☞ Note ytthat if there happen to bee in any equation either a fraction or surde quantity
or a Mechanichall one, (i:e: wchwhich cannot bee Geometrically computed, but is
expressed by yethe
To find in what proportion they unknowne quantitys increase or decrease doe thus. 1 Take two letters
yethe one (as ξ) to signify ytthat quantity, yethe other (as χ π) its motion of increase or
decrease: And making an equation betwixt ytthatyethe letter (ξ) & yethe quantity signifyed
by it, find thereby (by prop 7 if yethe Equation quantity bee Geometricall, or by some other
meanes if it bee mechanicall) yethe valor of yethe other letter (π). 2 Then substi
tuting yethe letter (ξ) signifying ytthat quantity, into its place in yethe maine Equation
esteeme ytthat letter (ξ) as an unknowne quantity & performe yethe worke of
seaventh proposition; & into yethe resulting Equation instead of those letters
ξ & π substitute theire valors. And soe you have yethe Equation required.
Example 1. To find p & q yethe motions of x & y whose relation is, . first suppose , Or . & thereby find π yethe motion of ξ, viz: (by prop 7) . Or . Secondly in yethe Equation , writing ξ in stead of , the result is , whereby find yethe relation of yethe motions p, q, &
π: viz (by prop 7)
. In wchwhich Equation instead of ξ & π writing theire valors, yethe result is, . wchwhich was required to.
[wchwhich equation multiplyed by , is . & in stead of , writing its valor , it is . Or . Which Which conclusion will also bee found by taking yethe surde quantity out yethe given Equation for both parts being squared it is . & therefore (by prop 7) , as before.]
☞ Note also ytthat it may bee more convenient (setting all yethe termes on one side of yethe Equation) to put every fractionall, irrationall & mechanicall terme, as also yethe summe of yethe rationall termes, equall severally to some letter: & then to find yethe motions corresponding to each letter of those letters yethe sume of wchwhich motions is yethe Equation required.
Example yethe 2d. If is yethe relation twixt x & y, whose motions p & q are required. I make ; ; & . & yethe motions of τ, φ, & ξ being called β, γ, & δ; yethe first Equation , gives (by prop 7) . yethe second , gives ; Or . & yethe Third , gives, ; Or . Lastly , is yethe Equation sought.
Example 3d. If . be=y. & yethe superficies abc=z suppose ytthat ax+xz−y3 , is yethe relation twixt x, y & z, whose motions are p, q, & r: & ytthat p & q are desired. The Equation gives (by prop 7), . Now drawing dh∥ab⊥ad=1−bh. I consider yethe superficies abhd=ab×bh=x×1=x, & abd=z doe increase in yethe proportion of bh to bc: ytthat is, 1∶
∷p∶r. Or . Which valor of r being substituted into yethe Equation , gives . wchwhich was required.
How to proceede in other cases (as when there are cube rootes, surde denominators, rootes within rootes (as &c: in the equation) may bee easily bee deduced from what hath been already said. 49 October 1666.
To resolve Problems by Motion these following Propositions are sufficient
1 If the body a in the Perimeter of yethe cirkle or sphære adc moveth towards its center b, its velocity to each point (b, c, d d, c, e,) of ytthat circumference is as yethe chords (ad, ac, ae) drawne from that body to those points are.
2 If yethe △s adc, aec, are alike viz ad = ec &c (though in divers plaines) & 3 bodys move from the point a uniformely & in equall times yethe first to d, the 2d to e, yethe 3d to c; Then is the thirds motion compounded of yethe motion of the firsts & second.s motion.
3. All the points of a Body keeping Parallel to it selfe are in equall motion velocity.
4. If a body move onely circularlyangularly about some axis, yethe motion velocity of its points are as their distance from that axis.
5. The motions of all bodys are either parallel or angular, Call these two simple motions, parallel & angular or mixed of ymthem both, after yethe same manner ytthat the motion towards c (Prop 2) is compounded of thatose towards d & e. And in mixed motion any line may bee taken for yethe axis (or if a line or superficies move in plano, any point in ytthat plane may bee taken for the center) of yethe angular motion
6 If yethe lines ae, ah being moved doe continually intersect; I describe yethe trapezium abcd, & its diagonall ac: & say ytthat, yethe proportion & position of these five lines ab, ad, ac, cb, cd, being determined by requisite data; shall designe yethe proportion & position of these five motions; viz: of yethe point a fixed in yethe line ae & moveing towards b, of yethe point a fixed in yethe line ah & moveing towards d; of yethe point a intersection point a moveing in yethe plaine abcd towards c, (for those five lines are ever in yethe same plaine, though ae & ah may chanch onely to touch that plaine in their intersection point a); of yethe intersection point a moveing in yethe line ae parallely to cb & according to yethe order of yethe letters c, b; & of yethe intersection point a moveing in yethe line ah parallely to cd & according to yethe order of the letters c, d.
Note ytthat one of yethe lines as ah (fig 3d & 4th) resting, yethe points d & a are coincident, & yethe point c shall bee in yethe line ah if it bee streight (fig 3), otherwise in its tangent (fig 4th)
7. Haveing an equation expressing yethe relation twixt two or more lines x, y, z &c: described in yethe same time by two or more moveing bodys A, B, C, &c. the relation of their velocitys may bee thus found p, q, r, &c may bee thus found, viz: Set all yethe termes on one side of yethe Equation that they become equall to nothing. And first multiply each terme by so many times as x hath dimensions in ytthat terme. Secondly multiply each terme by so many times as y hath dimensions in it. Thirdly (if there be 3 unknowne quantitys) multiply each terme by so many times as z hath dimensions in ytthat terme. (& if there bee still more unknowne quantitys doe like to every unknowne quantity). The summe of all these products shall bee equall to nothing. wchwhich Equation gives yethe relation of yethe velocitys p, q, r, &c Or thus. Translate all yethe termes to one side of yethe equation, & multiply them being ordered according to x by this progression, &c. or being ordered by yethe dimensions of y multiply them by this,: &c. The sume of these products shall bee equall to nothing, which equation gives yethe relation of their velocitys p, q, &c.
Or more Gera Generally yethe Equation may bee multiplyed by yethe terme of these progressions &c. And &c. (a & b signifying any two numbers whither rationall or irrationall
8. If two Bodys A & B, by their velocitys p & q describe yethe lines x & y. & an Equation bee given expressing yethe relation twixt one of yethe lines x, & y ratio of their motions q & p; To find yethe other line y.
Could this bee ever done all problems whatever might bee resolved. But by yethe following rules it may bee very often done. (Note ytthat ±m & ±n are logarithmes or numbers signifying yethe dimensions of x.)
ffirst get ytthat valor of . Which if it bee rationall & its Denominator consist of but one terme: Multiply ytthat valor by x & divide each terme of it by yethe logarithme of x in ytthat terme yethe quote shall bee yethe valor of y. As if
. Then is . Or if . Then is . (Soe if . Then is . soe ytthat y is infinite. But note ytthat in this case x & y increase in yethe same proportion ytthat numbers & thitheir logarithmes doe, y being like a logarithme added to an infinite number . But if x bee diminished by c, as if =, y is also diminished by yethe infinite number & becomes finite like a logarithme of yethe number x. & so x being given, y may bee mechanichally found by a Table of logarithmes, as shall bee hereafter showne.)
Secondly. But if yethe denominator of yethe valor of consist of more termes ynthan one, it may bee reduced to such a forme ytthat yethe denominator of each ꝑparte of it shall have but one terme, unlesse ytthat ꝑparte bee : Soe ytthat y may bee ynthen found by yethe precedent rule. Which reduction is thus performed, viz: 1st, If neither yethe numerrator nor neither yethe numerator nor denominator bee the denominator bee not a+bx, nor all the its termes of its termes not multiplyed in all their termes by x or xx, or x3, &c; Increase or diminish x untill yethe last terme of yethe Denominator vanish. 2dly, And when all yethe termes in yethe Denominator are multiplyed by x, xx, or x3 &c: Divide yethe numerator by yethe DenominrDenominator (as in Decimall numbers) untill yethe Quotient consist of such ꝑparts none of wchwhich whose DenominatorsDenominators are so multiplyed by x, x2 & begin yethe Division in those termes in wchwhich x is of its fewest dimensions unlesse yethe Denominator be a+bx
. If ynthen yethe termes in yethe Denominator valor of bee such as was before required yethe valor of y may bee found by yethe first ꝑparteparte of this Prop: onely it must bee so much diminished or increased as it was before increased or diminished by increasing or diminishing x. But if ty the denominator of any terme consist of two more termes ynthan one, in some of wchwhich x is of more ynthan one dimension unlesse ytthat terme bee
. First find those ꝑparts of y's valor wchwhich correspond to yethe other ꝑparts of its valor. & ynthen seeke ytthat of y's valor belon by yethe preceding reductions &c: seeke yethe ꝑparte of y's valor answering to this ꝑparte of its valor.
Example 1st. If . Then by Division tis . (as may appeare by multiplication.) Therefore (by 1st ꝑparte of this Prop:) tis . ( signifys ytthat ꝑparte of yethe valor of y wchwhich is correspondent to yethe terme of yethe valor of , wchwhich will may bee found by a Table of logarithmes as may hereafter appeare.)
Example 2d. If . I suppose x=a+c, & consequently . 50 = & by Division, .
Example 2d. If . I suppose x=z−a. Or . And by Division , (as may appeare by multiplication.) It consequently by yethe 1st ꝑparte of yethe Prop.) . And substituteing x+a into yethe place of z, tis
. And Therefore (by ꝑparte 1st of Prop 8) .
But sometimes The last terme of yethe Denominator cannot bee taken away, (as if yethe DenominrDenominator bee aa+xx. or a4+x4 or a4+bbxx+x4 &c) And then it will bee necessary to have in readinesse some examples wthwith such Denominators to wchwhich all other cases of like denomination may bee by Division reduced. As if
. Make bxx=z, Then is .
. Make bx3=z, Then is .
. Make bx4=z, Then is . &c. In Generall if . Make bxn=z, & ynthen is . Also if . Make & ynthen is . That is, if ; I make x=zz, & . Or if . Make , & □=CDV=y
Thirdly If yethe valor of is irrationall being a square roote, The simplest cases may bee reduced to these following examples.
1. If . Then .
2. If . Then
3. If . Then .
4. If . Then .
5. If . Then .
1. If . Then .
2. If . Then .
3. If . Then .
4. If . Then is .
5. If . Then .
1. If . Then .
1. If . Then is .
2. If . Then is .
3. If . Then .
4 . −630a3+
1. If . Then is .
2. . Then is .
3. If . Then is .
4. If . Then
12 If, . Make xn=zz. And ynthen is .
23. If, . Make xn=z. Then is .
34. If, . Make xn=z. Then, .
45. If . Make xn=z. ynthen is .
1 If . Make xn=zz. ynthen is .
1 If . Make . ynthen is
2 If . Make . ynthen is .
3 If . Make . ynthen is .
If . Make xn=z. Then is .
If . Make xn=z . Then is
If . Then is .
If . Make . Then is . 52
1.
●□::Note 2. ytthat●is to□::3.●□:: as yethe ordinately applyed line bc in some of yethe Conick sections: is to its corresponding superficies abc, yethe axis ab being in like manner related to z. But all those areas (& consequently □, □, □) may bee Mechanichally found either by a Table of logarithmes or signes & Tangents. And I have beene therefore hitherto content to suppose ymthem knowne, as yethe basis of most of yethe precedent propositions.
Note also ytthat if yethe valC Valor of consists of severall ꝑparts each ꝑpart must bee considered severally, as if:
. Then is □
. & □
. Therefore □.
Note also ytthat of yethe valor denominator of yethe valor of consist of both rationall & surde quantitys or of two or more surde quantitys First take those surde quantitys out of yethe denominator, & ynthen seeke (y) by yethe perecedent theoremes
But this eighth Proposition may bee ever thus resolved mechanichally. viz: Seeke yethe Valor of as if you were resolving yethe equation in Decimall numbers either by Division or Extraction of rootes or Vieta's Analyticall resolution of powers; This operation may bee continued at pleasure, yethe farther ther better. & from each terme ariseing from this operation may bee deduced a ꝑparte of yethe valor of y, (by ꝑparte yethe 1st of this prop).
Example 1. If
. Then by division is
&c. And consequently
&c.
Example 2. If
. Extract yethe roote & 'tis &c (as may appeare by squareing both ꝑparts). Therefore (by 1st ꝑparte of Prop 8) &c.
Example 3. If
But yethe Demonstracotions of hath beene said must not bee wholly omitted.
Prop 7 Demonstrated.
Lemma. If two bodys A, B, move uniformely yethe oneother from ab to c, d, e, f,g, h, k, l, &c: in yethe same time. Then are yethe lines ac,bg, & cd,gh, & de,hk, & ef,kl, &c: as their velocitys p.q. And though they move not uniformely yet are yethe infinitely little lines wchwhich each moment they describe, as their velocitys wchwhich they have while they describe ymthem. As if yethe body A wthwith yethe velocity p describe yethe infinitely little line (cd=)p×o in one moment, in ytthat moment yethe body B wthwith yethe velocity q, will describe yethe line (gh=)q×o. For p:q::po:qo. Soe ytthat if yethe described lines bee (ac=)x, & (bg=)y, in one moment, they will bee (ad=)x+po, & (bh=)y+qo in yethe next.
Demonstr: Now if yethe equation expressing yethe relation twixt yethe lines x & y bee x3−abx+a3−dyy=0. I may substitute x+po & y+qo into yethe place of x & y; because (by yethe lemma) they as well as x & y, doe signify yethe lines described by yethe bodys A & B. By doeing so there results x3+3poxx+3ppoox+p3o3−dyy−2dqoy−dqqoo=0−abx−abpo+a3. But x3−abx+a3−dyy=0 (by supp). Therefore there remaines onely 3poxx+3ppoox+p3o3−abpo−2dqoy−dqqoo=0. Or dividing it by o tis 3px2+3ppox+p3oo−abp−2dqy−dqqo=0. Also those termes are infinitely little in wchwhich o is ynthen those in wchwhich tis not. Therefore bomitting them there rests 3pxx−abp−2dqy=0. The like may bee done in all other equations.
Hence I observe. First ytthat those termes ever vanish in wchwhich o is not are not multiplyed by o, they being yethe propounded equation. Secondly those termes also vanish in wchwhich o is of more ynthan one dimension, because they are infinitely lesse ynthan those in wchwhich o is but of one dimension. Thirdly yethe rem still remaining termes, being divided by o will have ytthat forme wchwhich, by yethe 1st rule in Prop 7th, they should have (as may ꝑpartly appeare by Mr Oughtreds yethe second termes of Mr Oughtreds latter Analiticall table).
After yethe sàme manner may this 7th Prop: bee demonstrndemonstrated: there being 3 or more unknowne quantitys (x, y, z, &c
Prop 8th is yethe Converse of this 7th Prop. & may bee therefore Ab Analytically demonstrarted by it.
Prop 1st Demonstrated. If some body A move in yethe right line gafc from g towards c. From any point d draw df⊥ac. & call, df=a. fg=x, dg=y. Then is aa+xx−yy=0. Now by Prop 7th, may yethe proportion of (p) yethe velocity of ytthat body towards f; to (q) eits velocity towards d bee found viz 2xp−2yq=0. Or x∶y∷q∶p. That is gf∶gd∷ its velocity to d: its velocity towards f or c. & when yethe body A is at a, ytthat is when yethe points g & a are coincident then is ac∶ad∷ad∶af∷ velocity to c∶ velocity to d.
Prop 2d, Demonstrated.From yethe points d & e draw df⊥ac⊥ge. And let yethe firsts bodys velocity (it moveing to d) bee called ad, yethe seconds to e bee ae, & yethe 3ds toward c bee ac. Then shall yethe firsts velocity towards c bee af (by Prop 1): & The seconds towards c is ag, (prop 1). but af af=gc (for △adcsimilis=△aec, & △adf =△gec. by sup). Therefore ac=ag+gc =ag+af. That is yethe velocity of yethe third body towards c is compo equall to yethe summ of the velocitys of yethe first & second body towards c.
The former Theorems Applyed to Resolving of Problems.
Prob. 1. To draw Tangents to crooked lines.
Seeke (by prop 7th; or 3d, 4th & 2d, &c) yethe motions of those streight lines to wchwhich yethe crooked line is cheifely referred, & wthwith what velocity they increase or decrease: & they shall give (by prop 6t, or 1st or 2d) yethe motion of yethe point describing yethe crooked line; wchwhich motion is in its tangent.
Tangents to Geometricall lines.
Example 1. If yethe crooked line fac is described by yethe intersection of two lines cb & dc yethe one yethe one moveing parallely, yethe one viz: cb∥ad, & insisting upon ab, yethe other & dc∥ab; & insisting soe ytthat if ab=x, & bc=y=ad, Their relation is x4−3y+10ax3+ayxx−2y3x+a4−y4=0. To draw yethe tangent rhcr; I call p yethe velocity of cb ytthat is yethe velocity of yethe increase of ab & q yethe velocity of dc, ytthat is of yethe increasing of ad or cb. And so I have (by Prop 7) this Equation 4px3−9py+3opa+xx+2payx−2py3−3qx3+qaxx−6qyyx−4qy3=0. Now because yethe crooked line fac is described by yethe intersection point of yethe lines dc & cb, I consider ytthat yethe point c fixed in yethe line cb moves towards e parallely to ab (for so doth yethe line cb (by supp:) & consequently all its points): also yethe point c fixed in yethe line dc moves towards g parallely to ad (by sup): therefore I draw ce∥ab & cg∥ad, makeing ynthen & in such proportions as yethe motions they designe & so draw er∥cb, & gr∥dc, & yethe diagonall cr, (by Prop 6), ytthat is, if yethe velocity of yethe line cb, (ytthat is yethe celerity of yethe increasing of ab, or dc; or yethe velocity of yethe point c from d) bee called p, & yethe velocity of yethe line cd bee called q; I make ce∶gc∷p∶q(∷ce∶er∷hb∶cb.) & yethe point c shall move in the diagonall line cr (by prop 6) wchwhich is therefore the required tangent Now yethe relation of p & q may bee found by yethe foregoing Equation (p signifying yethe increase of x, & q of y) to bee 4px3−9pyxx+30paxx+2payx−2py3−3qx3+qaxx−6qyyx−4qy3=0. by Prop 7 And therefore . wchwhich determines yethe tangent hc
☞ Hence may bee observed this Generall Theorem for Drawing TangntsTangents to crooked lines thus referred to streight ones; ytthat is, to such lines in wchwhich y=bc is ordinately. applyed to x=ab at any given angle abc. viz: Multiply the termes of yethe Equation ordered according to yethe dimensions of y, by any ArimeticallArithmeticall progressiōon wchwhich product shall bee yethe Numerator: Againe change yethe signes of yethe Equation & ordering it according to x, multiply yethe termes by any Arithmeticall progression & yethe product divided by x shall bee yethe Denominator of yethe valor of (hb, ytthat is, of) x produced from y to yethe tangent hc.
As if
. 2.1.0.Then firstyy∗=0−rx0.-1.-2., produceth 2yy, or, 2rx−2xx. Secondly 2.1.0.Secondly+rx−yy produceth
. Therefore
. Or else
.
Example 2. If yethe crooked line chm bee describe by yethe intersection of two lines ac, bc circulating about their centers a & b, soe ytthat if ac=x, & bc=y; their relation is x3−abx+cyy=0. To draw yethe tangent ec I consider ytthat yethe point c moves to fixed in yethe line bc moves towards f in yethe line cf⊥bc (for yethe tangnttangent to a circle is perpendicular to its radius). also yethe point c fixed in yethe line ac moves towards d in yethe line cd⊥ac & from those lines cd & cf &I draw two others de∥cg & ef∥bc wchwhich must bee in such proportion one to another as yethe motions represented by ymthem (prop 6), ytthat is (prop 6) as yethe 54 as yethe motions. of yethe intersection point c moveing in yethe lines ca & cb to or from yethe centers a & b; ytthat is, as yethe velocity of yethe increase or decrease of ca=cx & cb=yn (yethe celerity of yethe increase or decrease of x being called p, & of y being q), de∶ef∷p∶q. Then shall yethe diagonall ce bee yethe required tangent. Or wchwhich is yethe same, (for △ecng=△ecd, & △ecf=△ecgn,) I produce ac & bc to g & n, so ytthat cg∶cn∷p∶q. & ynthen draw ne ne⊥bn, & ge⊥nag; & yethe tanegent diagonall ce to their intersection point e. Now yethe relation of p & q may bee found by yethe given Equation to bee, 3pxx−pab+2qcy=0 (by prop 7) Or 2cy∶ab−3xx∷p∶q∷cg∶cn, wchwhichdetermins yethe tangent ce.
But note ytthat if p, or q be negative cg or cn must bee drawn from c towards a or b, but from a or b if affirmative.
Hence tis easy to pronounce a sTheorems for Tangents in such like cases & yethe like may bee done in all other cases however Geometricall lines bee referred to streight ones.
Tangents to Mechanichall lines
Example yethe 3d. If yethe Quadratrix kbf is described by yethe intersection (b) of yethe two lines hb & ap, yethe one hp∥ma moving uniformly from k to a, whilest yethe other ap circulates from k to m about yethe center a. I dDraw yethe circle gbl wthwith yethe Rad. ab; & make ab⊥bd=bf bf= bl=bd⊥ab∥de; & to yethe intersection point e of yethe lines am & ed draw ytthat tan eb e wchwhich shall touch yethe Quadratrix in b. For suppose yethe motion of yethe point p fixed in yethe line ap, towards m to bee pm, ynthen yethe motion of yethe point b fixed in yethe line ab, towards d is bl=bd, (prop 4), & yethe motion of yethe line bh towards ca, & therefore of yethe point b fixed in it towards c (prop 3) is ha=bc (by supp): Also ce∥bh & ed∥ap (sup). Therefore (by Prop 6) is yethe diagonall eb yethe motion of yethe intersection point b of those two lines ap & hb, moves in yethe diagonall eb, & consequently eb toucheth yethe Quadratrix in b.
Example 4th
Prob 2d. To find yethe quantity of crookednesse of lines.
Lemma. The crookednesse of equall parts of circles are as their diameters reciprocally. For the crookednesse of a whole circle (acdea, bfghmb) amounts to 4 right angles. Therefore there is not more crookednesse in one whole circle acdea ynthan in another bfghmb. Suppose yethe perimeter acde=bfgh. Then tis, ar∶br∷ bfgh=acde∶bfghmb∷ crookednesse of bfgh∶crookednesse of bfghmb=crookednesse of acdea.
Lemma 2. That point of a Crooked Line's Perpendicular wchwhich is in least motion is yethe center required of yethe circle of equall whose crookednesse wthwith yethe given is required.
Resolution. This may bee done by drawing perpendiculars to 2 curved lines described by yethe motion of any two points fixed in yethe Tangent, wchwhich two perpendiculars shall intersect in yethe center of a Circle whose crookednesse was required. One of those curve lines & it perpendiculars may bee yethe given crooked line & yethe perpendicular to it at its given point.
But if y bee ordinately applyed to x at right angles yethe best way is
Resolution. Find That point of fixed in yethe crooked line's perpendicular wchwhich is ynthen in least motion, for it is yethe center of a circle wchwhich passing through yethe given point is of equall crookednesse wthwith yethe line at ytthat given point. To find wchwhich point I observe ytthat Now, since yethe crooked line's tangent & perpendicular &c: (at ytthat moment) circulate about ytthat center; I observe, 1st ytthat every point fixed in yethe TangntTangent or Perpendicular, or whose position to ymthem is determined, doth describe a curve line to wchwhich yethe right line drawne from ytthat center is perpendicular, & is also yethe radius of a circle of equall crookedness wthwith it: 2dly ytthat yethe motion of every such point is as its distance from ytthat center; & so are yethe motions of yethe intersection point, in wchwhich any radius drawn from ytthat center intersects two parallel lines.
Example 1. If cb=y is ordinately applyed to ab=x at a right angle abc, nc being tangent & mc perpendicular to yethe curve line ac: I draw seeke yethe motion of two points c & d fixed in yethe perpendicular cd; or (wchwhich is better & to yethe same purpose) I draw cg∥ab, & seeke yethe motions of yethe two intersection points c & d in wchwhich yethe perpendicular cd intersects those fixed lines cfg, & abdk: & ynthen draw cg & dk in such proportion as those motions are, & yethe line' gkm drawn by their ends shall intersect yethe perpendicular cd in yethe required center m: mc being yethe radius of a circle of equall crookednesse wthwith yethe curve line ac at yethe point c. Now, Making △cegfe= & like △ncdbc, suppose yethe motion of yethe line cb & consequently of yethe points c & b fixed in it & moveing towards f f & k, to bee p=nb=cf: Then is ce=cn yethe motion of yethe point c in wchwhich yethe lines ne & bc intersect, ytthat is bc intersects yethe tangent ne (by Prop 6), ytthat is of yethe point c infixed in yethe perpendicular cd, & moveing in yethe tangent ne: & therefore cg=nd==p+bd(=p+v if bd=v), is yethe motion of yethe intersection point c towards g in wchwhich point yethe perp: cd intersects cg (by Prop 6). If also yethe motion of yethe intersection point d from yethe point b (ytthat is yethe velocity of yethe increase of v) bee called r, ynthen is dk=nb+r=p+r soe ytthat, cg−dk∶cd∷cg∶cm; ytthat is, . Also . Lastly yethe motion of yethe point c from b, (ytthat is yethe velocity wthwith wchwhich y=cb increaseth) will bee q=cb=y.
As if yethe nature of yethe crooked line bee x3−axy+ayy=0. Then is (by examp: 1. Prob: 1.), & (for nb∶bc∷bc∶bd) Soe that 3xxy−ayy+−axv+2bayv=0. & therefore (by Prop 7) 6pxy−pavqay−2qay+3qxx+2qbav−rax+2rbay=0 & substituting yethe55 & substituting yethe , & , & y, into yethe places of p, v, & q in yethe this equation The product will bee . Or 6aax3yy . And therefore . Also . Soe ytthat . That is . Which Equation gives yethe point k & consequently yethe point m. for km∥abk.
☞ But in such cases where y is ordinately applyed to x at right angles, From yethe consideration of yethe Equation
; Or rather
: may yethe following Theoreme bee pronoundced. To wchwhich purpose let X signify the given Equation, ytthat is, all yethe algebraicall termes (expressing yethe nature of yethe given line) considered as equall to nothing & not some of ymthem to others. Let X· signify those termes multiplyed by ordered according to yethe dimensions of x & ynthen multiplyed by any arithmeticall progressiōon Let X· signify yethe those termes ordered according to yethe dimensions of y & ynthen multiplyed by any Arithmeticall progression. Let :X signify those termes ordered by x & ynthen multiplyed by any two arithmetiacall progressions one of ymthem being greater ynthan yethe other by a terme. Let X: signify those termes ordered by y & ynthen multiplied by any two Arith: Progr: one of ymthem being greater ynthan yethe other differing by a terme. Let ·X· signify those termes ordered according to x, & ynthen multiplyed by yethe greater of yethe progressions wchwhich multiplyed :X; & ynthen ordered by y & multiplyed by yethe greater progression wchwhich multiplyed X:. Then (observing ytthat all these progressions have the same difference & proceede yethe same way in respect of yethe dimension orof x & y) will yethe 3 Theorems bee
1st.
.
2d. .
3d.
radio circuli æqualis curvitatis cum curva ac in puncto c.
As if yethe line bee x3−axy+ayy=0. Then is ·X=3x3−axy. X·=−axy+2ayy. :X=6x3=3×2.1×0.0 x−1.
:X=6x3=x3−axy+ayy. 0x−1.1×0.2×1.X:=2ayy=x3−axy+ayy & 3×0.1×1.0×2.·X·=−axy=x3−axy+ayy. Which valors of ·X, X· &c being substituted into their places in yethe first rule, yethe result is;
. Which being conveniently reduced is
. As was found before.
.
Or suppose yethe line is a Conick section whose nature
. Then is ·X=rx+
210
=rx
xx+rx−yy
.
0312
X·=−2yy=
xx+rxyy∗−y
.
2×1.1×0.0x−1.
:X=xx=
xx+rx−yy
.
0x−1.0x−1.1×0.2×1.
X:=−2yy=
xx+rx∗−yy
.
2×01×00×2
·X·=0=
xx+rx−yy
. Which valors of ·X, ·X, :X, &c being substituted into their places in yethe first Theoreme, give
. Or (since 4qryy−4qrrx−4rrxx=0) tis,
. So by yethe second Theoreme tis
. Or
+
+2x+
. And so by yethe 3d Theoreme, tis
. Or
. Or
.
This Theoreme may bee thus Demonstrated
Note ytthat yethe curvity of any curve whose ordinates are applyed at inclined from right to oblique angles is to yethe curvity of the same curve whose ordinates are at right angles, as the curvity of an Ellipsis whose ordinates are in the sa a circle whose ordinates are in like manner inclined so as to make it becom an Ellipsis.
Prob: To find yethe points of curves where they have a given degree of curvity. 56
Prob 3d. To find yethe points distinguishing twixt yethe concave & convex portions of crooked lines.
Resolution. The lines are not crooked at those points: & therefore yethe radius cm determining yethe crookednesse at ytthat point must bee infinitely greate. To wchwhich purpose I put yethe denominator of its valor (in rule yethe 3d) to bee equall to nothing, & & so have this Theoreme ·X·XX:−2·XX··X·+X·X·:X=0. Or better ꝑperhaps
.
Example, was this point to be found in yethe Concha whose nature is
+bb
x4+2bx3+ccxx−2bccx−bbcc=0.
+yy
. Then is ·X=2x4+2bx3+2bccx+2bbcc. X·=2xxyy=X: :X=2x4−4bccx−bbbcc. ·X·=0. Which valors subrogated into yethe Theoreme, they produce 2x4+2bx3+2bccx+2bbcc+
. And subrogating
−bb
−x4−2bx3−ccxx +2bccx+bbcc yethe valor of xxyy into its stead & twice reducing yethe Equation by yethe divisor x+b. Tis x3+bcc+
. Or x4+4bx3+3bbxx−2bccx−2bbcc=0. Which being againe reduced by x+b=0. Tis x3+3bxx−2bcc=0. See Geometr: Chart: pag: 259.
Probl: 4. To find yethe points at wchwhich lines are most or least crooked.
Resol: At those points yethe afforesaid radius cm neither increaseth nor decreaseth. So ytthat yethe center m in ytthat moment doth absolutely rest, & therefore neither yethe line bk nor al doth increase or diminish, ytthat is, ck & bl doe soe much increase or diminish as xy & x (lb & ab) doe diminish or increase. Or in a word the point m resteth. Find therefore the motion of al or cm or lm & suppose it t nothing.
Thus in the Concha to find the point of least crookednes beyond yethe point of reflection, having substituted yethe valors of ·X, X· &c (exprest in the precedent problem) into this The computation is too tedious
Thus to find the point of least crookednesse in yethe curve x3=ccyy By the rule in prob 2 I make ·X=3x3. X·=−ccy. :X=6x3. X:=0 & ·X·equals;0 & thence obteine , or wchwhich is least when or . And then therefore happens the greatest crookednesse
In like manener if yethe curve be xxy=a3 The rule gives ·X=2a3 X·=a3 = :X=−6a3 X:;=0=·X· And thence which is least when
.
Soe if yethe curve bee x3=byy. Thein is ·X=3x3. X·=−2byy. :X=6x3 X:=−2yyc. ·X·=0. And therefore ck=3y+
, which hath no least when x=
nor the curve any least crookednesse. 57
Prob 5t. To find yethe nature of yethe crooked line whose area is expressed by any given equation.
That is, yethe nature of yethe area being given to find yethe nature of yethe crooked line whose area it is.
Resol. If yethe relation of ab=x, & abc=y bee given & yethe relation of ab=x, & bc=q bee required (bc being ordinately applyed at right angles to ab). Make de∥ab⊥ad∥be=1. & ynthen is □abed=x. Now supposing yethe line cbe by parallel motion from ad to describe yethe two superficies ae=x, & abc=y; The velocity wthwith wchwhich they increase will bee, as be to bc: ytthat is, yethe motion by wchwhich x increaseth being be=p=1, yethe motion by wchwhich y increaseth will bee bc=q. which therefore may bee found by prop: 7th. viz:
.
Example 1. If 4rx3=9yy
. Or −4rx3+9yy=0. Then is
. Or rx=qq & therefore abc is yethe Parabola whose area abc is
.
Example 2d. If x3−ay+xy=0. Then is
. Or
Example 3d. If
. Then is
. Or if axm=bxn; ynthen is
.
Note ytthat by this probleme may bee gathered a Catalogue of all those lines wchwhich can bee squared. And therefore it will not bee necessary to shew how this Probleme may bee resolved in other cases in wchwhich q is not ordinately applyed to x at right angles.
Prob 6. The nature of any Crooked line being given, to find other lines whose areas may bee compared to yethe area of ytthat given line.
Resol: Suppose yethe given line to be ac & its area abc=s, yethe sought line df & its area def=t; & ytthat bc=z is ordinately applyed to ab=x, & ef=v to de=y, soe ytthat ∠abc=∠def; & ytthat yethe velocitys of y wthwith wchwhich ab & de increase (ytthat is, yethe velocity of yethe points b & e, ofr of yethe lines bc & ef moving from a & d) bee called p & q. Then may yethe ordinately applyed lines bc & ef multiplyed by their velocitys p & q, (ytthat is pz & qv) signify yethe velocitys wthwith wchwhich yethe areas abc=s & def=t increase. Now yethe relation of yethe areas s & t (taken at pleasure) gives yethe relation of yethe motions pz & qv describing those areas, by Proposition yethe 7th.; Also yethe relation of yethe lines x & y ab=x & de=y (taken at pleasure) gives yethe relation of p & q, by Prop 7th: wchwhich two equations, together with yethe given Equation expressing yethe nature of yethe line ac, give yethe relation of de=y & ef=v ytthat is yethe desired nature of yethe line df.
Example 1. As if ax+bxx=zz is yethe nature of yethe line ac: & at pleasure I assume s=t &to be yethe relation of yethe areas abc & def; & x=yy to bee yethe relation of yethe lines ab & de. Then is pz=qv (prop 7), & p=2qy (by prop 7). Therefore 2yz=v (by yethe 2 last Equations), &
=zz=ax+bxx=ayy+by4. or vv=4ay4+4by6. &
: WchWhich is yethe nature of yethe line def whose area def is equall to yethe area abc, supposing
.
Example 2. If ax+bxx=zz as before; & I assume, as+bx=t, & x=yy. Then is apz+bp=qv (by prop 7), & p=2qy. 2azy+2by=v=2by+2ay
. Or v=2by+2ayy
; The required nature of yethe line def.
Example 3d. If ax+bxx=xx as if
: & at pleasure I assume
, & xx−a=y. Then is 4ccay+ccyy=aass+2aast+aatt, And (by prop 7) 4ccaq+8ccyq=2aaspz+2aatpz+2aasqv+2aatqv=
. & (by prop 7) 2px=q. Therefore
. But
. Therefore &
, Therefore
. That is 8 ca+2cy=2cy+
. Or
.
Example 4th. If ax+bxx=zz as before & I assume ss=t, & x=yy. Then is 2spz=qv, & p=2qy (by prop 7) Therefore
. Where note ytthat in this case yethe area line
is a Mechanicall one because s yethe area of yethe line ax+bxx=zz canott bee Geometrically found. The like is to bee observed in other such like cases.
Probl: 7. The Nature of any Crooked line being given to find its area when it may bee. Or more generally, two crooked lines being given to fingd the relation of their areas, when it may bee.
Resol: Reassuming yethe figure of the 5t probleme in wchwhich yethe given line is ac, & supposing its area y & xthe other area x to bee described by yethe line ce moving from a as was there taught: Suppose the motion describing x bee p=be=1, Then yethe motion describing y is cb=q=
. Now having yethe relation twixt ab=abed=x, & bc=q=
. (By supp:); I may (some times viz when it may bee) find
Resolution. In yethe figure of yethe fift probleme Let abc=y represent yethe area of yethe given line acf; cb=q yethe motion describing ytthat area; abed=x another area wchwhich is equall to yethe basis ab=x of ytthat given line acf, (viz: supposing ab∥de⊥be∥ad=1); & be=p=1 yethe motion describing ytthat other area. Now haveing (by supp) yethe relation twixt ab=x=abed, & bc=q= given, I seeke yethe area abc=y by yethe Eight proposition
Example 1. If yethe nature of yethe line bee,
. I looke in yethe tables of yethe Eight proposition for yethe Equation corresponding to this Equatiōon wchwhich I find to bee
, (For if instead of c, a, b, n I write a, aa, −1, 2, it will be
.) And against it is yethe equation . And substituting a, aa, −1, 2 into yethe places of c, a, b, n it will bee, yethe required area.
Example 2. If
. TBecause there are two termes in yethe valor of bc I consider them severally & first I find yethe area correspondent to yethe terme
, or
; To bee
, or
, by prop: 8. part 1. Secondly to find yethe area corresponding to yethe other terme I looke yethe Equation (in prop 8. part 3) corresponding to it wchwhich is , (for if instead of c, a, b, n, I write eeb, −1, a, −1, it will bee
, Or
): Against which is yethe Equation
. In 58 In which writing eeb, −1, a, −1, instead of c, a, b, n, the result will bee
; Or,
wchwhich is the the area corresponding to ytthat other terme. Lastly Now to see how these areas stand related one to another I draw yethe annexed scheme, in which is ab=x, bE=q=, adg yethe fkeg the given curved line, & adg yethe curve line described by one of its parts
ab=x.
.
. &
. Soe that
is yethe superficies corresponding to yethe first terme bd, wchwhich because it is affirmative must bee extended (or lye) from yethe side yethe from yethe line bd towards a. Also yethe other superficies correspondent to yethe 2d terme de, being negative must lye on yethe other side bd from a, wchwhich is therefore gde
. Lastly if x=ab=r Then is
. &
. And if aB=x=s, ynthen is
, And
. Soe ytthat,
. &
& substracting DdeE from bBDd there remaines bBEe
. wchwhich is yethe required Area of yethe given line &cfkeEg. Where note ytthat takging for yethe quantitys r=ab & s=aB taking any numbers you may thereby finde yethe area bBdD correspond to their difference bB.
Note ytthat sometimes one parte of yethe Area may bee Affirmative & yethe other negative. as if a2B=r, & ab=s. Then is b2B2Ee=kbe−k2B2BE==
=b2B2Ee=kbe−k2B2E. 59
Prob 89. To find such crooked lines whose lengths may bee found &also to find theire lengths.
Lemma 1. If to any crooked immovable line acg the streight line dcmσ moves to & fro perpendicularly every point of yethe said line cdmσ (as γ, θ, σ, &c) shall describe a curve line (as βγ, δθ, λσ, &c) all which will bee perpendicular to yethe said line cdmσ, & also parallel one to another & to yethe line acg.
2. If acg bee not a circle, there may bee drawne some curve line βδmλ, wchwhich yethe moving line cdmδ will always touch in some point or other (as at m) & to wchwhich therefore all yethe curve lines βγ, δθ, λσ, &c: are perpendiculars.
3. Soe that every point (γ, θ, m, σ, &c:) of yethe line cdmσ, when it begineth or ceaseth to touch yethe curve line βδmλ, doth ynthen move perpendicularly to or from it: & therefore yethe line γdmσ doth not at all slide upon yethe curve line βδmλ, but exactly measure it by applying it selfe to it point by point: & therefore yethe correspondent parts of yethe said lines are equall (viz: βm=γm. δm=θm. δλ=θσ. &c).
Resol. Take any Equation for yethe nature of yethe crooked line acg, & by yethe 2d probleme find yethe center m of its crookedness at c. That point m is a point of yethe required curve line βδmλ. For ytthat point m whereat cdmσ yethe perpendicular to acg doth touch yethe curve line βδmλ is lesse moved ynthan any other point of yethe said perpendicular, beeing as it were yethe hinge & center uponabout wchwhich yethe yethe perpendicular turnethmoveth at that moment.
Example 1. If acg is a Parabola whose nature is xx −rr=yy (supposing ab=x⊥bc=y) is al rx=yy. By Theoreme yethe 1st of Problem 2d I find cb+lm=y+
. By yethe 2d Theoreme, bl=2x+r. And by yethe 3d Theoreme . Soe ytthat supposing ab=r. bl=3x=z. . The relation twixt v & z will bee 27rvv=16z3, (for r4vv=16y6=16y3x3=) wchwhich is yethe nature of yethe required line βδmλ. And since cγ=aβ=r. Therefore . is yethe length of its ꝑparte βδm.
Example 2. Soe if aa=xy, is yethe nature of yethe line acg. By yethe afforesaid Theoreme I find, cb+lm=; &, ; & whereby yethe nature of yethe curve line βδmλ is determined. And lastly I find which determine its length.
Prob: 10. Any curve line being given to find other lines whose lengths may be compared to its length or to its area, & to compare ymthem.
Resolution. Take any Equation for yethe relation twixt ad & yethe perpendicular cd=y (whither ytthat relation bee expressed by an Geometricall Equation or whither it bee yethe same wchwhich some streight line beares to a curved one or to its superficies &c). Then (by prop 7) find yethe relation twixt yethe increarse or decrease (p & q) of yethe lines ad=x, & dc=y. & say (by prop 6) ytthat q∶p∷dc∶dn=
. And soe is yethe triangle dnc (rectanguled at c) given & consequently yethe Nature of yethe curve line acg to wchwhich dc is perpendicular, & cn a tangent.
Now yethe nature of yethe center (m) of yethe perpendiculars motion (wchwhich gives yethe nature & length of yethe required curve line βm) may be found as in yethe 2d or 9th Probleme, But more conveniently thus. Draw any fixed line he∥ad⊥ah=a=ef⊥ad. Also call fd=v. & yethe increase or decrease of (fd) call r. And yethe increase or decrease of yethe motions p & q, call γ β & γ Now considering ytthat yethe motion (p+r) of yethe intersection point e in yethe line he is to yethe motion (p) of yethe intersection point (d) in yethe line (ad), as (em) is to (dm) (see prob 2); That is ytthat yethe difference (r) of those motions is to yethe motion (p) of yethe point (b) soe is (ed) to (dm): First I find yethe valor of v. viz
∶q∷cn∶cd∷∷ef=a∶fd=v=
. Or aaqq+vvqq−vvpp=0. Secondly by this Equation I find yethe valor of r, viz: (by Prob 7), 2aaqγ+2vvqγ+2rvqq−2rvpp−2vvpβ=0. Thirdly
. Lastly supposing the motion p to bee uniforme ytthat its increase or decrease β may vanish, & also substituting ppzz yethe valor of aaqq+vvqq in its stead in yethe Denominator of dm, yethe result will bee
, if p=1. And to find yethe lines dl & ml, say −p∶q∷dm∶dl=
. &
Soe ytthat by the equation expressing yethe relation twixt x & y first I sefinde q & ynthen γ. wchwhich two give me
&c.
Example 1. If yethe relation twixt x & y bee supposed to bee yy−ax=0. Then (by Prop 7) I find, first 2qy−ap=0, Or 2qy−a=0. & secondly 2γy+2qq=0. And substituting these valors of
& γ=
in their stead in yethe Equatiotion
, &c: The results will bee
.
. &
. (And adding cd=−y to dm, & ad=x to dl =
to dl, yethe result is ad cm=
. & al=
.) WchWhich determine yethe nature & crookednesse of yethe line βm. (For if aβ=
& βl =z=al−ab=al−a aβ=
. βl=z. lm=v. The relation twixt z & v will bee 16z3=27rvv. The length of βm being
. I as before was found).
Example yethe 2d. If x=ad⊥dk=
, is yethe nature of yethe crooked line (yethe Hyperbola) gkw: And I would find other crooked lines (βm) whose lengths may bee compared wthwith yethe area sdkg (calling as=b⊥gs) of that crooked line gkw. I call 60 I call ytthat area sdkg=ξ, & its motion θ. Now since 1∶dk∷p∶θ, (see prob 5. & yethe Note on prop 7 in Example 3); Therefore θ dk×p=dk=θ=
. This being knowne I take at pleasure any Equation, in wchwhich ξ is, for yethe valor of cd=y.
As 1st suppose ay=ξ, That (by prop 7) gives aq=θ=
, Or qx=a; wchwhich also (by prop 7) gives γx+qp=0. Which valors of q=
, & γ=
; by helpe of yethe Theorems
&c: doe give
.
. &
=lm. wchwhich determines yethe nature of yethe required curve line βm. The length of ytthat portion of it wchwhich is intercepted twixt yethe point m & the curve line acg being −cm=cd+dm=
, Or mc=
.
☞ [Note ytthat in this case although yethe area sdkg=ξ cannot bee Geometrically found & therefore yethe line acg is a Mechanicall one yet yethe desired line βm is a Geometricall one. And yethe like will happen in all other such like cases, when in yethe Equation taken at pleasure to expresse yethe relation twixt x, y, & ξ; neither x, y, nor ξ doe multiply or divide one another, nor it selfe, nor is in any denominator or roote, except x wchwhich may multiply it selfe & bee in sometimes denominators & rootes, when y or ξ are not in those onefractions or rootes. & herein onely doth this excell the precedent 9th probleme. Such is this Equation aξ−aby+ax3+
. &c: But not this ξξ=a3y. nor ξ=xy. &c]
Secondly suppose ξ=xy. ytthat (by prop 7) give θ=qy+xq=
, Or xy+xxq=aa, & ytthat (by prop 7) gives y+qx+2qx+γxx=0. Which two valors of
, And
; by meanes of yethe Equatio Theoremes
,
, &
; doe determine yethe nature & length of yethe desired curve line βm.
Example yethe 3d. In like manner to find curve lines whose lengths may bee compared to yethe length gk of yethe said curve (Hyperbola) gkw. Call gk=ξ, & its motion θ. Now, drawing kh yethe tangent to gkw at k, I condsider that ad=x & gk=ξ doe increase in yethe proportion of dh to kh; ytthat is, dh∶kh∷p∶θ. Now finding (by prob 1) ytthat dh=−x, &
; therefore is
. Which being found, I take any equation, in wchwhich ξ is, for the valor of cd=y. & ynthen worke as in yethe precedent Example.
Note ytthat by this &or yethe Ninth Probleme may bee gathered a Catalogue of whatever lines, whose lengths can bee Geometrically found.
Prob 12. To find yethe Length of any given crooked line when it may bee.
Resolution. The length of any streight line to wchwhich yethe curve line is cheifly related being called (x), yethe length of yethe curve line (y), & theire motions (p & q) first (by prob 1) get yethe an equation expressing yethe relation twixt x & , & ynthen seeke yethe valor of (y) by yethe Eight proposition. [Or find a curve line whose area is equall to yethe length of yethe given line, by Prob 11. And then find that area by Prob 7.] 61
Prob 11. To find curve lines whose Areas shall bee Equall (or have any other given relation) to yethe length of any given Curve line drawn into a given right line
Resolution. The length of any streight line, to wchwhich yethe given curve line is cheifely referred, being called x, yethe length of yethe curve line y, & their motions of increase p & q. Get The valor of (tefound by yethe first probleme) being ordinately applyed at right angles to x, gives yethe nature of a curve line whose area is equall to (y) yethe length of yethe curve line.
And this thLine thus found gives (by prob 6) other lines. whose areas have any other given relation to yethe length (y) of yethe given curve line
Prob 13. To find yethe nature of a Crooked line whose length is expressed by any given Equation, (when it may bee done).
Resolution. Suppose ab=x, bc=y, ac=z. & their motions p, q, r. And let yethe relation twixt x & z bee supposed given. Then (by prop 8) finding the relation twixt p & r make
. (For drawing cd tangent to ac at c & de⊥cb⊥ab: yethe lines de, ec, dc, shall bee as p, q, r. but
, & therefore
). Lastly, yethe ratio twixt x & being thus knowne, seeke y (by prop 8). Which relation twixt ab=x & bc=y determines yethe nature of yethe crooked line ac=z. 62
Of Gravity.
DefinitnDefinition. 1. I call ytthat point yethe center of Motion in any Body, wchwhich always rests when or howsoever ytthat Body circulates wthwithout progressive motion. It would always bee yethe same wthwith yethe center of Gravity were yethe Rays of Gravity parallel & not converging towards yethe center of yethe Earth.
Lemma 1 Def: 2 And yethe right lines passing through ytthat point I call yethe axes of Motion or Gravity.
Lemma 1 The place & distance of Bodys is determinded by their center of Gravity. Which is yethe middle point of a right line circle &or Parallelogram:
Lemma 2 Those weights doe equiponderate whose quantitys are reciprocally proportionall to their distances from the common axis of Gravity, supposing their centers of Gravity were to bee in yethe same plaine wthwith ytthat common axis of Gravity.
Prob To find yethe center of Gravity in rectilinear plaine figures
1. In yethe Triangle acd make ab=bc, & cd=fd. & draw db, & af, their intersection point (e) is its center of Gravity.
In yethe Trapezium abdc, draw ad & cb. Ioyne yethe centers of Gravity e & h, f & g of yethe opposite triangles acb & dcb, bad & adc wthwith yethe lines eh, fg. Their intersection point n is yethe center of Gravity in yethe Trapezium. (And so of Pentagons, hexagones &c)
Propb: To find such plaine figures wchwhich are equiponderate to any given plaine figure in respect of an axis of Gravity in any given position.
Resol That yethe natures & positions of yethe given curvilinear plaine (gbc,) & sought plaine (bde) bee such ytthat they may equiponderate in respect of yethe axis (ak;) I suppose x=ab⊥bc=z, & y=ad⊥de=v to bee either perpendicular or parallel or coincident to yethe said axis ak: And yethe motions whereby x & y doe increase or decrease (i:e: yethe motions of bc, & de to or from yethe point a) I call p & q. Now yethe ordinatly applyed lines bc=z, & de=v, multiplyed into their motion p & q (ytthat is, pz, & qv) may signify yethe infinitly little parts of those areas (acb, & lde) wchwhich each moment they describe; wchwhich infinitly little parts doe equiponderate (by Lemma 1 & 2), if they multiplied by their distances from yethe axis ak doe make equall products. (ytthat is; pxz=qyv, in fig 1: pxz=qvv in fig 2: pzz=qv×fm, in fig 3; supposing dm=me. &) And if all yethe respective infinitly little parts doe equiponderate yethe superficies must do so too.
Now therefore, (yethe relation of x & yz being given by yethe nature of yethe curve line cg,) I take at pleasure any Equation for yethe relation twixt x & y, & thereby (by prop 7) find p & q, & so by yethe ꝑprecedent Theorem find yethe relation twixt y & v, for yethe nature of yethe sought are plaine lde.
Exam: 1. If cg (fig 2) is an Hyperbola, soe ytthat aa=xz. & I suppose 2x=y. ynthen is 2p=q (prop 7). & paa=pxz=qvv=pvv. Or aa=vv. or a=v=de. Soe ytthat le is a streight line, & lde a parallelogram, wchwhich equiponderates wthwith yethe Hyperbola cgkabc (infinitly extended towards gk) if 2ab=ad. al×al=ab×bc.
Example 2. If cg (fig 3) is a circle whose nature is,
. & I suppose at pleasure 3aax−x3=6aay. Then (by prop 7) I find 3aap−3xxp=6aaq. And therefore paa−pxx=
=qv×fm=(if fd=a,)
. Or
. Or 2aa=vv+av. Or
; . Soe that le is a streight line & alde a parallelogram
Example 3. If abcg is a parallelogram (fig 4) whose nature is, a=y a=z. & I suppose at pleasure x=yy−b. Then (by prop 7) tis p=2qy. Therefore aap=aaqy=pzz=qvy. Or aa=v. Soe ytthat aed is a parallelogram.
Or if I suppose at pleasure, x=y3−b. Then is (prop 7) p=3qyy. & therefore aap=
aaqyy=pzz=qvy. Or 3aay=2v.
Or if I suppose x=y4−b. ynthen is p=4qy3 & 2aayy=v. so that aed is a Parabola. [Soe if xx=y5. ynthen is 2px=5qy4. &,
=ap=pzz=qvy. Or 5aaqy
=4qvy. & 25a4y=16vv. soe ytthat aed is a pParabola.]
Example 34 If gbc (fig 1) is an Hyperbola whose nature is xx−aa=qy zz. & I suppose x=y+b. Then (by prop 7) is p=q. Therefore
=pxz=qyv=pyv. Or
. Or
in
. &c.
Or if I suppose xx=2y. Then is 2px=2q. Therefore
=pxz=qyv. Or (xx−aa=)2y−aa=yyvv.
Or if I suppose xx=yy+aa. Then (prop 7) is 2px+=2qy. Therefore qyz=pxz=qyv. Or
. & y=v; so ytthat aed is a triangle
Note ytthat This Probleme may bee resolved although the lines have a x, z, y, v, & ak have any other given inclination one to another, but the ꝑprecedent cases may suffice
Note also ytthat if I take a Parallelogram for yethe knowne superficies (as in yethe 3d Example I may thereby gather a Catalogue of all such curvilinear superficies whose weight in respect of yethe axis, may bee knowne.
Note also I might have shewn how to find lines whose wh weights in respect of any axis are not onely equall but have also any other given proportion one to another. And ynthen have made two Problems instead of this, as I did in Probl: 5 & 6: 9 & 10. 63
Prob 15. To find yethe Gravity of any given plaine in respect of any given axis, given in position when it may bee done.
Resol: Suppose ek to bee yethe Axis of Gravity, acb the given plaine, cb=y, & db=z to bee ordinatly applyed at any angles to ab=x. Bisect cb at m & draw mn⊥ek. Now, since [cb×mn] is yethe gravity of yethe line [cb], (by lem 1 & 2); if I make cb×mn=db=z, every line db shall designe yethe Gravity of its correspondent line cb, ytthat is, yethe superficies adb shall designe yethe Gravity of yethe superficies acb. Soe ytthat finding yethe quantity of ytthat superficies adb (by prob 7) I find yethe gravity of yethe supeerficies acb.
Example 1 If ac is a Parabola; soe ytthat, ab bc=4, & z=da∥k parallelTo;ak=axis rx=yy, & yethe axis ak is ∥ dcb. &, nb⊥ak, & ytthat, ab∶nb∷d∶e. Then is bc×nb=y×
. Or eerx3=ddzz, is yethe nature of yethe curve line ad. whose area (were abd a right angle would be
but now it) is
, (by prob 7) wchwhich is yethe weight of yethe area acb in respect of yethe axis ak.
Examp: 2 If ac is a Circle
Prob 16. To find yethe Axes of Gravity iof any Plaines
Resol. Find yethe quantity of yethe Plaine (by Prob 7) wchwhich call A & yethe quantity of its gravity in respect of any axis (by prob 15) wchwhich call B. & parallell to ytthat axis draw a line whose distance from it shall bee
. That line shall bee an Axis of Gravity of yethe given plain
Or If you y cannot find yethe quantity of the plane: Then find its gravitys in respect of two divers axes (AB & AC) wchwhich gravitys call B & C & D. & through (A) yethe intersection of those axes draw a line AD wthwith this condition ytthat yethe distances (DB & DC) of any one of its points (D) from the said axes (AB & AC), bee in such proportion as to the gravitys of the plane. That line (AD) shall bee an axis of gravity of yethe said plane EF.
Prob 17 To find yethe Center of Gravity of any Plaine, when it may bee
Resol Find two axes of Gravity &by the precedent Prop, & their common intersection is yethe Center of Gravity desired. If yethe figure have any knowne Diameter that may bee taken for one of its axes of Gravity.