The Solution of Kepler's Problem, by J. Machin, Astr. Prof. Gresh. and Secr. R. S.

X. The Solution of Kepler's Problem, by J. Machin, Astr. Prof. Gresh. and Secr. R. S. Many Attempts have been made at different times, but, if I mistake not, never any yet with tolerable Success, towards the Solution of the Problem proposed by Kepler: To divide the Area of a Semicircle into given Parts, by a Line from a given Point of the Diameter, in order to find an universal Rule for the Motion of a Body in an Elliptic Orbit. For among the several Methods offered, some are only true in Speculation, but are really of no Service. Others are not different from his own, which he judged improper: And as to the rest, they are all some way or other so limited and confined to particular Conditions and Circumstances, as still to leave the Problem in general untouched. To be more particular; it is evident, that all Constructions by Mechanical Curves are seeming Solutions only, but in reality unapplicable; that the Roots of infinite Series's are, upon account of their known Limitations in all respects, so far from affording an Appearance of being sufficient Rules, that they cannot well be supposed as offered for any thing more than Exercises in a Method of Calculation. And then, as to the universal Method, which proceeds by a continued Correction of the Errors of a false Position, it is, when duly considered, no Method of Solution at all in itself; because, unless there be some antecedent Rule or Hypothesis to begin the Operation, (as suppose that of an an uniform Motion about the upper Focus, for the Orbit of a Planet; or that of a Motion in a Parabola for the perihelion Part of the Orbit of a Comet; or some other such) it would be impossible to proceed one Step in it. But as no general Rule has ever yet been laid down, to assist this Method, so as to make it always operate, it is the same in Effect as if there were no Method at all. And accordingly in Experience it is found, that there is no Rule now subsisting but what is absolutely useless in the Elliptic Orbits of Comets; for in such Cases there is no other way to proceed but that which was used by Kepler: To compute a Table for some Part of the Orbit, and therein examine if the Time to which the Place is required, will fall out anywhere in that Part. So that, upon the whole, I think, it appears evident, that this Problem (contrary to the received Opinion) has never yet been advanced one Step towards its true Solution: A Consideration which will furnish a sufficient Plea for meddling with a Subject so frequently handled; especially if what is offered shall at the same time appear (as I trust it will) to contribute towards supplying the main Defect. Lemma I. Lemma I. The Tangent of an Arch being given, to find the Tangent of its Multiple. Let \( r \) be the Radius of the Circle, \( t \) the Tangent of a given Arch \( A \), and \( n \) a given Number. And let \( T \) be the Tangent of the Multiple Arch \( n \times A \) to be found. Then if \( \rho \rho \) be put for \(- rr \), and \( \tau \tau \) for \(- tt \): The Tangent \( T \) will be \[ \frac{r + \tau}{r - \tau} \frac{n}{n} \] Which Binomials being raised according to Sir Isaac Newton's Rule, the fictitious Quantities \( \tau \) and \( \rho \) will disappear, and the Tangent \( T \) will become equal to \[ nt \frac{n-1}{1} \frac{n-2}{2} \frac{t^3}{r^2} + \frac{n}{1} \frac{n-1}{2} \frac{n-2}{3} \frac{n-3}{4} \frac{n-4}{5} \frac{t^5}{r^4} \text{ &c.} \] \[ 1 \frac{n}{1} \frac{n-1}{2} \frac{tt}{rr} + \frac{n}{1} \frac{n-1}{2} \frac{n-2}{3} \frac{n-3}{4} \frac{t^4}{r^4} \text{ &c.} \] This Theorem (which I formerly found for the Quadrature of the Circle, at a time when it was not known here to have been invented before) has now been common for many Years; for which Reason I shall premise it, at present, without any Proof; only for the sake of some Uses that have not yet been made of it. Corollary 1. From this Theorem for the Tangent, the Sine (suppose) \( Y \), and Cosine \( Z \) of the Multiple Arch \( n \times A \), may be readily found. For if \( y \) be the Sine, and \( z \) the Cosine of the given Arch \( A \), then putting \( vv \) for \(- yy \), and substituting \( \frac{ry}{z} \) for \( t \), and \( \frac{rv}{z} \) for \( \tau \), and \( \frac{rT}{\sqrt{rr + TT}} \) for \( Y \): The Sine \( Y \) will be \( \frac{z + v}{2r^n} \). The Cosine \( Z \) will be \( \frac{z + v}{2r^{n-1}} \). Each of these may be expressed differently in a Series, either by the Sine and Cosine conjointly, or by either of them separately. Thus \( Y \), the Sine of the multiple Arch \( n \times A \), may be in either of these two Forms, viz. \[ \frac{z^n - 1}{r^{n-1}} y \text{ in } n - \frac{n-1}{2} \cdot \frac{n-2}{3} A \cdot \frac{y^2}{x^2} + \frac{n-3}{4} \cdot \frac{n-4}{5} B \cdot \frac{y^4}{x^4} - \&c. \] or \( = ny - \frac{nn-1}{2 \cdot 3 \cdot rr} A y^3 - \frac{nn-9}{4 \cdot 5 \cdot rr} B y^5 - \frac{nn-25}{6 \cdot 7 \cdot rr} C y^7 - \&c. \) Wherein the Letters A, B, C, &c. stand, as usual, for the Coefficients of the preceding Terms. The first of these Theorems terminates when \( n \) is any integer Number, the other (which is Sir Isaac Newton's Rule, and is derived from the former by substituting \( \sqrt{rr - yy} \) for \( z \)) terminates when \( n \) is any odd Number. The Cosine \( Z \) may, in like manner, be in either of these two Forms, viz. \[ \frac{z^n}{r^{n-1}} \text{ in } 1 - \frac{n}{1} \cdot \frac{n-1}{2} \cdot \frac{n-2}{3} \cdot \frac{n-3}{4} \cdot \frac{y^4}{x^4} - \&c. \] or \( = r - \frac{nn}{2 \cdot rr} A y^3 - \frac{nn-4}{3 \cdot 4 \cdot rr} B y^4 - \frac{nn-16}{5 \cdot 6 \cdot rr} C y^6 - \&c. \) The latter of which terminates when the Number \( n \) is even, and the other as before, when it is any Integer. Corollary 2. Hence the Sine, Cosine, and Tangent of any Submultiple Part of an Arch (suppose) \( \frac{1}{n} A \), may be determined thus: The Tangent of $\frac{1}{n} A$ will be $$\frac{r + \tau}{r - \tau} \left( \frac{1}{n} \right)^{\frac{1}{2}}$$ The Sine of $\frac{1}{n} A$ will be $$\frac{z + v}{z - v} \left( \frac{1}{n} \right)^{\frac{1}{2}}$$ For these Equations will arise from the Transposition and Reduction of the former for the Tangent and Sine of the Multiple Arch, upon the Substitution of $t, y, z$ and $A$; for $T, Y, Z$ and $n \times A$. Corollary 3. Hence regular Polygons of any given Number of Sides may be inscribed within, or circumscribed without, a given Arch of a Circle. For if the Number $n$ express the double of the Number of Sides to be inscribed within, or circumscribed about, the given Arch $A$; then one of the Sides inscribed will be the double of the Sine, and one of the Sides circumscribed the double of the Tangent of the Submultiple part of the Arch, viz. $\frac{1}{n} A$. Lemma II. To find the Length of the Arch of a Circle within certain Limits, by means of the Tangent and Sine of the Arch. Let $t$ be the Tangent, $y$ the Sine and $z$ the Cosine of the Arch $A$, whose Length is to be determined, and let $\varphi, \tau, v$ be expounded as before; then, if any Number $n$ be taken, the Arch of the Circle will be always less than $$\frac{r + \tau}{r - \tau} \left( \frac{1}{n} \right)^{\frac{1}{2}} \times n \varphi,$$ and bigger than $$\frac{z + v}{z - v} \left( \frac{1}{n} \right)^{\frac{1}{2}} \times n \varphi.$$ For if, by the preceding Corollaries, a regular rectilinear Polygon be inscribed within, and another without, the Arch $A$, each having half so many Sides as is expressed by the Number $n$; then will the former of these Quantities be the Length of the Bow of the circumscribed Polygon, (or the Sum of all its Sides) which is always bigger, and the latter will be the Length of the Bow of the inscribed Polygon, which is always less, than the Arch of the Circle; how great soever the Number $n$ be taken. Corollary 1. Hence the Series's for the Rectification of the Arch of a Circle may be derived. For by converting the Binomials into the Form of a Series, that the fictitious Quantities, $\rho$, $\tau$, $\nu$ may be destroyed; it will appear, that no Number $n$ can be taken so large as to make the inscribed Polygon so big, or the circumscribed so little as the Series. $$\frac{ry}{z} - \frac{ry^3}{3z^3} + \frac{ry^5}{5z^5} - \frac{ry^7}{7z^7} + \&c.$$ in one Case, or its Equal $$t - \frac{t^3}{3r^2} + \frac{t^5}{5r^4} - \frac{t^7}{7r^6} + \&c.$$ in the other Case. Wherefore since the Quantity denoted by the Sum of the Terms in either of these Series's is always bigger than any inscribed Polygon, and always less than any circumscribed, it must therefore be equal to the Arch of the Circle. Corollary 2. If, in the first of the above Series's, the Root $\sqrt{rr-yy}$, be extracted and substituted for $z$, there will arise the other Series of Sir Isaac Newton, for giving the Arch from the Sine; namely, $$y + \frac{y^3}{6r^2} + \frac{3y^5}{40r} + \frac{5y^7}{112r^6} + \&c.$$ or otherwise, $$= y + \frac{1}{1.2.3} \times \frac{y^3}{r^2} + \frac{3.3}{1.2.3.4.5} \times \frac{y^5}{r^4} + \frac{3.3.5.5}{1.2.3.4.5.6.7} \times \frac{y^7}{r^6} + \&c.$$ In like manner, as the Arches of the Polygons serve to determine the Arch of the Circle, so by comparing the Areas of the circumscribed and inscribed Polygons, $\frac{1}{2}nrT$ and $\frac{1}{2}nYZ$, the Area of the Sector of a Circle may be found. For if $T$, $Y$ and $Z$ are the Tangent, Sine and Cosine of the Arch $A$; then by the second Lemma the Area of the circumscribed Polygon will be found to be $\frac{1}{2}nr \times \frac{r + r}{r + r} = \frac{1}{2}nrT$. and the Area of the inscribed will appear to be $\frac{1}{2}nr \times \frac{z + z}{z + z} = \frac{1}{2}nYZ$. But upon the Expansion of these Binomials it will appear, that no Number $n$ can be taken so large as to make the one so big, or the other so little, as the Area denoted by the Series. $\frac{1}{2}r \int t - \frac{t^3}{3rr} + \frac{t^5}{5r^4} - \frac{t^7}{7r^6} + \&c.$ So that this Area being larger than any inscribed, and smaller than any circumscribed Polygon, must be equal to the Area of the Sector. It may further be observed, that as the Arch or Area is found from the Sine, Cosine or Tangent of the Arch, by means of the limiting Polygons, so may the Sine, Cosine or Tangent be found from the Length of the Arch by the same Method. Thus, if $A$ be the Arch whose Tangent $T$, Sine $Y$, and Cosine $Z$, are to be determined, then will the Tangent \[ \text{Tangent } T = \frac{A - \frac{1}{1.2.3} \times \frac{A^3}{r^2} + \frac{1}{1.2.3.4.5} \times \frac{A^5}{r^4}}{1 - \frac{1}{1.2} \times \frac{A^2}{rr} + \frac{1}{1.2.3.4} \times \frac{A^4}{r^4}} \] \[ \text{Sine } Y = A - \frac{1}{1.2.3} \times \frac{A^3}{r^2} + \frac{1}{1.2.3.4.5} \times \frac{A^5}{r^4} \] \[ \text{Cosine } Z = r - \frac{1}{1.2} \times \frac{A^2}{r} + \frac{1}{1.2.3.4} \times \frac{A^4}{r^3} \] For it may be made to appear, from the first Lemma; and its Corollaries, that if in any of these Theorems, as suppose in the First, the Quantity \( A \) stand for the Bow of the circumscribed Polygon, then will the Quantity \( T \) exhibited by the Theorem, be always bigger; but if for the Bow of the inscribed, always less than the Tangent of the Arch, how great soever the Number \( n \) be taken; and consequently, if \( A \) stand for the Length of the Arch itself, the Quantity \( T \) must be equal to the Tangent; and the like may be shewn for the Sine, and, mutatis mutandis, for the Cosine. These Principles, from whence I have here derived the Quadrature of the Circle, which is wanted in the Solution of the Problem in hand, happen to be upon another Account absolutely requisite for the Reduction of it to a manageable Equation. But I have enlarged, more than was necessary to the Problem itself, on the Uses of this sort of Quadrature by the limiting Polygons, because it is one of that kind which requires no other Knowledge but what depends on the common Properties of Number and Magnitude; and so may serve as an Instance to shew that no other is requisite for the Establishment of Principles for Arithmetick and Geometry. A Truth, which though certain in itself, may perhaps seem doubtful from the Nature and Tendency of the present Inquiries in Mathematicks. For among the Moderns some have thought it necessary, for for the Investigation of the Relations of Quantities, to have recourse to very hard Hypotheses; such as that of Number infinite and indeterminate; and that of Magnitudes in Statu fieri, existing in a potential Manner, which are actually of no Bigness. And others, whose Names are truly to be reverenced on account of their great and singular Inventions, have thought it requisite to have recourse even to Principles foreign to Mathematics, and have introduced the Consideration of efficient Causes and Physical Powers for the Production of Mathematical Quantities; and have spoken of them, and used them, as if they were a Species of Quantities by themselves. N.B. In the following Proposition I have, for the sake of Brevity, made use of a peculiar Notation for composite Numbers (or such Quantities as are analogous to them) whose Factors are in Arithmetical Progression. The Quantity expressed by this Notation has a double Index: that at the Head of the Root at the Right-hand, but separated by a Hook to distinguish it from the common Index, denotes the Number of Factors, and that above, within the Hook on the Left-hand, denotes the common Difference of the Factors proceeding in a decreasing or increasing Arithmetical Progression. Thus the Quantity $\frac{a}{n+a}$ denotes by its Index $m$ on the Right-hand, that it is a composite Quantity, consisting of so many Factors as there are Units in the Number $m$; and the Index $a$ above, on the Left, denotes the common Difference of the Factors, decreasing in an Arithmetical Progression, if it be Positive; or increasing, if it be Negative; and so signifies, in the common Notation, the composite Number or Quantity, $n+a, n+a-a, n+a-2a, n+a-3a$, and so on. For Example: $\frac{2}{n+5}$ is $=n+5, n+3, n+1, n-1, n-3, n-5$, consisting of six Factors whose common Difference is 2. After the same manner $\frac{2}{n+4}$ is $=n+4, n+2, n-2, n-4$, consisting of five Factors. According to which Method it will easily appear, that if $a$ be any Integer, then $\frac{2}{n+2a+1}(2a+2)$ will be $=nn-1, nn-9, nn-25$, continued continued to such a Number of double Factors as are expressed by $a + i$, or half the Index, which in this case is an even Number. So $$\frac{2}{n+2a} (2a+1)$$ will be equal to $n.n - 4.n.n - 16.n.n - 36$, and so on, where there are to be so many double Factors as with one single one ($n$) will make up the Index $2a + 1$, which is an odd Number. If the common Difference $a$ be an Unit, it is omitted: Thus, $\frac{n^6}{(n^6)} = n.n - 1.n - 2.n - 3.n - 4.n - 5$, containing 6 Factors. So $\frac{6}{6} = 6.5.4.3.2.1$, and the like for others. If the common Difference $a$ be nothing, then the hook is omitted, and it becomes the same with the Geometrical Power: $$\frac{m}{n+a} = \frac{1}{n+a} m$$ according to the common Notation. **Proposition I.** An Arch less than a Semicircle being given, with a Point in the Diameter passing through one of its Extremities; to find, by means of the Sine of a given Part of the Arch less than one half, the Area of the Sector subtended by the given Arch, and comprehended in the Angle made at the given Point. Let PNA be a Semicircle described on the Centre C, and Diameter AP, and let PN be the given Arch less than a Semicircle, and S the given Point in the Diameter AP passing thro' one of the Extremities of the Arch NP in P. Then taking any Number $n$ bigger than 2, let PK be an Arch in Proportion to the given Arch PN, as Unity to the Number $n$; and let it be required to find by means of the Sine of the Arch PK, the Area of the Sector NSP subtended by the given Arch NP, and comprehended in the Angle NSP made at the given Point S. From N and K let fall on the Diameter AP the Perpendiculars NM and KL, and join CN and CK. Then let \( t \) stand for CP the Semidiameter of the Circle; \( f \) for CS the Distance of the given Point S from the Centre; \( p \) for SP the Distance of it from the Extremity of the Arch through which the Diameter AP passes; and \( y \) for KL the Sine of the Arch KP in the given Circle. These Substitutions being presupposed, the Problem is to be divided into two Cases; one when SP is less, and the other when it is greater than the Semidiameter CP. **CASE I.** If SP be less than CP, then take an Area \( H \) equal to the Sum of the Rectangles expressed by the several Terms of the following Series continued ad libitum: \[ \frac{py}{1} + \frac{t^2}{3!} + \frac{t^4}{5!} + \frac{t^6}{7!} + \cdots \] And the Area \( \frac{1}{2} n \times H \) will determine the Area of the Sector NSP ad libitum. For the Sector PSN, being the Excess of the Sector NCP above the Triangle NCS, will be the Difference of two Rectangles: \( \frac{1}{2} CP \times PN - \frac{1}{2} CS \times NM \); but PN is the Multiple of the Arch PK, namely \( n \times PK \); and NM is the Sine of that multiple Arch: Wherefore if for CP be put \( t \), for CS, \( f \), according to the Supposition; and if for PK be substituted: \[ \frac{y}{1} + \frac{1}{3} \times \frac{1^3}{2} + \frac{9}{5} \times \frac{1^5}{4} + \frac{9 \times 25}{7} \times \frac{1^7}{6} + \&c. \text{ by Cor. 2. Lem. 2;} \] and for NM: \[ \frac{ny}{1} - \frac{n \cdot n + 1}{3} \times \frac{1^3}{2} + \frac{n \cdot n + 3}{5} \times \frac{1^5}{4} - \frac{n \cdot n + 5}{7} \times \frac{1^7}{6} + \&c. \text{ accord-} \] ing to Cor. 1. Lem. 1. the Area of the Sector will appear in a Series, as is above determined. But since the Number \( n \) is greater than 2, and the given Arch PN is less than a Semicircle, and consequently KL or \( y \), the Sine of the Submultiple Arch PK, is less than the Semidiameter CP or \( t \); it may thence be easily proved, that the Series will approximate to the just Quantity of the Area, ad libitum. Corollary 1. Hence, if the Number \( n \) be taken equal to \( \sqrt{\frac{5}{5}} + \sqrt{\frac{25}{f}} \), the Sector NSP will be \[ = \frac{1}{2} npy + \frac{n^3 t - n \cdot n - 1 \cdot p}{12tt} y^3 + \&c. \] For the Numerator of the Coefficient of the third Term in the Series, that determines the Area \( H \), namely, \( \frac{9t - n + 3}{3} \times f \) is equal to \( 9t - n \cdot n - 1 \cdot n - 9 \cdot f \), which, according to the above Determination of the Number \( n \), will become nothing; wherefore, if for \( t - p \) be put \( f \) in the second Term, and the Value of \( n \) be substituted for \( n \) in the Third and Fourth, the Series for the Area will appear upon Reduction to be as is here laid down. Corol. Corol. 2. Hence the Area of the Sector NSP may be always defined nearly by the Terms of a Cubic Equation. For the Number \( n \), as constructed in the former Corollary, is always greater than the square Root of 10, and consequently \( \frac{y}{t} \) is always less than the Sine of one third Part of the given Arch; so that the fourth Term \( \frac{n^3}{1120t^4} \), with the Sum of all the following Terms of the Series, can never be more than a small Part of the whole Sector. Corol. 3. If \( R \) stand for 57,2957795, &c. Degrees, (or the Number of Degrees contained in an Angle subtended by an Arch of the same Length with the Radius of the Circle) and \( M \) be the Number of Degrees in an Angle which is to 4 right Angles, as the Area NSP to the Area of the whole Circle; then will \( M \) be \( \frac{np}{t} \times \frac{Ry}{t} + \frac{n^3t - n.nn - 1.p}{6t} \times \frac{Ry^3}{t^3} \), nearly. For \( \frac{M}{R} \times \frac{t}{2} \) will appear by the Construction to be equal to the Sector NSP. CASE II. If SP be greater than CP, then take an Area \( H \) equal to the Sum of the Terms in the following Series: \[ \frac{py}{x} + \frac{t-n+1}{3} \times f \times \frac{y^3}{t^2} + \frac{9t+n+3}{5} \times f \times \frac{y^5}{t^4} + \frac{9\times 25t-n+5}{7} \times f \times \frac{y^7}{t^6} + \&c. \] and the Area \( \frac{1}{2} n \times H \), will be the Sector, as before. For the Point S being on the contrary Side of the Centre to what it was before, it will easily appear, that the Change of $+f$ into $-f$, must reduce one Case to the other, without any other Proof. Corollary. Hence, if the Number $n$ be taken equal to $\sqrt{\frac{t+f}{f}}$ or in this Case $\sqrt{\frac{E}{f}}$, then the Series for the Sector will want the second Term, as in the former it wanted the Third. Definition. The Angle called by Kepler the Anomalia Eccentri; is a fictitious Angle in the Elliptic Orbit of a Planet, being analogous to the Area described by a Line from the Centre of the Orbit, and revolving with the Planet from the Line of Apsides; in like manner as the Mean Anomaly is a fictitious Angle, analogous to the Area described by a Line from the Focus. Otherwise, if C be the Centre, S the Focus of an Elliptic Orbit described on the transverse Axis AP, and the Area NSP in the Circle be taken in Proportion to the whole, as the Area described in the Ellipsis about the Focus, to the whole: Then is the Arch of the Circle PN, or the Angle NCP, that which Kepler calls the Anomalia Eccentri. This Angle may be measured either from the Aphelion, or from the Perihelion; in the following Proposition it is supposed to be taken from the Perihelion. PROPOSITION II. The mean Anomaly of a Comet or Planet revolving in a given Elliptic Orbit being given; to find the Anomalia Eccentri. The Solution of this Problem requires two different Rules; the first and principal one serves to make a Beginning for a further Approximation, and the other is for the Progression in approximating nearer and nearer ad libitum. I. The Rule for the first Assumption: Let $t$, $f$, and $p$, stand as before, for the Semi-transverse Axis of the Ellipsis, the Semi-distance of the Foci, and the Perihelion Distance; then taking the Number $n$ equal to $\sqrt{5 + \frac{2p}{f}}$, let $T$ stand for $\frac{2t}{nnt - nn - 1.p}$; and $P$ for $\frac{2p}{nnt - nn - 1.p}$ (or $\frac{p}{t}T$); which constant Numbers, being once computed for the given Orbit, will serve to find the Angle required nearly by the following Rule: Let $M$ be the Number of Degrees in the Angle of mean Anomaly to the given Time, reckoned from or to the Perihelion; and supposing $R$, as before, to stand for $57,2957$, &c. Degrees, take the Number $$N = \sqrt{\frac{3T}{nR}M}$$ and let $A$ be the Angle whose Sine is $$\frac{3}{N} \left( \sqrt{\frac{1}{2}} + \sqrt{\frac{1}{4}} + \frac{P^3}{N^6} \right) + \frac{3}{N} \left( \sqrt{\frac{1}{2}} - \sqrt{\frac{1}{4}} + \frac{P^3}{N^6} \right),$$ then the Multiple Angle $n \times A$ will be nearly equal to the Anomalia Eccentri. The Truth of which will appear from the Resolution of the Cubic Equation in the last Corollary to the preceding Proposition. Corollary I. If the Quadruple of the Quantity \( \frac{P^3}{N^6} \) be many times greater or many times less than Unity; or, which amounts to the same, if the mean Anomaly \( M \), be many times less, or many times greater, than the Angle denoted by the given Quantity \( \frac{2np}{3t} R \sqrt{P} \) (one or the other of which two Cases most frequently happens in Orbits of very large Eccentricity) then the Theorem will be reduced to a simpler Form near enough for Use. Case I. If \( M \) be many times less than \( \frac{2np}{3t} R \sqrt{P} \), then the Angle \( A \) may be taken for that whose Sine is \( \frac{t \times M}{np \times R} \). Case II. If \( M \) be many times greater than \( \frac{2np}{3t} R \sqrt{P} \), then let \( A \) be the Angle whose Sine is \( N - \frac{P}{N} \); and the Multiple Angle \( n \times A \), according to its Case, will be nearly equal to the Angle required. Corollary II. In Orbits of very large Eccentricity, the Perihelion Distance \( p \) is many times less than the Semi-distance of the Foci \( f \), and the Number \( n = \sqrt{5 + \sqrt{25 + \frac{9p}{f}}} \) is always nearly equal to \( \sqrt{10} \) or to the Integer 3, either of which may be used for it without any material Error in the Orbits of Comets. II. The II. The Rule for a further Correction ad libitum. Let $M$ be the given mean Anomaly, $t$ the Semi-transverse Axis, as before; and let $B$ be equal to or nearly equal to the Multiple Angle $n \times A$ before found, then if $\mu$ be the mean Anomaly, and $x$ the Planet's Distance from the Sun, computed to the Anomalia Eccentri $B$; the Angle $B$ taken equal to $B + \frac{t}{x} \times M - \mu$, will approach nearer to the true Value of the Angle sought; and by Repetitions of the same Operation, the Approximation may be carried on nearer and nearer, ad libitum. This last Rule being obvious, the Explication of it may be omitted at present. SCHOLIUM. In this Solution, where the Motion is reckoned from the Perihelion, the Rule is universal, and under no Limitation: But had the Motion been taken from the Aphelion, the Problem must have been divided into two Cases: One is, when the Eccentricity is less than $\frac{9}{16}$; the other is, when it is not less, but is either equal to, or more than in that Proportion. If the Eccentricity be not less than $\frac{9}{16}$, then the same Rule will hold, as before, only putting the Aphelian Distance, suppose ($a$) instead of the Perihelian Distance ($p$), and substituting $-f$ for $+f$ in the Rule for the Number $n$. If the Eccentricity be less than $\frac{9}{16}$, then take the Number $n$ equal to $\sqrt{\frac{a}{f}}$, and $\frac{t}{n} \times \frac{M}{R}$ will be nearly equal to the Sine of the Submultiple Part of the Anomalia Eccentri denominated by the Number $n$, as before. It is needless to observe, that the like Rules would obtain in Hyperbolic Orbits, *mutatis mutandis*. But that which perhaps may not appear unworthy of being remarked, concerning this sort of Solution from the Cubic Root, is, that although the Rule be altogether impossible, upon a total Change of the Figure of the Orbit either into a Circle, or into a Parabola; yet it will operate so much better, and stand in need of less Correction, according as the Figure advances nearer in its Change towards either of those two Forms. That the Use of the Method may better appear, it may not be amiss to add a few Examples. I have given two for the Orbits of Planets, one the most, and the other the least Eccentric; but which are more to shew the Extent of the Rule, than to recommend the Use of it in such Cases; for there are many other much better and more expeditious Methods in Orbits of small Eccentricity. The other two Examples are adapted to the Orbits of two Comets, whose Periods have been already discovered by Dr. Halley; the one is to shew the Use of one of the Rules in the first Corollary, and the other is to explain the Use of the other Rule. **Example I.** *For the Orbit of Mercury.* If an Unit being put for the Semi-transverse Axis (*t*), the Eccentricity 0.20589 will become (*f*), and the Perihelion Distance (*p*) will be 0.79411; wherefore by means of the Number *R* given as before, the constant Numbers for this Orbit will appear to be, \[ n = 3.56755, \quad T = 0.5857271, \quad P = \frac{p}{t} T = 0.4651319, \text{ and hence } \] \[ \frac{3T}{n \times R} = 0.0085965. \] *Example.* Example. Suppose $M$ the mean Anomaly from the Perihelion to be $120^\circ.00'.00''$, to which it is required to find the Anomalia Eccentri. Here, since the mean Anomaly $M$ is not many times more than the limiting Angle $\frac{2np}{3t}R\sqrt{P}$, (which in this Orbit is about 74 Degrees) recourse must be had to the general Rule in the Proposition. The Number $N$ then, which is $\sqrt[3]{\frac{3T}{nR}}M$ will be $=1,0104195$; which found gives $$N\sqrt{\frac{1}{2}+\frac{1}{4}+\frac{P^3}{N^6}} = 1,0389090; \text{ and also}$$ $$N\sqrt{\frac{1}{2}-\frac{1}{4}+\frac{P^3}{N^6}} = -0,4477126.$$ Wherefore the Sum of both (under their proper Signs) viz. $0,5911964$ will be the Sine whose Arch $36^\circ,24195$ is the Angle $A$; the Multiple whereof $n\times A = 129^\circ,295503$, will be the Angle to be first assumed for the Anomalia Eccentri. For a further Correction; this Angle, now called $B$, whose Sine is suppose $y$, and its Cosine $z$, gives, by a known Rule, $x + \frac{f}{t}z = 1,1304$ for $x$ the Planet's Distance from the Sun; and by another known Rule $B-\frac{fR}{t}y = 120^\circ,16568$ for $\mu$, the mean Anomaly to the Anomalia Eccentri $B$. Wherefore the correct Angle $B = B + \frac{t}{x}\times M - \mu$ will be $129^\circ,14846 = 129^\circ.08'.54''.5$, erring, as will appear from a further Correction, about $\frac{1}{10}$ of a Second. This Angle, being thus determined, will give by the common Methods $137^\circ.48'.33''\frac{1}{2}$, for the true Anomaly or Angle at the Sun: The Sine of the true Anomaly being in Proportion to the Sine of the Anomalia Eccentri, as the Semi-conjugate Axis to the Planet's Distance from the Sun. So that the Equation of the Centre in this Example is $17^\circ 48'.33''\frac{1}{2}$. **Example II.** *For the Orbit of Venus.* Supposing, as before, the mean Distance $t$ to be Unity, and the Eccentricity $f$ to be $0.0069855$; the constant Numbers for this Orbit will be, $p=0.9930115$; $n=6.4116$; $T=1.562134$; $P=0.1551217$; $\frac{3T}{nR}=0.0127571$; and the limiting Angle $\frac{2np}{3t}R\sqrt{P}$, will appear to be about 303 Degrees. *Example.* Let $M$ be $120^\circ.00'.00''$, as in the former Example. Then, since the mean Anomaly is, in this Case, not many times less than the limiting Angle, the general Rule must be used as before; according to which the Number $N$ will appear to be $1.152585$; the Sine of $A$ will be $0.3217917$; the Angle $A$, $18^\circ.77132$; and the Multiple $n\times A$, or Angle $B$; for the first Assumption of the Anomalia Eccentri will be $120^\circ.35416$. This Angle $B$ will give, by the Method before explained, the Angle $B=120^\circ.34555$; or $120^\circ.21'.44''$ fere, for the Anomalia Eccentri correct; the Error of which will appear, upon Examination, to be but a small Part of a Second. In this Example the true Anomaly is $120^\circ.41'$. $25''$, and consequently the Equation of the Centre no more than $41'.25''$. **Example III.** *For the Orbit of the Comet of 1682.* To know the mean Anomaly of this Comet to any given Time, it is to be premised, that it was at the Perihelion in the Year 1682, on the 4th Day of September, at 21 Ho. 22 Min. equated Time to the Meridian of Greenwich, and makes its Revolution about the Sun, as Dr. Halley has discover'd, in $75\frac{1}{3}$ Years. The Perihelion Distance $p$ is, according to his Determination, $0.0326085$ Parts of the mean Distance $t$. So that the constant Numbers for the Orbit will be, $n = 3.1676061; T = 0.2054272; P = 0.00669867$; and the limiting Angle $\frac{2np}{3t}R\sqrt{P}$ will be about 19 Minutes or $\frac{1}{3}$ of a Degree. In the Orbits of Comets, the Rule for the first Assumption of the Anomalia Eccentri is generally sufficient without Correction. Thus, suppose the mean Anomaly $M$ to be $0.072706$, (as it was at the time of an Observation made at Greenwich on the 30th of August 1682, at 7h. 42'. Æq. T.) then the general Rule (which must be here used, since the Angle of mean Anomaly is not above 4 or 5 times less than the limiting Angle) will give $nxA$ or $B = 2^\circ 12'.48''$, erring about $\frac{7}{10}$ of a Second from the true Anomalia Eccentri. But in these Orbits the Rules in the first Corollary to the second Proposition most frequently take Place, especially the last; and the Calculation may also be further abbreviated, by putting the square Root of 10, or the Integer 3, for the Number \( n \). **Example.** Suppose the mean Anomaly to be \( 0^\circ 006522 \), or \( 23''4792 \): Here, since \( M \) is 50 times less than the limiting Angle, the Rule in the first Case of the first Corollary may be used; that is, to take the Sine of the Angle \( A = \frac{t \times M}{n \times R} \). Wherefore, if the Number 3 be put for \( n \), the Sine of \( A \), which is \( \frac{t M}{3 p R} \), will be \( = 0.00116367 \); and consequently the Angle \( A \) will be \( 4'.00''.011 \); and the multiple Angle \( n \times A \) to be assumed for the Anomalia Eccentri will be \( 12'.00''.033 \), the Error of which will be found to be about \( \frac{1}{30} \) of a Second. **Example IV.** *For the Orbit of the great Comet of the Year 1680.* This Comet, according to Dr. Halley, performs its Period in 575 Years; and was in its Perihelion on the 7th of December 1680, at \( 23^h\ 09' \) Æq. T. at London; the perihelion Distance \( p \) is \( 0.000089301 \), in Parts of the mean Distance \( t \): Wherefore supposing the Number \( n \) to be \( \sqrt{10} \), the constant Numbers for the Orbit will be \( T = 0.2000161 \); \( P = 0.000017862 \), and the limiting Angle \( \frac{2 \times p}{3 \times R} \sqrt{P} \) will be about \( \frac{1}{6} \) of a Second. **Example.** Suppose the mean Anomaly to be \( 3'.31''4478 \) or \( 0^\circ 05873541 \), (as it was at the Time of the first Observation made on it in Saxony, on November November the 3d, at 16h. 47' Aeq. T. at London.) here, since the mean Anomaly is many times greater than $\frac{1}{6}$ of a Second, the Rule in the second Case of the first Corollary may be used; that is, by taking the Sine of $A = N - \frac{P}{N}$. But the Number $N$ or $\sqrt{\frac{3T}{nR}} M$ is $= 0.05794134$; and $\frac{P}{N}$ will be $= 0.0030827$; wherefore $(N - \frac{P}{N}) = 0.05763307$, will be the Sine whose Arch $3^\circ 30' 39"$ is the Angle $A$; and the multiple Angle $n \times A = 10^\circ 26'. 53'' 05"$, will be the Angle to be first assumed for the Anomalia Eccentri, the Error of which will be found to be less than a Second. The true Anomaly, computed from this Angle according to the Rule in the Example for Mercury, will appear to be $171^\circ 38'. 24''$. from the Perihelion. By these Examples it appears, that the Solution is universal in all respects; for the two first, compared with the two last, serve to shew that it is not confined to any particular Parts of the Orbit, but extends to all Degrees of mean Anomaly: And by comparing the second with the last, it sufficiently appears to be universal with respect to the several Degrees of Eccentricity; since in one the Equation of the Centre for the Reduction of the Mean to the true Motion is not so much as the $\frac{1}{170}$th Part of the whole; whereas in the other it amounts to almost 3000 times as much as the mean Motion itself. POSTSCRIPT. UPON reviewing the Reflections on the Quadrature of the Circle in Page 212, I believe it may be necessary for me, to prevent any Mistake that may arise from the different Opinions that obtain about the Nature of Mathematical Quantity, to explain myself a little upon that head; as also to add a few Words to shew how the Method of Quadrature by limiting Polygons, takes place in other Figures as well as the Circle. I take then a Mathematical Quantity, and that for which any Symbol is put, to be nothing else but Number with regard to some Measure which is considered as one. For we cannot know precisely and determinately, that is, mathematically, how much any thing is, but by means of Number. The Notion of continued Quantity, without regard to any Measure, is indistinct and confused; and although some Species of such Quantity, considered physically, may be described by Motion, as Lines by Points, and Surfaces by Lines, and so on; yet the Magnitudes or Mathematical Quantities are not made by that Motion, but by numbering according to a Measure. Accordingly, all the several Notations that are found necessary to express the Formations of Quantities, do refer to some Office or Property of Number or Measure; but none can be interpreted to signify continued Quantity as such. Thus some Notations are found requisite to express Number in its ordinal Capacity, or the Numerus Numerans, as when one follows or precedes another, in the first, second or third Place from that upon which it depends; as, the Quantities \(x\), \(x'\), \(x''\), referring to the principal one \(x\). So, in many Cases, a Notation is found necessary to be given to a Measure as a Measure; as for Instance, Sir Isaac Newton's Symbol for a Fluxion \(x\); for this stands for a Measure of some kind, and accordingly he usually puts an Unit for it, if it be the principal one upon which the rest depend. So some Notations are expressly to shew a Number in the form of its Composition, as the Index to the Geometrical Power \(x^n\) denoting the Number of equal Factors which go to the Composition of it, or what is analogous to such. But that there is no Symbol or Notation but what refers to discrete Quantity, is manifest from the Operations, which are all Arithmetical. And hence it is, there are so many Species of Mathematical Quantity as there are Forms of composite Numbers, or Ways in the Composition of them; among which there are two more eminent for their Simplicity. Simplicity and Universality than the rest: One is the Geometrical Power formed from a constant Root; and the other, though well known, yet wanting a Name as well as a Notation, may be called the Arithmetical Power; or the Power of a Root uniformly increasing or diminishing, and is that whose Notation is designed in Page 213: The one is only for the Form of the Quantity itself, the other is for the Constitution of it from its Elements. Now from the Properties of either of these it would be easy to shew how the Quadratures of simple Figures are deducible from the Areas of their limiting Polygons. I shall just point out the Method from the Arithmetical Power, as being the shortest and readiest at hand. Let \( z, \dot{z}, \ddot{z} \) &c. or \( z, \dot{z}, \ddot{z} \), &c. be Quantities in Arithmetical Progression, diminishing or increasing by the common Difference \( \dot{z} \), and let, as before explained, \( \dot{z}^{m} \) signify the Arithmetical Power of \( z \), denominated by the potential Index \( m \), namely, \( z \times \dot{z} \times \ddot{z} \), &c. whose first Root is \( z \) and last \( z - m = 1 \times z \); which being supposed, the Element of the Arithmetical Power will be \( m z \times \dot{z} \), that is, the Product made from the Multiplication of the two Indices, and the next inferior Power of the next Root in Order. For the first Arithmetical Power \( z \) is \( z \times \dot{z} \), and the next \( \dot{z} \) is \( \dot{z} \times \ddot{z} - m \dot{z} \), wherefore the Difference will be as is explained. And consequently, since the Sum of these Elements or Differences, taken in order from the first to the last, do make up the Quantity according to its termini; hence, if \( x \) be the Abscissa of a curvilinear Figure whose Ordinate \( y \) is equal to \( m z^{m-1} \); a Demonstration might easily be made that the [Form of the Quantity for] the Area will be \( z^{m} \); that is, the same Multiple of the next superior Power of \( z \) divided by the Index of that Power. For since the Arithmetical Powers do both unite and become the same with the Geometrical Power, when the differential Index \( \dot{z} \) is supposed to be nothing; the Magnitude of the Geometrical Figure will be implied from the Magnitudes of the two Polygons made up of Rectangles, one from the increasing Arithmetical Power, the other from the diminishing, although it be true, that the Elements of the Polygons cannot be summed up, when \( z \), the Measure of the Abscissa \( z \), is supposed to be nothing. In like manner, in any other Case where \( z \) and \( z' \) are two Abscisses whose Difference as a Measure is \( z \); and \( y, y' \) the two Ordinates; the Magnitude of the Figure will be implied by the Magnitudes of the two Polygons which are made from the Sum of the inscribing and circumscribing Elements \( z, y \) and \( z', y' \), although the Figure itself is not to be resolved into any such primogenial rectangular Elements. And thus, I think, the Symbol \( z \), considered as a component Part of the Rectangle \( z, y \), may bear a plain Interpretation; viz. that it is the Measure according to which the Quantity \( z \) is measured; nor can I see that any other Interpretation need to be put upon a Symbol, which, like a Measure, is used only to make other things known, but is of itself for nothing but a Mark. And what is said of the Elements of the first Resolution, is easily applied to those of a second or third, and so on; the last may always be considered as the Measure of the former and indivisible, although, in respect of the following, it be taken as the Part according to which the Measure was made, and therefore divisible. The candid Reader is desired to strike out a Remark of mine, about the Comet of 1556, subjoined to an Observation of the late Comet made at Lisbon, printed in Page 123 of the last Transactions, No 446. The Note was there inserted by some Accident, without my Intention; for I had soon afterwards informed the Society, that the Remark was ill-founded. According to Mr. Bradley's Observations at Oxford, which were not then communicated, the Place at the Time mentioned ought to have been in Long. \( \text{H}^\circ \). 13\(\frac{1}{2}\). 21\(\frac{1}{2}\). Lat. 0\(\circ\). 29\(\frac{1}{2}\). South. So that in all probability there happened some Mistake in making this Observation. ERRATA. P. 213. L. 31. for \( n.n-2 \) read \( n.n-2 \). P. 222. L. 26. for being read be. Printed for T. Woodward, at the Half-Moon, between the Two Temple-Gates in Fleetstreet; and C. Davis, the Corner of Pater-noster-row, next Warwick-lane; Printers to the Royal Society. M.DCC.XXXIX.