<2r>

Of the extraction of Pure Square Cubick. Square-square & square-cubick rootes &c.

Let the number whose roote is to bee extracted bee pointed makeing the first point under the {unite} & comprizeing soe many numbers under each point as the number hath dimensions as if the number be square-cube tis thus pointed 5708.63524.10802.

Then out of the figures of the first point next the left hand extract the greatest roote proper to the power of the number & set that downe in the Quotient which is the first side & is called A. (as the roote quintuplicate of 5708 is (5) , & (5) quintuplicate is 3125 ) then takeing that roote duely multiplied out of the number (as 3125 out of 5708 ) with the rest of the numbers to the next point. seeke the seacond side which is found by divideing that number by another number made out of the first side (which is called the Divisor) & this second side I name E. (thus by divideing 258363524 by 5Aqq+10Ac+10Aq+5A after such a maner that 5 Aqq E + 10 Ac Eq + 10 Aq Ec + 5 A Eqq + Eqc may be conteined in the number the product of that division shall be E =

<2v>

The extraction of the square roote

The square to be resolved29.16.(54 The Product The square of  ye  first side250be taken away. The rest of  ye  sqare to be0416resolved. The divisor for finding  ye  seacond010sidie. which is  ye  first side doubled The rectangle by 2A & E04 The square of E0 00g 16q } to be substracted The sum¯e of  ye  rectangles40 16to be subducted 0 00The remainder

The extraction of the cube roote

The cube to be resolved157. 464.(54 The cube to be subducted125 whose roote isA=5 The remainder foryefinding32 464of E The divisors foryefinding of (E)yeseacond side. { 0 7 0 0 5 0 15 0 3Aq 3A The sume ofyedivisors07 650 Sollids to be substracted { 0 30 0 2 0 0 0 0 0 40 0 064 3AqE 3AEq Ec The sume of those32 464sollids The remainder00 000

The extraction of the square square roote

The square-square33. 1776.(24 The square-squ: to be subduc:16 =Aqq Remainder.17 1776 Divisors for findingye seacond side E. { 0 3 0 0 0 0 2 0 24 0 008 0 4Ac 6Aq 4A Theire sum¯e03 4480 Squ-Squares to be sub= =ducted { 12 3 0 0 8 84 5120 0256 4AcE 6AqEq 4AEc Eqq Theire Sum¯e17 1776

<3r>

The extraction of the Square-Cube roote

The squ: cube to be resolved79. 62624.(24 Substract32 Aqc Remainde47 62624 Divisors { 5q 8 0 0 0 0 q 0 05q 80 Ac 0400 Aq 0010 A 5Aqq 10Ac 10Aq 5A The Sume ofyedivisors08 84100 Plano-Sollids to be substracted { q 32 q 12 q 02 q 00 q 00 0 q 80 q 5600 q 25600 q 01024 q 5AqqE 10AcEq 10AqEc 5AEqq Eqc. Theire Sum¯e47 62624 Remainder00 00000

Note that the 3d 4th 5th & other figures are found by the same manner that the seacond figure is found onely makeing all the figures found to stand for A the first side & the figure sought for e or the 2d side

And if roote is found inexpressible in whole numbers then adding ciphers & pointing them from the unite towards the right {kind} as was before explained & soe hold on the worke in decimalls.

As for the Divisors they are easily found by the 2d Table of Powers from a Binomial roote.

If the Number bee of 6.7.8.9.10 &c dimensions The roote may be extracted after the same manner

<4r>

Of the Extraction of Rootes in Affected powers.

The manner of the extraction of rootes in pure & affected powers is very much alike, especially when the affected powers are decently prepared, that is, when theire affections are not over large & those altogether either affirmative or negative, & the power affirmative, affirmations & negations so mixt that there be noe ambiguity & all fractions & Asymmetry taken away

All the figures in the coefficients & affected power are to be pointed (after the manner before explained in the Analisis of pure powers) according to the degree of theire dimensions & the worke onely differs from that in pure powers in that the coefficients enter into the divisors

Let the first side be called A. the 2d be called E. the Roote of the equation {L} the coefficients B . Cq . Dc . Fqq . Gqc . Hcc &c the Power P . Pq . Pc . Pqq &c & the Operation follows

The analysis of Cubick Equations.

The equation supposed Lc* + 30L = 14356197 . Lc + CqL = Pc
0.The square coëfficient 0.The cube affected to be 0 0. 1 4. 0. 0 3 3 5 6. 0. 0. 0. 1 9 7. (243 Sollids to bee substracted { 0 8 0 0 0 0 0 0 0 6 0 0 0 0 0 0 =Ac =ACq Theire sum¯e 0 8 0 0 6 0 0 0 Rests 0 6 3 5 0 1 9 7 for findingye2dside The extraction ofyeseacond side 0.Coëfficient 0.The rest ofyecube to be 0. 0 0. 6 0 0. 0 3 5 0. 3 0. 0. 1 9 7. or superior divisor0. resolved0. The inferior divisors { 0 1 0 0 2 0 0 0 6 0 0 0 0 0 0 0 3Aq 3A Theire sum¯e 0 1 2 6 0 3 0 0 Sollids to be sub= stracted { 0 4 0 0 0 0 0 0 8 0 0 9 6 0 0 6 4 0 0 1 0 0 0 0 0 0 0 0 0 2 0 0 =3AqE =3AEq =Ec =ECq Theire sum¯e The superior part ofyedivisor 0 5 0 8 2 5 0 2 0 0 0

<4v>

The superior part ofyedivisor The remainder for finding 0 0 0 0 0 0 0 5 2 4 0 3 0. 9 9 7. oryesquare coefficient yethird side The inferior part ofye divisor { 0 0 1 7 2 0 0 0 8 0 0 7 2 0 3Aqthat is3×24×24 3Aor3×24. The sum¯eofyedivisor 0 0 1 7 3 5 5 0 Sollids to be taken away { 0 0 0 0 0 0 0 0 5 1 8 0 0 6 0 0 0 0 0 0 4 0 0 4 8 0 0 2 7 0 0 9 0 0 3AqE 3AEq Ec ECq Theire Sum¯e 0 0 5 2 4 9 9 7 Remaines 0 0 0 0 0 0 0 0

But the Coëfficient maybe greater than the Power soe that it cannot be substracted from it which argues that the Cube more propperly affects than is affected. In this case the coëfficient must descend towards the unite soe many points untill it may be substracted, & soe many points as the coëfficient is devolved soe many pricks must be blotted out towards the left hand in the power affected. As the Example shews
Lc+95400L =1819459 .

Sollids to be substracted { 0 9. 0 1. 5. 4 0 8 1 9. 0. 0. 0. 4 5 9. Coefficiens0. The Power0. Since9 is greateryn1make a devolution thus. 0.T 0.The Quote (19 0. 0 0. 1. 9. 5. 4. 8 1 9. 0 0. 0. 4 5 9. The Coefficient0. The affected power0. Sollids to be substracted { 0 0 0 0 9 5 4 0 0 1 0 0 0 0 0 0 ACq Ac Sum¯a 0 0 9 5 5 0 0 0 substrahenda Divisoru¯superior pars 0 0 0. 9 5 4 0 0. Coefficiens Planum 0 0 8 6 4 4 5 9 Potestas reliqua Divisoru¯pars inferior { 0 0 0 0 0 0 0 0 0 0 3 0 0 0 3 0 3Aq 3A Divisoru¯   sum¯a 0 0 0 9 5 7 3 0 Sollida ablativa { 0 0 0 0 0 0 0 0 8 5 8 0 0 2 0 0 2 0 0 0 6 0 0 7 0 0 4 3 0 7 2 9 ECq 3AqE 3AEq 3Ec Eoru¯  Summa 0 0 8 6 4 4 5 9 Restat 0 0 0 0 0 0 0 0

<5r>

To place the unite of the coefficient in its right place in respect of the power make so many pricks above as there are under the power begining at the unit, & if the coefficient be one dimension lesse than the power make a prick on every figure if 2 dimensions les than every other figure of 3 dimensions lesse make it one each third figure &c

If there be many coefficients in the equation each must be placed according to this rule.

Sometimes the coefficient is under a negative sine as Lc10L=13584 & the Analysis is as follows

Coëfficiens planum0 10.0sublaterale Cubus resolvendus+13. 584.(24 Solida ablativa { +08 0 0 20 Ac ACq Sum¯a+07 800 Restat+0.5 784.resolvendum Divisoru¯p_ssuperior+00 -10coëfficiens planum Divisoru¯p_sinferior { +01 +01 20 06 +3AA +3A Suma divisoru¯+01 250 Solida ablativa { +04 +0 +0 0 0 800 960 064 040 3AAE 3AEE EEE ECC Eoru¯  sum¯a+05 784

But sometimes the square coëfficient hath more paires of figures than the cube to be analysed, hath & then there is præfixing so many ciphers to the cube as figures are wanting, the first side will not much differ from the square roote of the coefficient. as Lc116620L=352947

1 1. 6 6 2. 0. 0. 0. Coefficiens planu¯ Cubus resolvendus + 0 0. 3 5 2. 9 4 7. (343 Sollida Ablativa { + 2 7 3 4 0 0 0 9 8 6 0 0 0 Ac ACq Restat auferendu¯ 0 7 9 8 6 0 0 0 Reliquum resolvendi 0 + 8 3 3 8 9 4 7 Cubi ________________________________________________________

<5v>

Divisoru¯p_ssuperior Coeff: 0 1. 1 6 6. 2 0. 0. planum. Reliquu¯  resolvendi cubi 0 + 8. 3 3 8. 9 4 7. negative affecti Divisoru¯p_sinferior { + + 2 + + 0 7 0 0 0 9 0 0 0 0 3AA. 3A. Sum¯a Divisorum { + * * + + 1. * * * 6 2 3. * * * 8 0. 0 ************ 3AA + 3A + Cq Sollida ablativa { + 1 0 0 + 1 0 + 0 0 4 8 0 0 4 4 0 0 6 4 6 6 4 0 0 0 0 0 0 0 0 0 8 0 0 3AAE 3AEE EEE CCE Eorum summa + + 7 6 3 9 2 0 0 Restat Resolvend¯ + + 0. 6 9 9. 7 4 7. pro3olatere Divisorum¯p_ssuperior ~ + 0. 1 1 6. 6 2 0. CC Divisoru¯p_sinferior { + + 0 + + 0 3 4 6 0 0 1 8 0 0 0 2 0 3AA 3A *********** * * * * * * * * * * Eorum Summa 0 0 0 2 3 1 2 0 0 =3AA+3A+CC Sollida ablativa { 0 + 1 0 + 0 0 + 0 0 0 0 4 0 0 0 9 0 0 0 3 4 9 4 0 0 1 8 0 0 2 7 8 6 0 3AAE 3AEE EEE ECC Eorum Summa 0 + 0 6 9 9 7 4 7 ________________________________________________________

Sometimes though there be as many 2 figures in the coefficient as 3 figures in the cube affected yet the coëfficient may be so greate as to deceive an unwary Analist As in this Lc6400L=153000 . where the roote of 64 is 8 which cubed is 512 which added to 153 makes 665 then whose roote the number immediately greater is 9 which is the first side =A .

But if the coefficient had beene affirmative, then not the aggregate of the facts but the difference must be taken as in this. Lc+64L=1024 .

Since the roote of 64 is 8 . which cubed is 512 . & 1024512= 512 . the roote of which is 8=A . The like is observable in equations of higher powers

If the Cube be affected with a negative sine as 13,104LLc=155,520 . Then the Equation is expressible of 2 rootes: whereof the square of one is <6r> lesse & the square of the other is greater then 131043 . & therefore one roote is lesse the other greater then 15552013104 . & in this equation 27755LLqq=217944 are two rootes whereof one is greater the other lesse then 21794427755 .

Suppose in the former cubick equation the lesse roote be 12. then 15552012=12960 . or else 1310412×12=12960 . & Lq+12L=12960 . where L=108 is the greater roote.

And in the latter equation if the greater roote be 27. & 21794427=8072 , c. or 27×27×27 + 27755=8072 . 27×27=729 . If there be 4 cubes continually proportionall whose greate extreame is 27c=19683 . & the aggregate of the 3 rest is 8072 & Lc the lesse extreame, therefore Lc+27Lq+729 L =8072 . the roote of which .is 8 the other roote of the equation

Or haveing one roote of an equation the Equation may be lessened by division thus 13104llc = 155520 or l3 13 1 04l+155520=0 . & one roote is 12 . therefore divide this equation by l12 & the Quote is an equation conteining the other roote viz: lq+12l=12960 .

<7v>

Figure

<8r>

Propositiones Geometricæ. Franc: Vietæ.

prop 1

Figure

abaccebd

prop 2

& if abacacbd: then acababce

prop 3. If ab×ac=bd×ce . then bdacacababce

prop 3. To find two meane proportionalls {twixt} Bc & IK . On the center a with the radius ai describe Figure the circle ibck . inscribe b c =cd . draw da through the center & bg parallel to it. draw hk through A soe that gh=ab=(=ai) . & ikhbhbhihibc . ∺

Figure

Prop: 4

If ad=db=cb . then the Angle c be is tripple to the Angle abd .

Prop 5

If ab=bd=Rad. 3Ang:bad=cde

Figure Figure

Prop 6

Figure

If 3rpq=spq:recto . that is If 2qr=pr . then 3or×or=sp×sp+op×op+px×px

Prop 7

Figure

If ad=dc=ce=ef . then Figure ecf=efc=3dac=3dca & AC3=3AC×ad2+Cf×ad2 . Z3=aaz+b3.~ ~ ~

Figure

prop 8. If op=pr=qr=qs . then, prt=3qsr &c and sr3=3sr×qr2or×qr2 . Z3=aazb3

Figure

prop 9 If ah=hb=bf=fd . & ch=2eh or ceh=3hce . then ac3=3ac×ah2db×ah2&cb3=3cb×ah2db×ah2 }Z3=aazb3.

<8v>

Prop 10

Figure

If de=ea & dbdaab×abdc×dc . then be is a side of a 7 equall sided & angled figure. or 7eab=4right angles .

Figure

prop 10

If ac=ef . & aef a right angle & ab passes through the center then cddededfdfdb . And if cddededfdfdb then ae is perpendicular to ef . 2hd is the difference of the extreames & 2do is the difference of the meanes. which given the proportionall lines may be found &c.

Figure

prop 11

Pseudomesolabium wherby To find 2 meane proportionalls. If, aeececededeb . they be inscribed in the circle acbd the diam : being ae+eb . If twixt g i & i h two meane proportionalls are sought on the same center f with the Rad : gi+ih2 describe gkhl & inscribe a line kl parallel to cd cutting ab in the point i & gikikiililih .      Examine it.

[1] Figure

prop 12

If do=dh & ac bisected in b & bd bee drawne rd is the side of a pentagon which may be inscribed in defcro

prop 13.

Figure

If rd be the side of a {octogondecagon} & pd the side of an {hexagonoctogon} the arch rp divided in o , od will be the side of an {heptagonenneagon} to be inscribed in the circle ord & the arch RP is rightly divided by Bisecting the Line ac . Examine it

<9r>

Of Angular sections.

Figure

prop 14

If ead=cab . Then abab2cbeb×adae×dbacae×ad+eb×db or, abaq×abopeb×anae×nqaoae×an+eb×nq . But the angles anq , aop are right ones and e an=oap = eabdab

prop 15

If the angle cab+dab=eab . or naq+oaq=eab . & anq , aop are right angles then Abab2Ebad×bc+ac×dbeaad×acdb×cb . or the triang: unequall. abap×aqeban×op+ao×nqeaan×aonq×op

prop 16.

In 2 rectang: triang: acb & aed , if the first have an acute angle cab submultiple to the acute angle eab of the 2d triang aeb the sides of the seacond have this proportion. Suppose the Hypoten of the first tri: be z . the base b . the Cathetus c .

If  ye  acute angle ofye  seacond triangle beto  ye  acute angle ofye  first triangle ina proportion{   Hypoten:BasePerpendicular Duple,Z2.B2.-C2.2BC. Triple,Z3.B3.-3DDC.3BBC.-C3. Quadruple,Z4.B4.-6B2C2.+C4.4B3C.-4BC3. Quintuple,Z5.B5.-10B3C2.+5BC4.5B4C.-10B2C3.+C5.

Figure

Prop 17. If {} ab=bc=ce=eg &c: & ac=cd . & ae=ef &c then abacacadaeaf &c & ed=ab & ac=gf &c. {nam} triangle cde & cba , efg & eac &c: = & sim.

Prop.18.

Figure

If bd=dg=gh=hk=pq=pw &c Then alak pepc eddo pdpc+do rggs rqqo qgqo+gs &c & if ( lf2 = ) from ʒ to the center be drawne cʒ then alak didv iqqx dqdv+qx gzgx wzwx &c Ergo acak ab +adadab+agagad+ahahag+ak &c.

<9v> Figure

Prop 19

If fa=ab=be=eh &c. &. af+ab+be+eh are greater than the semiperiphery: & dh is the greatest, db the least line drawn from d to these points a , b , e , h . then rad dh db dade .

Prop 20

Out of the 18th & 19th Propositions To divide An angle into any number of points in the figure of the 18th prop: al=diam= 2 z . ah is the greatest of the inscribed lines =B : now z ∶ Bah+ 2z. therfore bb = ahinz + 2z2 . & bb 2 zz z =ah . And zB b22z2z b+ag . therfore b32zzb zz b = ag Likewise b44zzbb2+2z4z3 = ad . & B55zzB3+5z4B z4 = ab B6 6zzB4 + 9z4BB 2z6 zzzzz =
B77zzB5+14z4B37z6B z6 = to a seaventh line
B88zzb6 + 20z4b4 16z6bb + 2z8 z7 = to an eight line 00 00 B99zzb7 + 27z4b5 30z6b6 + 9z8b z8 = a nineth line 00 00 B10 10zzb8 + 35z4b6 50z6b4 + 25z8bb 2z10 z9 = tenth &

Prop 21

out of the 17th Theor.: in the figure whereof if ab {:} the least inscribed line =z . & ac the next line bee B . then zB Bz+ae . & bb zz z = ae & B3 2z2B zz = ag & B4 3z2bb + z4 z3 = to a fift line. B5 4zzb3 + 3z4b z4 z4 = a sixt. & b6 5zzb4 + 6z4bb z6 z5 = seaventh b7 6zzb5 + 10z4b3 4z6b z6 = to an eight line
B8 7zzb6 + 15z4b4 10z6bb + z8 z7 = to a nineth line B9 + 8zzb7 + 21z4b5 20z6b3 + 5z8b z8 = to a tenth line B10 9zzb8 + 28z4b6 35z6b4 + 15z8bb z10 z9 = eleventh .

<10r> Figure

Prop 22.

If aq=ab=bd=dc=ch=hk=kl=lf . Then GK Rad kl kl el (=alah) hl hm (=qhqd= akac) dl do (=qdqa=acab) lc cn (=qcqb= ahad &c ) Soe that the Periph: divided into any number of points. gl lk lk al ak lh ak ac lc ah ad dl ac ab &c. & gl lk ah lc lk ac ld lh ad lb lc ab al ah &c.

hence Prop 23.

In the former scheame If al =2x =hypotenusa . kl=b . xb b2xah & bb+2xx x = ah / xb bb+2xx x 2bxxb3 xx (=lc b ) therefore 3bxxb3 xx = lc . & 2x4 4bbxx + b4 x3 = ad the base of the 4th triang: & 5bx4 5b3xx + b5 x4 = the perpendicular ( bl ) of the 5t triangle & 2x6 9x4bb + 6xxb4 b6 x5 = base of the 6t triang. 7x6b 14x4b3 + 7x2b5 b7 x6 = perpendicular of the 7th triangle 2x8 16x6bb + 20x4b4 8xxb6 + b8 x7 = base of the 8th tri. 9x8b 30x6b3 + 27x4b5 9xxb7 + b9 x8 = perp: of the 9th tri:

<10v> Figure

Prop 24:

If bd =dh=hk=kl= b g &c: then bh=gd & bk=gh & bl=hm &c: & then
xtbx acag ceeh eiem iook opon pqql qrqy rssx stsf baad therefore xtbx btdg+hm+kn+ly+xf . againe xtbt abbd accg cech eiim &c Therefore xtbt btbd+gh+mk+ml+yx+ft . & since, as xtbt bxdg+hm+kn+ly+xf Therefore xtbt bx+bt bd+dg+gh+hm+mk+kn+ +nl+ly+yx+xf+ft . And xtbt bx+bt+xt to all the perpen dicular & transverse line + bt . that is
( 5 ) xtbt xt+bt+bx 2bd + 2bh + 2bk + 2bl + 2bx + 2bt .

Prop 24

Figure

If in the circle cfgh Figure be inscribed the helix bedc & ac touch it in the point c then ab = to the circumference.

Prop 25

Figure

If apcr be les than halfe the circle. & vt=tp . & vo= to vrap : then rq×po 2 = 4 times the section rapc

Prop 26

Figure

If ab=bd=ad & bh perpendicular to ad from the angle b . ce=ed . then aed=adi=3dae . & ed is the side of a heptagon

Prop 27.

If a line be cut by extreame & meane proportion the lesse segment almost is to the whole line as the diameter is to 5 times the periphery divided by 6 .

Prop 28

Si secetur linea per extremam & mediam proportionem erit proximè, ut tota linea plus minori segmento ad bis totam lineam, ita quæ potest quadrato sesquialterum semidiametri, ad latus quadrati circulo equalis.
linea secta sit 100,000 . minus segmentum 38,197 . Semidiametrum 100,000 , quæ potest quadrato sesquialterum semidiametri paulo maior est quam 122,474 . Radix Peripheriæ, 177,245 .

<11r>

Prop 28.

Figure

If er=rh=or . & ao=fc= to the side of a decagon; & fn parallell to cd then en shall be almost equall to the fourth parte of a circle for ef is divided in extreame & meane propor in the point c . & ecef ef512 Perimeterhbkfa hr524Perim 610ef(=de) 14Perim ; by the 27th prop: & ecef ed en ( =14 Perim ) .

Prop 29.

Figure

If os=2cp . & co is divided by extreame & meane proportion in r . & od parallell to rp then db is the side of a square = to the area of the circle. for by the 28th prop: As br(=to line+less segm¯) bo(=twice  ye  line) bp ( ( =32of the square of  ye   semidiameter ) bd(=to  ye  roote of a square equall to  ye  area of a circle .

Prop 30

Figure

If the line dc touch the helix in the line ag . & the line hf toucheth the beginning of it in the center a & 4ac = af then 2ad shall bee equall to perim: asr . & ac being the Diameter: the area of the triang acd = to the area of the circle asr

Prop 31

Figure

If bed be a square of one revolution of an helix & the angle dbe=dba & through the points a , d , in the helix be drawne the line adk & through the points e , d in the Helix be drawne edg . & the angle kdg bisected by dh ; then dh shall almost touch the helix in d . & it shall be soe much the nigher a touch line by how much the angles
ebd dba are lesser.

<11v>

Prop 32

Figure

If many Polygons be inscribed in a circle the number of theire sides increaseing in a double proportion. & theire apotomies, or the base of a tri: whose cathetus is a leg of the Polygon & hypotenusa is the Diameter (as the apotome of the Polygon cgp is ce . of pacegi is ae &c) if the Apotome of the sides of the first Polygon be called b . of the 2d = c . of the 3d = d . of the 4th = f . of the 5t = g of the Sixt = h . & the diameter be z And the first Polygon be = p . the 2d = q . the 3d = r = abcdefghiopq . the fourth = s . the 5t = t the sixt = v . the 7th = w &c then
pq bz . & pr bc zz . & ps bcd zzz . & pt bcdf z4 . & pv bcdfg z5 . & pw bcdfgh z6 &c



To know how many divers ways things, whereof some of them are equall, may bee ordered. . as of . a.b.b.c.c.c.d.d. doe thus { a bb ccc dd 11 × 2×31×2 × 4×5×61×2×3 × 7×81×2 =112 } { 1a1 × 2b×3b1×2 × 4c×5c×6c1×2×3 × 7d×8d1×2 =112 } the number of changes, in order.

To know how many elections may bee made doe thus 2a × 3bb × 4ccc × 3dd × 5eeee = 360 = 2a × 2b×3b2 × 2c×3c×4c2×3 × 2d×3d2 × 2e×3e×4e×5e2×3×4 therefore there are 359=3601 elections in abbc3dde4 .

<12r>

Propositiones Geometricae Ex Schootenij
Sectionibus miscellaneis.

Sectio 1ma

To know how many changes 6 Bells, abcdef or how divers conjuctions the 7 planets can make . or how many divisors abcde hath, or how man{y} divers compositions the 24 letters can make &c the examples following show.
1. a1
2. b . ab3
4. c . cb . cab . ac . 7
8. d . da . db . dab . dc . dac . dcb . dcab . 15
16. e . ea . eb . eab . ec . eac . ecb . ecab . ed . eda . edb . edab . edc . edac . edcb . edcab . 31
32. f . fa . fb . fab . fc . fcb . fac . fcab . fd . fda . fdb . fdab &c 63.
64. g . ga . gb . gab . gc . gcb . gac . gcab . gd . gda . gdb &c 127
which shows that in 7 letters 127 elections may be made. that 7 Planets may be conjoyned 120 divers ways. that abcdefg . hath 128 divisors for an unite is one of {them}& 1×2×3×4×5×6.=720 ; are the number of changes in six bells.

Sec 2

To know how many things & of what sort they are which may be chosen 15 ways. 15+1=16 . 162=8 . 82=4 . 42=2 . 22=1 . & 21 . 21 . 21 . 21 . = 1 . 1 . 1 . 1 . = 4 . that 4 things all unequall may be varyed 15 ways. also. 164=4 . 42=2 . 22=1 41 . 21 . 21 = 5 &. 5 things whereof 3 are equall viz: a . a . a . b . c . & 164=4 . 44=1 . 41 . + 41 = 3+3 = 6 . & 6 things whereof 3 & 3 are equall as aaabbb . may be varied 15 ways. & 168=2 22=1 . 81 . 21 = 8 = 7+1 . & 8 things whereof 7 are = may be varyed 15 ways. as aaaaaaab . 1616=1 . 161 = 15 . 2 wherefore 15 alike things &c as a 15. 2 what things vary 23 ways. 23+1 = 24 24 admitts a 7 fold divisor therefore the multitude of things sought may be 7 fold but since 43 is a primary number (viz which cannot bee divided) 42+1 = 43 . 4343=1 431 = 42 . therefore onely 42 like things can be varyed 42 ways as a42 .

<12v>

Sec 3

Every quantity hath one divisor more that it hath aliquote parts (that is parts of whole numbers.). How to find a quantity haveing a given multitude of divisors or aliquote parts: suppose its aliq: parts must be 15. 15+1 = 16 & soe by the former section abcd . a3bc . a3b3 . a7b . a15 . may be varyed 15 ways. therefore they shall have 15 aliquote parts & 16 divisors. but since onely 42 like things (as a42 ) can be varyed 42 ways therefore onely a42 hath 42 aliquote parts & 43 divisors. &c

Sec 4

To find the least numbers haveing a given multitude of divisors & aliquote parts instead of soe many letters in the former sec: put soe many least primary numbers & take the least result from them. as from the former example: abcd . a3bc . a3b3 . a7b . a15 . that is 2 . 3 . 5 . 7 . or 2 . 2 . 2 . 3 . 5 . &c. now. 2×3×5×7 . = 210 . & 2×2×2×3×5 = 120 . &c therefore 2×3×5×7 . =210 is the least number haveing 16 divisors.

Sec: 5 conteines a table of Primary numbers.

Sec 6

To find progressions constituteing rectangular triangles with sides rationall the examples following shew. take two numbers as 1 . 2 . then 1×2 = 2 since the product is eaven double it viz: 2×2 = 4 . & 4 is the numerator then 1+2 = 3 & since 3 is od multiply it by the difference of the termes: 1×3 = 3 & 3 is the denominator. & the first terme 43 . then since (1) the difference of the termes is od multiply it by 4 . 4×1 = 4 & 4× per 2 majorem terminum. 4×2 = 8 8+4 (the former numerato{r}) = 12 = numerator 2d. then 3 (the former denom) added to. 2 (the double square of the diff: of the termes because the square (1) is odd) =5 the 2d denominator. I ad another example take 1 . 3 . then 1 × 3 = 3 = 1st numerator. then 1+3 = 4 & since 4 is eaven 4×22 (diff: of the termes) =4 & the first denom is 4. the first terme 34 . then becaus the diff of the termes is eaven 2 2×2 = 4 & 4×3 = 12 & 12+3 = 15 . then 2×2 = 4 . 4+4 = 8 . & 158 the 2d terme & now termes may be had by Arithmeticall proportion. thus. 43 . 125 or 113 . 225 . 337 . 449 . 5511 . 6613 7715 8817 9919 101021 &c & 34 . 158 or 34 178 . 21112 . 31516 . 41920 . 52324 . 62728 . 73132 . 83536 &c thus may other progressions be obteined. For the use take the numerator for one leg & the denom for another & the Hypoten: will be rationall as in 225 or 125 144+25 = 169 = 13 . & in this 178 or 158 225+64 = 17 . Figure

<13r>

If the suposed numbers be 2 . 5 . then 2×5 = 10 . 10+10 = 20 . & 2+5 = 7 . 3×7 = 21 . so that 2021 . then 4×3 = 12 . 12×5 = 60 . 60+20 = 80 . & 3×3 = 9 . 9doubled = 18 . 18+21 = 39 . & the 2 first termes 2021 . 8039 or 2239 . Againe, if the numbers be 3 . 4 3×4 = 12 . 12×2 = 24 . & 3+4 = 7 . 1×7 = 7 . therefore 247 . then 4×1 = 4 . 4×4 = 16 16+24 = 40 . & 1×1 = 1 . 2×1 = 2 . 7+2 = 9 therfore 409 is the 2d & the progres may be continued, as 2021 . 2239 . 3557 . 4875 . 51193 . & 337 . 449 . 5511 . 6613 &c.

Sec 7

To find a {number} which divided by 7 leaves 2 . by 11 leaves 1 . by 13 leaves 9 . the least common divisor of 7 . 11 . 13 . is 7 × 11 × 13 × = 1001 . divide 1001 twice by each & consider the remainder of the seacond division thus.

1 Since more than 1 is left (viz 3) multiply 3 till it divided by 7 leavs 1 . 5×37 = 217 therfore 5×143 = 715 the multiplier | 10017 ( 1437 ( 2037 .

2 Since more than 1 is left (viz: 3 ) 3×411 = 1111 therfore 4×91 = =364 the multipl: | 100111 ( 9111 ( 8311 .

3 If but 1 had beene left 77 had beene divisor but now 12×12 11 = 13111 . therfore 12×77 = 924 is multiplyer. 100113 ( 7713 ( 51213 . now the number sought is thus found.

< insertion from the center right of f 13r >

Divisor.Reliq:Multip. 70.2×715=1430. 11.1×364=0364. 13.9×924=8316. The Sum¯e10110.

< text from f 13r resumes >

Lastly divide
by the least com. divis: 101101001 ( 101001001 wherefore 100 the number left is the number sought.

Sec 8.

Touching the Method of weights suppose a man have weights of 1.2.4.8.16.32 pounds &c by them all intermediate pounds may be thus weighed 1. 2 3. 4 5 6 7 8 9 10 11 12 13 14 1. 2 1+2 4 1+4. 2+4. 1+2+4. 8 8+1 8+2. 8+1+2. 8+4. 8+4+1. 8+4+2 &c or if his weights be 1.3.9.27.81. all weights may be supplyed thus. 1. 2. 3. 4. 5 6 7 8 9 10 11 12 13 14 1 31. 3. 3+1. 913. 93. 9+13. 91. 9. 9+1. 9+31. 9+3. 9+3+1. 27931. &c Note that weight marked with − signifie the weight to be put in the opposite ballance.

<13v>

Sec. 9.

To find numeri amicabiles that is 2 numbers whose aliquote parts are mutually equall to theire wholes. take this Des-Cartes his rule

If (2), or any other number produced out of 2 as 2×2. 2×2×2 &c (viz 2.4.8.16.32 &c ) bee such a number that 1 taken out of it triple there rests a primary number{,} & that if 1 taken from it sextuple there rests a primary number, & if 1 taken from its square octodecuple a primary number rests: then multiply this last prime number by the assumed number doubled & the product is one amicable number & the aliquote points of it make the other Example. if 2 be taken. 2×3 1 = 5 numero primario primo. 2×6 1 = 11 numero primario secundo. 2×2×18 1 = 71 numero primario tertio. 4×71 = 284 , one amicable number, & the 2 former prime numbers × one another & the product ×4 the double of the assumed number viz 5×11 = 55 55×4=220 . Thus from 8 . & 64 &c. may be deduced amicable numbers.

Sec 10

To find triangles whose sides, segments of theire bases, & Perpendiculars are expressible by rationall numbers
Figure 1st if the perpendic: is without the tri: let ac=z . bd=x cd=y . ad=z+y . ad=y+b . xx + yy = yy + 2by + bb . y = xx bb 2b . & cd = z+y+a . xx + zz + 2yz + yy = zz + yy + aa + 2zy + 2za + 2ay . 2ay = xx aa 2za = axx abb b . bxx baa 2zab = axx abb . bxx baa axx + abb 2ab = z . puting any numbers for a , b , & x ; y & z may be found. then ad = z+y = = xx + bb 2b . cd = z+y+a = xx + aa 2a . which reduced to the common denominator 2ab ; & that cast away. cd = bxx + baa . ad = axx + abb . de = 2abx . ae = axx abb . ce = bxx baa . ac = bxx axx + abb aab .

Figure

In like manner if the perpendicular fall within side. ab = bxx + baa . bd = 2abx . ad = bxx baa . dc = axx abb . bc = axx + abb . ac = bxx + axx abb baa .

Also by the conjunction & disjunction of 2 triangles it may be found that ab = bbx + aax ad = bbx aax . ac = bbx aax aax + abb . bc = axx + abb . db = 2abx . dc = axx abb . For if bd=x dc = xx bb 2b . bc = xx + bb 2b . that is bd = 2bx . dc = xx bb . bc = xx + bb . Likewise bd = 2ab . ad = bb aa . ab = bb + aa . 2abx the least quantity divisible by 2bx & 2ab , being divided by them, leaves a & x which must multiply the bases & hypotenusas. If the perpendic: fall without the legs may be thus exprest cd = acc + ayy . da = yyc + aac <14r> ca = acc ayy + cyy aac . ae = yyc aac . ce = acc ayy . ad = 2acy .

Sec 11

To make that two such tri: be of the same base & altitude. Suppose an equation twixt the bases & perpendiculars of the 2 last tri: as 2abx = 2acy . x = cy b . xx = ccyy bb . bbx aax axx + abb = acc ayy + yyc aac or bbcy aacy b accyy bb + abb = acc ayy + yyc aac & yy = + b3cy + aabbc aabc + ab4 abbcc bbc + acc abb . Suppose aabbc + ab4 = abbcc . or a = c bb c . let c=3 greater than b=2 . a=53 . y=2261 . x=3361 & consequently Figure

Sec 14 differs not from Cap 19: prob 18 Oughtred.

Sec: 15 Of Polygons or multangular numbers

The summe of all the tearmes in an arithmet: progres: increasing from an unite by 1 composeth triangles. by 2 , composes squares. by 3 , composes pentangles. by 4 , hexang: &c as 1. 2. 3. 4. 5. 6. compose the triangles 1 3 6 10 15 &c likewise 1. 3. 5. 7. 9. compose 1 4 9 16 25 &c So 1. 4. 7. 10. 13. compose the quintangles 1. 5. 12. 22. 35. 51. 70. &c. If a=1= the first term{e} the excess of the progression =x. The summe of the termes =z= to the polygon the multitude of the termes =t= to the side of the Polygon. Suppose t given to find z. z = 1tt+1t 2 or z = tt+t 2 in trigons. z=tt in 4gons. z = 3ttt 2 in 5gons. z = 2ttt in 6gons z = 5tt3t 2 in 7gons. z = 3tt2t in 8gons. z = 7tt5t 2 in 9gons. &{c} & z given t is found thus t = 1+1+8z 2 in tri. t = 0+16z 4 in 4gons t = +1+1+24z 6 , in 5gons. { t = +2+4+32z } { t = +2+4+32z 8 } in 6gons &c. As the side 12 of a tri given. the tri = z = 12×12+12 2 = 78 &c & if z=21 be octangled. t = +4+16+48z 12 = 4+16+48×21 12 t = 4+1024 12 = 3 .

<14v>

July 4th 1699. By consulting an accompt of my expenses at Cambridge in the years 1663 & 1664 I find that in the year 1664 a little before Christmas I being then Senior Sophister, I bought Schooten's Miscellanies & Cartes's Geometry (having read this Geometry & Oughtred's Clavis above half a year before) & borrowed Wallis's works & by consequence made these Annotations out of Schooten & Wallis in winter between the years 1664 & 1665. At which time I found the method of Infinite series. And in summer 1665 being forced from Cambridge by the Plague I computed the area of the Hyperbola at Boothby in Lincolnshire to two & fifty figures by the same method.                 Is. Newton

<15r>

Annotations out of Dr Wallis his Arithmetica infinitorum.

1 A primanary series of quantitys is arithmetically proportionall, as 0, 1, 2, 3, 4 . & its index is 1

A Secundanary series are those whose rootes are arithmetically proportionall; as 0, 1, 4, 9, 16 . & its index is 2

A Tertianary, quartanary, quintanary series of quantitys are those whose cube, square square, square cube rootes are Arithmetically Proportionall as 0, 1, 8, 27, 64 . / 0, 1, 16, 81, 156 . / 0, 1, 32, 243, 624 . &c Their indices being 3, 4, 5 &c.

3 Subsecundanary, subtertianary, series &c: are those whose squares, cubes, &c are arithmetically proportionall, as 0, 1, 2, 3 . . c:0, c:1, c:2, c:3 &c. Theire indices being 12, 13, &c.

2 Primary Secundanary, tertianary series &c are said to bee reciprocally proportionall ( that is to the same se increasing) which continually decrease as. 10, 11, 12, 13, 14, . 10, 11, 14, 19, 116, . 10, 11, 18, 127, 164, . Their indices being negative as -1, -2, -3 .

4 The indices of compound or mixt of rationall & irrati{onall} series, by multiplying or dividing the indices of the simple series may bee found as in a subsecundanary progression cubed 0, 1, 8, 27, 64 the index is 12×3 = 32 . So in the cube rootes of a secundanary progression, c:0, c:1, c:4, c:9 &c. the index is 13×2 = 23 . so in irrationall reciprocal progressions
qq:10, qq:11, qq:12, qq:13 , the index is -1×14 = -14 .

<15v> Figure

Now suppose the line ac be divided into an infinite number of equall parts ad, de, ef, fg &c, from each of which are drawne parallels nd pe qf &c. which increase continually in some of the foregoeing progressions or in some progression compounded of them, all those lines may be taken for the surface bqnac , & to know what proportion that superficies hath to the superficies ambc that is what proportion all those lines have to soe may equal to the greatest of them, I say as the index of the progression increased by an unite is to an unite soe is the square abcm to the area of the crooked line. As if abc is a parabola the lines ad, pe, qf, &c are a subsecundanary series (for y=rx) whose index is 12 which added to an unite is 1+12 = 32 Therefore 32 1 3 2 so is the square ambc to the area of the Parab. (the names of the lines are (ad), ae, af &c =x . dn, pe, qf &c =y . ac=p. bc=q.) The case is the same if abc bee supposed a sollid, as suppose it a Parabolicall conoides. then since the nature of it is rx = yy . yy designes the squares nd, pe, qf &c: all which taken together are equivalent to the Sollid. & those squares increase in the same proportion which rx . or x doth. that is they are a primanary series whose index is 1 to which (according to the rule I ad an unite & tis 2. Therefor 1 2 soe are all the squares of the Primary series to soe many squares equall to the greatest of that series. & soe is the conoides to a cilinder of the same altitude.

<16r>

Also if a superficies be compounded of 2 or more of these series, Their area is as easily found as if the nature of the line bee , y = aa xx , or y = a4 2aaxx + x4 or y = a6 3a4xx + 3aax4 x6 . &c. Their areas will bee to the parallelograms {about} them as 2 to 3 , as 8 to 15 , as 48 to 105 &c. but if I put in the intermediate termes in these last named lines their order will bee y = aa xx , y = aa xx , y = aa xx aa xx ¯ , y = a4 2aaxx + x4 . y = a4 2aaxx + x4 aa xx ¯ . y = a6 3a4xx + 3aax4 x6 ; &c: & since these lines observe a geometricall progression their areas must observe some kind of progression. of which every other terme is given viz 1. . 23. . 815. . 48105. . 384945. . 384010395 . Twixt which termes if the intermediate termes . can bee found the 2nd square will give the area of the line y = aa xx , the circle. Soe likewise in this progression of lines y =1 . y = ax xx . y = aa xx y = ax xx ax xx ¯ . y = aaxx 2ax3 + x4 &c: the progression of their areas is 1: : 16: : 130: : 1140: 1630: &c. the 2nd if it can bee found gives the area of the circle for as its denominator to its numerator so is the square of the diameter to the area of a semicircle. If this last progression bee multiplyed by the respective termes in the progression 1. 2. 3. 4 & it may bee diminished the result being 1. 2. 12. 4. 16. 6. 120. 8.170 soe that in this progression 1, b, 12, c, 16, d, 120, e, 170, f, &c: if <16v> b can be found then, the square of the diameter to the area of the circle is as the denominator of b to its numerator. Likewise the 1st series of areas may be diminished by multiplying each terme by its correspondent terme in this progression 1, 2, 3, 4, 5, 6, &c: & it will become, 1, a, 2, b, 83, c, 4815, d, 384105, e, 3840945 . &c. In which if a can bee found then as the denominator of a to its numerator: so the square of the Radius is a semicircle, that making the radius =q . 2aq= . The same kinds of changes may bee performed by any other progressions, as by division by the geometricall progression 1, 2, 4, 8, 16, & the first series of areas becomes 1, g, 16, h, 130, k, 1140, l, 1630, &c viz the same with the 2d series. Also these changes may be done by addition or substraction of mutuall termes in 2 proportions. Soe that the most convenient way may be chosen, wherby to reduce any series of proportions to the most convenient forme.

Now if it be propounded to find these middle termes, It will bee convenient to find how the given proportion may bee deduced from an Arithmeticall, Geometricall, or some other familiar proportion, viz whose meane termes may be found, as this progression 1. 23. 815. 48105 deduceth its originall from this A × 0×2×4×6×8 1×3×5×7×9 & in which A is an infinite number =10.

It will also be convenient to find what relation all the other meanes have to the first soe that if the first bee had all the other may bee deduced thence. As in this case suppose the 1st meane to bee a . The progression will bee <17r> 12a: 1: a: 32: 4a3: 158: 8a5: 10548: 64a35: 945384: 640a315: deducing its originall from A × 0×2×4×6×8×10 1×3×5×7×9×11 & from this A(=12) × 2a × 4a × 6a × 8a × 10a 1×3×5×7×9 . &c {+} (note that the proportions of these meane termes to one another, or to (a), are found by finding the proportion of the circle y = aa xx to the line y = aa xx aa xx ¯ &c).

In this case to find the quantity a : Naming the termes in the progress: 12a: 1: a: 32: 4a3: 158: 8a5: 3516: bcdefghk. 1st observe that db=2 . ec=32 . fd=43 . ge=54 . hf=65 &c the proportions still decreasing & therefore that in cb . dc . ed . fe . gf . hg . kh &c: the latter terme is lesse than the former; & therefore dd cc { is less  yn  dc×cb=db=2. greater  yn  dc×ed=ec=32. or a= d is { less  yn  1×2=1+11 greater  yn  1×32=1+12 . Also
ff ee is { lesse  yn  fe×ed=fd=43 greater  y¯  fe×gf=ge=54 } = a: 8a9 = 2×4a3×3 .
Therefore a is { lesse  yn 3×3 2×4 43 greater  yn 3×3 2×4 54 . And So by the same reasoning. a is { less  yn 9×25×49×81×121×169 2×16×36×64×100×144×14 1413 . greater  yn 3×3×5×5×7×7×9×9×11×11×13×13 2×4×4×6×6×8×8×10×10×12×12×14 1514 . &c. Thus Wallis doth it. but it may bee done thus. a is { greater  yn   1 . less then 32 . | 4a3 is { greater then 32 less then 158 Therefore <17v> a is { greater  yn   3×3 2×4 lesse then 3×3×5 2×4×4 . 8a5 is { greater  yn   158 lesse  yn   3516 that is a is { greater then 3×3×5×5 2×4×4×6 = 3×5×5 2×4×4×2 lesse then 3×3×5×5×7 2×4×4×6×6 = 5×5×7 2×4×4×4 &c. By the same reasoning
a is { greater  yn   32×34×54×56×76×78×98×910 = 5×49×81 2×16×4×64×2 . lesse  yn   32×34×54×56×76×78×98×910×1110 = 49×81×11 2×16×4×64×4 . Or
a is { greater  y¯   11×169×225×289×369×441 2×16×4×64×4×144×4×256×4×400×2 lesse  yn   169×225×289×369×441×23 2×16×4×64×4×144×4×256×4×400×4 . Note that a is greater than 12 these two summes.

<18r> Figure

Having the signe of any angle to find the angle or to find the content of any segment of a circle

Suppose the circle to be aec its semidiameter ap=pc=1 . the given sine pq=x , viz: the signe of the angle epa . the segment sought eapq . abcp the square of its Radius. & that, qi: qk: qg: qf: qe: qd: qh: ql: qr: &c are continually proportionall. Then is eq = 1 xx . fq = 1 xx . gq = 1 xx in 1 xx ¯ . hq = 1 2xx + x4 . iq = 1 2x2 + x4 1 xx dq=1 . kq = 1 1 xx . lq = 1 1 xx . rq = 1 1 xx in 1 xx . &c & since all the ordinately applyed lines in these figures abcq , aecq , afcq , agcq , &c are geometrically proportionall their areas adqp , aeqp , afqp , agqp , ahpq &c will observe some proportion amongst one another. To find which proportion, 1st adqp = 1×xx = x . 2dly afc is a parab: therefore afqp = x x34 . also since tis qh = 1 2xx + x4 , therefore ahqp = x 23x3 + 15x5 . Also vq = 1 3xx + 3x4 x6 , therefore avqp = x x3 + 35x5 17x7 . & by the same proceeding the proportion may bee still continued after this manner <18v> x : x 13 x3 : x 23 x3 + 15 x5 : x 33 x3 + 35 x5 17 x7 : x 43 x3 + 65 x5 47 x7 + 19 x9 : x 53 x3 + 105 x5 107 x7 + 59 x9 111 x11 : x 63 x3 + 155 x5 207 x7 + 159 x9 611 x11 + 113 x3 : x 73 x3 + 215 x5 357 x7 + 359 x9 2111 x11 + 713 x13 115 x15 . &c.
And if the meane termes be inserted it will bee
x : x : x 13 x3 : x 36 x3 + : x 23 x3 + 15 x5 : x 56 x3 + 25 x5 The first letters x run in this progression 1. 1. 1. 1. 1. &c. the 2d x3 in this -13. 03. 13. 23. 33. 43. 53 &c the 3d x5 in this 6. 3. 1. 0. 0+1=1 . 1+2=3 3+3=6 . 6+4=10 . 10+5=15 . the 4th x7 this 1st . +x×1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 2d . x33×0 . 0+1=1 . 1+1=2 . 2+1=3 . 3+1=4 . 4+1=50 . 6 . 7 . 8 . 9 . 10 . 3d . +x55×0 . 0+0=0 . 0+1=1 . 1+2=3 . 3+3=6 . 6+4=10 . 15 . 21 . 28 . 36 . 45 . 4th . x77×0 . 0+0=0 . 0+0=0 . 0+1=1 . 1+3=4 . 4+6=10 . 20 . 35 . 56 . 84 . 120 . 5th . +x99×0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+1=1 . 1+4=50 . 15 . 35 . 70 . 126 . 210 . 6th . x1111×0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+1=10 . 6 . 21 . 56 . 126 . 252 . 1st. .2d. .3d .4th .5th .6th 1 . 7 . 28 . 84 . 210 . | 7th . 1 . 8 . 36 . 120 . | 8th 1 . 9 . 45 . | 9th 1 . 10 . | 10th 1 . | 11th . Now if the meane termes in these progressions can bee calculated the first of them gives the area aeqp. Which is thus done 1st +x×1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 2d . x33×0 . 12 . 1 . 32 . 2 . 52 . 3 . 72 . 4 . 92 . 5 . 112 . 6 . 3d . +15x5×0 . 18 . 0 . 38 . 1 . 158 . 3 . 358 . 6 . 638 . 10 . 998 . 15 . 4 . 17x7×0 . +116+ . 0 . 116 . 0 . 516 . 1 . 3516 . 4 . 10516 . 10 . 23116 . 20 . 5 . +19x9×0 . 5128 . 0 . 3128 . 0 . 5128 . 0 . 35128 . 1 . 315128 . 5 . 1155128 . 15 . 6 . 111x11×0 . 7256 . 0 . 3256 . 0 . 3256 . 0 . 7256 . 0 . 63256 . 1 . 693256 . 6 . 7 . 113x13×0 . 211024 . 0 . 71024 . 0 . 51024 . 0 . 71024 . 0 . 211024 . 0 . 2311024 . 1 . 30031024 . Soe that 1×x 12×13×x3 18×15×x5 116×17×x7 5128×19×x9 &c. is the area, apqe that is 00×x 00×12×13x3 00×12×14×15x5 00×12×14×36×17x7 1×3×5x9 2×4×6×8×9 &c: The progression may be deduced from hence 0× 1× -1× 3× -5× 7× -9× 11 0× 2× 4× 6× 8× 10× 12× 14 . &c <19r> Figure Soe that if the given sine bee pq=te=x . & if the Radius pc=1 . Then is the superficies ape = x x1xx 16x3 140x5 x7112 5x91152 7x112816 &c: And the area ade = x36 + x540 + x7112 + 5x91152 + 7x112816 + 21x1313312 + 11x1510240 + 429x17557056 + 715x191245184 + 2431x215505024 &c. By which meanes the angle ape is easily found for aecpa apc=90 ape ape .

The same may bee thus done.

adp = x2 . Or 2adp = x . 2afp = x+x33 . 2ahp = x+2x333x55 . And 2avp = x + x3 9x55 + 57x7 . &c. as in this order x . x + 13x3 . x + 23x3 35x5 . x + x3 95x5 + 5x77 . x + 4x33 18x55 + 20x77 79x9 . x + 53x3 305x5 + 50x77 35x99 + 9x1111 . x + 6x33 45x55 + 1007x7 105x99 + + 54x1111 11x1313 . x + 7x33 63x55 + 175x77 245x99 + 18911x11 77x1313 + 13x1515 . &c Which progression with their intermediate termes may bee thus exhibited. By which it may appeare that if pe=1. pq=x. then aep = 12x + x312 + 3x580 + 5x7224 + 35x92304 &c. And the area aep given gives the angle ape for apce apc=90d ape ape Likewise the angle ape given its sign may bee found hereby &c +911x11in0 2×adpa= 2×aep= 2afp= 2agp= 2ahp= 2aip= 2avp= +xin1 . 1 . 1 . 1 . 1 . 1 . 1 . x33in0 . 12 . 1 . 32 . 2 . 52 . 3 . 35x5in0 . 18 . 0 . 38 . 1 . 158 . 3 . +57x7in0 . 116 . 0 . 116 . 0 . 516 . 1 . 79x9in0 . 5128 . 0 . 3128 . 0 . 5128 . 0 . +911x11in0 . 7256 . 0 . 3256 . 0 . 3256 . 0 . 2×adpa= . 2×aep= . 2afp= . 2agp= . 2ahp= . 2aip= . 2avp= . Note that 1xx = x2 2x36 5x580 7x7224 45x92304 77x115632 &c that is 1xx = x2 x33 x516 x732 5x9256 7x11512 21x132048 &c. According to this progression 12 × 12 × 14 × 3×5×7×9×11×13×15×17 6×8×10×12×14×16×18×20 &c. Note also that the segment ae = x312 + 3x580 + 5x7224 + 35x92304 . &c. aep = x2 + x312 + 3x580 + 5x7224 + 35x93304 + 63x115632 + 231x1326624 + 143x1520480 + 6435x171114112 + 12155x192490368 + 46189x2111010048 .

<19v> Figure

If pq=a. qd=x. pc=1=pb. db= 1 aa 2ax xx then the areas of the lines in this progression. (supposeing also b 2ax xx _ }32 . bb 4abx 2bxx + 4ax3 + x4 + 4aa . b3 6abbx 3bbxx + 8abx3 + 3bx4 6ax5 x6 + 12aabx2 8a3x3 12aax4 + 4abx3 . &c

<20r>

To square the Hyperbola.

Figure

So if nadm is an Hyperbola. & cp=1=pa. pq=x . qd , qe , qf , qg &c =y . & 11+x = y = dq . 1 = y = eq . 1+x = y = qf . 1+2xxx = y = qg . 1+3x+3xx+x3. 1+4x+6x2+4x3+x4. &c. Their squares are. . x . x + xx2 . x + 2xx2 + x33 . x + 3xx2 + 3x33 + x44 . x + 4xx2 + 6x33 + 4x44 + x55 . x + 5xx2 + 10x33 + 10x44 + 5x55 + x66 . &c As in the following table. By whose first terme is represented the square of the Hyperbola, viz that it is -0.0 0×-=. =. =. =. =. =. =. =. =. 0 -0.0 x×a. a. a. a. 1. 1. 1. 1. 1. -0.0 x×p. p. p. p. 1. 4. 10. 20. 35. -0.0 x×q. q. q. q. 0. 1. 5. 15. 35. -0.0 x×d. e. f. g. 0. 0. 1. 6. 21. =. =. =. =. 0. 0. 1. 6. 21. -0.0 x×1. 1. 1. 1. 1. 1. 1. 1. 1. -0.0 x22×- 1. 0. 1. 2. 3. 4. 5. 6. 7. -0.0 x33×1. 0. 0. 1. 3. 6. 10. 15. 21. -0.0 x44×- 1. 0. 0. 0. 1. 4. 10. 20. 35. -0.0 x55×1. 0. 0. 0. 0. 1. 5. 15. 35. -0.0 x66×- 1. 0. 0. 0. 0. 0. 1. 6. 21. -0.0 x77×1. 0. 0. 0. 0. 0. 0. 1. 7. -0.0 0×-=. =. =. =. =. =. =. =. =. -0.0 0000. 00000. 00000. -0.0 3333. 33333. 33333 . 33333. 33333. 3 -0.0 0000. 00000. 00000 . 00000. 00000. 0 -0.0 8571. 42857. 14285 . 71428. 57142. 8 -0.0 1111. 11111. 11111 . 11111. 11111. 1 -0.0 9090. 90909. 09090 . 90909. 09090. 9 0 0 xx22+x33x44+x55x66+ +x77x88+x99x1010. &c. 0 As ifx=110. orcq=1110=1,1 . yn isdapq=1101200+13000 140000+150000016000000. &c yt0is0apqd=0,10000.00000.000000. 13x3=+0,00033.33333.333333. 15x5=+0,00000.20000.000000. 17x7=+0,00000.00142.857142. 19x9=+0,00000.00001.111111. 111x11=+0,00000.00000.009090. 113x13=+0,00000.00000.000076. 000115x15= +0,00000.00000.000000.¯ 0,10033.53477.310755. Summa000000000 . -0.0 9230. 76923. 07692 . 30769. 23076. 9230769230. -0.0 6666. 66666. 66666 . 66666. 66666. 6666666666 . -0.0 0058. 82352. 94117 . 64705. 88235. 2941176470 . -0.0 0000. 52631. 57947 . 36842. 10526. 3157947368. -0.0 0000. 00476. 19047 . 61904. 76190. 4761904761 . -0.0 0000. 00004. 34782 . 60869. 56521. 7391304347 . -0.0 0000. 00000. 04000 . 00000. 00000. 0000000000 . 0000000000. 00000000 -0.0 0000. 00000. 000 ==. =====. =====. 0000000000. 0000000000. 00000000 -0.0 0000. 00000. 000 37. 03703. 70370. 3703703703 . 0000000000. 00000000 -0.0 0000. 00000. 00000. 34482. 75862. 0689655172. 4137931034. 48275862 -0.0 0000. 00000. 00000. 00 ===. =====. = 000000000. 0000000000. 00000000 -0.0 0000. 00000. 00000. 00322. 58064. 5161290322. 5806451612. 90322580 -0.0 0000. 00000. 00000. 00003. 03030. 3030303030. 3030303030. 30303030 -0.0 0000. 00000. 00000. 00000. 02857. 1428571428. 5714285714. 28571428 -0.0 0000. 00000. 00000. 00000. 00027. 0270270270. 2702702702. 70270270 -0.0 0000. 00000. 00000. 00000. 00000. 2564010256. 4010256401. 02564010 -0.0 0000. 00000. 00000. 00000. 00000. 0024390243. 9024390243. 90243902 -=. 8063. 57265. 52007. 40736. 63159. 415063 &c = summæ -0.0 0000. 00000. 00000. 00000. 00000. 0000000000. 0000000000. 00000000 -0. 00500. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 0000 2. 50000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 0 1666. 66666. 66666. 66666. 66666. 66666. 66666. 66666. 66666. 6. -0. 00000. 000 12. 50000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 00000 10000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 00000 000 83. 33333. 33333. 33333. 33333. 33333. 33333. 33333. 3. -0. 00000. 00000 00000 71428. 57142. 85714. 28571. 42857. 14285. 71428. 5. -0. 00000. 00000 00000 00 630. 55555. 55555. 55555. 55555. 55555. 55555. 5. -0. 00000. 00000 00000 00000 0 5045. 45454. 54545. 45454. 54545. 45454. 5. -0. 00000. 00000 00000 00000 00000 41666. 66666. 66666. 66666. 66666. 6. -0. 00000. 00000 00000 00000 00000 00 384. 61538. 46153. 84615. 38461. 5. -0. 00000. 00000 00000 00000 00000 0000 3. 57142. 85714. 28571. 42857. 1. -0. 00000. 00000 00000 00000 00000 00000 0 3364. 58333. 33333. 33333. 3. -0. 00000. 00000 00000 00000 00000 00000 00000 29411. 76470. 58823. 5. -0. 00502. 51679. 26750. 72059. 17744. 28779. 27385. 30147. 14044. 12586. 0.

<20v>

cui addendum -277.77777.77777.7 00-2.63157.89421.0 000.-02523.80952.4 000.00000.-22727.3 000.00000.00-217.4 that is 000.00000.0000-2.1 -280.43459.71098.0 that is And so the summe will bee +0.10033.53477.31075.58063.57265.52007.40736.63159.41506.3 -0.00502.51679.26750.72059.17144.28779.27385.30427.57503.8 -0.09530.01798.04324.86004.40121.23228.13351.32731.84002.5 which is the quantity of the area adpq . If cpab=1. & cp= ab=10pq & qde||ap||bc=ap . In like manner if I make x= 1100=pq . The opperation followeth. +0,01000.03333.33333.33333.33333.33333.33333.33333.33333.30 +15x5+17x7= 33333.20001.42857.14285.71428.57142.85714.28571.40 19x9+111x11= ==.=====.=====.=====.=====.=0202.00 19x9+111x11= 11.11202.02020.20202.02020.20202.00 113x13+115x15= 769.23076.92307.69230.70 117x17= 6666.66666.66666.60 119x19= 58823.52941.10 121x21= 5.26315.70 47.60 +0,01000.03333.53334.76201.58821.07551.40422.38870.97309.30 -0,00005.00025.00166.66666.66666.66666.66666.66666.66666.6. 18x8110x10112x12= -1250.10000.83333.33333.33333.33333.3. 114x14= -7.14285.71428.57142.8. 116x16118x18= -62.50555.55555.5. =====.=====.=. 120x20= -5000.4. -0,00005.00025.00166.67916.76667.50007.14348.21984.17699.0. +0,00995.03308.53168.08284.82153.57544.26074.16886.79610.0.
which is the quantity of the area apqd if 100p=cp . and abcp=1

<21r>

y=db . x=ba aay=x3 . b + z =y z=bc aab+aa z =x3. b − z=y= z=bf. aabaa z =x3. z− b=y= z=bg aa z− a ab=x3. x=d+ξ ξ=ah. aayd33ddξ3dξξξ3=0 aab+aaz aabaaz aazaab } = d3+3ddξ+3dξξ+ξ3 . x=bξ . aq=ξ aay aab±aaz aazaab } = d33ddξ+3dξξξ3 . x=ξb . ak=x aay aab±aaz aazaab } = ξ33dξξ3ddξd3 .

na=ξ . nd=z . Or. ξ3a2z+bbξ aban ab . &. abnb ac then ξ3 = b3az bbcaξ . ξ=d+ζ ζ=mn d3+3d2ζ +bbcζ + 3dζζ + ζ3 b3az + bbcda

<21v> <22r>

db=x . ba=y . aax=y3 . x=b+z. z=bc. y=c+ξ. ξ=ah aab + aaz = ( x=bz.z=bf. ) aab aaz = ( x=zb.z=bg ) aaz aab = } ξ3 + 3cξ2 + 3ccξ + ξ3 . y=cξ . ξ=aq . c3 3ccξ + 3cξ2 ξ2 . y=ξc . ξ=ak . ξ3 3cξ2 + 3ξc2 c3 . na=y . d n =x . aban ab . abnb ac . & y3 = b3ax bbcay . Or d2x = εεy + y3 . ( x=z+o.z=nc ) ξ3 d2z + d2o = ξ3 ( x=zo.z=gn ) ξ3 ddz ddo = ξ3 ( x=oz.z=nf ) ξ3 ddo ddz = ξ3 ddz } εεy + y3 . ( y=n+ξ.ξ=na ) ξ2n + εεξ + n3 + 3nnξ + 3nξ2 + ξ3 . ( y=nξ.ξ=as ) εεn εεξ + n3 3nnξ + 3nξ2 ξ3 . ( y=ξn.ξ=av. ) εεξ εεn + ξ3 3nξ2 + 3ξn2 n3 . ddz al=y . dl=x . ab al ab &. albl bc . whence y3 = xb3 + yb2c a or y3 εεy = ddx &c: as before onely varying the signes at εεn & εεξ . ao=y. do=x. ab bddo . bc doob . a3bx = y3 3cxyyb + 3ccxxybb c3x3 bb .

<22v>

Dr Wallis in a letter to Sr Kenelme Digby promiseth the squareing of the Hyperbola by finding a meane proportion twixt 1 , & 56 in the progression 1, 56, 3130, 209140, 1471630, 106252772 &c.

<23r>

The resolution of cubick equations out of Dr Wallis in his dedication before Meibomius confuted

suppose x=ae. then x3 = a3 3aae 3aee e3 . or x3 = + 3aex a3 e3 . that is making a3 + e3 = q . . & 3ae = p . then x3= + px q . Againe suppose x = a e . then x3 = a3 3aae + 3aee e3 . that is making a3 e3 = q , & 3ae = p , then x3 = px q .

Then in the first of these p = 3ae . or p3e = a . or p327e3 = a3 = qe3 . Therefore e6 = qe3 p327 . & e3 = 12q 14qq p327 . & by the same reason a3 = 12q O 14qq p327 where the irrationall quantitys have. divers signes otherwise a3 + e3 = q would bee false. Soe that
x = a e = c: 12q O 14qq 127p3 c: 12q 14qq 127p3 . is a rule for resolving the equation x3 p O q = 0 , when it hath but one roote that is when it may be generated according to the supposition x = a e . &c. By the same reason x3 + px O q . may be resolved by this rule x = a e = c: 12q 14qq + 127p3 c: 12q O 14qq + 127p3 .

But here observe that Dr Wallis would Argue that since in the first of these two cases sometimes (viz when the equation hath 3 reall rootes) the rule faileth as it were impossible for the equation to have rootes when yet it hath, therefore the fault is in Algebra. & therefore when Analysis leads us to an impossibility wee ought not to conclude the thing absolutely imposible, untill wee have tryed all the ways that may bee.

But let me answer that the fault is not in the Analysis in this example, but in his opperation. for when the equation x3 + p x q = 0 , hath 3 roots hee supposeth it to have but one roote viz x = a e . but since the Equation cannot be then generated according to that supposition it is impossible it should be resolved by it.

<23v>

In like manner hee sayeth that Algebra representeth a thing possible when tis not so as in this example, in the triangle abc, make ab=1 . bc=2 ac=4. FigureThen to find dc=x, worke thus, ad = 4x . bd×bd = 116+8xx2 = 4x2 therefore 8x = 19 . or x = 198 . In which opperacon all things proceede as possible though they are not soe for ac is greater than ab+bc .

yet I answer that if the opperation & conclusion be compared together the absurdity will appeare. for in the equation bd×bd = 4xx = 436164 = 25636164 or bd×bd = -1058 . but it is impossible that a square number should be negative.

Thus x=-b is impossible. square it & tis xx=-b. Againe, & tis x4 = bb . Extract the roote & tis xx = b or x=b. which is possible. The reason of this Event is that x4bb=0 hath two possible rootes viz x=b . x=-b. & two impossible viz: x=-b. x=--b.

Thus the valors of x8 a8 = 0 are x=a , -a , -aa, --aa , 4:-a4, - 4:-a4, --a4, ---a4.

<24r>

Dr Wallis in a letter to Sr Kenelme Digby teacheth how to find the center of gravity in divers lines first when their position is as in Figure this figure.

Suppose ad the Axis, a their vertex Then saying, as 1 to the index of the line increased by an unite (vide pag 2dam) so cd to ca Then c is their center of gravity.

The Demonstration.

Let p bee the index of the series according to which the odinately aplyed lines (parallell to db) increase, then 1p+1 area of the line ∶ to nmbq . the distances of those ordinate lines from the vertex a are equall to the intercepted diameters & therefore a primanary series (whos index is 1. & since supposing a the center of the ballance the whole weight of the surface or figure is composed of its magnitude & distance from the center and therefore the index of all its moments or whole weight is p+1, viz: the aggregate of the other two. Therefore as all its moments (or the weight of the figure in its site in respect of the center a are to soe many of the greatest (or to the weight of the rectangle nmbq hung on the point d ) soe is 1, to p+2. and if apad 1p+2 , then nmbq hung on the point q shall counterballance the figure in its site &c therefore if accd p+11 , c shall be the center of gravity of those figures.

Figure

Also as the figure is now put extending infinitely towards δ if -2p + 1 - {p } +1 amac . m being the center of qnbd then c shall bee the center of gravity of the whole figure qndbδ .

Demonstration

<24v>

since the lines parallell to aδ increase in series reciprocally proportionall their index is -p & since the halfes of those lines increase in the same proportion their index is -p. whose extremitys or middle points of the whole lines (suposing a the center of the ballance) are theire centers of gravity, their distances from a being proportionall to the lines whose centers they are & consequently their index is -p & since all the moments (or whole weight of the figure) increase in a proportion compounded of the proportion of the magnitudes & distances of the lines from the center a, they will be in a duplicate proportion of the lines magnitudes that is a reciprocall series whose index is -2p. Therefore the figure is to the inscribed parallelogram as 1 to 1p. & all its moments or whole weight in this its site to the weight of the parallelogram as 1 to 12p. Therefore if, amap 12p 1 , the parallelogram hanging on the point p shall counterballanc{e} the whole figure in its site &c: whence the point c may be found easily, viz amac 12p 1p .

<26r>

Of Refractions.

1 If the ray ac bee refracted at the center Figureof the circle acdg towards d & abbegced . Then suppose abed de . See Cartes Dioptricks

2 If there be an hyperbola Figure the distance of whose foci bd are to its transverse axis hf as d to e . Then the ray acbd is refracted to the exterior focus (d). See C: Dioptr

3 Having the proportion of d to e, or. bdhf. The Hyperbola may bee thus described.

1 Upon the centers a, b Figure let the instrument adbtec bee moved in which instrument observe that ad de c et & that the beame cet is not in the same plane with adbe but intersects it at the angle tev soe that if tvev, then de ettv . Or de Radsine of ∠ tev . Also make de=q2, i.e half the transverse diamet{er.} Then place the fiduciall side of plate chm in the same plaine with ab . & moving the instrument adbect to & fro its edge cet shall cut or weare it into the shape of the desired Parabola. Or the plate chm may bee filed away untill the edge cet exactly touch it everywhere.

2 By the same proceeding Des=Cartes concave Hyperbolicall wheele may bee described by beeing turned with a chissell d tec whose edge is a streight line inclined to the axis of the mandrill by the ∠ tev which angle is found by making de ettv Radsine of etv .

3 By the same reason a wheele may be turned Hyperbolically concave the Hyperbola being convex. Or a Plate may bee turned Hyperbolically concave

<26v>

Figure Also Des=Cartes his Convex wheele B may be turned or {ground} trew a concave wheele A being made use of instead Figure of a patterne

5 In turning the concave wheele A it will perhaps bee best to weare it with a stone p & let the streight edged chissell d serve for a patterne. And it may bee convenient to grind the stone (or iron &c ) p into the fashion Figure of a cone S That may fit the hollow of the wheele A. The angle of which cone being

<27r>

9 Halving such a cone smoothly pollished within & without, by the helpe of a square set the plate perpendicular to one side hae the fiduciall edge being distant from the vertex the length of ae = edde3 dd+ee & if the edge of the plaine every where touch the cone, tis trew.

10 The exact distance (ae) of the plate from the vertex of the cone neede not bee much regarded for that changeth onely the bigness not the shape of the figure.

[By the broken lookinglasse I find in glasse refraction, that de 4328 1000651+ 1536 1000. These are almost insensibly different from truth de 2013 10065 1531000 . Or de 2315 100652+ de 6643 100651,5151+ . Or

For the Ellipsis ddee d x + eedd dd xx = yy

<27v>

The former {descriptionspropositions} demonstrated.

Lemma. If in the Opposite Hyperbolas abc Figure edf (one of which are to bee described) supposing bd=d. hf=e. gh=x gc=y. gcghd & gc terminated by the hyperbola Then is ddee ee xx + ddee e x = yy . b h =de2 . dh=d+e2 . b g=2xd+e2. gd=2x+d+e2. dc2= 4xx + 4dx + 4ex + dd + 2ed + ee + 4yy 4=gd×gd +gc×gc. bc2= 4xx 4dx + 4ex + dd 2ed + ee + 4yy 4=gb2 +gc2. And since dc=bc+hf. Or dc2 = bc2 + 2bc×hf + hf2 Therefore 2dx + ed ee = e 4xx 4dx + 4ex + dd 2ed + ee + 4yy . Both parts of which squared & ordered the result is 4ddxx 4eexx + 4eddx 4e3x 4eeyy = 0 . That is ddee e x + ddee ee xx = yy .

Description the 1st demonstrated Synthetically. See that Scheame

Nameing the quantitys ed=dh=e2. gh=x. gc=be=y. dg=x+e2=nc. cgdhg. cd2 = x2 + ex + ee4 + y ce2 = xx + ex + yy . eg2 = xx + ex . Also de ettv ceeg , therefore ddxx + ddex = eex2 + e3x + e2y That is ddee e x + ddee ee xx = yy . As in the lea

The Same demonstrated Analytically.

Nameing the quantitys, de=dh=a . gh=x . gc=y. dg=a+x dc2 = aa + 2ax + xx + yy . ce2 = 2ax + x2 + yy . eg2 = xx + ex . Supose that bc ettv ceeg .

<28r>

Then is bbxx + bbex = 2ccax + ccxx + ccyy . That is bbcccc xx + bbe2ccacc x = yy . Therefore the line chm is a Conick Section & since (bb) is greater than (cc) tis an Hyperbola, which that it may bee the same with that in the lemma, Their correspondent termes are to bee compared together & soe I find that bbcccc xx = ddeeee xx . & bbe2ccacc x = ddeee x by the 1st equation bb = ccdd ee . Or b=cde. that is bc de . by the 2nd ccee ccdd + bbee = 2ccea . And by substituting ccdd ee into the place of bb And ordering it tis ccee = 2ccea . Or e2 = a . Therefore if I take e2 = a = de . & de bc cttv . then shall chm bee the Hyperbola desired Q:E:D.

The 2d 3d 4th & 5th Propositions are manifest from this

Figure

Instead of the 6th & 7th Descriptions which are false use these

6 Draw 2 concentrick circles (na & cd) with the Radij e & d. Then from the common center b draw 2 lines bc & {babd} at the given angle {bae=abcaed=cbd} of {sectionyeCone} then draw a line cad from c by the end of the Rad {babd} & to the intersection of that line with the circle {cdna} draw {bdba} & so the angle of {yeCone,hek=cbdsection,eab=abc} is found.

<28v>

Or which is the same make ab=e. bd=d & then if that cone is sought the angle cba being given, make ac=a. Then is cd = dd ee + aa a . & soe the ∠ cbc=aed is knowne & also ae = ed = d3dee dd ee + aa , & ad= dd ee a . But if the ∠ bae=abc of the section is sought the cone being given than make cd=2b. And it will bee ac=b+eedd+bb . & soe ∠ abc=bae is given also ad=beedd+bb . & ae = dbdeedd+bb 2b

In generall observe that in any cone cut any ways bd = be+ea = d . & ba=e.

7. DesCartes his wheele thus described cut by any plaine produceth one of the Conick=Sections.

Description the 6th Demonstrated. Synthetically.

Call, bd=d. ba=e. cp=pd=a. bp=ddaa. ag=x ap=eedd+aa. ac=a+eedd+aa. ad=aeedd+aa. baac aggh= ax + xeedd+aa e . baad bggk= ea+ax e exe eedd+aa . gk×gh=gm2=y2 . Therefore ddee ee xx + ddee e x = yy . by ordering the result of gk×gh. which is like that in the lemma.

The 7th Proposition may be easyly demonstrated after the same manner.

If the two equall cones bad bcd intersect the Figure one the other soe that ab=bc their intersection (bf) shall bee one of the Conick sections as they had each beene intersected by the plane bf .

<29r>

To describe the Parabola (& other figures after the same manner) pretty exactly.

Figure

Take a squire cbe , soe that cb=r2 (for then the circle described by (bc) will bee as crooked as the Parabola at the vertex d ). Divide the other leg (be) of the Squire into any number of points, Then get a plate of Brasse &c: lkfd streight & eaven. And taking one point d for the vertex of it & another point c for the Squire to moven soe that cd= cb=r2 , & weareing away the edge of the plate untill (the Squire being erected) ab= qd. the squire touching the plate at a . thus shall the edge adf become Parabolicall. the Rad: ab describe a circle it may bee knowne when ab=a. Instead of the leg be a circle may be used Demonstracon. qd=x. cd=r2=cb. cq=r2x. aq=rx=y. ac2=rr4+xx ab2=rr4+xxrr4 & ab=x. Q.E.D.

Another description of the Parabola with the compasses. Make ab= bc=r4. Make ce=cd Figure & cebd. Make af=ae, & bf=bd then shall f be a point in the Parabola.

Another. Make ab = r+x2 = ac . eb=xce & the point c shall bee in the parabola. This like the first by calculation may bee made use of in other lines.

<29v>

The manner whereby any kind of little lines may be described very accurately. And that the same Instrument serve for all lines (though never so small) differing in quantity but not in quality.

Figure Make the plate d of the figure required (by some of the former meanes) the larger the better. Then hold the streight steele staffe b against the center a & {roule}{route} it to & fro it shall grind c into the same figure but soe much lesse as ac is lesse than ad.

Figure Soe if the glass c bee fastened upon the mandrill f, it may be ground acording to the sollid figure d by the helpe of a stick of steele (as a cone) whose cuspis is in the hole a upon which it is moved as on a center. when the cone b leanes uppon the vertices of d & c it must be perpendicular to the mandrill f. Perhaps it may be convenient to cause the cone b to turne about its axis. Or it may bee better instead of the nutt at a with a hole in it to make a sharpe pointed nutt, & instead of the cone b to make use of a broad plate to cover a, c & d & move every way upon them

<30r>

Another way to describe lines on plates

Figure

Suppose the plate bee abc, whose edge boc is to be made into the fashion of a given crooked line suppose (o) is its vertex & that a circle described with the Radius eo would bee as crooked as the given line at its vertex. Againe suppose two streight rulers mn & pq to bee very trew & steddyly fastened together Figure which must a very little incline the one to the other, soe as that being produced they would meete at ar. Then are the lines pn=a , & pr=b given.

Suppose then the point d in the crooked line is to bee found then is dc given by supposition, & consequently (supposing dk to bee a tangent) dg=y. gc=x. fg=v. fd=s ec=c. fk=v+yyv . ef=v+xc. ek=cx+yyv. & (if ehdkdf) then is eh= cv xv + yy vv + yy = d . (eh) being thus found, supposing that pn=a=ec , then I take re=bda. that is pe=bbda . haveing thus found the point e lay the plate twixt the two rulers so that the point of it, fall upon the point e then should the line mn touch the plate in d. But note that pnmn.

In both telescopes & microscopes tis most convenient to have a convex glasse next the eye for by that meanes the angle of vision will bee much greater than it will bee with a concave one (though both doe magnifie alike). If the convex glasse be Hyperbolicall (&c) make it soe bigg that the penecilli may crosse in the pupill; that is, the exterior focus will be as far distant from the vertex as the eye is. let the glass bee as thinn as may bee that the eye bee not too far from the vertex that it should bee about as thick as the distance of the interior focus from the vertex.

And by this meanes also, (the focus of the objectglasse being within the telescope twixt the glasses) there may bee placed at that focus the edge of <30v> a steele ruler accurately divided into equall parts (to measure the diameters or distances of starrs &c) which should bee soe made that by a pinne or handle it may be placed in any posture & in any parte of the focus, without otherwise altering the Telescope in observations.

Note that were not the glasses faulty they would not onely magnify objects but render vision more distinct; each of the penicilli passing through (perhaps but) the 10th, 20th or 100th parte of the pupill must bee more exactly refracted to one point of the Tunica Retina than in ordinary vision in which each of the penicilli spreads over all the pupill.

Note also that Figurethat the glasse a may be ground Hyperbolicall by the line cb, if it turne on the mandrill e whilst cb turnes on the axis rd being inclined to it as was shewed before. If the edge (cb) bee not durable enough, inough instead thereof use a long small cilinder: which I conceive to bee the best way, of all. For a Cilinder of all sollids is most easily made exact (being Figure turned, as in the figure, by a gage untill its thicknesse bee every where equall). 2 the Cilinder may bee made to slip up & downe & turne round whereby it will not onely grinde the glase crosse wise to take of all hubbes, but also the glasse & cilinder will grinde the one the other truer & truer. All the difficulty is in placing the axis rd perpendicular to the Mandrill ae & vertex to vertex, which yet may bee done exactly severall ways. & untill then the glasse & Cilinder will not fit. & should the axis not intersect the glasse would bee still Hyperbolicall except a point at the vertex of it. The same instrument may also serve for severall glasses onely making df longer or shorter. Let the Cilinder han{g} over the glasse.

<31r>

To Grinde Sphæricall optick Glasses

If the glasse (bc) is to bee ground sphærically Figure hollow: naile a steele plate to the beame (fg), on the upper side: In which make a center hole for the steele point (f) of the shaft (def): to which shaft fasten a plugg (a) of stone or leade or leather &c: (with which you intend to grinde the glasse (bc)): which shaft & plugg being swung to & fro upon the center f will grind the glasse bc sphærically hollow.

The manner whereby Figure glasses may bee ground sphærically convex may appeare by the annexed figure (being the former way inverted). Also the plugg (a), in the firstsecond figure, is ground sphærically convex.concave.

But if this way bee not exact enough yet hereby may bee {grownd}{ground} plates of mettall well nigh sphæricall, And by those plates may bee ground glasses after the usual manner; If a circular hoope of steele (abc) bee put about the edge of the glasse (d) to keepe it Figure from grinding away at the edges faster than in the middle.

But the best way of all will bee to turne the glass circularly upon a mandrill whilest the plate is steadily rubbed upon it or else <31v> to turne the plate upon a mandrill whilest the glasse is rubbed upon it or let sometimes the one, sometimes the other bee turned.: & by this meanes they will either of them weare the other to a truely sphericall forme. but however let there bee a hoope or of some mettall which weares more difficultly then glasse to defend the glasse from wearing more at its edges then in the middle. Perhaps it may doe well first to weare the plate sphæricall by the hoope alone without the glasse.

The same meanes may bee used for grinding plaine glasses.

Let not an object glasse bee ground sphærically convex on both sides, but sphaerically convex on one side & plane or but a little convex ~ on the other, & turne the convexest side towards the object.

<32r>

<32v> Figure

If the Glasses of a Telescope bee not truely ground Theire errors may bee thus found.

Because an error is much more easily discernable in the object glasse than in the eye glasse let us first suppose the eye glasse to bee ground true towards its center, (tis exact enough if it be sphericall, & not Hyperbolicall), & so wee may find & rectifie the errors of the object glasse.

First make a thin plate (A) of brasse & in the center of it a Small hole (whose diameter perhaps may bee about the 50th or 100dth parte of an inch. With which plate cover the eye glass the center of it respecting the center of the glasse.

Secondly make two other plates the one B with two holes as neare to its edge as may bee their{e}{} distance being about the 5th parte of an inch or lesse, & the other C with one hole close to the midst of its edge. Let the diameters of these 3 holes bee about a 20th parte of an inch or lesse. And theire edges must bee true that they may slide one upon another, & yet not let the suns rays passe through, to which purpose make them oblique. with these two plates cover the object glasse (first stopping the hole of C the holes of the other plate respecting the center of the glasse & looke at a stare (or the edge of the sunne &c) & if the object appeare double (like two starrs &c) make the Tube longer or shorter untill it appeare single. Then open the hole of C , & the plate B being fixed, slide the plate C up & downe still looking at the starre, When then appeares <33r> but one starre that part of the glasse under the hole of C is truely ground in respect of the 2 parts of the glasse under the two holes of B. But {no} when the starre appeares double. And the position of the starre caused by the hole of C in respect of the starre caused by the holes of B, shews which way the glasse under the hole of C is erroneously inclined; the distance of the two starres giving the quantity of that error.

Thus the errors of the object glasse bein{g} found in every place of it they may bee all rectified, & found againe, & againe rectified, untill they almost or altogether vanish.

Then may the eye=glasse bee rectified much after the same manner, in every parte of it, & if it bee necessary the object glasse may bee againe rectified & againe the eye=glasse untill the Telescope bee as perfect as the workeman can make: Whome perhaps experience may teach by this & the former rules to make telescopes as perfect as men can hope to make them.

These glasses may also bee rectified whilst on the Mandrill by observing the images made by reflection from the vertex & all other parts of the glasse what proportion they have one to another & how much they are longer than broader in one place then another. &c.

<33v>

BLANKThe sines measuring refractions are in Aere42 water56 Glasse65 christall70 The proportions of the motions of theextreamely heterogeneous rays are in} 39,4.40,4. 7038.7138. 095110.096110 11013.11113 The proportions ofyesines of refraction of the extreamely hetero-00 geneous rays into aire out of Their common sine of incidence00 Which substracted the difference is00 000Ty000 c00 000gf000 00 00 000Ty000 9023.9123. 000Ty000 6813 2213.2313. 000Ty000 068.0006900 000Ty000 4414 02334.02434 000Ty000 06145.06245. 000Ty000 3645 024.00025.00 The like proportions for refrac-tions made into water out of 27545.27645 23825 03725.03825 19613.19713 15749 03919.04019

of the method of infinite series

I Newton

<35r>

Theoremata varia. Circa angulorum æqualitates.

Figure

si ang DAB & DAE bisecentur a rectis FH et IG et ducatur quævis KLMN . Erit
1. AK.AM KL.LM KN.MN Euclid 6 3
2. AK×AM = ALq + KL×LM = KL×LM ALq . Scho{o}ten de {concis} {æqu{is}}
3. AM+AK . PK AQ.AL posito AP=AM .

Figure Si in angulo quovis PAQ inseribantur æquales AB, BC, CD, DE, EF, FG, GH &c anguli BA P erit angulus CBQ duplus DCP tripl, ED Q quadr FEQ quint, GFQ sext, HGP sept. IHQ oct &c. Horum vero angulorum posito radio AB sinus erunt Bβ , Cχ &c cosinus AB , Bχ , Cδ &c. Ergo si AB=r, & AB=x erit AC=2x =2xxr. . AD = (2AB) = 4xxrrr . = 4x3rrxrr &c

<163v>

In Generall.

mm+8mn+15nn¯ × ddx2n1 mm2mn¯ eex1 in dxm+n+exm = y . 1=v . &, 2m+6n¯ × ddx2n + 2ndexn 2m4n¯ × ee in dxm+n+exm = z.

<40v>

VII. 14

{illeg} Geomet. {illeg}. Frn Vieta Schooten {illeg} July 4th 1699 {illeg}

<41r>

To find the sume of the squares cu{bes} &c. of the rootes of an equation

If a , bg , c , d , e , f &c be the rootes of the equation x6 + px5 + qx4 + rx3 + sxx + tx + v = 0 . then is a + b + c + d + e + f = p(=g) a2 + b2 + c2 + d2 + e2 + f2 = pp2q . (= pg 2q =h) a3 + b3 + c3 + d3 + e3 + f3 = p3 3pq + 3r . (= ph qg + 3r = k ) a4 + b4 &c = p4 4ppq + 4pr + 2qq 4s . (= pk qh + rg 4s = l ) a5 &c = p5 5p3q + 5pqq + 5ppr 5ps 5qr + 5t . (= pl qk + rh sg + 5t = { m) } a6 &c: = p6 6p4q + 9ppqq + 6p3r 12pqr 6pps + 6pt 2q3 + 3rr + 6qs 6v .

<80r> Figure

ag=a . ab=x . bh=dxc . bc=y . bg=xxaa gh = ddxxeexx+eeaaee . dg=b. ce = ydx ddxxeexx+eeaa = fg . cf=dx+eydxxxaa . df = b ydx ddxxeexx+eeaa . z2=dc2={ xxaa+2eyxd2eyaadx+bb+yy 2bydx ddxxeexx+eeaa 2bydx ddxxeexx+eeaa = xx+yyzz+bbaa+2eyxx2eyaadx . ddx6 + 4deyx5 + 2ddyyx4 4deaayx3 4bbddyyxx 4dey3aax + 4e2a4y2 2ddzzx4 + 4dey3 + 4bbeeyy + 4deyz2a2 + 2ddbb 4deyzz 8aaeeyy 4deybba2 2ddaa + 4deybb + 2ddy4 + 4deya4 4bbe2a2y2 + 4eeyy 4deyaa

Ad constructionem Canonis angularis.

905gr = 18gr . 185gr = 3gr + 36 . Et 603gr = 20gr . 203gr = 6gr + 40 6gr + 40 2 = 3gr + 20 . 3gr + 36 3gr 20 = 16 . 162=8 . 82=4 . 42=2 22=1 .

If r= radius. Then
yesine of 0 0 0 78degr is, r5r+r30+65 8 . 66degr is, r5r+r3065 8 . 42degr is, 5rr+r+30rr+6rr5 8 . 06degr is, 30rr6rr55rrr 8 .

<80v>

Suppose gh=x . nh=r . Then gh=x. unisectio ab×r=2rrxx. bisectio hb2×r2=3rrxx3. trisectio ab3×r3=2r44rrxx+x4.quadriseco. hb4×r4=5r4x5rrx3+x5.quintusecto. ab5×r5=2r69r4x2+6rrx4x6. hb×r6=7r6x14r4x3+7rrx5x7. ab×r7=2r816r6x2+20r4x48rrx6+x8. hb×r8=9r8x30r6x3+27r4x59rrx7+x9. ab×r9=2r1025r8x2+50r6x435r4x6+10rrx8x10. hb×r10=11r10x55r8x3+77r6x544r4x7+11rrx9x11.

< insertion from the top right of f 80v >

As on the other leafe excepting some signes here changed.

< text from f 80v resumes >

If gh=x . bh=y .Then y=bh. duplicatio angulihag yyxx=x×hb2. triplicatio angulihag. y32xxy=xx×hb3. quadruplicatio. y43xxyy+x4=x3×b4h. quinto y54xxy3+3x4y=x4×hb5. sexto y65xxy4+6x4yyx6=x5×hb. septo y76xxy5+10x4y34x6y=x6×hb. octo y87xxy6+15x4y410x6yyx8=x7×hb. nonc y98xxy7+21x4y520x6y3+5x8y=x8×hb. dec y109xxy8+28x4y635x6y4+15x8y2x10=x9×hb. und, y1110xxy9+36x4y756x6y5+35x8y36x10y=x10×hb. duod

<81r> Figure

Of Angular sections

Suppose ab=q . ah2=r . & ag=x. & that the arches hg, gb, bb are equall. By the following Equations an angle bah may bee divided into any number of partes. x=q. unisectio. x22rr=rq. bisectio. x33rrx=rrq. trisectio. x44rrxx+2r4=r3q. quadrisectio. x55rrx3+5r4x=r4q. quintusectio. x66rrx4+9r4xx2r6=r5q. sextusectio. x77rrx5+14r4x37r6x=r6q. septusectio. x88rrx6+20r4x416r6x2+2r8=r7q. x99rrx7+27r4x530r6x3+9r8x=r8q. x1010rrx8+35r4x650r6x4+25r8x22r10=r9q. x1111rrx9+44r4x777r6x5+55r8x311r10x=r10q. x1212rrx10+54r4x8112r6x6+105r8x436r10x2+2r12=r11q. x1313rrx11+65r4x9156r6x7+182r8x591r10x3+13r12x=r12q. x1414rrx12+77r4x10210r6x8+294r8x6196r10x4+49r12x2&c x1515rrx13+90r4x11275r6x9+450r8x7318r10x5+140r12x3&c x1616rrx14+104r4x12352r6x10+660r8x8672r10x6+336r12x4&c x1717rrx15+119r4x13442r6x11+935r8x91122r10x7+714r12x5&c x1818rrx16+135r4x14546r6x12+1287r8x101782r10x8+1386r12x6&c x1919rrx17+152r4x15665r6x13+1729r8x112717r10x9+2508r12x7&c x2020rrx18+170r4x16800r6x14+2275r8x124604r10x10+4290r12x8&c

This scheame is the former inversed. Figure

<81v>

Suppose the perifery bgh to bee a & the whole perifery to bee p. The line bh subtends these arches. a. pa. p+a. 2pa. 2p+a. 3pa. 3p+a. 4pa. 4p+a. 5pa. 5p+a. 6pa. 6p+a. &c: All which are bisected, trisected, quadrisected, quintusected &c after same manner. As for example

The rootes of the equation hb2 × rr = 3rrx x3 . are 3. The first whereof subtends the arches a3 . 3pa3. 3p+a3. 6pa3. 6p+a3. 9pa3. 9p+a3 &c. The second subtends the arches pa3. 2p+a3. 4pa3. 5p+a3. 7pa3. &c. The 3d p+a3. 2pa3. 4p+a3. 5pa3. 7p+a3 &c.

Soe the rootes of the equation ~ hb×r4=5r4x5rrx3+x5 , doe the first subtend the arches a5 . 5pa5. 5p+a5 &c: the 2d pa5. 4p+a5. 6pa5. the 3d p+a5. 4pa5. 6p+a5 &c. the 4th 2pa5. 3p+a5. &c the 5t 2p+a5. 3pa5. 7p+a5. &c.

Hence may appeare the reason of the number of rootes in these equations & that the points of the circumference to which they are extended æquidistant. & by the lower scheme may bee known which rootes are affirmative & which negative.

The numerall cöefficients of the afforesaid equations may bee deduced from this progression (if 1n .) 1 × 0+n × 1+n 1 × 1n × n2 × n3 2 × 2n × n4 × n5 3 × 3n × n6 × n7 4 × 4n × n8 × n9 5 × 5n × n10 × n11 6 × 6n &c. As if n=10. the progression 1 × 10 × 72 × 107 × 12 × 225 × 0 . And the coefficients 1 . 10 . +35 . 50 . +25 . 2 .

<82r>

1663 /4 January.

All the parallell lines which can be understoode to bee drawne uppon any superficies are equivalent to it, as Figure all the lines drawne from (ao) to (co) may be used instead of the superficies (aco.)

If all the parallell lines drawne uppon any superficies be multiplied by another line they produce a Sollid like that which results from the superficies drawne into the same line Figure as if either all the lines in the superficies (oac) or if the superficies oac be drawne into the line (b) they both produce the same sollid (d) whence All the parallell superficies which can bee understoode to bee in any sollid are equivalent to that Sollid. And If all the lines in any triangle, which are parallell to one of the sides, be squared there results a Pyramid. if those in a square, there results a cube. If those in a crookelined figure there results a sollid with 4 sides terminated & bended according to the fashion of the crookelined figure{.}

If each line in one superficies bee drawne into each correspondent line in another superficies as in aebk, & omnc Figure if ae×dh. bk×cn. qv×wx. &c. they produce a sollid whose opposite sides are fashioned by one of the superfic as Sollid fpsrg. where all the lines drawne from fr to ps are equall to all the correspondent lines drawne from ow to mx. & those drawne from fg to fr are equall to the correspondent lines drawne from qz to vz.

<82v>

Theorema. 1

If in the Circle abcdeP there be Figure inscribed any Poligon abcde with an odd number of sides, & from any point in the circumference P there bee drawne lines Pe, Pa, Pb, Pc, Pd to every corner of the Polygon: the summ of every other line is equall to the summ of the rest, Pa+Pb+Pc =Pd+Pe. & soe are their cubes Pa3+Pb3+Pc3=Pd3+Pe3 . unless the figure be a Trigon

Theorema 2

If from the points of the Polygon Figure then bee drawne perpendicular ap, br, ct, ds, eq to any Diameter pt: the summe of the Perpendiculars on one side the Diameter is {equall}{equal} to their summe on the other ap+br+ct=eq+ds . & soe is the summe of their cubes (unlesse when the figure is a Trigon), ap3+br3+ct3= eq3+ds3 . & of theire square cubes (except when the figure is a Trigon or Pentagon. &c.

Theorema 3

If the 2 circles (fig 1 & 2) be equall with like Poligo{illeg}{ns} inscribed, & Pa in fig 1 be assumed double to pa in fig 2. then are all the other corresponding lines in fig 1 double to those in fig 2 viz Pb=2rb, Pc=2tc, Pd=2sd, Pe=2qe.

<83r>

To square the Parabola

In the Parabola cae suppose the Figure Parameter ab=r. ad=y. dc=x. & ry=xx or xxr=y. Now suppose the lines called x doe increase in arithmeticall proportion all the x's taken together make the superficies dch which is halfe a square let every line drawne from cd to hd be square & they produce a Pyramid equall to every xx=x33. which if divided by r there remaines x33r= =yx3 equall to every xxr equall to every (y) or all the lines drawne from ag to accc equall to the superficies ag c equall to a 3d parte of the superficies adcg & the superficie acd=2yx3.

Otherwise. suppose ce=b. co=x. to=y. & ry = bxxx the lines x increasing in arithmeticall proportion every x is equall to 4 times the superficies cdh=bb2 which drawne into b produceth the sollid b32 but if every x be squared they produce a pyramid equall to b33. wherefore every bxxx=b36 equall to every ry equall to the superficies adce drawne into r & b36r= to cade as before.

<83v> Figure

* = pd = q+y5 & qy+yy5 = xx

<84r>

To Square the Hyperbola

In the Hyperbola eqaw. suppose ef=a. fa=b. ap=rq=y {=d} pq=ar=x. ad=q=5oa=5ac=5 r=. & da+ar ar arrq . xx = dy+ yx . In which equation Every x taken together is equall to the triangle aβb equall to aa2 & every xx taken together is a pyramid = a33 . Every y taken together is equall to the superficies eba=mkt If then gh=lm=ns==d . every dy is equall to the solid nglmhs. If the angle mhk is a right one & if mh=gl=ba=ef=a that is if the triangle mhk=abβ. every y x will be equall to the sollid mhstk Joyne these two sollids together as in lmtng=a33 . *

⊛ Againe Suppose every x taken together to be equall to the superficies aef , the line squared is 4xx every 4xx composeth a Sollid like (♊♉♌) an eighth parte whereof (which is equall to every xx2) being like ♊♉♓o= xyzv; xv will be equall to ♊ ♓ =ef =a=st=xz=o♓=hm. & vz=♊♓ =km . whence the convexe superficies xyv of the figure xyvz will fitly joyne with the concave superficies mst of the figure shmkt . If every x is equall to the superficies aef , every y shall be equall to the triangle afπ=bb2. every yy=bbb3 every qy=qbb2 & therefor the Sollid yxzv=b330+qbb20=2b3+3qbb60. Joyne the Sollid shmkt to yxvz & there resulteth shmztk= =aab2 from which againe substract xvzy=2b3+3qbb60 & there remaines the sollid mhstk=30aab2b33qbb60 which substract from the sollid ntmlg=a33 & there remaines nglmhs=20a3+2b3+3qbb30aab60 which being divided by θ=5rr4. there remaines 40a3+4b3+6qbb60aabr5= to the superficies abe

<85r>

The squareing of severall crooked lines of the Seacond kind.

Figure In any two crooked lines I call the Parameter or right side of the greater. (r). but of the lesse (s) Transverse side (q). the right axis as cf (x) or ef=v . y Transverse axis as fe y, or fd z.

Suppose in the Parab: ddc: ac=r. & in eec: bc=s rx=zz=df2. sx=yy=fe2. rxsx=de= =p. rx=sx+pp+2psx. rxsxpp=2psx rrxx2rsxx+ssxx2pprx2ppsx+p4=4ppsx. Or p42rxpp6sxpp+rrxx2rsxx+ssxx=0. if p=y. xx=+2ryy+6syyxy4rr2rs+ss. make cf=a. fd=b. fe=c. ceeff=2ac3. & cddeff=2ab3 therefore 2ab2ac3=cddee the square of the crooked line cdd (when the line cee is supposed too close with the line cf ) whose nature is exprest by the foregoing Equation.

Figure 2 lk=b. li=x. qi=y. in=z. +bxxx =ry bx xx =sz. xx+bxr=y. xx+bxs=z qn=sxxs+bsx+rxxrbrxrs=v=y or, rrxsxx+bsxbrxrsy=0. Or xxbxrsyrs=0

Figure 3 tg=x. dg=z. gp=y. rxrz=dp2. rxrz+zz=y2. zz=rzrx+yy

<86r>

ra rxrz zz . zz = arxarz . zz = az + ax . z = 12a + 14aa+ax . axaz = rz rx + yy Or yy + rx + ax = 12aa 12ar + ar14aa+ax . 18631 35459 944

y4 2rx yy + rrxx + 14a4 + 2aarx 2ax + 2arxx + 12a3 + arrx aa + aaxx + 14aarr + a3x ar = 14a4+a3x+14rraa+rrax 0 y4 2rx yy + rrxx = 0 2ax + 2arxx aa + aaxx ar + 12a3r + 2aarx . . xx + 2aar x + y4 2ryy aayy 2ayy aryy + 12a3r rr+2ar+aa =0 .

17548875.(5849625 51)358(7,019608. 00)357¯(0,000-58. 00)00100¯(0,0-00. 00)000490(0,-00. 00)000459¯(0,-00. 00)0000310(,-00. 00)0000306¯(,-00. 00)000000400(,-. 0000000 51)197(3,862745 00)153¯(0,00+392 000)440(0,0+000 000)408¯(0,0+000 0000)320(0,0+00 0000)306¯(0,0+00 00000)140(0,0+0 00000)102¯(0,0+0 000000)380(0,0+ 000000)357¯(0,0+ 0000000)230(0,+ 0000000)204¯(0,+ 00000000)26(0,+ 00)93¯(0000000. 53)372(7018868. 00)371¯(000+682. 0000)100¯(000+0. 00000)470(00+0.00000)424¯(00+0. 000000)460(00+.000000)424¯(00+. 0000000)360(0+. 0000000)318¯(0+. 0000000)420(+. 000 212)819¯(3863207 00)636(0000070 00)1830(000000 00)1696¯(000000 000)1340(00000 000)1272¯(00000 000000)680(0000 000000)636(0000 0000000)44(0000 0000000)424¯(000 00000000)160(00

<86v>

0·0·9·7·5·4·5·5·5·9·7·7·7·0·0·3·0·0· 0·8·9·0·0·0·2·0·0·2·0·0·

<87r>

4 In the Parabola Figure cb=a . be=x . 2aa 2ax aa + 2ax xx = ed2 aa xx = ed2 = yy . aa xx r=y . cp=cb eb×df = fg { ×c } = zc. aax x3 r + xx =zc x3 rxx aax + rcz = 0 . Since all eb×df = 18 all co2 = 14ab×ab×r . ab=b. all eb2 = a33 therefore bgpf 14bbr + a33 .

<87v> Figure <88r>

ab= b e=y. bd=x. bq=dg=b. nb=c. ybxyc = bxc = ef Then shall bq&c: be the axis of gravity in feb&c & bqgd.

<88v> Figure <89r>

In the 1st figure.

gccd cfedckhg = cd×cfed gc . ac=gc. xz zaxy . zza x = ckhg . or xxy z = cdef . Suppose cdca acbc the swiftnesse of de ∶ to the swiftnesse of gh . de×its swiftnesgh×its swiftness gccd . de×cd gh×ca de×ac gh×bc gc×cd .

Fig 2d. 3d.

c θ ca acbc nmam swiftnessθeswiftnessegh . de× its swiftnesse gh× its swiftness ∷ c k×degh × ac de×ac gh×bc de× k gh×ac de×ac gh×bc de×nm gh×am gccd . de×ck×cd = gh×ac×gc . de×cd = gh×bc . de×nm = gh×gc .

Fig 4

ckca motion of the point a from c ∶ motion of the point a from m ∷ ckca increasing of ac=gc∶increasing of cd ∷motion of gh ∶ motion of de. &c as before.

These are to find such figures cghk, cfed, as doe equiponderate in respect of the axis acfk.

<89v> Figure
<90r>

Reasonings concerning chance.

If

1 If p is the number of chances by one of which I may gaine a, & q those by one of which I may gaine b, & r those by one of which I may gaine c; soe that those chances are all equall & one of them must necessarily happen: My hopes or chance is worth pa+qb+rcp+q+r=A. The same is true if p, q, r signify any proportion of chances for a, b, c.

2. If I bargaine for more than one chance (viz: that after I have taken the gaines by my first chance, from the stake a+b+c; I will venter another chance at the remaining stake &c) my second lott is worth A AAa+b+c = A AAa+b+c = B . My third lot is worth A AA AB a+b+c = C . My Fourth lot is worth A AA AB AC a+b+c = D . My Fift lot is worth A AA AB AC AD a+b+c = E . My sixt lot is worth A A × A+B+C+D+E a+b+c . &c

As if 6 men ( 1. 2. 3. 4. 5. 6. ) cast a die soe that he gaines a who throws a cise first: since there is but one chance to gaine a & 5 to gaine nothing at each cast, I make b=0=c=r. p=1 & q=5. Therefore by the <90v> The first mans lot is worth a6 The seconds is worth a6 a36 = 5a36 . The thirds is worth 5a36 5a216 = 25a216 . The fourths is 25a216 25a1296 = 125a1296 The fifts lot is worth 125a1296 25a7776 = 625a7776 . The Sixts lot is 625a7776 625a46656 = 3125a46656 . &c. Soe that their lots are as 7776: 6480: 5400: 4 5 00 : 3 950:3 1 25 .

Soe that if I cast a die two or more times tis 1. to 5 that I cast a cise at the first cast & 11 to 25 that I throw it at two casts, & 91 to 125 that I cast it at thrice, & 671 to 625 that I cast it once in 4 trialls, & 4651 to 3125 that I cast it once in 5 times. &c

3. If I bargaine to cast severall sorts of lots successively at the same stake the valor of each lot is thus found viz: The first prop: gives the valor of the first lot; which valor being destructed from the stake, the remainder is the stake of the 2d lot which therefore may bee also found by the first prop: &c.

As if I gaine a by throwing 12 at the first cast, or 11 at the 2d or 10 at the 3d &c with two dice. Since at the first cast there is but one chance for a (viz 12 ) & 35 for nothing Therefore its valor is a36 (by Prop 1). & the stake for the 2d cast is a a36 = 35a36 . Now since there are two chances for it (viz: ⚅⚄ & ⚄⚅) & 34 for 0 at the 2d cast therefore its valor is 2×35a 36×36 = 35a 648 . as the stake for the 3d lot is 595a 648 for which there are 3 chances (viz ⚄⚄, ⚅⚃, ⚃⚅) & 33 for nothing Therefore its valor is 595a7776.

<91r>

4 If I bargaine with one or two more to cast lots in order untill one of us by an assigned lott shall win the stake a: Since the chances may succede infinitly I onely consider the first revolution of them The valor of each mans whole expectation being in such proportion one to another as the valors of their lots in one revolution. & the valors of each mans first lot being to the valor of his whole expectation as the summe of the valors of their first lots to the stake a.

As if I contend with another that who first throws 12 with 2 dice shall have a, I haveing the dice. My first lot is worth a36 (by prop 1), The 2d his first lot is worth 35a36×36. And a3635a36×36 3635 my expectationto his. for the two first lots make one revolution because I have the same lot If I throw a 2d time that I had at the first. Therefore ( 36+35=71 a 3636a71 ) 36a71 is my interest in the stake.

If our bargaine bee soe that there is some lott at the beginning of our play which returnes not in the after revolutions, detract the valor of those irregular lotts from the stake & the rest shall bee the stake of the lots which follow & revolve successively. As if I contend with another that who first casts 11 must have a , onely I have {the} first cast for 12. My first lot is worth a36. & the stake for our after throws is 35a36. his firts lot being 35a648. & my next lot 595a11664. soe that his share in the stake 35a36 is to mine as 35a648 595a11664 1817 . Soe that my share in it is 17a36. To which adding the valor of my first lot viz: a36, the summe is 18a36 = a2 , my interest in the stake a at the begining.

5 If the Proportion of the chances for any stake bee irrationall the interest in the stake may bee found after the same manner. As if the Radij ab , ac , divide the horizontall circle bcd into two points <91v> abec & abdc in such proportion as 2 to Figure 5 . And if a ball falling perpendicularly upon the center a doth tumble into the portion abec I winn (a): but if into the other portion, I win b . my hopes is worth 2a+b5 2+5 .

Soe if a die bee not a Regular body but a Parallelipipedon or otherwise unequall sided, it may bee found how much one cast is more easily gotten then another.

6 Soe that the facility of the chances & the stake belonging to each chance being knowne the worth of the lott may bee ever found by the precedent precepts. And if they bee not both immediatly known they must bee sought before the valor of the lott can bee found.

As if I want two games at Irish & my adversary three to win a , & I would know my interest in the stake (a.) my first lot can gaine me nothing but the advantage of another lot, & therefore to know its vallue I must first find the value of that other lot &c. First therefore if wee each wanted one lot to win a our interest in it would bee equall viz my lot worth a2. Secondly If I want one game & my adversary two, & I gaine the next game then I gaine a but if I loose it I onely gaine an equall lot for a at the next game which is worth 12a, Therefore my interest in the stake is a+12a 2 = 3a 4 . Thirdly If I want one game & my adversary three & I gaine the next game I get a; but if I loose it, then I want one game & my adversary but two, that is I get 3a4: Therefore (there being one chance for a & one for 3a4) my interest in the stake is a+3a4 2 = 7a 8 . Fourthly If I want 2 games & my adversary 3; & I win I get 7a8. but if I loose I get 12a for our chances <92r> will then bee equall; Therefore my interest in the stake is 11a16. Soe if I want 1 games & my adversary 4 my interest in a is 15a16. If I want two and hee 4 , it is 13a16. If I want 3 and hee 4 it is 21a32. If I 1 and hee 5 , it is: 31a32. If I 2 and hee 5 it is 57a64. If I 3 and hee 5 it is 99128a. If I 4 and hee 5 , it is: 163256a. (The like may bee done if 3 or more play together. (as if one wants one game, another 3 a third 4: Their lots are as 616:82:31 . &c.) As also if their lots bee of divers sorts.)

By this meanes also some of the precedent questions may bee resolved. as if I have two throws for a cise to win a , with one die; If I have missed my first lot already, I have at my second cast five chances for nothing. & one for a . therefore that cast is worth a6. Soe that in my first cast I had five chances for 16a & one for a , which therefore (with my 2d cast) is worth 1136a. That is tis 11 to 25 that I cast a cise once in two throws. as before

By this meanes also my lot may bee known if I am to draw 4 cards of severall sorts out of 40 cards 1 0 of each sort.

Or if out of two white & 3 black stones I am blindfold to chose a white & a black one.

<92v>

Figure Symbol (equal subscript u) in text Equation

An equation given; if both x , y , have divers dimensions, try if the roote of one of them may be extracted: & If a quantity wherein y is not is divided by x in the line equall to x . that crooked cannot be squared.

<93r>

The line cdf is a Parab. 4 ac=2 ad=r=2 na=2 ge. ce=x. ef=y. rx=yy. g e ef 12r rx eoap . eo=z. ap=a. 12ra = z rx . 14rraa = rzzx . raa4 = zzx . or supposeing ea= y. x = y+14r & raa4 = zzy + 14 zzr. which shews the nature of the crooked line po. now if dt=ap. then drst=eoap. for supposeing eo moves uniformely from ap , rs moves from dt with motion decreaseing in the proportion that the line eo doth shorten. Suppos aq=ap=r2=a & eq= y . x= y14r . then r316 = zzy 14zzr . suppose z=a+y. then r3=aax+2ayx+yyx. Or aax + 2ayx + yyx + 14aar + 18ayr + 14yyr = r316 . Or aax + 2ayx + yyx 14aar 18ayr 14yyr = r316 . Or suppose z=ya . then r316 = aax 2ayx + yyx . Or aax 2ayx + yyx + 14aar 18ayr + 14yyr = r316 Or aax 2ayx + yyx 14aar + 18ayr 14yyr = r316 . Or, if x=a r . r316 = zza zz x . &
a3 + 2 aay + ayy aax 2ayx y yx = r316 . Or a3 2aay + aay aax + 2ayx yyx = r3 16 . 2= 2z2. mp=ʒ . mq = a + ʒ =. mo = y = a + ʒ 2z2 pq(=a) aq(=12r) ƅq(=x+z) = y + 2zz ay+a2zz = 12rx + 12rz . 2ay+2z2aa r 12z = x = r3+4zzr 16zz . 32zzay + 32z3a2 8z3r r4 4zzrr = 0 mv = ξ = a + ʒ z2. ξaʒ2 = z . ξ22ξa2ξʒ+aa+2aʒ+ʒ2 2 = zz . ξ33ξ2a3ξ2ʒ+ 3aaξ+6aʒξ+3ʒʒξa33aaʒ3aʒʒʒ3 8 = z3 322aa 8r = c cz3+32aξzzr4 4rr = 0 <93v> Figure <94r> cξ3 3caξ2 + 3caaξ caʒ + 32aξ3 3cʒξ2 + 6caʒξ 3ca2ʒ + 3cʒʒξ 3caʒ2 64aaξ2 + 32a3ξ cʒ3 64aʒξ2 + 64aaʒξ 4rraa 4rrξ2 + 32aʒʒξ 8rraʒ + 8rraξ 4rrʒʒ + 8rrʒξ r4 = 0 Or
ξ3 deʒξ2 + ggξ fʒ d + hʒ ʒm2 + igʒʒ nʒ2 onʒ3 = 0 .

Let ed=a. ae=x ab=y se=v. sb=s. x3 2vx2 +vv ss +aa x + a3 = 0 2 1 0 1

aax = yyx + yya . y2 = ss vv + 2vx xx aax = ssx vvx + 2vxx x3 + ssa + 2avx axx avv . x3 2v +a x2 ss + vv 2av + aa x ssa + avv = 0 . x22ex+ee×x+f x3 2e +f x2 +eefx +2eex +eef =0 2v+a = f 2e . f = a+2e 2v. eef=avvass eea 2e3 + 2eev + avv a = ss . vvx2vvx+x3 vva2vax+ax2 +aax = a x } 2eev + aav 2e3 eea a

<94v>

To square those lines in which is y onely

If y is in but one terme onely of the Equation (as xx=ay. or, a3=xxy) resolve the Eq: into the proport ya (as ya xxaa . or, ya aaxx .) If the line hath Assymptotes
x3=aay . v = 3x5 a4 + x .

<95r>

+a +x a x vv 2vx.2 2vax 2vee. 2xaeev +x.3 +ax.2 +aa.x +2e.3 +ee.a +2xae.3 +xee. =0 v= 4x3 + 2axx + aa.x + 2x4a 4xx + 2ax + 2x3a 0 v= 4ax2 + 2aax + a3 + 2x3 4ax + 2aa + 2xx 0 sa= +4ax2 + 2x3 + 2aax + a3 4ax2 2x3 2aax 4ax + 2aa + 2xx 0 sa= a3 2x2 + 4ax + 2aa 2a4x x+a 2a6 4xx + 8ax + 4aa 2aax 2a6 8x3 + 24ax + 24aax + 8a3 2 x3 + 2ax2 ad. 2a4x2+2a5x 8x3& a4x 4xx + 8ax + 4aa 2x3+4ax2+2aaxa3=0 oe=p aaxa+x a32x2+4ax+2aa pz z2aax a+x = a6pp 4x4+16ax3+8aaxx+16a3x+4a4 +16aa a5pp+a4xpp=4zzx5+16zzax4+24aazzx3+16zza3x2+4a4z2x divided by x+a it produceth. 4zzx4+12zzax3+12zzaax2+4zza3x a4pp=0

<95v> Figure

By the Squares of the simplest lines to square lines more compound. 1st those whein y.

find the valor of y. If the number of the termes in the denominator thereof be neither 1.3.6.10.15.21.28. &c. the line cannot be squared If it have but one terme tis squared by finding the square of each particular terme in the valor of y & then adding all those squares together. Example 1st. 3x4+a4=yaxx . & y=3x4+a4axx. Then makeing y equall to each particular terme. 3xxa=y. a3xx=y or 3xx=ay whose square is x3a. & a3=xxy. whose square is a3x Add these 2 squares together & they (viz: x4+a4ax) are the square of the line 3x4+a4=ayxx. Againe 2a72 b x6+x7=a3x3y. Or y = 2a72bx6+x7 a3x3 . then disjoynting the valor of y. y=2a4x3 . y=x4a3. y=2bx3aaa Or x3y=2a4 , whose square is a4xx . ya3=x4 , whose square x55a3. y a3 =2bx3, whose square x4b2a3. which 3 squares (viz 10a7 + 2x7 5bx6 10a3xx ) taken together are the square sought for. And these lines may bee ever squared unless in the valor of y there bee found aax, abx, cc+dex, &c. for the Squareing of that line depends on the squareing of the Hyperbola. As in the line x xxy = x4 +a3x+a4 .

<96r>

Secondly. If it have 3 termes See if it may be reduced to {one orfewer} dimensions by adding or subtracting a knowne quantity to or from x . Example. 2bax + axx = bby + 2bxy + xxy . which (makeing x+b = z ) is thus reduced
zzy= bba+azz . Or bba+azzzz=y

<97r>

aax+bby=y3 . x=vssyy . aavaassy2+bby= y3 aav+bbyy3=aassy2 .
a4v2+2aavbby2aavy3+b4y22bby4+y6 a4ss +a4y2 013246 6y58bby3+2b4y+2a4y 2aabb+6aayy =v vx= 00 00 2b4y+6bby36y5 +2b4y+2bby3+6y5 +2a4y8bby3 vx=2a4y6aay22aabb vx=aay3y2bb . bby+ aa | y aay 3y2 bb y3bby aa & ad. aay4aabbyy 3aay3aabby y3bby 3y2bb .

aay 3y2bb = y3bby aa. a4 = 3y4 4bbyy + b4
y4 = 4bbyyb4+a4 3 a=b y2 = 4bb 3 . y = 2b 3 = dm = dv . 8b3 33 2bbb 3 aax . 2b 33 = dc = ds . y aay 3y2bb 2b33 z . yz = 2aaby 9y233bb3 9yyz3bb z=2aab3 An Equation expressing the nature of the line ns .

<98r>

aax+byx=y3 . x=vssyy.
aav+byvy3= aa by ssyy
a4v2+2aav2by2aavy3+bbvvy22bvy4+y6 a4ss +a4yy+bb bbss 21+1024 01+3246 ss = 2a4vv + 2aavvby + aavy3 + 4bvy4 2bby4 4y6 2a4 12a4y6 16a4bvy4 + 4a4bby4 + 4bba4vvy4 12a6vy3 + 4a6vvby +4a8yy 12a4y6 16a4bvy4 +4a6vvby 8b3vy6 + 8a4bby4 12a6vy3 + 4a8yy +4b4y62aabbvy54aavvb3 } 0 +8y8bb¯ +4a6by +4aab3y3 { vv= 0 0 12a6y3av8a4bby4 2aab2y58y8bb 16a4by44b4y6 8b3y612a4y6 4a6by 4aab3y3 v= 6a6y2 + a2b2y4 + 8a4by3 + 4b3y5 4a6b 4aab3y2 36a12y4+8aab5y9 + 12a8b2y6+64a8b8y6 + 16a10by5+16bby10b4 + 24a8b3y7 + a4b4y8 b5 16a12b232a8b4y2 +16a4b6y4 . 8a4bby3b0 8y7bbb0 4b4y5 + 12a4y5b3 + a4b4y8 b5 + 4a6b 4aab3y2

<99r>

a3x=y4 . x=vssyy . a3vy4=a3ssyy abv22a3vy4+y8+a6yy abss 0482 . v=4y6+a64a3y2 vx=4y6+a64y64a3y2=a34y2 .
y · a34y2 y4a3 a3y44a3y3 = y4

<100r>

aaxaay=y3 . aavaayy3=aassyy a4v22a4vy2aavy3+2a4y2+2aay4+y6 a4ss 013246 v=3y5+4aay3+2a4ya4+3aayy
vx = 3y.5 + 4aa.y3 + 2a4y aa.y3 3y.5 ya.4 3aa.y3 a4 + 3aayy vx = aay aa+3yy . y aay a2+3y2 a2y+y3 aa a4y2+a2y4 a4y+3a2y3 = aay+y3 aa + 3yy

<101r>

aax = by2+y3 . aav by2 y3 = aa ss yy a4v2 2aabyyv 2aavy3 + bvy4 + 2by5 + y6 a4ss + a4 023456 v = 3y4+5by3+2bby2+a4yy 2aab+3aay vx = 3y.4+5by.3+2bby.22bbyy3by.32by33y.4+a4 2aab+3aay vx = a4 2aab+3aay a=b . a3 + 3aay = byy + bby aa+3ayyyay = 0 yy = 2ay + aa y = a + aa + aa . y = a + a2 = dm = vd . a3 + 3a32 + 6a3 + 2a32 a3 + 2a32 + 2a3 aa = 10a + 7a2 = dc y y3+by2 bb+3by 10b+7b2 z . yz = 10y3+10by2+7y32 +7by22 b+3y bz + 3yz = 10yy + 10by + 7yy2 + 7by2 .

<102r>

axy = y3 + b3 . ayv y3 b3 = ay ss yy . aavvy2 2avy4 2avb3y + y6 + 2b3y3 + b6 aass + aa 021412 v = 4y6 + 2b3y3 + 2aay4 2b6 4ay4 2ab3y vx = 4y.6 + 2b3y.3 + 2aay4 2b.6 4y.6 + 2b3y.3 4b3y.3 + 2b.6 4ay4 2ab3y vx = ay3 2y3b3 . y ay3 2y3b3 y3+b3 ay y4+b4 2y3b3

<102v> Figure <103r>

a3=xyy . a=dg=dp . go=x . oa=y . cg=v . ca=s . ss = y2 + xx 2vx + v2 . a3 xss + x3 2vx2 + vvx=0 1 0 2 1 0 2x3 a3 2x2 = v . a3 x 4x6 4a3x3 + 4a6 4x4 x . { Ad } 4x6 4a3x3 + a6 4xxa3 . xv = a3 2x2 . a3x a32xx p z zza3 x = a6pp 4x4 4x3zz = a3pp go=y . oa=x . &c: a3=xyy . s s = xx + yy + vv 2vy xx = ss yy + 2vy + vv . a6 ssy4 + y6 2vy5 vvy4 = 0 4 0 2 1 0 v = 2y6 4a6 2y5 yv = 2a6 y5 a3y2 2a6y5 p z z = 2a3p yyy . zy3 = 2a3p . which shewes the nature of another crooked line that may be squared.

<104r>

axx=y3 . x = v ss yy . x2 = v2 2v ss y2 + ss y2 . av2 + ass ay2 y3 = 2 a v ss yy .
a2v4 2av2y3 + aay4 2ay5 + y6 + aas4 2aassy2 2assy3 2a2v4ss + 2vvaay2 +3 1 0 1 2 3 3a2s4 2a4y2ss + 3a2v4 &c + 2v2a2y2 a2y4 + 4ay5 3y6 6aavv ss = +2yy3 +6vv3 + 4y49v4+y43+y62ay53aa ss = 2yy+6vv3 4y.4a2+24aayy.vv+36aav4 9aav46aayy.vv12ay.5 +3aay.4+9y.6 9aa ss = 2y2+6v2 3 9y612ay5+7aay4+18aavvy2 +27aav4 9aa

<104v>

Figure

a = bA = CB = Dd = eD = FE = Gf = 1,117313 . Aa = dC = ffF = 0,921787 . Bb = Ee = 0,706724 . A = ba = dB = DC = fE = GF = 2,039100 . BA = Cb = ED = Fe = 1,824037 . ed = aaf = 2,234626 . b = DB = eC = GE = aaF = AAf = 3,156413 . CA = Ed = FD = 2,941350 . Ba = db = fe = 2,745824 B = Ca = dA = Db = EC = fD = Ge = 3,863137 . eB = aaE = bbf = 4,273726 . Fd = 4,058663 . A2F = 4,078200 . C = DA = eb = EB = FC = fd = GD = aae = B2f = 4,980450 A2E = bbF = 5,195513 . da = 4,784924 . d = Da = fC = A2e = B2F = 5,902237 . bbe = 6,312826 . eA = FB = Gd = aaD = C2f =6,097763. Eb = 5,687174 . D = ea = fB = GC = A2D = bbe = B2E = C2F = ddf = 7,01955 EA = Fb = 6,804487 . aad = 7,215076 . e = GB = aaC = A2d = bbD = C2E = D2f = 8,136863 Ea = fb = B2e = 7,726274 . FA = 7,921800 . ddF = 7,941337 . E= Fa = fA = Gb = B2D = C2e = 8,843587 . A2d = ddE = D2F = 9,058650 . aaB = bbd = eef = 9,254176 . F = GA = aab = B2d = C2D = E2f = 9,960900 . A2B = bbC = D2E = eeF = 10,175963 . fa = dde = 9,765374 . f = Ga = A2b = B2C = ddD = D2e = E2F = 10,882687 . aaA = C2d = F2f = 11,078213 . B2C = eeE = 11,293276 .

This table shews the distance of any two notes As the distance of C & E is B , or a third, or 3,863137 halfe notes. Of B & E tis a fourth, or 4,98045 halfe notes. of B & F tis 6,097763 halfe notes, or greater than a fifth ♭, by 0,095526 halfe notes &c.

<105r>

aa = xy . eg = eh = a . ga = x . ad = y . d c = s . cg = v y2 = ss xx + 2vx vv . a4+x2s2x4+2vx3=0 x2v2 +2021 2x42a4 2x3 = v xv = a4x3 . aa x a4x3 p z . zxx = a2p .

8th5t = 4th = GD = C = 6t3d = EB
5t4th = 5t+5t8th = 2d = A .
4th+4th = 8th+8th5t5t = 7th = F .
4th3d = 2d = a
8th3d = 6t = e
4th+3d = 6t = E
5 3d = 3d = 8th6t = b
3d+5t = 7th = f
7th4th = 2d+3d = 5t = d = 3d+5t4th = 2d+5t .

By the helpe of concordant notes all the notes in the Gam ut may bee thus tuned viz:

First tune the eighths, G , G2 , G3 , G4 &c.

Seacondly tune fifts to them both above them D , D2 , D3 , D4 . & below them C2 , C , C2 , C3 .

Thirdly tune thirds to them both above them B , B2 , B3 , B4 , & below them E2 , E , E2 , E3 .

Fourthly from each B , B2 , B3 , B4 rise a fift for F , F2 , F3 , F4 & fall a fift for E2 , E , E2 , E3 .

Fiftly from E2 , E , E2 , E3 rise a fift for B B2 , B3 , B4 . & fall a fift for A2 , A , A2 , A3 .

Sixtly from D, D2 , D3 , D4 . rise a fift for A2 , A3 , A4 A5 . & from C2 , C , C2 , C3 fall a fift for F3 , F2 , F , F2 .

Seaventhly from each F , F2 , F3 , F4 . rise a fift for D2 , D3 , D4 , D5 . The rest as A , D are supplyed by eighths viz to A2 , D2 &c.

<105v>

November 20. 1665.

0000000000000000000 0000000000000000000 0000000000000000000 0000000000000000000 0000000000000000 00 1200 036000 2,55630247 2,5563025 360,00000,0 12,00000.,0 G 81500 038400 2,58433118 2,5813883 381,40667,8. 10,88268,7 f 91600 040500 2,60745497 2,6064742 404,08640,6. 9,96090,0. F 3500 043200 2,63548369 2,6315600 428,11458,1. 9,84358,7. E 5800 045000 2,65321247 2,6566458 453,57157,8. 8,13686,3. e 2300 048000 2,68124123 2,6817317 480,54236,7. 7,01955,0. D 324500 051200 2,70926992 2,7068175 509,11688,24543. 5,90223,7. d 3400 054000 2,73239371 2,7319033. 539,39055,9. 4,98045,0. C 4500 057600 2,76042244 2,7569892 571,46447,4. 3,86313,7. B 5600 060000 2,77815121 2,7820750 605,44546,7. 3,15641,3. b 8900 064000 2,80617993 2,8071608 641,44697,3. 2,03910,0. A 151600 067500 2,82930373 2,8322467 679,589514,9. 1,11731,3. a 100 072000 2,857332447. 2,8573325 720,00000,0. 0,00000,0. G How ye string 1 or 720 is to bee di= =videdytit may sound allye musicall notes & halfe notes in an eight The proportion wch those musicall notes &12notes beare yeone toyeother (viz yelogarithmes ofye string sounding them) Twelve exact or equidistant12notes (oryelogarithmes of a cord divided into 12 geome tricall partes)yedista¯ce of each12note being 0,025085833333&c. A just note being 0,050171666666&c. A string (720) divided into 12 (geometrically progressionall) parts,yt it may soundye12 exact12notes in an eight 509,11688,24543. The proportion of all ye12musicall12notes in a eight; An exact halfe note being a unite. 00 0000000000000000000 0000000000000000000 0000000000000000000 0000000000000000000 0000000000000000 00

<106r>

2,635483 69 a4 = xy3 . ad=x. a8+y6ssy8+2vy7vvy6=0 +60210 2y86a8 2y7 = v . vy = 3a8 2y7 . a4 y3 3a8 2y7 pz za4 y3 = 3a8p 3y7 . zy4 = 3a4p 2

0,921787 = Aa = dC = fF . 2,039100 = A = bA = dB = DC = fE = bF 0,706724 = Bb = Ee . Fe = 1,824037 = BA = Cb = ED . ed = 2,234626 aaf = 2,234626 2,745824 = Ba = db = fe . aaE = bbf = = 4,273726 = eB . Fd = 4,058663 2941350 = CA = Ed = FD . 4,078200 = A2F 3,156413 = b = DB = eC = GE = aaF = AAf . bbe = 6312826 3863137 = B = Ca = dA = Db = EC = fD = Ge 4980450 = C = DA = eb = EB = FC = fd = GD = aae = B2f 4784924 = da 5,195513 = A2E = bbF 5,902237 = d = Da = fC = A2e = B2F . 7,9218 = FA 6,097763 = eA = FB = Gd = aaD = C2f 7941337 = ddF 5,687174 = Eb 6,804487 = EA = Fb 7,019550 = D = ea = fB = GC = A2D = bbe = B2E = C2F = ddf 5,215076 = aad . 7,726274 = Ea = fb = B2e 8,136863 = e = GB = a2C = A2d = bbD = C2E = D2 f 9,254176 = aaB = bbd = eef . D2F = 9,058650 A2d = ddE . 9,765374 = fa = dde 10,175963 = A2B = bbC = D2E = eeF 11,078213 = aaA = C2d = F2f 11,293276 = B2C = eeE perhaps d = Eb is better than d = 3d+5t4th . 0.1,1.2.3,1.3,9.5.5,9.7.8,1.8,9.10.109.12. G.a.A.b.B.C.d.D.e.E.F.f.G. 0.1,2.2.3,2.3,8.5.5,8.7.8,2.8,8.10.108.12 0.6.10.16.19.25.29.35.41.44.50.54.60.

<106v>

00 00 00 00 00 00 6445=14222&c00 2,00000g0002=2,0000. 1,8877400158=1,8750. 1,7818a00169=1,777777&c 1,681800053=1,66666&c 1,5874b00085=1,6000 1,4983c00032=1,5000 1,414213600107=1,428571428571&c 1,3349d00043=1,33333&c 1,259900054=1,2500. 1,18920e00065=1,2000. 1,12245f00098=1,1250. 1,05946001615=1,066666&c 1,00000g0001=1,0000. A string divided in a geometricall progres =sion 0 0 0 0 0 0 0 0 0 g A string divided by a musicall progressi¯ By this table it may appeare that a Second minor Secondminor¯ Third minor } is higher byye { 17th+ 51th+ 13th Thirdminor¯ Fourthminor¯ } is lower byyeo { 15th 102th Fiftmineis lower byye20th12 or higher byye20th12 Fiftminor¯ Sixtminor } is higher byye { 102th 15th Sixt Seaventh Eight minor00 } is lower byyeo { 13th 51th 17th } parte of a note then it would bee wereyemusicall chord divided in geometricall progression. { 12notes, notes =1,058522. =2,02. =3,07. =3,93. =4,99. =5,95. =6,05. 0 0 0 0 0 0 0 1 , 00001516,2425,2021, 000089,78,910 56 , 45 , 34 , 1825573245,4564,7102536 23 , 58 , 35 , 0000916,47,59, 0000815,25482140 12 0 0 12 , 1532 , 49 , 512 , 25 , 38 , 1645 , 13 , 516 , 310 , 932 , 415 , 14 14 , 1564 , 29 , 524 , 15 , 316 , 845 , 16 , 532 , 320 , 964 , 215 , 18 17 A3dmineis highery¯ just, byye564+thpteof a note A Fift minor is higherynjust by ye241thpteof a note

<107r>

a5=xxy3

Figure By this table may bee knowne the distance of any two notes whither a {trew} second of the lesse, second, third f the lesse, a third fourth &c: As to know the distance twixt A re & B sol re I follow the pricked stroke from A to D or from D to A where I find it crossed by a black crooked line & against it, a Fourth written, therefore I conclude A re & D la sol distant a true fourth.

<107v>

And Thus to find the distance of B mi & D la sol re I follow the prick line from the top B to the right hand side thence to the bottom B thence towards the left hand side untill I come {over} D. Or (which is the same) I follow the prick{t} line from the top D to the left hand side thence to the bottom D, thence toward the right hand side untill I come just over B, where I find the pricked line to be crossed by a Symbol (roundAngle) in text stroke & against it to bee written on the upper line a tenth , on the lower Figure a ninth therefore tis a tenth minor exactly. But if it

<108r>

a3 = xxy . x = v + ssyy . a3 v y y = yy ssyy a62a3vyy+vvy4+y6=0 ss 4202 2y6+4a6 2a3y2 . xv = 2y62a6 2a3y2 . 2a3y3 2y62a6 p z 2a3y3z = 2y6p2a6p py6 a3y3 a6p

Figure 0. 5. 9. 14. 16. 21. 25. 30. 35. 37. 42. 46. 51. 0. 5. 9. 14. 17. 22. 26. 31. 36. 39. 44. 48. 53. g. a. A. b. B. C. d. D. e. E. F. f. G. 0. 4. 7. 11. 13. 17. 20. 24. 28. 30. 34. 37. 41. 0. 3. 5. 8. 9. 12. 14. 17. 20. 21. 24. 26. 29. 0. 4. 6. 10. 11. 15. 17. 21. 25. 26. 30. 32. 36. 0. 11. 20. 31. 39. 50. 59. 70. 81. 89. 100. 109. 120. 0. 4. 2. 6. 1. 5. 3. 7. 11. 6. 10. 8. 12. 0 0. 2. 5. 7. 8. 10. 13. 15. 17. 18. 20. 23. 25. 0. 1. 4. 5. 7. 8. 11. 12. 13. 15. 16. 19. 20.

<108v> Figure

0. a. . b. b+a. 0. a. a+b. a+b+c. 2a+b+c. 2a+2b+c.

The notes The proportion of their or thus or thus How they maybee otherwisedistinguished byfigures The notes soundsThe proportion of their or thus or thus thus thus orthus or thus or thus G536125910010036291212.00 f4855590893226102310,900 F44508823024101000 E394517226218238,900 e364154069211777 00 D313585925208138,100 d26301494817145235,900 C22254414115125500 B171971931301193233,900 b1416128291083133,100 A91041818652200 a5571011431131,100 G00000000000

<109r>

ao=a=ad . dc= p . ai=x. oi=y. aaxx=y2 in=z. xx aaxx pp zz . zzxx = aapp p2x2 . id=x . oi = 2axxx . aa2ax+x2 2axx2 pp zz zzx22azzx+aazz=0 pp2app .

Figure

The 3 meanes are best there being an imperfect fift in the outward extreame & a tritonus in the inmost.
1. 6. 4. 3. 2. 5. / 10. 7. 2. 5. . 6. 8. 4. 6. 11. 8. 3. 10. 2. 3. 9. 10. 12. 7. 1. 6. 11. 8. 4. 3: 9: 10: 12. 7. 2. 5.

<109v> Figure Figure <110r>

In the Hyperbola dm . suppose ak=a=kh ao=x. od=y. dc= a secant. ca=v og=z. xy=aa. y y=ssxx+2vxvv
a4+xxssx4+2vx3=0 vvxx +2021 double route equall x4a4x3=v . xv=a4x3=oc. od oc kh og . aa x a4 x3 a z . aaz x = a5 x3 . zxx=a3 . which equation continues the nature of the crooked line gh. Now supposeing the line og always moves over the same superficies in the same time, it will increase in motion from kh in the same proportion that it decreaseth in lenght & the line ne will move uniformely from (mq), soe that the space mqen=gokh. suppose ok=a. ao=2a. od=a2=nm. & mqen = 12aa = ogkh .

Modi 1010110101101 0202022020220 3030303303033 4404040440404 055050505505 6066060606606 0707707070770 8080880808088 9909099090909 01010010010100100100 110111101101111011011 01201212012012120120 1.6.

<110v> Figure Figure

In the order of the musicall tones the 2 halfe notes may not be together 1st because every note would then bee distant 3 tones from some other which is most ungratefull Secondly whole notes ought to bee interposed to moderate their harshnesse. Thirdly since there must bee a Fift to the ground: these 12 notes must bee either next the ground or its Fift which would make them harsh & that wee could not gradually passe to or from them.

Neither ought they to be distant but one tone for the second reason {afforesd} & because they will bee more consonant by the absense of more 3 tones &c if they be distant 2 tones yet perhaps they may not bee wholly uselesse. See the last modes.

A catalogue of the 12 Musicall modes in theire order of gratefulnesse. __ __ _ __ _ _ __ _ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ 1 G . a . b c . d . e f . g 3 c . d . e f . g . a . b c 2 d . e f . g . a . b c . d 4 a . b c . d . e f . g . a 5 e f . g . a . b c . d . e 6 f . g . a . b c . d . e f b c . d . e f . g . a . b . b c . d . e f . g . a . . a . b c . d . e f . g . . d . e f . g . a . b c . . g . a . b c . d . e f . . e f . g . a . b c . d . __ __ _ __ _ _ __ _ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ 0 0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10 . 11 . 12 __ __ _ __ _ _ __ _ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ 2 G . . a . b c . d e . f . g 1 c . d e . f . g . a . b c 3 d e . f . g . a . b c . d __ __ _ __ _ _ __ _ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ __ _ 0.9.17.22.31.39.44.53.f 0.9.17.22.31.40.48.53.f 0.9.14.22.31.40.45.53.f 0.9.14.22.31.36.45.53f 0.5.14.22.31.36.45.53.f 0917.26.31.40.48.53f f f f f f f __ 0 __ G . a . b . cde . f . g de . f . g . a . b . cd f . g . a . bc . de . f .

<111r>

suppose the line last found to be md . mk=kh=a=ka ao=x. od=y. dc=s. ca=v. yxx=a3. yy = ss vv + 2vx xx . ssx4+2vx5x6a6=0 vv 012+4 v = x62a6 x5 . xv = 2a6 x5 . to find at what point do=oc: a3xx = 2a6 x5 . x3=2a3. x = a c:2 = af . mq = fh = a . og=z. od oc fh og . a3xx 2a6 x5 a z . a3zx5 = 2a7xx . zx3 = 2 a4 . which shews the nature of the line (gh). & mneq=gofh or nbpe=gokl. suppose ko=ka=a. oa=2a= x od = a3xx = a34aa = a4 . bn = 3a 4 . bpen = 3aa 4 = gokl .

kl=ka=a=kb. Suppose kl=ka=kb=a ak=x=a. bk=y=a. bs=s. as=v. yy=ssvv2vxxx. yyxxxx=a6. ssx42vx5x6a6 vv 012+4 . 2x6+4a62x5=v v+x=2a6x5=ks=2a. ksbk klfh . 2aa aa2 . fh=a2=ne=mq=rp. mn=3a4=qe 3aa8=mneq=lkog. ao=2a=x. a34aa=do do=a4. oc=2a632a5=a16. a4a16 a2og=a8 2a6x5 a3xx z a2 . a4=zx3.

<111v>

a,b,c=12 tone max; medî: minimus. a+a=d , a+b=e, a+c=f~tone {maj me:mi.} a+a+b=a+e=3d
a+a+b+c=e+f=3d.
3a+b+c=a+e+f=f+3d=4th

Figure

2r+s+t=5t. 3dmaj=r+s.r+t=3dmi 6tmin=2r+s+2t 6thmaj=2r+2s+t 4th=r+s+t hath 8 Fifts

r.s.t.r.s.t.r,rstrstr. 12345fifts 12thirdmajs2 12third min. 12.6tsmaj 112sixt minors 12345forths srtrstr,s.r.t.rst 12345fourths3dmode 12345fifts 1233dmaj.1 123third min 123Sixt maj. 123Sixt min. rstrrts,rstrrts. 1234fourths 12third maj.4 12third minor srtrrts,srt. 12fourths5 13dmaj 1233dmin. r.r.t.sstr,rrts 12fourths6 03dmaj 123dmin rrtsrts,rrts 12345fourths 12third maj.3 12third min. 6t.mode rsrtsrt,rs 12345.4ths 1233dmaj 1.2.3.3dmin srrtrst,s ac,ab,a,ab,ac,a,ab:ac,ab ac,ba,aaaba ab.ac.aab:acaba.a ...b.c.b.c.b abaacabaacab,a abacabaacaba 0 0 0 0 0 0 0 9×4ths.7×3ds.6×3ds.

Figure <112r>

Suppose againe the last line whose nature is comprised in this equation y x3=a4. ak=bk=lk=a ao=x. ac=v. do=y. dc=s. og=z. ssx6+2vx7x8a8=0 vv 012+6 v=2x86a82x7 xv=3a8x7. to find where do=dc
a4x3=3a8x7 a4x4=3a8. x4=3a4. x=qq:3 af=aqq:3: bk=y=a. bs=s as=v ak=x=a x+v=3a8x7= x+v=3a. 3a a ki(=a) fh = a3 = mq = ne oa=x= 2 a. a4x3 = do = y = a48a3 = a8 = do . mn=7a8=qe. mqen=7aa24=lkog=7aa24.

0000000000000000000000000 rstrstr;rs. 2×3d.2×3d. 5×4th. no4th 1stmode2dmode.3d.0 0000000000000000000000000 srtrstr;sr 3×3d.3×3d. 5×4th. 6t,4th&5tm¯ode 3dm¯ode.1stm¯ode.no2d 0000000000000000000000000 rstsrtr,rs 2×3d.2×3d. 3×4th. 0000000000000000000000000 srtsrtr,sr 3×3d.3×3d. 5×4th.no1st 5t.4thm¯ode.3dmode 6tm¯ode.2dm¯ode 0000000000000000000000000 rrtsstr;rr. 2×3d.0×3d. 2×4th. 0 0d 0000000000000000000000000 sstrrtr;ss 3×3d.1×3d. 2×4th. 0 0d 0000000000000000000000000 rstrrts;rs 2×3d.2×3d. 2×4th. 0 2dmod. 0000000000000000000000000 srtrrts;sr. 3×3d.1×3d. 2×4ths. 0 0d 0000000000000000000000000 rrtrsts;rr. 2×3d.2×3d. 3×5t. 0d 0d 0000000000000000000000000, rrtsrts,rr 2×3d.2×3d. 5×4ths. 2d.noe3d. 4thmode.6tmodeno2d

<112v>


1.ef0g0a0bc0d0e5 2.a0bc0d0ef0g0a4 3.d0ef0g0a0bc0d2 4.g0a0bc0d0ef0g1 5.c0d0ef0g0a0bc3 6.f0g0a0bc0d0ef6

r=ton. maj. s=ton min. t=semit maj. v=semit min.

1st. gd=cg=da=5t=2r+s+t. dg=gc=ad=r+s+t. cd=ga=r. ac=s+t. ca=3r+s+t. ab=s. bc=t. dc=2r+2s+2t. de= s. ef=t. per sup. et fg 345 strrstr rstrstr rstrrst Modus Symbol (largeAsteriskEightBars) in text harum vocum respectu fundamenti.

2d. gd=da=ae=5= 2 r+s+t. dg=ad=ea=r+s+t. ga=dadg= de=aead=r. ge=3r+s+t=gd+de. eg=s+t. ef=t. fg=s. And if ab=s. bc=t. Then cd=r & the voyces in respect of their {ground} are best 234 strrtsr. rtsrstr. rstrrts. Symbol (largeAsteriskSixBars) in text

If in the 1st case de=r. then 3.4.5. rtsrstr. rstrrst. rrstrst. Symbol (largeAsteriskSixBars) in text If in the 2d case ab=r then 2.3.4. rtsrtsr. rtsrrts. rrtsrts. Symbol (largeAsteriskEightBars) in text

2.strrtsr.0 3.rtsrstr. 4.rstrrts. 5.rrtsrst. 1.tsrrtsr.0 2.rtsrtsr. 3.rtsrrts. 4.rrtsrts. 3.strrstr.0 4.rstrstr. 5.rstrrst. 6.rrstrst.

<113r>

Likewise supposing the line yx4=a5. xv=4a10x9=oc. af= a qc:4. ka(=x)+v=4a fh=a4. do=a16. mq=ne=a4. 15a16 15aa64. &c whence supposeing x to be a line increasing in arithmeticall proportion from the quantity of the line (a) untill it be as long as b. the superfices resulting out of a3xx.a4x3 &c is found as follows.
a3xx=aaa3b. a6x5=aa4a64b4
a4x3=aa2a42bb. a7x6=aa5a75b5.
a5x4=aa3a53b3. a8x7=aa6a86b6. &c

1.trsrtsr0 2.rtsrtrs 3.rtrsrts

1 Of the Key or Ground sound. Secondly, Of its Eighths. Thirdly, of their divisions into Fifts & Fourths Sixts & Thirds, illustrated by the division of a corde. Fourthly, The order of the concords in respect of gratefulnes deduced thence & from other considerations. Fifthly the degrees deduced thence & of the proportion of the concords & degrees i.e. the logarithmes of their strings. 6 Of the various ordering of the degrees & distance of the halfe notes , the keys fift being onely stable 7 Of the moodes ariseing thence & their dignity; explained by one line, o.p. qr.s.tv.o.p.qr.s.tv.o.p. &c. Eighthly, How the tones major & minor are best ordered in every Moode. Ninthly of passing from one moode to another explained by 3 lines g.a.bc.d.ef.g c.d.ef.g.a.bc f.g