proportion among themselves. Let the angles FGH, GHI, be so far increased that the right lines FG, GH, HI, may lie in directum ; and byconstructing the Problem in this case, a right line fghi will be drawn,whose parts fg, gh, hi, intercepted between the four right lines given byposition, AB and AD, AD and BD, BD and CE, will be one to anotheras the lines FG, GH, HI, and will observe the same order among themselves. 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 toGH ; and DL to BD as GI to FG;and join KL meeting the right lineCE in i. Produce iL to M, so asLM may be to iL as GH to HI ;then draw MQ, parallel to LB, andmeeting the right line AD in g, andjoin gi cutting AB, BD in f, h ;I M*say, the thing is done.For let MO- cut the right line AB in Q, and AD the right line KL iuII^52 THE MATHEMATICAL PRINCIPLES [BOOK I.S, arid draw AP parallel to BD, and meeting iL in P, and -M to Lh (g。to hi, Mi to Li, GI to HI, AK to BK) and AP to BL, will be in the sameratio. Cut DL in 11, so as DL to RL may be in that same ratio; and because ffS to g~M, AS to AP. and DS to DL are proportional; therefore(ex ceqit.o) as gS to LA, so will AS be to BL, and DS to RL ; and mixtly.BL RL to Lh BL, as AS DS to gS AS. That is, BR is toEh as AD is to Ag, and therefore as BD to gQ. And alternately BR isto BD as 13/i to g-Q,, or asfh to fg. But by construction the line BLwas cut in D and R in the same ratio as the line FI in G and H ; andtherefore BR is to BD as FH to FG. Wherefore fh is to fg as FH toFG. Since, therefore, gi to hi likewise is as Mi to Li, that is, as GI toHI, it is manifest that the lines FI, fi, are similarly cut in G and H, gand //.. Q.E.F.In the construction of this Corollary, after the line LK is drawn cuttingCE in i, we may produce iE to V, so as EV may be to Ei as FH to HI,arid then draw V/~ parallel to BD. It will come to the same, if about thecentre i with an interval IH, we describe a circle cutting BD in X, andproduce iX to Y so as iY may be equal to IF, and then draw Yf parallelto BO.Sir Christopher Wren and Dr. Wallis have long ago given other solutions of this Problem.PROPOSITION XXIX. PROBLEM XXI.To describe a trajectory given in kind, that may be cut by four rightlines given by position, into parts given in order, kind, and proportion.Suppose a trajectory is to be described that may besimilar to the curve line FGHI, and whose parts,similar and proportional to the parts FG, GH, HI ofthe other, may be intercepted between the right linesAB and AD, AD, and BD, BD and CE given by position, viz., the first between the first pair of those lines,the second between the second, and the third betweenthe third. Draw the right lines FG, GH, HI, FI;and (by Lem. XXVII) describe a trapezium fghi thatmay be similar to the trapezium FGHI, and whose angles/, g, h, i, may touch the right lines given by position AB, AD, BD, CE, severally according to their order. And then aboutbins trapezium describe a trajectory, that trajectory will be similar to thecurve line FGHI.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 makeSEC. VI OF NATURAL PHILOSOPHY. 153Elthe angles CAK. DAL equal tothe angles PGH, VFH. LetAK, AL meet the right lineBD in K and L, and thencedraw KM, LN, of which letKM make the angle AKM equalto the angle CHI, and be itselfto AK as HI is to GH ; and letLN make the angle ALN equal to the angle FHI, and be itselfto AL as HI to FH. But AK, KM. AL, LN are to be drawntowards those sides of the lines AD, AK, AL, that the lettersOA.KMC, ALKA, DALND may be carried round in the sameorder as the letters FGHIF ; and draw MN meeting the right vline CE in L 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 maywith the right line ADE contain an angle PQE equal to the angle FIG,and may meet the right line AB in /, and join fi. But PE and PQ arctobe drawn towards those sides of the lines CE, PE, that the circularorder of the letters PEtP and PEQP may be the same as of the lettersFGHIF ; and if upon the line/i, in the same order of letters, and similarto the trapezium FGHI, a trapezium /^//.i is constructed, and a trajectorygiven in kind is circumscribed about it, the Problem will be solved.So far concerning the finding of the orbits. It remains that we determine the motions of bodies in the orbits so found.SECTION VI.How the motions are to be found in given, orbits.PROPOSITION XXX. PROBLEM XXII.To find at any assigned time the place of a body moving in, a givenparabolic trajectory.Let S be the focus, and A the principal vertex ofthe parabola; and suppose 4AS X M equal to theparabolic area to be cut off APS, which either wasdescribed by the radius SP, since the body s departurefrom the vertex, or is to be described thereby beforeits arrival there. Now the quantity of that area tobe cut off is known from the time which is proportional to it. Bisect AS in G, and erect the perpendicular GH equal to3M, and a circle described about th j centre H, with the interval HS, willcut the parabola in the place P required. For letting fall PO perpendicular on the axis, and drawing PH, there will be AG2-f- GH2(=.= HP2 -_AO^TAGJ* + PO GH|2) = AO2 + PO2 2CA > ?G!I f POA G S154 THE MATHEMATICAL PRINCIPLES [BOOK IAG* + GH2. Whence 2GH X PO ( AO2 + PO2 2GAO) = AOJPO2-f | PO2. For AO2 write AO X ; then dividing all the terms by2PO; and multiplying them by 2AS, we shall have ^GH X AS (= IAOthe area APO SPO)| = to the area APS. But GH was 3M, andtherefore ^GH X AS is 4AS X M. Wherefore the area cut off APS isequal to the area that was to be cut off 4AS X M. Q.E.D.Con. 1. Hence GH is to AS as the time in which the body describedthe arc AP to the time in which the body described the arc between thevertex A and the perpendicular erected from the focus S upon the axis.COR. 2. And supposing a circle ASP perpetually to pass through themoving body P, the xelocity of the point H is to the velocity which thebody had in the vertex A as 3 to 8; and therefore in the same ratio isthe line GH to the right line which the body, in the time of its movingfrom A to P, would describe with that velocity which it had in the vertex A.COR. 3. Hence, also, on the other hand, the time may be found in whichthe body has described any assigned arc AP. Join AP, and on its middlepoint erect a perpendicular meeting the right line GH in H,LEMMA XXVIII.There is no oval figure whose area, cut off by right lines at pleasure, can,be universally found by means of equations of any number of finiteterms and dimensions.Suppose that within the oval any point is given, about which as a polea right line is perpetually revolving with an uniform motion, while inthat right line a mov cable point going out from the pole moves alwaysforward with a velocity proportional to the square of that right line within the oval. By this motion that point will describe a spiral with infinitecircumgyrations. Now if a portion of the area of the oval cut off by thatright line could be found by a finite equation, the distance of the pointfrom the pole, which is proportional to this area, might be found by thesame equation, and therefore all the points of the spiral might be foundby a finite equation also ; and therefore the intersection of a right linegiven 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 number of points ; and the equation by which any one intersection of two linesis found at the same time exhibits all their intersections by as many roots,and therefore rises to as many dimensions as there are intersections. Because two circles mutually cut one another in two points, one of those in8FC. Vl.J OF NATURAL PHILOSOPHY. 155terscctions is not to be found but by an equation of two dimensions, fowhich the other intersection may be also found. Because there may b(-four intersections of two conic sections, any one of them is not to be founduniversally, but by an equation of four dimensions, by which they may bi>all found together. For if those intersections are severally sought, because the law and condition of all is the same, the calculus will be thesame in every case, and therefore the conclusion always the same, whichmust therefore comprehend all those intersections at once within itself, andexhibit them all indifferently. Hence it is that the intersections of theconic se"f ions with the curves of the third order, because they may amountto six, (。,me out together by equations of six dimensions ; and the intersections 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 necessarily happen, we might reduce all solid to plane Problems, and thosehigher than solid to solid Problems. But here i speak of curves irreducible in power. For if the equation by which the curve is defined may boreduced to a lower power, the curve will not be one single curve, but composed of two, or more, whose intersections may be severally found by differentcalculusses. After the same manner the two intersections of right lineswith the conic sections come out always by equations of two dimensions ; thethree intersections of right lines with the irreducible curves of the thirdurder by equations of three dimensions ; the four intersections of rightlines with the irreducible curves of the fourth order, by equations of fourdimensions ; and so on in iitfinitum. Wherefore the innumerable intersections of a right line with a spiral, since this is but one simple curveand not reducible to more curves, require equations infinite in r- .imber ofdimensions and roots, by which they may be all exhibited together. Forthe law and calculus of all is the same. For if a perpendicular is let fallfrom the pole upon that intersecting right line, and that perpendiculartogether with the intersecting line revolves about the pole, the intersections of the spiral will mutually pass the one into the other ; and thatwhich was first or nearest, after one revolution, will be the second ; aftertwo, the third ; and so on : nor will the equation in the mean time bechanged but as the magnitudes of those quantities are changed, by whichthe position of the intersecting line is determined. Wherefore since thosequantities after every revolution return to their first magnitudes, the equation will return to its first form ; and consequently one and the sameequation will exhibit all the intersections, and will therefore have an infinite number of roots, by which they may be all exhibited. And thereforethe intersection of a right line with a spiral cannot be universally found byany finite equation ; and of consequence there is no oval figure whose area,cut off by right lines at pleasure, can be universally exhibited by an^such equation.1 56 THE MATHEMATICAL PRINCIPLES [BOOK 1By the same argument, if the interval of the pole and point by whichthe spiral is described is taken proportional to that part of the perimeterof the oval which is cut off, it may be proved that the length of the perimeter cannot be universally exhibited by any finite equation. But here Ispeak of ovals that are not touched by conjugate figures running out ininfinitvm.COR. Hence the area of an ellipsis, described by a radius drawn fromthe focus to the moving body, is not to be found from the time given by afinite equation ; and therefore cannot be determined by the description olcurves geometrically rational. Those curves I call geometrically rational,all the points whereof may be determined by lengths that are definableby 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 tothe tenth Book of Elements) are arithmetically rational or irrational.And therefore I cut off an area of an ellipsis proportional to the time inwhich it is described by a curve geometrically irrational, in the followingmanner.PROPOSITION XXXI. PROBLEM XXIII.Tofind the place of a body moving in a given elliptic trajectory at anyassigned time.Suppose A to bethe principal vertex,S the focus, and Othe centre of theellipsis A PB ; andlet P be the place ofthe body to be found.Produce OA to G soas OG may be to OAas OA to OS. Erectthe perpendicular GH; and about the centre O, with the interval OG, describe the circle* GEF ; and on the ruler GH, as a base, suppose the wheelGEF to move forwards, revolving about its axis, and in the mean time byits point A describing the cycloid ALL Which done, take GK to theperimeter GEFG of the wheel, in the ratio of the time in which the bodyproceeding from A described the arc AP, to the time of a whole revolutionin the ellipsis. Erect the perpendicular KL meeting the cycloid in L ;then LP drawn parallel to KG will meet the ellipsis in P, the requiredplace of the body.For about the centre O with the interval OA describe the semi-circleAQB, and let LP, produced, if need be, meet the arc AQ, in Q, and joinSEC. VI. OF NATURAL PHILOSOPHY. 157SQ, OQ. Let OQ meet the arc EFG in F, and upon OQ let fall theperpendicular Sll. The area APS is as the area AQS, that is, as tliedifference between the sector OQA and the triangle OQS, or as the difLienceof the rectangles *OQ, X AQ, and -J.OQ X SR, that is, because .>,_is given, as the difference between the arc AQ and the right line Sll : ai.;ltherefore (because of the equality of the given ratios SR to the sine of thearc AQ,, OS to OA, OA to OG, AQ to GF; and by division, AQ Siito GF sine of the arc AQ) as GK, the difference between the arc C 1and tlie sine of the arc AQ. Q.E.D.SCHOLIUM.But since the description of this curveis difficult, a solution by approximationwill be preferable. First, then, let therebe found a certain angle B which maybe to an angle of 57,29578 degrees,which an arc equal to the radius subtends,as SH, the distance of the foci, to AB,the diameter of the ellipsis. Secondly, a certain length L, which may be tothe radius in the same ratio inversely. And these being found, the Problemmay be solved by the following analysis. By any construction (or evenby conjecture), suppose we know P the place of the body near its trueplace jo. Then letting fall on the axis of the ellipsis the ordinate PRfrom the proportion of the diameters of the ellipsis, the ordinate RQ ofthe circumscribed circle AQB will be given ; which ordinate is the sine ofthe angle AOQ, supposing AO to be the radius, and also cuts the ellipsisin P. It will .be sufficient if that angle is found by a rude calculus innumbers near the truth. Suppose we also know the angle proportional tothe time, that is, which is to four right a iules as the time in which tliebody described the arc A/?, to the time of one revolution in the ellipsis.Let this angle be N. Then take an angle D, which may be to the angleB as the sine of the angle AOQ to the radius ; and an angle E whichmay be to the angle N AOQ -fD as the length L to the same lengthL diminished by the cosine of the angle AOQ, when that angle is lessthan a right angle, or increased thereby when greater. In the nextplace, take an angle F that may be to the angle B as the sine of the angle1OQ H- E to the radius, and an angle G, that may be to the angle NAOQE -f F as the length L to the same length L diminished by thecosine 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 f- E 4- G to theradius; and an angle I to the angle N AOQ E G -f- H, as the58 THE MATHEMATICAL PRINCIPLES jB(OK 1.length L is to the same length L diminished by the cosine of the angleAOQ -f- E + G, when that angle is less than a right angle, or increasedthereby when greater. And so we may proceed in infinitum. Lastly,take the angle AOy equal to the angle AOQ -f- E 4- G + I -。-} &c. andfrom its cosine Or and the ordinatejor, which is to its sine qr as the lesseraxis of the ellipsis to the greater, 。。 e shall have p the correct place of thebody. When the angle N AOQ, -f D happens to be negative, thesign -|- of the angle E must be every where changed into , and the signinto +. And the same thing is to be understood of the signs of the anglesG and I, when the angles N AOQ E -f F, and N AOQ EG + H come out negative. But the infinite series AOQ -f- E -f- G -|- I +,&c. converges so very fast, that it will be scarcely ever needful to proceed beyond the second term E. And the calculus is founded uponthis Theorem, that the area APS is as the difference between the arcAQ and the right line let fall from the focus S perpendicularly upon theradius OQ.And by a calculus not unlike, the Problemis solved in the hyperbola. Let its centre beO, its vertex A, its focus S, and asymptoteOK ; and suppose the quantity of the area tobe cut off is known, as being proportional tothe time. Let that be A, and by conjecturesuppose we know the position of a rij;ht i neSP, that cuts off an area APS near the truth.Join OP, and from A and P to the asymptoteT A Sdraw AI, PK parallel to the other asymptote ; and by the table of logarithms 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 2A PS, the double difference ofthe area A that was to be cut off, and the area APS that is cut off, to theline SN that is let fall from the focus S, perpendicular upon the tangentTP, we shall have the length of the chord PQ. Which chord PQ is tobe inscribed between A and P, if the area APS that is cut off be greaterthan the area A that was to be cut off, but towards the contrary side of thepoint P, if otherwise : and the point Q will be the place of the body moreaccurately. And by repeating the computation the place may be foundperpetually to greater and greater accuracy.And by such computations we have a generalanalytical resolution of the Problem. But the particular calculus that follows is better fitted for astronomical purposes. Supposing AO, OB, OD, tobe the semi-axis of the ellipsis, and L its latus rectum, and D the difference betwixt the lesser semiSEC. VII.] OF NATURAL PHILOSOPHY. J 59axis OD, and -,L the half of the latus rectum : let an angle Y be found, whosesine may be to the radius as the rectangle under that difference J), andAO 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 rectangle under the distance of the foci SH and that difference D to triplethe square of half the greater semi-axis AO. Those angles being oncefound, the place of the body may be thus determined. Take the angle Tproportional to the time in which the arc BP was described, or equal towhat is called the mean motion ; and an angle V the first equation of thrmean motion to the angle Y, the greatest first equation, as the sine ofdouble the angle T is to the radius ; and an angle X, the second equation,to the angle Z, the second greatest equation, as the cube of the sine of theangle T is to the cube of the radius. Then take the angle BHP the meanmotion 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 V, thedifference of the same, if that angle T is greater than one and less thantwo right angles ; and if HP meets the ellipsis in P, draw SP, and it willcut off the area BSP nearly proportional to the time.This practice seems to be expeditious enough, because the angles V andX, taken in second minutes, if you please, being very small, it will be sufficient to find two or three of their first figures. But it is likewisesufficiently accurate to answer to the theory of the planet s motions.For even in the orbit of Mars, where the greatest equation of the centreamounts to ten degrees, the error will scarcely exceed one second. Butwhen the angle of the mean motion equated BHP is found, the angle oithe true motion BSP, and the distance SP, are readily had by the knownmethods.And so far concerning the motion of bodies in curve lines. But it mavalso come to pass that a moving body shall ascend or descend in a rightline : and I shall now go on to explain what belongs to such kind ofmotions.SECTION VII.Concerning the rectilinear ascent and descent of bodies,PROPOSITION XXXII. PROBLEM XXIV.Supposing that the centripetal force is reciprocally proportional to thtsquare of tlie distance of the places from the centre ; it is requiredto define the spaces which a body, falling directly, describes in giventimes.CASE 1. If the body does not fall perpendicularly, it will (by Cor. I160 THE MATHEMATICAL PRINCIPLES [BOOK IProp. XIII) describe some conic section whose focus is Aplaced in the centre of force. Suppose that conic section to be ARPB and its focus S. And, first, if thefigure be an ellipsis, upon the greater axis thereof ABdescribe the semi-circle ADB, and let the right lineI) PC pass through the falling body, making right angleswith the axis; and drawing DS, PS, the area ASD will cbe proportional to the area ASP, and therefore also tothe time. The axis AB still reaiaining the same, let thebreadth of the ellipsis be perpetually diminished, and sthe area ASD will always remain proportional to thetime. Suppose that breadth to be diminished in, in fruitum ; and the orbitAPB in that case coinciding with the axis AB, and the focus S with theextreme point of the axis B, the body will descend in the right line AC1.and the area ABD will become proportional to the time. Wherefore thespace AC will be given which the body describes in a given time by itsperpendicularfall from the place A, if the area ABD is taken proportionalto the time, and from the point D the right line DC is let fall perpendicularly on the right line AB. Q,.E.I.CASE 2. If the figure RPB is an hyperbola, on thesame principal diameter AB describe the rectangularhyperbola BED ; and because the areas CSP, CB/P,