Principia Mathematica (Motte Translation, 1848) — part 11 of 45

Under the preceding Propositions are comprehended those Problems wherein either the centres or asymptotes of the trajectories are given. For when points and tangents and the centre are given, as many other points and as many other tangents are given at an equal distance on the other side of the centre. And an asymptote is to be considered as a tangent, and its infinitely remote extremity (if we may say so) is a point of contact. Conceive the point of contact of any tangent removed in wjinitiun, and the tangent will degenerate into an asymptote, and the constructions of the preceding Problems will be changed into the constructions of those Problems wherein the asymptote is given.

After the trajectory is described, we may find its axes and foci in this manner. In the ^^--^r~~-~^

construction and figure of Lem. XXI, let those , /^/ \

legs BP, CP, of the moveable angles PBN, / / K p

PCN, by the concourse of which the trajec- \jJ\ / J ''\ \ tory was described, be made parallel one to |i"^"^^"^-^|~^ ^ \ the other; and retaining that position, let cV-— — «1^^-^^' them revolve about their poles B, G, in that \

figure. In the mean while let the other legs ^rjj^ ^-^

CN, BN, of those angles, by their concourse

K or Ic, describe the circle BKGC. Let O be the centre of this circle ; and from this centre upon the ruler MN, wherein those legs CN, BN did concur while the trajectory was described, let fall the perpendicular OH •meeting the circle in K and L. And when those other legs CK, BK meet in the point K that is nearest to the ruler, the first legs CP, BP will be parallel to the greater axis, and perpendicular on the lesser ; and the con-

148

THE MATHEMATICAL PRINCIPLES

[Book 1.

• V 
  

— ^ ,

T^/fx

^^^'

\0\ ^

"" '/

~

^^^B

M ^^«^

■ -^

trary will happen if those legs meet in the remotest point L. Whence if the centre of the trajectory is given, the axes will be given ; and those be- ing given, the foci will be readily found.

But the squares of the axes are one to the other as KH to LH, and thence it is easy to describe a trajectory given in kind through four given points. For if two of the given points are made the poles C, B, the third will give the moveable angles PCK, PBK ; but those being given, the circle BGKC may be described. Then, because the trajectory is given in kind, the ratio of OH to OK, and and therefore OH itself, will be given. About the centre O, with the interval OH, describe another circle, and the right line that touches this circle, and passes through the concourse of the legs CK, BK, when the first legs CP, BP meet in the fourth given point, will be the ruler MN, by means of which the trajectory may be described. Whence also on the other hand a trapezium given in kind (excepting a few cases that are impossible) may be inscribed in a given conic section.

There are also other Lemmxas, by the help of which trajectories given in kind may be described through given points, and touching given lines. Of such a sort is this, that if a right line is drawn through any point given by position, that may cut a given conic section in two points, and tlie distance of the intersections is bisected, the point of bisection will touch another conic section of the same kind with the former, and having its axes parallel to the axes of the former. But I hasten to things of greater use.

LEMMxi XXVI.

To place the three angles of a triangle, given both in kind and magjii- tude, in respect of as many right lines given by position, provided they are not all parallel among themselves, in such 'manner that the several angles m.ay tonch the several lines.

Three indefinite right lines AB, AC, BC, are given by position, and it is required so to place the triangle DEF that its angle D may touch the line AB, its angle E the line AC, and its angle F the line BC. Upon DE, DF, and EF, describe three segments of circles DRE, DGF, EMF, capable of angles equal to the angles BAC, ABC, ACB respectively. But those segments are to be de- scribed towards such sides of the lines DE, DF, EF, that the letters

Sec. Y.]

OF NATURAL PHILOSOPHY

149

the thing is done.

DRED may turn round about in the same order with the letters BACB ; the letters DGFD in the same order with the letters ABCA ; and the letters EMFE in the same order with the letters ACBA ; then, completing those segments into entire circles let the two former circles cut one the other in G, and suppose P and Q, to be their centres. Then joining GP, PQ, take Ga to AB as GP is to PQ. ; and about the centre G^ with the interval Ga, describe a circle that may cut the first circle DGE in a. Join aD cutting the second circle DFG in b, as well as c/E cutting the third circle EMF in c. Complete the figure ABCdef similar and equal to the figure abcDFiF : I say,

For drawing Fc meeting aD in n, and joining aG, bG, (iG, Q.D, PD, by construction the angle EaD is equal to the angle CAB, and the angle acF equal to the angle ACB ; and therefore the triangle avc equiangular to the triangle ABC. Wherefore the angle anc or F?iD is equal to the angle ABC, and conse- quently to the angle FbD ; and there- fore the point ?h falls on the point b. Moreover the angle GPQ., which is half the angle GPD at the centre, is equal to the angle GaD at the circumference ; and the angle GQ.P, which is half the angle GQD at the centre, is equal to the complement to two right angles of the angle GbD at the circum- ference, and therefore equal to the angle Gba. Upon which account the triangles GPQ, Gab, are similar, and Ga is to ab as GP to PQ, ; that is (by construction), as Ga to AB. Wherefore ab and AB are equal ; and consequently the triangles abc, ABC, which we have now proved to be similar, are also equal. And therefore since the angles D, E, F, of the triangle DEF do respectively touch the sides ab, ac, be of the triangle abc, the figure ABCdef may be completed similar and equal to the figure abcDFF, and by completing it the Problem will be solved. Q.E.F.

Cor. Hence a right line may be drawn whose parts given in length may be intercepted between three right lines given by position. Suppose the triangle DEF, by the access of its point D to the side EF, and by having the sides DE, DF placed in directimb to be changed into a right line whose given part DE is to be interposed between the right lines AB, AC given by position ; and its given part DF is to be interposed between the right lines AB, BC, given by position ; then, by applying the preceding construction to this case, the Problem will be solved.

^^

150 THE MATHEMATICAL PRINCIPLES [BoOX I.

PROPOSITION XXYIIL PROBLEM XX.

To describe a trajectory given both in kind and magnitude, given parts of luhich shall be interposed between three right lines given by position. Suppose a trajectory is to be described that

may be similar and equal to the curve line DEF^

and may be cut by three right lines AB, AC,

EG, given by position, into parts DE and EP,

similar and equal to the given parts of this / \ \ ^""^.,^

curve line. -^ -^^

Draw the right lines DE, EF, DF : and

place the angles D, E, F, of this triangle DEF, so

as to touch those right lines given by position (by

Lem. XXVI). Then about the triangle describe

the trajectory, similar and equal to the curve DEF.

Q.E.F.

LEMMA XXYII.

To describe a trapezium given in kind, the angles ivhereof may be so placed, in respect of four right lines given by position^ that are neither all parallel among themselves, nor converge to one coinmon point, that the several angles m^ay touch the several lines. Let the four right lines ABC, AD, BD, CE, be given by position ; the first cutting the second in A, the third in B, and the fourth in C : and suppose a trapezium /^/ii is to be described that may be similar to the trapezium FGHI, and whose angle /, equal to the given angle F, may touch the right line ABC ; and the other angles g, h, i, equal to the other given angles, G, H, I, may touch the other lines AD, BD, CE, re- spectively. Join FH, and upon FG. FH, FI describe ^ as many segments of circles FSG, FTH, FVI, the first of which FSG may be capable of an angle equal to the angle BAD : the second FTH capable of an angle equal to the angle CBD : and the third FVI of an angle equal to the angle ACE.^ But the segments are to be described towards those sides of the lines FG, FH, FI, that the circular order of the letters FSGF may be the same as of the letters BADB, and that the letters FTHF may turn about in the same order as the letters CBDC, and the letters FYIF in the same order as the letters ACEA. Complete the segments into entire cir- cles, and let P be the centre of the first circle FSG, Q the centre of the second FTH. Join and produce both ways the line PQ,, and in it take QR in the same ratio to PQ. as BC has to AB. But QR is to be taken towards that side of the point Q, that the order of the letters P, Q, R.

Sec. v.]

OF NATURAL PHILOSOPHY.

151

may be the same as of the letters A^ B, C ; and about the centre R with the interval RF describe a fourth circle FNc cutting the third circle FVI in c. Join Fc cut- tino- the first circle in a, and the second in b. Draw aG, bH, cl, and let the figure ABC/j^/a-be made similar to the figure abcFGBl] and the trapezium fghi will be that which was required to be de- scribed.

For let the two first circles FSG, FTH cut one the other in K ; join PK, GIK, RK, aK, hK, cK, and produce QP to L. The anoies FaK, YbK, FcK at the circumferences are the halves of the ano'les FPK, FQ.K, FRK, at the centres, and therefore equal to LPK, LQKj LRK; the halves of those angles. Wherefore the figure PQ,RK is equiangular and similar to the figure abcK, and consequently ah is to he as PQ, to Q,R, that is, as AB to BC. But by construction,, the angles fAg,fBh,fCi, are equal to the angles FaG, FbR, Fcl. And therefore the fio'urc ABCfghi may be completed similar to the figure abcFGHl. Which done a trapezium fghi will be constructed similar to the trapezium FGHI, and which by its angles/, g, h, i will touch the right lines ABC, AD, BD, CE. aE.F.

Cor. Hence a right line may be drawn whose parts intercepted in a given order, between four right lines given by position, shall have a given proportion among themselves. Let the angles FGH, GHI, be so far in- creased that the right lines FG, GH, HI, may lie i7i directmi%; and by constructing the Problem in this case, a right line fghi will be drawn, whose parts /g", gh, hi, intercepted between the four right lines given by position, AB and AD, AD and BD, BD and CE, will be one to another as the lines FG, GH, HI, and will observe the same order among them- selves. But the same thing may be more readily done in this manner.

Produce AB to K and BD to L, . _^

so as BK may be to AB as HI to GH ; and DL to BD as GI to FG; and join KL meeting the right line CE in i. Produce ih to M, so as LM may be to iL as GH to HI ; then draw MQ, parallel to LB, and meeting the right line AD in g, and join gi cutting AB, BD m f h] I say, the thing is done.

For let Mg cut the right line AB in Q., and AD the right. line KL.in

IVU---'

H

152

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[Book I.

S, and draw AP parallel to BD, and meeting ih in P, and gM. to hh (gi to hi, Mi to hi, GI to HI, AK to BK) and AP to BL, will be in the same ratio. Cut DL in R, so as DL to RL may be in that same ratio ; and be- cause gS to gM, AS to AP, and DS to DL are proportional; therefore {ea: cequo) as gS to LA, so will AS be to BL, and DS to RL ; and mixtiy, RL — RL to LA — BL, as AS — DS to gS — AS. That is, BR is to Bh as AD is to Ag, and therefore as BD to gGl. And alternately BR is to BD as BA to gO,, or as /A to fg. But by construction the line BL was cut in D and R in the same ratio as the line FI in G and H ; and therefore BR is to BD as FH to FG. Wherefore /A is to fg as FH to FG. Since, therefore, gi to hi likewise is as Mi to hi, that is, as GI to HI, it is manifest that the lines FI, /i, are similarly cut in G and H, g and A. Q.E.F.

In the construction of this Corollary, after the line LK is drawn cutting CE in i, we may produce iE to V, so as EV may be to Ei as FH to HI, and then draw Yf parallel to BD. It will come to the same, if about the centre i with an interA^al IH, we describe a circle cutting BD in X, and produce iX to Y so as iY may be equal to IF, and then draw Yf parallel

toBD.

Sir Christopher Wren and Dr. Wallis have long ago given other solu- tions of this Problem.

PROPOSITION XXIX. PROBLEM XXL

To describe a trojectory given in kind, that may be cut by four right lines given by 'position, into parts given i?i order y kind, and proportion. Suppose a trajectory is to be described that may be similar to the curve line FGHI, and whose parts, similar and proportional to the parts FG, GH, HI of the other, may be intercepted between the right lines AB and AD, AD, and BD, BD and CE given by po- sition, viz., the first between the first pair of those lines, the second between the second, and the third between the third. Draw the right lines FG, GH, HI, FI; and (by Lem. XXYII) describe a trapezium fghi that may be similar to the trapezium FGHI, and whose an- gles/, g, A, i, may touch the right lines given by posi- tion AB, AD, BD, CE, severally according to their order. And then about this trapezium describe a trajectory, that trajectory will be similar to the curve line FG-HI.

SCHOLIUM.

This problem may be likewise constructed in the following manner. Joining FG, GH, HI, FI, produce GF to Y, and join FH, IG, and make

Sec. VI.l

OF NATURAL PHILOSOPHY.

153

the angles CAK, DAL equal to

the angles FGH, YFH. Let

AK; AL meet the right line

BD in K and L, and thence

draw KM, LN, of which let

KM make the angle AKM equal

to the angle GHI, and be itself

to AK as HI is to GH ; and let

LN make the angle ALN equal to the angle FHI, and be itself

to AL as HI to FH. But AK, KM, AL, LN are to be drawn

towards those sides of the lines AD, AK, AL, that the letters

CAKMC, ALKA, DALND may be carried round in the same

order as the letters FGHIF ; and draw MN meeting the right ^

line CB in i. Make the angle iEP equal to the angle IGF, '^^

and let PE be to Ei as FG to GI ; and through P draw PQ/ that may

with the right line ADE contain an angle PQ,E equal to the angle FIG,

and may meet the right line AB in /, and join fi. But PE and PQ. are

to be drawn towards those sides of the lines CE, PE, that the circular

order of the letters PEiP and PEGIP may be the same as of the letters

FGHIF ; and if upon the line fi, in the same order of letters, and similar

to the trapezium FGHI, a trapezium /«^/w is constructed, and a trajectory

given in kind is circumscribed about it, the Problem will be solved.

So far concerning the finding of the orbits. It remains that we deter- mine the motions of bodies in the orbits so found.

SECTION VL

Uoio the motions are to be found in given orbits.

PROPOSITION XXX. PROBLEM XXII.

To find at amj assigned time the place of a body Qnoving in a given parabolic trajectory. Let S be the focus, and A the principal vertex of the parabola; and suppose 4AS X M equal to the parabolic area to be cut off APS, which either was described by the radius SP, since the body's departure from the vertex, or is to be described thereby before its arrival there. Now the quantity of that area to be cut off is known from the time which is propor- tional to it. Bisect AS in G, and erect the perpendicular GH equal to 3M, and a circle described about the centre H, with the interval HS, will cut the parabola in the place P required. For letting fall PO perpendic- ular on the axis, and drawing PH, there will be AG^ + GH^ (= HP^ = AO — AGl^ + PO — GH|2) = A02 + PO^ — 2GA0 — 2GH + PO h

A G

154 THE MATHEMATICAL PRINCIPLES [BoOK I.

AG2 + GW. Whence 2GH X PO (= AO^ + PO^ — 2GA0) = AO"

• I POl For AO^ write AO X j-^; then dividing all the terms by

3P0, and multiplying them by 2AS, we shall have |GH X AS (=.. JAO

V Dr. , lAo T^r^ AO + 3AS ^^ 4A0— 3S0

X PO + |AS X P0= ^ X PO = X PO = to

the area APO — SPO)| = to the area APS. But GH was 3M, and therefore ^GH X AS is 4AS X M. Wherefore the area cut off APS is equal to the area that was to be cut off 4AS X M. Q.E.D,

Cor. 1. Hence GH is to AS as the time in which the body described the arc AP to the time in which the body described the arc between the vertex A and the perpendicular erected from the focus S upon the axis.

Cor. 2. And supposing a circle ASP perpetually to pass through the moving body P, the velocity of the point H is to the velocity which the body had in the vertex A as 3 to S ; and therefore in the same ratio is the line GH to the right line which the body, in the time of its moving from A to P, would describe with that velocity which it had in the ver- tex A.

Cor. 3. Hence, also, on the other hand, the time may be found in which the body has described any assigned arc AP. Join AP, and on its middle point erect a perpendicular meeting the right line GH in H.

LEMMA XXYHL

There is no oval figure lohose area, cut off by right lines at pleasure, can be universally found by means of equations of any number of finite terms and dimensions.

Suppose that within the oval any point is given, about which as a pole a right line is perpetually revolving with an uniform motion, while in that right line a moveable point going out from the pole moves always forward with a. velocity proportional to the square of that right line with- in the oval. By this motion that point will describe a spiral with infinite circumgyrations. Now if a portion of the area of the oval cut off by that right line could be found by a finite equation, the distance of the point from the pole, which is proportional to this area, might be found by the same equation, and therefore all the points of the spiral might be found by a finite equation also ; and therefore the intersection of a right line given in position with the spiral might also be found by a finite equation. But every right line infinitely produced cuts a spiral in an infinite num- ber of points.; and the equation by Avhich any one intersection of two lines is found at the same time exhibits all their intersections by as many roots, and therefore rises to as many dimensions as there are intersections. Be- cause two circles mutually cut one another in two points, one of those in-

Sec. VL] of natural philosophy. 155

tersections is not to be found but by an equation of two dimensions, by which the other intersection may be also found. Because there may be four intersections of two conic sections, any one of them is not to be found universally, but by an equation of four dimensions, by which they may be all found together. For if those intersections are severally sought, be- cause the law and condition of all is the same, the calculus will be the same in every case, and therefore the conclusion always the same, which must therefore comprehend all those intersections at once within itself, and exhibit them all indifferently. Hence it is that the intersections of the conic sections with, the curves of the third order, because they may amount to six, come out together by equations of six dimensions ; and the inter- sections of two curves of the third order, because they may amount to nine, come out together by equations of nine dimensions. If this *did not ne- cessarily happen, we might reduce all solid to plane Problems, and those higher than solid to solid Problems. But here I speak of curves irreduci- ble in power. For if the equation by which the curve is defined may be reduced to a loAver poAver, the curve will not be one single curve, but com- posed of two, or more, whose intersections may be severally found by different calculusses. After the same manner the two intersections of right lines with the conic sections come out always by equations of two dimensions ; the three intersections of rio-ht lines with the irreducible curves of the third order by equations of three dimensions ; the four intersections of right lines with the irreducible curves of the fourth order, by equations of four dimensions ; and so on m wjinitum. Wherefore the innumerable inter- sections of a right line with a spiral, since this is but one simple curve, and not reducible to more curves, require equations infinite in number of dimensions and roots, by which they may be all exhibited together. For the law and calculus of all is the same. For if a perpendicular is let fall from the pole upon that intersecting right line, and that perpendicular together with the intersecting line revolves about the pole, the intersec- tions of the spiral will mutually pass the one into the other ; and that which was first or nearest, after one revolution, will be the second ; after two, the third ; and so on : nor will the equation in the mean time be changed but as the magnitudes of those quantities are changed, by which the position of the intersecting line is determined. Wherefore since those quantities after every revolution return to their first magnitudes, the equa- tion will return to its first form ; and consequently one and the same equation will exhibit all the intersections, and will therefore have an infi- nite number of roots, by which they may be all exhibited. And therefore the intersection of a right line with a spiral cannot be universally found by any finite equation ; and of consequence there is no oval figure whose area, cut off by right lines at pleasure, can be universally exhibited by any such equation.

156

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[Book I.

By the same argument^ if the interval of the pole and point by which the spiral is described is taken proportional to that part of the perimeter of the oval which is cut off; it may be proved that the length of the peri- meter cannot be universally exhibited by any finite equation. But here I speak of ovals that are not touched by conjugate figures running out in infinitum.

Cor. Hence the area of an ellipsis, described by a radius drawn from the focus to the moving body, is not to be found from the time given by a finite equation ; and therefore cannot be determined by the description of curves geometrically rational. Those curves I call geometrically rational, all the points whereof may be determined by lengths that are definable by equations ; that is, by the complicated ratios of lengths. Other curves (such as spirals, quadratrixes, and cycloids) I call geometrically irrational. For the lengths which are or are not as number to number (according to the tenth Book of Elements) are arithmetically rational or irrational. And therefore I cut off an area of an ellipsis proportional to the time in which it is described by a curve geometrically irrational, in the following maimer.

PROPOSITION XXXI. PROBLEM XXIII.

To find the place of a body moving in a given elliptic trajectory at any

assigned time.

Suppose A to be the principal vertex, S the focus, and O the centre of the ellipsis APB; and let P be the place of the body to be found. Produce OA to G so as OG may be to OA as OA to OS. Erect the perpendicular GH ; and about the centre O, with the interval OG, de- scribe the circle GEF ; and on the ruler GH, as a base, suppose the wheel GEF to move forwards, revolving about its axis, and in the mean time by its point A describing the cycloid ALL Which done, take GK to the perimeter GEFG of the wheel, in the ratio of the time in which the body proceeding from A described the arc AP, to the time of a whole revolution in the ellipsis. Erect the perpendicular KL meeting the cycloid in L ; then LP drawn parallel to KG will meet the ellipsis in P, the required place of the body.

For about the centre O with the interval OA describe the semi-circle AQ,B, and let LP, produced, if need be, meet the arc AQ, in Q, and join

Sec. YI.

OF NATURAL PHILOSOPHY.

157

SO,, OQ. Let 00. meet the arc EFG in F, and upon OQ, let fall the perpendicular SR. The area APS is as the area AQ.S, that is, as the difference between the sector OQ.A and the triangle OQ,S, or as the differ- ence of the rectangles ^OQ. X AQ, and |0a X SR, that iS; because |0a is o-iven, as the difference between the arc AQ, and the right line SR : and therefore (because of the equality of the given ratios SR to the sine of the arc AQ; OS to OA, OA to OG, AQ, to GF; and by division, AQ,— SR

to GF sine of the arc AQ,) as GK, the difference between the arc GF

and the sine of the arc AQ,. O.E.D.

H

SCHOLIUM.

But since the description of this curve is difficult, a solution by approximation will be preferable. First, then, let there be found a certain angle B which may be to an angle of 57,29578 degrees, which an arc equal to the radius subtends, as SH, the distance of the foci, to AB, ^' s B.^' the diameter of the ellipsis. Secondly, a certain length L, which may be to the radius in the same ratio inversely. And these being found, the Problem may be solved by the following analysis. By any construction (or even by conjecture), suppose we know P the place of the body near its true place p. Then letting fall on the axis of the ellipsis the ordinate PR from the proportion of the diameters of the ellipsis, the ordinate RQ, of the circumscribed circle AQ,B will be given ; which ordinate is the sine of the angle AOQ, supposing AO to be the radius, and also cuts the ellipsis in P. It will be sufficient if that angle is found by a rude calculus in numbers near the truth. Suppose we also know the angle proportional to the time, that is, which is to four right angles as the time in which the body described the arc Ap, to the time of one revolution in the ellipsis. Let this angle be N. Then take an angle D, which may be to the angle B as the sine of the angle AOQ, to the radius; and an angle E which may be to the angle N — AOQ, -fD as the length L to the same length L diminished by the cosine of the angle AOQ., when that angle is less than a right angle, or increased thereby when greater. In the next place, take an angle F that may be to the angle B as the sine of the angle AOQ, -I- E to the radius, and an angle G, that may be to the angle N — AOQ, — E + F as the length L to the same length L diminished by the cosine of the angle AOQ, + E, when that angle is less than a right angle, or increased thereby when greater. For the third time take an angle H. that may be to the angle B as the sine of the angle AOQ, r E + G to the radius; and an angle I to the angle N — AOQ. — E — G -}- H, as the

158

THE MATHEMATICAL PRINCIPLES

[Book 1.

lengtli L is to the same length L diminislied by the cosine of the angle AOQ, -f- E + Gj when that angle is less than a right angle, or increased thereby when greater. And so we may proceed i?i wfinitum. Lastly, take the angle AOq equal to the angle AOQ,-fE+ G + I+j <S^c. and from its cosine Or and the ordinate pr, which is to its sine qr as the lesser axis of the ellipsis to the greater, we shall have p the correct place of the body. When the angle N — AOQ, + D happens to be negative, the sign -f of the angle E must be every where changed into — , and the sign — into +. And the same thing is to be understood of the signs of the angles G and I, when the angles N — AOQ. — E + F, and N — AOQ — E — G + H come out negative. But the infinite series AOQ + E + G -j- I +, &c. converges so very fast, that it will be scarcely ever needful to pro- ceed beyond the second term E. And the calculus is founded upon this Theorem, that the area APS is as the difference between the arc AQ and the right line let fall from the focus S perpendicularly upon the radius OQ.

And by a calculus not unlike, the Problem is solved in the hyperbola. Let its centre be O, its vertex A, its focus S, and asymptote OK ; and suppose the quantity of the area to be cut off is known, as being proportional to the time. Let that be A, and by conjecture suppose Ave know the position of a right line SP, that cuts off an area APS near the truth. Join OP, and from A and P to the asymptote draAv AI, PK parallel to the other asymptote ; rithms the area AIKP will be given, and equal thereto the area OPA, which subducted from the triangle OPS, will leave the area cut off APS. And by applying 2APS — 2A , or 2A — 2APS, the double difference of the area A that was to be cut off, and the area APS that is cut off, to the line SN that is let fall from the focus S, perpendicular upon the tangent TP, we shall have the length of the chord PQ. Which chord PQ is to be inscribed between A and P, if the area APS that is cut off be greater than the area A that was to be cut off, but towards the contrary side of the point P, if otherwise : and the point Q will be the place of the body more accurately. And by repeating the computation the place may be found perpetually to greater and greater accuracy.

And by such computations we have a general analytical resolution of the Problem. But the par- ticular calculus that foUoAVS is better fitted for as- tronomical purposes. Supposing AO, OB, OD, to be the semi-axis of the ellipsis, and L its la.tus rec- tum, and D the difference betwixt the lesser semi-

O T ^ S

and by the table of

Sec. VII.l OF natural philosophy. 159

axis O D, and -L the half of the latus rectum : let an angle Y be found, whose sine may be to the radius as the rectangle under that difference D, and AO 4- OD the half sum of the axes to the square of the greater axis AB. Find also an angle Z, whose sine may be to the radius as the double rec- tano-le under the distance of the foci SH and that difference D to triple the square of half the greater semi-axis AO. Those angles being once found; the place of the body may be thus determined. Take the angle T proportional to the time in which the arc BP was described, or equal to what is called the mean motion ; and an angle V the first equation of the mean motion to the angle Y, the greatest first equation, as the sine of double the angle T is to the radius ; and an angle X, the second equation, to the ano-le Z, the second greatest equation, as the cube of the sine of the ano-le T is to the cube of the radius. Then take the angle BHP the mean mo'tion equated equal, to T + X + V, the sum of the angles T, V, X, if the angle T is less than a right angle ; or equal to T + X — Y, the difference of the same, if that angle T is greater than one and less than two right angles ; and if HP meets the ellipsis in P, draw SP, and it will cut off the area BSP nearly proportional to the time.

This practice seems to-be expeditious enough, because the angles Y and X, taken in second minutes, if you please, being very small, it will be suf- ficient to find two or three of their first figures. But it is likewise sufficiently accurate to answer to the theory of the planet's motions. For even in the orbit of Mars, where the greatest equation of the centre amounts to ten degrees, the error will scarcely exceed one second. But when the angle of the mean motion equated BHP is found, the angle of the true motion BSP, and the distance SP, are readily had by the known methods.

And so far concerning the motion of bodies in curve lines. But it may also come to pass that a moving body shall ascend or descend in a right line ; and I shall now go on to explain what belongs to such kind of motions.

SECTION YII.

Concerning the rectilinear ascent and descent of bodies,

PROPOSITION XXXII. PROBLEM XXIY.

Supposing that the centripetal force is reciprocally proportional to the sqnare of the 'distance of the places from the centre ; it is required to define the spaces lohich a body, falling directly, describes in given times. Case 1. If the body does not fall perpendicularly, it will (by Cor. 1.

160

THE MATHEMATICAL PRINCIPLES

[Book L

Prop. XIII) describe some conic section whose focus is a placed in the centre of force. Suppose that conic sec- tion to be ARPB and its focus S. And, first, if the figure be an ellipsis, upon the greater axis thereof AB describe the semi-circle ADB, and let the right line DPC pass through the falling body, making right angles with the axis ; and drawing DS, PS, the area ASD will ^ be proportional to the area ASP, and therefore also to the time. The axis AB still remaining the same, let the breadth of the ellipsis be perpetually diminished, and the area ASD will always remain proportional to the time. Suppose that breadth to be diminished in wjinituni ; and the orbit APB in that case coinciding with the axis AB, and the focus S with the extreme point of the axis B, the body will descend in the right line A C, and the area ABD will become proportional to the time. Wherefore the space AC will be given which the body describes in a given time by its perpendicular fall from the place A, if the area ABD is taken proportional to the time, and from the point D the right line DC is let fall perpendic- ularly on the right line AB. Gt.E.I.