171006 m \\ + & 204630 & + & 3669006 \\ + & 1225230 & + & 36774606 \\ + & 228718446 \\ &c \end{array} \\ \begin{matrix} The 4^{th} \\ diff. \end{matrix} & \begin{array}{r} &c \\ 24 e & + & 120 f & + & 480 g & + & 1680 h & + & 5544 i & + & 17640 k & + & + & 168960 m \\ 24 e & + & 0 & + & 120 g & + & 0 & + & 504 i & + & 0 & + & 2040 l & + & 0 \\ 24 e & - & 120 f & + & 480 g & - & 1680 h & + & 5544 i & - & 17640 k & + & - & 168960 m \\ - & 186480 & - & 3498000 \\ - & 1020600 & - & 33105600 \\ 191943840 \\ &c \end{array} \\ \begin{matrix} The 5^{t} \\ diff \end{matrix} & \begin{array}{r} &c \\ 120 f & + & 1080 g & + & 6720 h & + & 35280 i & + & 168840 k & + & + & 3329040 m \\ 120 f & + & 360 g & + & 1680 h & + & 5040 i & + & 17640 k & + & + & 168960 m \\ 120 f & - & 360 g & + & 1680 h & - & 5040 i & + & 17640 k & - & + & 168960 m \\ 120 f & - & 1080 g & + & 6720 h & - & 35280 i & + & 17640 k & - & + & 3329040 \\ + & 29607600 \\ + & 15883824 \\ &c \end{array} \\ \begin{matrix} The 6^{t} \\ Diff \end{matrix} & \begin{array}{r} &c \\ 720 g & + & 5040 h & + & 30240 i & + & 151200 k & + & + & 3160080 m \\ 720 g & + & 0 & + & 10080 i & + & 0 & + & + & 0 \\ 720 g & - & 5040 h & + & 30240 i & - & 151200 k & + & - & 3160080 m \\ - & 26298560 \\ - & 129230640 \\ &c \end{array} \\ \begin{matrix} The 7^{th} \\ diff \end{matrix} & \begin{array}{r} &c \\ + 5040 h & + & + & 514080 k & + & + & 23118480 m \\ + 5040 h & + & 20160 i & + & 151200 k & + & + & 3160080 m \\ + 5040 h & - & 20160 i & + & 151200 k & - & + & 3160080 m \\ + 5040 h & - & + & 514080 k & - & + & 23118480 m \\ + & 10295208 \\ &c \end{array} \\ \begin{matrix} The 8^{th} \\ diff. \end{matrix} & \begin{array}{r} &c \\ 40320 i & + & 514080 & + & + & 19958400 m \\ 40320 i & + & 0 & + & + & 0 \\ 40320 i & - & 362880 & + & - & 19958400 m \\ - & 79833600 m \\ &c \end{array} \\ \begin{matrix} The 9^{th} \\ diff \end{matrix} & \begin{array}{r} 362880 k & + & + & 19958400 m \\ 362880 k & - & + & 19958400 m \\ + & 59875200 m \end{array} \\ \begin{matrix} The 10^{th} Diff. \end{matrix} & \begin{array}{l} + \\ + \end{array} \\ \begin{matrix} The 11^{th} Diff. \end{matrix} & \begin{array}{l} 39916800 m \end{array} \end{aligned}$

The use of these differences is for composing rules to find the differences of {y^e terms of a table w^ch {illeg} to be interpoled by y^e 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 $\{\begin{cases} columnis \\ seriebus \\ ordinibus \end{cases}$ \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, $l = \frac{h + h2}{2}$ {&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= $\frac{h + h2}{2}$ , m=g2, $n = \frac{f2 + f3}{2}$ , o=e³, $p = \frac{d3 + d4}{2}$ , {illeg} q=c4, $r = \frac{b4 + b5}{2}$ 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, $\frac{b4 + b5}{2} = l$ , c4=m, $\frac{d3 + d4}{2} = n$ , e3=0, $\frac{f2 + f3}{2} = p$ , g2=q, $\frac{h + h2}{2} = r$ {illeg}i=s. Et erecta ordinatim applicata PQ die A5P=x, due terminus hujus pr{illeg}|ogr|essionis $1 \times \frac{x^{2}}{2} \times \frac{x x - 1}{3, 4} \times \frac{x x - 4}{5, 6} \times \frac{x x - 9}{7, 8} \times \frac{x x - 16}{9, 10} \times \frac{x x - 25}{11, 12} \times \frac{x x - 36}{13, 14} \times$ &c in se continuò {illeg}|&| orientur coefficientes termini $1 . \frac{x x}{2} . \frac{x^{4} - x x}{24} . \frac{x^{6} - 5 x^{4} + 4 x x}{720} .$ $\frac{x^{8} - 14 x^{6} + 49 x^{4} + 36 x x}{40320}$ &c $1 \times \frac{x}{1} \times \frac{x}{2} \times \frac{x x - 1}{3 x} \times \frac{x}{4} \times \frac{x x - 4}{5 x} \times \frac{x}{6} \times \frac{x x - 9}{7 x}$ $\times \frac{x}{8} \times \frac{x x - 16}{9 x} \times \frac{x}{10} \times \frac{x x - 25}{11 x} \times \frac{x}{12} \times \frac{x x - 36}{13 x}$ &c. \in se continuo/ Et orientur termini 1. x. $\frac{x x}{2} . \frac{x^{3} - x}{6} . \frac{x^{4} - x x}{24} . \frac{x^{5} - 5 x^{3} + 4 x}{120} . \frac{x^{6} - 5 x^{4} + 4 x x}{720} . \frac{x^{7} - 14 x^{5}}{5040} +$ $\frac{+ 49 x^{3} - 36 x}{5040}$ &c per quos si termini seriei k, l, m, n, o, p &c resspectivè multiplicentur, prodibunt aggregatum factorum $k + x l + \frac{x x}{2} m$ $+ \frac{x^{3} - x}{6} n + \frac{x^{4} - x x}{24} o + \frac{x^{5} - 5 x^{3} + 4 x}{120} p$ &c erit longitudo ordinatim applicatǽ PQ

Cas 2.

In casu posteriori sint A{illeg}, A5 \A4B4, A5B5/ duo medij termini, hoc est $\frac{A4 B4 + A5 B5}{2}$ =k, b4=l, $\frac{c3 + c4}{2} = m$ , d3=o|n|, e2+e3={illeg}|{o}|, f2=p, $\frac{g + g2}{2} = q$ & h=r{,|.|} Et erecta ordinatim applicata PQ, biseca A4|5|A5 in O {et} d{illeg} OP=x due terminos hujus progressionis $1 \times \frac{x}{1} \times \frac{x x - \frac{1}{4}}{2 x} \times \frac{x}{3} \times \frac{x x - \frac{9}{4}}{4} \times \frac{x}{5}$ $\times \frac{x x - \frac{25}{4}}{6} \times \frac{x}{7} \times \frac{x x - \frac{49}{4}}{8}$ &c in se continuo. Et orientur termini $1 . x . \frac{4 x x - 1}{8}$ $\frac{4 x^{3} - x}{24} . \frac{16 x^{4} - 40 x x + 9}{384}$ &c per quos si termini seriei k, l, m, n, o, p &c respectivè multiplicentur, aggregatum factorum $k + x l + \frac{4 x x - 1}{8} m$ $+ \frac{4 x^{3} - x}{24} n + \frac{16 x^{4} - 40 x x + 9}{384} o$ &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 y^e Equations ax³+bxx+cx+d=0 & fx³+gxx+hx+k=0.

I feighe y^e 3 valors of x in y^e first Equation to bee -r, -s, & -t. [{illeg} is {illeg}|[| $r + s + t = \frac{b}{a}$ . $r r + s s + t t = \frac{b b}{a a} - \frac{2 c}{a}$ . $r^{3} + s^{3} + t^{3} = \frac{b^{3} - 3 a b c + 3 a d d}{a^{3}}$ . $r s + r t + s t = \frac{c}{a}$ . & that is] {illeg} y^e summe of y^e rootes is $\frac{b}{a}$ {;} of their squares is $\frac{b b - 2 a c}{a a}$ of their cubes is {illeg} $\frac{b^{3} - 3 a b c + 3 a a d}{a^{3}}$ ; of their rectangles is $\frac{c}{a}$ &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. r^3⁴=b³−3bc+3d. rrst⁴=bd. rrss=cc-2bd. r³s⁴=bbc-2cc-bd. r^4⁴=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. r³s³t^3⁴=d³. ^*r³s^3⁴=c³−3bcd+3dd.

Now supposing k (or any other quantity of y^e second Equation) to bee {illeg} i|a|n unknow{ne} quantity, it must have 3 severall valors by reason of y^e 3 valors of x in y^e first Equation, & therefore x being taken away, \h will bee of three dimensions in/ y^e 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 y^e 3 fold valor of k: out of w^ch equation I take {illeg} out r, s, t by writing b for their summe c for y^e sum̄e of their rectangles rs+rt+st. bc−3d for y^e summe of all theire rectangles of this for{me {illeg}|rrs|} (viz: for rrs+rrt+{illeg}|rss|+rtt+rtt+sst+stt) &c as in y^e Table. W^ch substitution may bee most briefly done in y^e said multiplication, thus; writing a to make up six dimensions $\begin{matrix} \begin{array}{l} k - h r + g r r - f r^{3} \\ k - h s + g s s - f s^{3} \\ k - h t + g t t - f t^{3} \end{array} & \begin{matrix} Produceth, \end{matrix} & \begin{matrix} \begin{array}{l} a^{3} h^{3} - a a b h k k + a b b g k k - 2 a a c g k k - b^{3} f k k + 3 a b c f k k \\ - 3 a a d f k k + a a c h h k - a b c g h k + 3 a a d g h k + b b c f h k \\ - 2 a c c f h k - a b d f h k + a c c g g k - 2 a b d g g k - b c c f g k \\ + 2 b b d f g k + a c d f g k + c^{3} f f k - 3 b c d f f k + 3 a d d f f k \\ - a a d h^{3} + a b d g h h - b b d f h h + 2 a c d f h h - a c d g g h \\ + b c d f g h - 3 a d d f g h - c c d f f h + 2 b d d f f h + a d d g^{3} \end{array}\} =0 \end{matrix} \end{matrix}$

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

$\begin{matrix} \begin{array}{r} A \\ A + x \end{array} & \begin{array}{r} a \\ a & + & b x & + & c x x & + & d x^{3} + & e x^{4} & + & f x^{5} & &c \end{array} \\ \begin{array}{r} A + p \\ A + q \\ A + r \\ A + s \\ A + t \end{array} & \begin{array}{r} a & + & b p & + & c p p & + & d p^{3} + & e p^{4} & + & f p^{5} & = & α \\ a & + & b q & + & c q q & + & d q^{3} + & e q^{4} & + & f q^{5} & = & β \\ a & + & b r & + & c r r & + & d r^{3} + & e r^{4} & + & f r^{5} & = & γ \\ a & + & b s & + & c s s & + & d s^{3} + & e s^{4} & + & f s^{5} & = & δ \\ a & + & b t & + & c t t & + & d t^{3} + & e t^{4} & + & f t^{5} & = & ε \end{array} \\ \begin{array}{r} p - q & ) & α - β & = \\ q - r & ) & β - γ & = \\ r - s & ) & γ - δ & = \\ s - t & ) & δ - ε & = \end{array} & \begin{array}{r} b & + & c \times \bar{p + q} & + & d \times \bar{p p + p q + q q} & + & e \times \bar{p^{3} + p p q + p q q + q^{3}} & + & f \times p^{4} & &c & = & ζ \\ b & + & c \times \bar{q + r} & + & d \times \bar{q q + q r + r r} & + & e \times \bar{q^{3} + q q r + q r r + r^{3}} & + & f \times q^{4} & &c & = & η \\ b & + & c \times \bar{r + s} & + & d \times \bar{r r + r s + s s} & + & e \times \bar{r^{3} + r r s + r s s + s^{3}} & + & f \times r^{4} & &c & = & θ \\ b & + & c \times \bar{s + t} & + & d \times \bar{s s + s t + t t} & + & e \times \bar{s^{3} + s s t + s t t + t^{3}} & + & f \times s^{4} & &c & = & ι \end{array} \\ \begin{array}{r} p - r & ) & ζ - η & = \\ q - s & ) & η - θ & = \\ r - t & ) & θ - ι & = \end{array} & \begin{array}{r} c & + & d \times \bar{p + q + r} & + & e \times \bar{p p + p q + q q + p r + q r + r r} & + & f p^{3} & &c & = & λ \\ c & + & d \times \bar{q + r + s} & + & e \times \bar{q q + q r + q s + r r + r s + s s} & + & f q^{3} & &c & = & μ \\ c & + & d \times \bar{r + s + t} & + & e \times \bar{r r + r s + r t + s s + s t + t t} & + & f r^{3} & &c & = & ν \end{array} \\ \begin{array}{r} p - s & ) & λ - μ & = \\ q - t & ) & μ - ν & = \end{array} & \begin{array}{r} d & + & e \times \bar{p + q + r + s} & + & f \times \bar{p p + p q + p r + p s + q q + q r + q s + r r + r s + s s} & = & ξ \\ d & + & e \times \bar{q + r + s + t} & + & f \times \bar{q q + q r + q s + q t + r r + r s + r t + s s + s t + t t} & = & π \end{array} \\ \begin{array}{r} p - t & ) & ξ - π & = \end{array} & \begin{array}{r} e & + & f \times \bar{p + q + r + s + t} & = & σ \end{array} \end{matrix}$

<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 $\frac{A B - A2 B2}{A A2} = b$ , $\frac{A2 B2 - A3 B3}{A2 A3} = b2$ , $\frac{A3 B3 - A4 C4}{A3 A4} = b3$ , $\frac{A4 C4 - A5 C5}{A4 A5} = b4$ , $\frac{A5 C5 - A6 C6}{A5 A6} = b5$ {,} $\frac{A6 C6 + A7 C7}{A6 A7} = b6$ . $\frac{- A7 C7 - A8 B8}{A7 B8} = b7$ . Deinde $\frac{b - b2}{A A2} = c$ . $\frac{b2 - b3}{A2 A4} = c2$ . $\frac{b3 - b4}{A3 A5} = c3$ &c. Tunc $\frac{c - c2}{A A4} = d$ . $\frac{c2 - c3}{A2 A5} = d2$ . $\frac{c3 - c4}{A3 A6} = d3$ &c. Tunc $\frac{d - d2}{A A5} = e$ . $\frac{d2 - d3}{A2 A6} = e2$ . $\frac{d3 - d4}{A3 A7} = e3$ &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, $\frac{b3 + b4}{2} = l$ . c3=m, $\frac{d2 + d3}{2} = n$ , {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} $1 \times \bar{x - O A4} \times \bar{x - \frac{4 O A3 + O A5}{2}} \times \frac{\bar{x - O A3} \times \bar{x - A 05}}{x - \frac{O A3 + O A5}{2}} \times x - \frac{+ O A2 + O A6}{2} \times$ &c \et {illeg} ortam progressionem asserva./ Vel quod perinde est duc terminos hujus progressionis $1 \times \bar{x - O A4} \times \bar{\bar{x - O A3} \times \bar{x - O A5}}$ $\times \bar{\bar{x - O A2} \times \bar{x - O A6}} \times \bar{\bar{x - O A} \times \bar{x - O A7}} \times$ &c in se gradatim et terminos ex ortos duc \respective/ in terminos hujus progressionis $1 . x - \frac{+ O A3 + O A5}{2} . x - \frac{+ O A2 + O A6}{2}$ $x - \frac{+ O A + O A7}{2}$ . &c et orientur termini intermedij: {illeg} tota progressione existente $1 . x - O A4 . x x - \frac{+ O A3 + 2 O A4 + O A5}{2} x + O A3 \times O A4 + O A5 \times O A4$ &c Vel dic OA=α, OA2=β, OA3=γ{,} OA4=δ, OA5=e, OA6=ζ, OA7=η $\frac{O A3 + O A5}{2} = θ$ , $\frac{O A2 + O A6}{2} = μ$ . $\frac{O A + O A7}{2} = λ$ . Et ex progressione $1 \times \bar{x - δ} \times$ $\bar{\bar{x - γ} \times \bar{x - ε}} \times \bar{x - β} \times \bar{x - ζ} \times \bar{x - α} \times \bar{x - η}$ 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 $\frac{A4 B + A5 B5}{2} = k$ {.} b4=l. $\frac{c3 + c4}{2} = m$ . d3=n. $\frac{e2 + e3}{2} = o$ . f2=p{illeg} Et {illeg} \{illeg} k, m, {illeg}, o, {illeg} &/ coefficientes orientur \en/ multiplicatione terminorum hujus progressiones {illeg} $1 \times \bar{\bar{x - O A4} \times \bar{x - O A5}} \times \bar{\bar{x - O A3} \times \bar{x - O A6}} \times \bar{\bar{x - O A2} \times \bar{x - O A7}}$ $\times \bar{\bar{x - O A} \times \bar{x - O A8}}$ &c et ubiquorum coefficientes en multiplicatione {illeg} {illeg} {c}ujus progressionis $x - \frac{\bar{+ O A4 + O A5}}{2}$ . $x - \frac{\bar{+ O A3 + O A6}}{2}$

<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 w^ch y^e 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.

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

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

Or thus If their sume is {illeg}|−|p=\−/a. Then is y^e 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

$x - \frac{+ O A2 + O A7}{2}$ . $x - \frac{+ O A + O A8}{2}$ &c Hoc est eritk+x−OA4+ $k + \bar{x - \frac{+ O A4 + O A5}{2}} \times l + \bar{x x - \bar{O A4 - O A5} \times x + O A4 \times O A5} \times m$ &c=Ordinatim applicatæ $PQ /= \begin{array}{l} k & \begin{matrix} \begin{array}{l} + x \\ - \frac{1}{2} O A4 \\ - \frac{1}{2} O A5 \end{array}× & l \end{matrix} & \begin{matrix} \begin{array}{l} + x \\ - O A4 \end{array}× & \begin{array}{l} x \\ - O A5 \end{array}× & m \end{matrix} & \begin{matrix} \begin{array}{l} + x \\ - O A4 \end{array}× & \begin{array}{l} x \\ - O A5 \end{array}× & \begin{array}{l} x \\ - \frac{1}{2} O A3 \\ - \frac{1}{2} O A6 \end{array}× & n \end{matrix} \end{array}$ $\begin{matrix} \begin{array}{l} + x \\ - O A3 \end{array}× & \begin{array}{l} x \\ - O A4 \end{array}× & \begin{array}{l} x \\ - O A5 \end{array}× & \begin{array}{l} x \\ - O A6 \end{array}× & o \end{matrix}$ . {illeg} Sive dic $x - \frac{+ O A4 + O A5}{2} = π$ . $\bar{x - A4} \times \bar{x - A5}$ =ρ. $ρ \times \bar{x - \frac{+ O A3 + O A6}{2}} = σ$ . $ρ \times \bar{x - O A3} \times \bar{x - O A6} = τ$ . $τ \times x - \frac{+ O A2 + O A7}{2}$ =ν. $τ \times \bar{x - O A2} \times \bar{x - O A7} = φ$ . $φ \times x - \frac{+ O A + O A8}{2} = χ$ . $φ \times \bar{x - O A} \times$ $\bar{x - O A8} = ψ$ Et erit k+πl+ρm+σn+τo+νp+φq χ{illeg}r+ψs=PQ.

<84v>

Prob{l}

<85r>

Of Equations,

Every equation hath soe many roots as dimensions of w^ch 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 y^e signes of y^e Equations termes: Namely there are soe many true rootes as variations of signes & soe many false ones as successions of y^e same signes. When any termes a{illeg}|re| wanting supply their{e} voyd places w^th ±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 w^ch in y^e one case appeare true, |&| in a|y^e|nother false are neither, but imaginary; & that of y^e other two roots one is true {y^e} other false.

Hence it appeares y^t to know y^e particular constitution of any Equation it is {chi{illeg}fely} necessary to understand\{illeg}/ w^t 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 y^e rules would bee more in{illeg}|tr|icate & laborious than y^e to put in practise then to solve the Equations either by lines or numbers. Soe y^t y^e 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 w^thout much labour.

First then {illeg} if you see any terme betwixt

First yⁿ if in any three termes together y^e two extreame termes having y^e 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 y^e terme betwixt them, conclude there are two imaginary roots. Thus \{in}/{illeg} +aaax−b³{illeg} x³−2|3|xx+2x{\{illeg}/}−4=0 has two roots imaginary because neither 3 nor 4 are lesse then 2. And y^e 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 y^e 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 y^e 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. Symbol (two surmounted by circle) in text If Thus x⁴−2x+3=0 {illeg} have all foure roots reall & therefore must have two imaginary. Symbol (one surmounted by circle) in text 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 w^ch {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 y^e product y³−2yy+3y−2=0 hath \{times}/ soe many reall roots, halfe true halfe false, the other shal & four times soe imaginary ones y^e 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/ y^e dimension of the terme{illeg} above {illeg} it for its {illeg} & number {illeg} {denominating} y^t terme first second third &c for its denominator & \{illeg} if/ {illeg} {illeg} any {illeg} together so y^t the \rectangle of the/ first {illeg} have {illeg} {illeg} multiplyed {viz the fraction} {illeg} first {illeg} square of the {illeg} terme multipl{illeg} by {y^e fraction} {illeg} conclude there are two imaginary roots {illeg} if {illeg} {illeg} one {illeg} <85v> in all throughout y^e 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 y^e said squ/ces of y^e 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 y^e 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\ y^e progre{illeg}|ss| is $\frac{3}{1} . \frac{2}{2} . \frac{1}{3}$ .\ & because $- 3 x x - 2 \times \frac{2}{2} (6) < 4 \times 4 \times \frac{1}{3} (\frac{16}{3})$ , I conclude there are two roots imaginary at least.

$\begin{matrix} Th u s in & x^{3} & - & 6 x^{2} & + & 6 x & - & 2 & = & 0. & because - 6 x - 2 \times \frac{2}{2} (12) = 6 \times 6 \times \frac{1}{3} (12) but \\ \frac{3}{1} & . & \frac{2}{2} & . & \frac{1}{3} & . & 1 \times 6 × \frac{3}{1} (18) < - 6 \times - 6 \times \frac{2}{2} (36 I conclude \end{matrix}$ there are two imaginary roots.

$\begin{matrix} T hus in & x^{3} & - & 6 x x & + & 12 x & - & 8 & = & 0 & because 1 \times 12 \times 3 (36) = - 6 \times - 6 \times 1 (36) and also \\ 3 & . & 1 & . & \frac{1}{3} & . & - 6 × - 8 \times 1 (48) = 12 \times 12 × \frac{1}{3} (48) I conclude that \end{matrix}$ all the 3 roots are equall.

$\begin{matrix} Bu t in & x^{3} & + & 6 x x & + & 12 x & - & 8 & = & 0, & because 1 \times 12 \times 3 (36) = 6 \times 6 \times 1 but \\ 3 & . & 1 & . & \frac{1}{3} & 6 × - 8 \times 1 (- 48) < 12 \times 12 × \frac{1}{3} (+ 48) I \end{matrix}$ conclude two rootes are imaginary.

$\begin{matrix} In & x^{4} & - & x^{3} & + & 2 x x & - & 2 x & + & 3 & = & 0, & because 1 \times 2 \times \frac{4}{1} (8) < - 1 \times - 1 \times \frac{3}{2} \\ \frac{4}{1} & \frac{3}{2} & \frac{2}{3} & \frac{1}{4} & (\frac{3}{2}), I conclude there are two \end{matrix}$ imaginary roots at least, also by the three last termes $\frac{2 \times 3}{} \times \frac{2}{3} (4) < - 2 \times - 2 \times \frac{1}{4}$ (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 $- 1 \times - 2 \times \frac{3}{2} (3) < 2 \times 2 \times \frac{2}{3} (\frac{8}{3}$ & soe can conclude but two rootes imaginary.

$\begin{matrix} In & x^{4} & - & x^{3} & + & 3 x x & - & 2 x & + & 3 & = & 0 & because 1 \times 3 \times \frac{4}{1} (12) < - 1 \times - 1 \times \frac{3}{2} (\frac{3}{2}) and also \\ \frac{4}{1} & . & \frac{3}{2} & . & \frac{2}{3} & . & \frac{1}{4} & . & 3 \times 3 \times \frac{2}{3} (6) < - 2 \times - 2 \times \frac{1}{4} (1), but not - 1 \times - 2 \times \frac{3}{2} (3) < \end{matrix}$ $< 3 \times 3 \times \frac{2}{3} (6) .$ Therefore I conclude all four roots imaginary.

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

Had it been

Thus in litterall Equations, if $\begin{matrix} x^{3} & - & p x x & + & 3 p p q & - & q^{3} & = & 0 \\ \frac{3}{1} & \frac{2}{2} & \frac{1}{3} \end{matrix}$ , because 1×3ppx{illeg} $(9 p p) < - p x - p \times \frac{2}{2} (p p)$ \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 y^e termes of y^e equation set a series of fractions each having y^e dimensions of the terme under it for its numerator, & the number denominating y^e 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 y^e factus of the {terme} before & after it multiplyed by y^e fraction over y^e terme before it. If greater write y^e signe + underneath; if equal or lesse write {illeg} the signe − underneath y^e middle terme {illeg} & lastly set + under y^e first terme of y^e 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| y^e roots bee equall|

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

<86r>

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

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

$\begin{matrix} \frac{4}{1} & . & \frac{3}{2} & . & \frac{2}{3} & . & \frac{1}{4} & . \\ Thus if & x^{4} & - & x^{3} & + & 2 x x & - & 2 x & + & 3 & = & 0. & becaus - \frac{23}{4} \times - 3 (\frac{69}{4}) < - \frac{1}{3} \times - \frac{1}{3} (\frac{1}{9}) I change \\ + \frac{4}{1} & - & \frac{23}{4} . & - & \frac{1}{3} . & - & 3 . & the signe of - \frac{1}{3} & soe the signes +.−.+.− \\ + & . & - & . & + & . & - & . & shew all foure roots imaginary. \end{matrix}$

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

Any Equation being propounded, set down a series of \so many/ fractions as y^e Equation hath dimensions, whose {illeg}|nume|rators are a prog & denominators are a progression of units backward & forward. Divide each fraction by y^t p^eceding it & set the quotes in order overal |y^e| middle termes of the Equation. Then observe whither of every {thee term such} of every middle terme whither it square multiplyed by y^e fraction over it bee greater equall or lesse yⁿ y^e rectangle of y^e two termes on either hand. If greater write + underneath, if equall or lesse write {illeg} −. Lastly set + under y^e 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 y^t all y^e roots are equall, for &c:|

Thus if x³−6|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 y^e middle termes of the equation thus $\begin{matrix} \frac{1}{3} . & \frac{1}{3} . \\ x^{6} & - & 3 x x & + & 6 x & - & 4 & = & 0. \\ + & - & - & + \end{matrix}$ . Then I observe in y^e 2^d terme that $- 3 \times - 3 \times \frac{1}{3} (3)$ is lesse then {illeg}4|1×6| & therefore I write − under it. so in y^e 3^d terme I find $6 \times 6 \times \frac{1}{3} (12) = - 3 \times - 4 (12)$ , therfore I write − under it. Lastly seting + under y^e first & last terme I find two changes of sines & so conclude there are

Thus if x⁵−\5/{illeg}x⁴+5x³− xx− x⁵−2|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 y^e middle termes of the equation th{illeg}|{illeg}s| $\begin{matrix} \frac{2}{5} . & \frac{1}{2} . & \frac{1}{2} . & \frac{2}{5} . \\ x^{5} & - & 4 x^{4} & + & 4 x^{3} & - & 2 x x & - & 5 x & - & 4 & = & 0. & Then in the second time I find - 4 \times - 4 \times \frac{2}{5} (\frac{32}{5}) \\ + & + & - & + & + & + & < 1 \times 4 (4) : therefore I set + under it. In \end{matrix}$ the third $4 \times 4 \frac{1}{2} (8) = - 4 \times - 2 (8)$ : therefore I write −. In the 4^th $- 2 \times - 2 \times \frac{1}{2} (2) < 4 \times - 5$ (-20) therfore I write +. In y^e 5^t $- 5 \times - 5 \times \frac{2}{4} (10) < - 2 \times - 4 (8)$ therefore I w{illeg}|rit|e +. lastly under y^e first & last terme I w{illeg}|rit|e +. And soe finding two changes of termes I conclude two roots to bee impossible.

$\begin{matrix} \frac{3}{8} & \frac{4}{8} & \frac{3}{8} \\ T hus in & x^{4} & + & 3 x^{3} & - & 6 x^{2} & - & 3 x & - & 2 & = & 0. & two roots are impossible \\ + & +. & + & - & + \end{matrix}$ $\begin{matrix} \frac{1}{3} & \frac{1}{3} \\ In & x^{3} & + & 0 x x & + & p p x & - & q^{3} & = & 0 & two roots are impossible \\ + & - & + & + \end{matrix}$ $\begin{matrix} \frac{1}{3} & \frac{1}{3} \\ x^{3} & - & 6 x x & + & 12 x & - & 8 & = & 0 & All the roots are equall. \\ + & - & - & + \end{matrix}$

<86v>

Sometimes there may bee impossible not by this meanes discovered, w^ch if you suspect, augment or diminish y^e 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 y^t 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 w^ch noe/ impossible appeare I put x−p{illeg} x=y−p & the result is y³−3pyy−p³=0 in w^ch two appeare, Or if I put x=y−2p the result is y³−6pyy+9ppy−5p³=0 in w^ch also two appeare.

Thus |if| $\begin{matrix} x^{5} & + & x^{4} & - & 4 x^{3} & + & 5 x x & - & 2 x & + & 1 & = & 0 \\ + & + & + & + & - & + \end{matrix}$ .|,| I set y^e signes + & − under it as before and find two imaginary roots & to try if it have any more I suppose x=y+1 & y^e results is x^4|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 y^t terme \also set y^e signe + on either side the term {wanting}/. As in this $\begin{matrix} x^{7} & + & 0 x^{6} & + & 2 x^{5} & - & 2 x^{4} & + & 3 x^{3} & - & 4 x x & + & 6 x & - & 2 & = & 0 \\ + & - & + & - & - & + \end{matrix}$ . In w^ch 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 w^th a negative{illeg} only end w^th an affirmative/ /if the terms on either hand have contrary signes\. As in {illeg} $\begin{matrix} x^{5} & - & a x^{4} & * & * & * & + & a^{5} & = & 0 \\ - & + & - \end{matrix}$ so in $\begin{matrix} x^{5} & + & a x^{4} & * & * & * & - & a^{5} & = & 0. \\ - & + & + \end{matrix}$ The first shows 4, y^e last two roots imaginary: Soe{.} in x⁹ * x⁸ * $\begin{matrix} x^{10} & * & + & x^{8} & * & * & - & 3 x^{5} & * & * & * & * & - & 6 & = & 0. \\ + & - & + & - & + & + & - & + & - & + & + \end{matrix}$ w^ch hath {illeg}|{8}| roots imaginary

<88r>

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

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

Or yet better. Do not add {illeg}|&| subduct y^e divisors from 9, 4, 1, 0, 1, 4, 9 but try if of those in {illeg}|y|^e first & last rank y^e difference of any two eaven or two odd ones be divisible by 6. Call $\frac{1}{2}$ that difference G & $\frac{1}{2}$ y^e summ of y^e same terms H. Then {illeg}|try| in y^e middle collumn there be any terms w^ch 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 y^e middlemost w^ch has least /fewest\ divisors there be any te{rm} {illeg} w^ch added |to| or subducted from $\frac{1}{3} G$ produces a number divisible by 9] Then making & y^e summ or difference {illeg}K, & putting $\frac{1}{9} K - 3 = a$ , & G+ /−\=b. Try if you can find 8−4a+2b−c {illeg}|on| y^e {illeg} |2^d| rank next above y^e middle one or 84|+|4a+2b+c in y^e \2^d/ rank 2^d next \ra{illeg}/ below {illeg}|it|. If so {illeg}|tr|y y^e 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 |y^e| three ranks \of the fewest terms/ & in them those numbers \one in each/ {b}by w^ch get y^e valor of a, b, & c. & those gotten will for y^e give you numbers to be sought in the other ranks, w^ch 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 y^e last term of y^e æquation divided by {c} be p y^e last but one q, Then if c & $\frac{q}{c}$ {must {illeg}ust} have a common divisor w^ch divides {illeg}|{n}|ot q, that c is to be rejected. Also if $\frac{1}{4} b \times \frac{p}{c}$ is greater then {+} {×} y^e grea or $\frac{1}{4} a \times \frac{p}{c}$ be greater then {illeg} {illeg} y^e greatest term of y^e æquation y^t b or a is to be rejected.

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

Or thus best. L|I|f t{illeg}|he| Let y^e numbers w^ch 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 y^e numbers arising be F. {illeg}|& M{illeg}| but if I end in 5 {illeg}|or| 0 increase y^e root of & H or K do not, increase \or decrease/ y^e root of y^e æquation or little by an unit. Do so also if {H} be an eaven number and H or K an o{illeg}|dd| one w^th 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/ y^e divisors of \suppose/ y^e resulting terms, w^ch 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 y^e 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 y^e divisor of Q from y^e divisor of F not y^t of F from y^t of Q so y^t S will be negative if y^e divisor of Q be y^e greater ✝ < insertion from the left margin > ✝ And if y^e æquatiō be not of more yⁿ {{illeg}}5 dimensions so y^t {{illeg}} it must be divisible (if at all) by a divisor of 2 dimensions set down xx∓Sx $\mp \frac{1}{2} R - 25$ to be tried for such a divisor. Where R & S must have y^e same signes if s {illeg} y^e divisor of F{illeg} was greater then y^e divisor of K otherwise contrary signes. < text from f 88v resumes > Quadruple y^e divisors of L. L {F} Take one & if {illeg}|th|e two last figures of any one be the same w^th y^e two last figures of 2R, take it from 2R. Let y^e residue {be T} divided by 100, be T. Or if y^e two last figures of any one added to y^e two last figures of 2R make 100 add it to 2R & let y^e summ \divided by 100/ be {illeg} T & let y^e number \whose quadruple is/ added to or subducted from 2R be a And [if y^e 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 y^e 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 y^e following operations.|

But if y^e æquation be of more then 7 dimensions then look in y^e columns{illeg} among y^e divisors of K for a number w^ch 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 w^ch gave it call β And set down $\begin{array}{l} x^{4} & + & v x^{3} & + & T x x & + & S x^{} & + & a \\ - & 1 & - & 25 & - & 25 v \\ - & 25 \end{array}$ to be tried for a divisor \if the æquation be not of more then 9 dimensions/. Where {illeg}|{v}| must be negative if y^e 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 y^e æquation be of more then 9 dimension then Cook in y^e column among the divisors of M for a number w^ch added to or subducted from −S{illeg}\+/T+a gives a number divisible by 24. This number divided by 24 call W & y^e divisor w^ch gave it {γ}. And set down $\begin{array}{l} x^{5} & + & \frac{1}{2} w & x^{4} & + & \frac{1}{2} v & x^{3} & + & T & x x & + & S & x & + & a \\ + & \frac{1}{2} v & - & \frac{1}{2} w & - & \frac{25}{2} v & - & \frac{25}{2} v \\ - & 26 & - & \frac{25}{2} w & + & \frac{25}{2} w \\ + & 25 \end{array}$ to be tried for a divisor if y^e 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 y^e divisor be a+bx+cxx+dx³+ex⁴+fx⁵ &c And observe if {illeg} be among y^e divisors of L{,} if {illeg}|th|is Divisor ascend but to {two} dimensions{;} & a+b{illeg}{+c+{illeg}+1}+c+1 among y^e 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 y^e divisors of I if it ascend not to more then 5 dimensions & a−2b+4c{illeg} \−d+e−f+1/ among y^e divisors of N if it ascend to no more then six dimensions & a{illeg}+3b+9c+27d+81e+243f|+|&c among y^e divisor{illeg}s of H if it ascend not {illeg}|to| more then 7 dimensions, & so in infinitum In all w^ch put c=1 & d, e, f = 0 if y^e 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 y^e divisors w^ch may be found by this {rale}, {If}|{&}| rejected those w^ch will not hold this trial: if there remain more or if y^e æquation be not reducible divisible by \any of/ those w^ch remaine, you may conclude y^e {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} $\frac{SB \times SQ}{SC} ∷ {AO}^{q} . {B0}^{q}$ , 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 RC^q 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 $BK \times KE . AK \times KC ∷ \frac{BH \times HE}{DH} . HI ∷ FG . \frac{AF \times FC}{FD}$ , 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 BP^q.GQ^q ∷PV.QV. et erit PV diameter V vertex et $\frac{{BP}^{q}}{PV}$ latus rectum figuræ {illeg}.

^[61] His præmissis nihil aliud restat agendum in solutio compositio loci solidi quam ut quin puncta quæramus per quæ figur{a} transibit. Id quod in Problemat{illeg}|{e}| Veterum facillimum est{.} Su{illeg} AP, AT, ST, AG, SG quatuor positione datæ rectæ et ad hac ducendæ sint \{illeg} s{um}{illeg}/ in datis angulis a puncto aliquo com{illeg}muni C {a}liæ quat{uor} CD, CF, CB, CH ea lege ut rectangulum duarum primarum CD ×CF dat{a}m habeat rationem ad rectangulum e|r|eliquarum CB×CH {illeg} curva in qua punctum C perpetim reperitur transibit per intersectiones datarum A, G, S, T, nam ubi FC nulla est, rectangul{u}m FC×CD nullum erit adeo et rectangulum CD×CH, {illeg} et {illeg} rectarum CD, CH. Si CD{;} punctum C incidet adT, si CH{;} ad S. At ita ubi CD nulla est punctum C incidet {vel in A vel in G.} Dantur ita quatror puncta A, G, S, T {illeg} et sola restat quinta investiganda. Id /quod\ < insertion from f 90r > facillimum est. Nam per punctum A ag{illeg}|at|ur recta quævis AC {illeg} {illeg} et in ea quaratur punctum C quod Problemati satisfaciet. {Et ergo} Jam adeo ratio DC ad BC, et proinde etiam ratio CH ad FC siquidem ratio DC×CH ad BC×FC detur. Age ergo rectam SC ea lege ut sit CH ad FC in ratione ista data, et hæc secabit rectam AC in puncto quæsito C. {illeg} Eadem lege unum era {puncto} invenire possunt sed uno aliquo invento {habebimus} {illeg} C quin puncta quæ Prob {Conicam Sectionem jaxta} præcedentia determinando Problemati satisfaciunt.

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.

Junge AB. Eam secent refringentes superficies in C et E. {illeg} Et posito d ad e ut sinus incidentiæ ex aere in medio ACD in medium EDC ad sinum refractionis, cape B produc AB ad G {illeg}|ut| sit BG.CE∷d−e.e, & AD ad H ut sit AH=AG, et DF ad K ut sit DK.DH∷e.d. Junge KB et centro D radio DH describe circulum occurrentem KB productæ in L. Ipsi DL parallelam age FBF et erit F punctum superficiei EF quæ radios omnes ADF convergere faciet ad punctum B. Nam |de| fluxio DF est ad fluxionem AD+BF ut e ad d: adeo CE−DF.AD+BF−AC−BE ∷e.d.

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 AB^q æquales sit rectangulo sub posteriore abscissa ac data DC×E

Vel sic. Si AB^q=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 AC^q−BC^q punctum C est ad rectā

7 Si dentur puncta duo A, B et summa quadratorum AC^q+ BC^q 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.

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^[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] $\{\begin{cases} 3 Detur positione DA et punctum P. Et si detur AC (Locus \\ Conchoid) d iff AC ± AP (loc \{\begin{matrix} circ. \\ linear. \end{matrix}) proportio PA ad AC (loc. rect.) rectangulum \\ APC (circ) rectang P AC (lin) rectang PCA (lin.) {AP}^{q} \pm {PC}^{q} . \\ {AP}^{q} \pm {AC}^{q} . {PC}^{q} \pm {AC}^{q} (lin.) \\ 4 Detur circ AD, punctum P et si detur etiam APC \\ vel \frac{AP}{AC} (erit C in loc. Pla n .) \end{cases}$

<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 $PR = \frac{PD \times QR}{AD - QR}$ 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 =EC^q−EF^q PR.PQ Et $BC . \frac{PE \times RQ}{PR} - EF (FC) ∷$ $\frac{PE \times RQ}{PR} + EF (DC) . \frac{PE \times BC}{PR} (PC) .$ Seu BQ|C|,PR.PE,PR|RQ|−PR,EF∷ $\frac{PE \times RQ}{} + PR, EF . PE, BC . {BC}^{q}, PE, PR = {PE}^{q}, {RQ}^{q} - {PR}^{q}, {EF}^{q}$ Si PR infinitum{Q} Ergo Q locatur in Conica sectione cujus diameter PR, ordinata RP|{PR}|. Sit RQ=0, erit PR=0 et $\frac{- {BC}^{q} PE}{{EF}^{q}}$ In EP producta cape ergo $PS . \frac{1}{2} PE ∷ {BC}^{q} . {EF}^{q}$ et erit S centrum et P vertex figuræ. Pone PR infinitus et erit PE^q,RQ^q=PR^q,EF^q, 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. $\frac{{AC}^{q} - {BC}^{q}}{AB} = AD$ 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=EM^q=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 MG^q+FG^q−KF erit AM×BM{−} AG−MG×BG−{illeg}G(AM×BM)=MG^q+FG^q−KF^q {illeg} est AG×BG−FG^q+KF^q=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= $\frac{{AD}^{q}}{AB}$ $= \frac{{NQ}^{q}}{NF} - \frac{{AD}^{q}}{AB}$ . Dividendo|a| est ita \data/ AH in D ita ut $\frac{{DH}^{q}}{NF} - \frac{{AD}^{q}}{AB}$ dato{illeg} æquale sit, nempo dato AB(Hb)+DQ−HF, seu bk. {AD^q}DH^q×AB AD =AB, HK, bL. DH=A{D}{illeg} AH−AD.DH^q=AH^q−2DAH+AD^q.AD×{illeg}K^q{−2A}{illeg}{AH} −2DAH+AD^q−{illeg} $\frac{HK \times {AD}^{q}}{AB} = HK, bk$ . [A{B}{illeg}−HK(bK).AH∷A{illeg}|{B}|.AV.AH^q−2DA{H} $\frac{{AD}^{q}, AH}{AV} = HK \times BL$ ] $AH - 2 DA + \frac{AB - HK}{BAH} {AD}^{q} = \frac{HK, bk}{AH}$ . \Fact/ AB−HK.{illeg}AH∷BA.AV. & AH. $HK ∷ bk . Hp . AH - 2 DA + \frac{{AD}^{q}}{AV} = Hp$ . AD^q−DA{illeg}V+D{illeg} AV^q=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 CD^q+CK^q(DK^q)+GD^q=GK^q Aufer BG^q seu CK^q et restabit CD^q+GD^q=BK^q 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 y^e Ellipsis to be a circle first & then reducing it to y^e 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 1^gr.46′6″. α, β, γ &c loca Comet{æ}

$\begin{array}{l} Feb 25 hora 8 \frac{1}{2} vesp. Dist Aα 2^{gr} .17′.42″. Ang BAα. \\ F \end{array}$

$\begin{array}{l} Dist. C ometæ stella A. & Angulus \\ Die Feb 25 hora 8 \frac{1}{2} vesp & Aα = 104, \frac{7}{12} part = 2^{gr} . 17'. 42 ″ & BAα 9^{gr} . 17' \\ Feb 25 hora 9 \frac{1}{2} & Aβ = 103 \frac{3}{4} = 2^{} . 16'. 36 ″ & BAβ 9^{} . 27' \\ Feb 27 hora 8 \frac{1}{4} & Aγ = 77 - & BAγ \\ Mart 1 hora 11 & Aδ = 52 = 1^{} . 8'. 30 ″ & BAδ 31^{} . 40' \\ Mart 2 hora 8 & Aε = 43 = 0^{} . 56'. 37 ″ & BAε 43^{} . 8' \\ Mart 5 hora 11 \frac{1}{2} & Aζ = & BAζ \\ Mart 7 hora 9 \frac{1}{2} & Aη = & BAη \\ Mart 9 h o ra 8 \frac{1}{2} & Aθ = 78 \frac{7}{12} = 1^{} . 43'. 30 ″ & BAθ 143^{} . 00' \\ Mart 9 hora 12 & Aι = 80 \frac{1}{3} = 1^{} . 45'. 46 ″ & BAι \end{array}$

<98r>

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

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

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 ≏ 23^gr cum lat Aust. 1^gr 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. 5^gr

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

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

<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 ♍ 12^gr cum lat. b|B|oreali 2^gr. Locus ☉is ♏ 29^gr. 53′

Romæ per Marcum Antonium Cellium observationes hæ factæ sunt.

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

Romeæ per Galletium hæ

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

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

Ejusdem posterioris Cometæ Observationes Grenovici habitæ

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

<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 ♑ 26^gr 43′ cum lat bor. 38^gr 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^tæ et 6^tæ magnitudinis, δ et ζ in tribulo sagittæ, quasi 40′ ad occidentem, et ultra per 3^grad 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^tæ 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. 15^gr 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, 72^gr Conica{illeg} existens circiter. Decem 28 {illeg} orta {illeg} fortior et clarior apparuit {illeg} sed 56^gr. {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{m}. P{enulti}{ma in Ala} austrina Cygni. (qu{illeg}|æ| tertiæ magnitudinis est et in Tabulis Bayeri {illeg} {ζ} diatur) llunat per caudam quarta parte latitudinis cauda{illeg} a latere australi ejus distans. erminus ejus {illeg} habebut multitudinem seu distantiam ab horizonte cum {stellas} quibus {illeg} extrema cauda cygni, Bayero dictis ♊. De{illeg} {igitur} {illeg} ♓ {illeg} lat bor {illeg} 52^gr. 20′.

♂ Dec 21 In cauda stella nulla apparuit sed {cauda} incuvata {illeg} versus {illeg} ad {illeg} onminò {illeg} in loco qui {illeg} <99v> seu pectore Cassiopeiæ et alia tertiæ magnitudinis stella in summa fere cathedra prope brachium dextrum (Bayero β dicta) triangulum æquilaterum constituit, tantum ab utra distans quantum {utra} ab invicem, (Flamsted. epist. {illeg}|{s}|{ubs.}) adeo in long. ♓ 23^gr.54′ lat 47^gr 24′ \desinebat. Cauda jam 70 C||adas \fere/ longa duas Cata tendebat versus intervallum inter {schedir} et lucidam cathedræ({illeg} Epist. 2)/. In Astrolabio vero Flamstedij, Ar{is} caudæ productus secabat ab intervallo inter caput & pectus Cassiopeiæ tertiam ejus partem versus pectus, desinebat autem e regione schedir. Transibat axis ille per caput Delphini, dimidio gradu a stellis duab{us} orientalibus in capite equiculum versus distans. Dein a distantia ultimarum duarum in ala austrina cygni auferebat quin nonas partes distantiæ illius versus stellam ultimam in ala. Postea a spatio inter terminum catenæ Andromadæ & stellam proximam in capite Cephei auferebat tertiam partem distantiæ \illius/ versus terminum caten{æ}

♀ Decem 24 Cauda transibat per medium intervallum stellarum duarum borealium in manu s{illeg}|u|periori Andromedæ et vix ultra Schedir extendebat (Flamst. Epist. post.) Desinebat igitur in long. ♉ 4 vel 5^gr, lat. $43 \frac{2}{3}$ ^gr. In Astrolabio Flamstedij cauda desinebat e regione pectoris Andromadæ. Transibat \autem (sed on rect{æ})/ per medium punctum inter genu dextrum seu australe Pegasi et stellam illam informem ad pedem dextrum quartæ magnitudinis cujus Long ♒ 29 55 lat bor 36. 11. Dein per pr{illeg}|{æ}|dictas st{illeg}|ell|as duas in manu superiori Andromedæ. {Dein per} Juxta Astrolabium erat stellæ duæ {γ}, δ invictu equiculi, et Cometa trian{illeg}|gul|um rectangulum constituebant Angulus rectus erat ad stellam occidentaliorem γ. Cometa boream versus distabat ab hac stella tertia parte distantiæ stellarum.

☉ Dec 26 \{illeg}/ Genu {illeg} sinistrum Pegasi (quæ stella tertiæ magnitudinis est et {sic} Bayero dicitur {η} erat inmedio caudæ \Flamst. Epist. {ult}.)/. Sed hac nocte et præc{æ}denti|{e}|bus caudæ terminus ob Lunæ splendorem haud satis definiri potuit (Flamst. epist. 2.) Unde die 24 gradus unus forte et alter ad caudæ longitudinem addi debet. Cauda vero hactenus semper curva apparuit, sed non valde curva. Convera sui parte austr{illeg}|{um}| respiciebat: qua etiam parte lucidior et {illeg} distinctiùs terminata apparuit quam altera.

♂ Decem 28 Cauda 56^grad longa distantiam inter Alamac et lucidam in femure Cassiopeiæ bisecans ultra pergebat ad us Persei Caput (Observat Hamburgens.)

☿ Decem 29 Cauda tangebat Scheat \sitam/ ad sinistam & intervallum stellar{is} in pede boreali Andromadæ \accur{a}te complebat/ (Flamsteed {illeg} \epist ult/ stella {φ} in femure boreali Andromade erat in medio caudæ, et \Decem 30, hora $8 \frac{1}{2}$ / Situs erat humerus Pegasi seu scheat in latere australi caudæ ita ut per caudam l{a}ceret, a termino caudæ quinta circiter vel sexta \quarta/ \vel quinta sexta/ parte latitudinus caudæ distans. Implebat autem cauda quasi \ $\frac{1}{3}$ vel/ $\frac{2}{5}$ intervalli inter Scheat & genu sinistrum Pegasi (Bayero {illeg}|{η}|) Stella φ in femure boreali Andromedæ erat in medio caudæ. B|b|isecabat axis caudæ intervallum stellarum \{illeg}/ in pede Borel|a|li Andromadæ, {illeg}|&| cauda intervallum illud plusquam implebat. Desinebat vero in medio loco inter stellam τ quintæ magnitudinis in capite Persei & extemam |in| borealis pede Andromadæ, sive inter stellam γ tertiæ magnitudinis in humero {illeg} boreali Persei & punctum \qu{a}/ distantia duarum in pede boreali Andromadæ bisecantur. Un(Ego.) Unde caudæ longitudo tota erat $53 \frac{1}{4}$ ^gr Deflectio caudæ ab oppositione ☉^is, seu angulus quem linea jungens caput et extremitatem caudæ effecit cum {Cabe} linea jungente solem & cometam, 5^gr. Latitudo caudæ juxta duas {illeg}|λ, μ| in {illeg}|pectore| Pegasi (hoc est 5^gr a capite Cometæ) erat {dimidium} distanti{illeg}|{a}| humeri illius duarum illarum stellarum {illeg} una cum triente distantiæ (nempe 1^gr {30′} circiter). Ejusdem juxta humerum Pegasi (seu $10 \frac{1}{2}$ gr a capite) latitudo erat dimidium distantiæ humen illius et \Pegasi/ genu {illeg} sinistri orientalis, Pegasi adeo 2^gr 30′ circitem. Ejusdem inter {illeg}put Andromadæ et annulum qui est in termino catenæ, (hoc est $21 \frac{1}{4}$ ^gr a capite) latitudo caudæ erat quinta par{illeg} distantiæ stellarum illarum \feré/ adeo{illeg} \{illeg}/ {illeg} caudæ latitudo adhuc {illeg} {illeg} <100r> aliquantulum us ad extremitatem fere, ita ut tandem evaderet 5^gr vel paullo major (Ego).

♃ Decem 30 \hor {8} {& hor}/ Scheat sita erat e latere caudæ ad dextram, et australior duarum in boreali pede Andromedæ erat in medio caudæ. {ad dextram} (Flamsteed. ^[79] & Ego \Idem et ego observabam hora 9./) Ultra vero hanc stellam australiorem cauda quasi ad 7^gr vel $7 \frac{1}{2}$ ^gr extendebat circiter.|(|Ego)

☾ Jan 3 Hor $11 \frac{1}{2}$ Cauda transibat per medium intervalli inter Alamac et australiorem in pede boreali Andromadæ & $\frac{4}{9}$ partes distantiæ stellarum (id est $3 \frac{1}{2}$ ^grad) ibi \(hoc est 30^gr a capite)/ lata erat. Tendebat ver{d}|u|s lucidam in latere Persei sed magis accuratè ve{illeg}{d}|rs|us stellam {ι} quartæ magnitudinis in dorso Persei lucidæ proximam \quæ tamen sex vel decem minutis circiter distabat ab axe austrum versus./ Desinebat verò {illeg} quasi \cauda e regione/ {in}|{a}| medij{illeg} luci|e| inter stellam illam quarta magnitudinis, et aliam ejusdem magnitudinis in humero dextro seu clypeo Persei quæ Bayero θ dicitur. Desinebat igitur in ♉ 22^gr. 27′ & lat. bor. 30^gr {illeg} 50′. Si borealiorum duarum μν in angulo Andromadæ distantia dividatur in tres partes \æquales/ & una pars sumatur versus mediam {trium} {illeg} trium in angulo μ, ibi erat medium caudæ (hoc est in ♈ 25^g. 4′ lat 30^gr 52′) et ibi \hoc est 18^gr a capite)/ latitudo ejus æquabat distantiam stellarum illarum, vel paullo superabat adeo erat✝^[80] 2^gr 6′ circiter. Ex his colligitur caudam curvam fuisse & convexo sui latere austrum respexisse concavo boream. Cauda jam haud multò lucidior erat quam partes lucidiores viæ {illeg}|ha|cteæ, si partes capiti proximas excipias, et quidem per ultimos duodecim vel quindecim gradus non erat illis {illeg}|u| lucidior. Caput jam multo magis conspicuum erat quam cauda at Decemb 15 cauda maximè conspicua erat caput vero instar stellæ adeò exiguæ apparuit (a crepusculo; scilicet et luce lunari obscuratum) ut nudis oculis ne quidem videre possem quamvis adstantes {illeg}digitum ad eam intenderent. Longitudo caud{illeg}|æ| 41^gr. Distantia circul termini caudæ a circulo solem et cometas jungente 4^gr 30′ {illeg}′. Delinatio caudæ ab oppositione ☉^is 7^gr.

♂ Jan 4 Hor {illeg}|{9}| Cauda juxta caput Cometæ tendebat versus lucidam in eductione cruris sinistri Persei, sed postea vergebat ad lucid{illeg}|am| in latere Persei et ubi aer admodum {illeg} \defæcatus erat, et meo et \{illeg}{sinetiū} judicio// exte{illeg}|nd|ebat ad \us/ stellam {illeg}|{ι}| in dorso Persei. \Axis caudæ non transibat per stellam {illeg}|{ι}|, sed paucis minutis australior existens, dirigebatur \{illeg} accurate/ versus/. Media {illeg}rium in ci{illeg}|ngulo| /lucidam in eductione cruris Persei Alge{illeg}b. dictam. vel potius versus punctum 5′ aut 6′ australius.\ Andromadæ erat in media|o| caudæ. Latitudo caudæ e regione capitis Andromadæ erat $\frac{4}{5}$ partes distantiæ medij caudæ a capite Andromadæ: Inter Alamac et lucidiorem et|in| altero pede Andromadæ æquabat $\frac{1}{2}$

Newton's Waste Book (Part 3)

How to Draw Tangents to Mechanicall Lines

Lemma.

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

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

To draw a tangent to ye Ellipsis

To draw a Tangent to ye Concha.

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

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

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

The same done otherwise.

Resolution.

Demonstration.

By helpe of ye preceding probleme divers others may bee readily resolved.

[28] 1. To draw tangents to crooked lines (however they bee related to sreight {sic} ones).

Resolution

[31] 2. Hitherto may bee reduced ye manner of drawing tangents in mechanicall lines. see Fol 50.

[32] 3. To find ye quantity of crookednes in Geometricall lines.

{illeg} Resolution

Of the construction of Problems.

Theoremata Optica.

Hujus autem Theorematis inventio totis est.

Probl.

Probl.

Telescopij novi delineatio

Some Problems of Gravity & levity &c

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

A Method Whereby to Square lines Mechanichally.

Octob. 1676.

Prob.

Cas 1

Cas. 1.

Cas 2.

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

Cas 1

Cas. 2.

Of the nature of Equations

Prob{l}

Of Equations,

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

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

Veterum Loca solida componere restituta.

Problema

De resolutione Quæstionum circa numeros.

Loca plana.

De Loco rectil{í}neo.

Quæstionum solutio Geometrica.

Quæstionum solutio Geometrica

Quæstionum solutio Geometrica. Prob 1

Prob. 2

Prob. 3

Prob 4

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

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

Observationes ejusdem cometæ habitæ a R. P Ango in Fleche

Observationes Venetijs habitæ a M. Montenaro.

Observationes Hevelij destituti instrumentis

Observationes Cometæ Mense Novembri anni 1680

Observationes Parisijs habitæ Cometæ subsequentis 1680 & 1681.

Ejusdem posterioris Cometæ Observationes Grenovici habitæ

Observationes de Cauda Cometæ prioris

De cauda Cometæ posterioris

To resolve Problems by motion y^e 6 following prop: are necessary & suffcient.

To draw a tangent to y^e Ellipsis

To draw a Tangent to y^e Concha.

To find y^e point c w^ch distinguisheth twixt y^e concave & convex portion of y^e Concha.

By helpe of y^e preceding probleme divers others may bee readily resolved.

^[28] 1. To draw tangents to crooked lines (however they bee related to sreight {sic} ones).

^[31] 2. Hitherto may bee reduced y^e manner of drawing tangents in mechanicall lines. see Fol 50.

^[32] 3. To find y^e quantity of crookednes in Geometricall lines.

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

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

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

Quæstionum solutio Geometrica.
Prob 1

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