PROB

PROB. I. To determine the momentum of libration corresponding to any position of the earth respecting the sun, that is, to determine the accumulated energy of the disturbing forces on all the protuberant matter of the spheroid.

Let B and $b$ be two particles in the ring formed by the revolution of AQ, and so situated, that they are at equal distances from the plane NM; but on opposite sides of it. Draw BD, $b\; d$, perpendicular to NM, and FLG perpendicular to LT.

Then, because the momentum, or power of producing rotation, is as the force and as the distance of its line of direction from the axis of rotation, jointly, the combined momentum of the particles B and $b$ will be $f\; \backslash cdot\; BD\; \backslash cdot\; DC\; =\; f\; \backslash cdot\; bd\; \backslash cdot\; dc$, (for the particles B and $b$, are urged in contrary directions). But the momentum of B is $f\; \backslash cdot\; BF\; \backslash cdot\; DC\; +\; f\; \backslash cdot\; FD\; \backslash cdot\; DC$, and that of $b$ is $f\; \backslash cdot\; bG\; \backslash cdot\; dC\; =\; f\; \backslash cdot\; dG\; \backslash cdot\; dC$; and the combined momentum is $f\; \backslash cdot\; BF\; \backslash cdot\; DC\; +\; f\; \backslash cdot\; FD\; \backslash cdot\; DC\; +\; f\; \backslash cdot\; dG\; \backslash cdot\; dC\; =\; 2\; f\; \backslash cdot\; BF\; \backslash cdot\; LF\; -\; 2\; f\; \backslash cdot\; LT\; \backslash cdot\; TC$.

Because $m$ and $n$ are the sine and cosine of the angle ECS or LCT, we have $LT\; =\; m\; \backslash cdot\; CL$, and $CT\; =\; n\; \backslash cdot\; CL$, and $LF\; =\; m\; \backslash cdot\; BL$, and $BF\; =\; n\; \backslash cdot\; BL$. This gives the momentum $=\; 2\; f\; m\; n\; BL^2\; -\; CL^2$.

The breadth AQ of the protuberant ring being very small, we may suppose, without any sensible error, that all the matter of the line AQ is collected in the point Q; and, in like manner, that the matter of the whole ring is collected in the circumference of its inner circle, and that B and $b$ now represent, not single particles, but the collected matter of lines such as AQ, which terminate at B and $b$. The combined momentum of two such lines will therefore be $2\; m\; n\; f\; \backslash cdot\; AQ\; \backslash cdot\; BL^2\; -\; CL^2$.

Let the circumference of each parallel of latitude be divided into a great number of indefinitely small and equal parts. The number of such parts in the circumference, of which Q $q$ is the diameter, will be $\backslash pi\; \backslash cdot\; QL$. To each pair of these there belongs a momentum $2\; m\; n\; f\; \backslash cdot\; AQ\; \backslash cdot\; BL^2\; -\; CL^2$. The sum of all the squares of BL, which can be taken round the circle, is one half of as many squares of the radius CL: for BL is the sine

of an arch, and the sum of its square and the square of its corresponding cosine is equal to the square of the radius. Therefore the sum of all the squares of the sines, together with the sum of all the squares of the cosines, is equal to the sum of the same number of squares of the radius; and the sum of the squares of the sines is equal to the sum of the squares of the corresponding cosines: therefore the sum of the squares of the radius is double of either sum. Therefore $\backslash int\; \backslash pi\; \backslash cdot\; QL\; \backslash cdot\; BL^2\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; \backslash cdot\; QL\; \backslash cdot\; QL^2$. In like manner the sum of the number $\backslash pi\; \backslash cdot\; QL$ of $CL^2$'s will be $=\; \backslash pi\; \backslash cdot\; QL\; \backslash cdot\; CL^2$. These sums, taken for the semicircle, are $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash cdot\; QL\; \backslash cdot\; QL^2$, and $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash cdot\; QL\; \backslash cdot\; CL^2$, or $\backslash pi\; \backslash cdot\; QL\; \backslash cdot\; \backslash frac\{1\}\{2\}\; QL^2$, and $\backslash pi\; \backslash cdot\; QL\; \backslash cdot\; \backslash frac\{1\}\{2\}\; CL^2$: therefore the momentum of the whole ring will be $2\; m\; n\; f\; \backslash cdot\; AQ\; \backslash cdot\; QL\; \backslash cdot\; \backslash pi\; (\backslash frac\{1\}\{2\}\; QL^2\; -\; \backslash frac\{1\}\{2\}\; CL^2)$: for the momentum of the ring is the combined momenta of a number of pairs, and this number is $\backslash frac\{1\}\{2\}\; \backslash pi\; \backslash cdot\; QL$.

By the ellipse we have $OC\; :\; QL\; =\; EO\; :\; AQ$, and $AQ\; =\; QL\; \backslash cdot\; \backslash frac\{EO\}\{OC\}\; =\; QL\; \backslash cdot\; \backslash frac\{d\}\{b\}$; therefore the momentum of the ring is $2\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{b\}\; \backslash cdot\; QL\; \backslash cdot\; \backslash pi\; (\backslash frac\{1\}\{2\}\; QL^2\; -\; \backslash frac\{1\}\{2\}\; CL^2)\; =\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{b\}\; \backslash cdot\; QL^2\; \backslash cdot\; \backslash pi\; (\backslash frac\{1\}\{2\}\; QL^2\; -\; CL^2)$: but $QL^2\; =\; b^2\; -\; x^2$; therefore $\backslash frac\{1\}\{2\}\; QL^2\; -\; CL^2\; =\; \backslash frac\{1\}\{2\}\; b^2\; -\; \backslash frac\{1\}\{2\}\; x^2\; -\; x^2\; =\; \backslash frac\{1\}\{2\}\; b^2\; -\; \backslash frac\{3\}\{2\}\; x^2\; =\; \backslash frac\{b^2\; -\; 3x^2\}\{2\}$; therefore the momentum of the ring is $m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{b\}\; \backslash cdot\; \backslash pi\; (b^2\; -\; x^2)$

$(\backslash frac\{b^2\; -\; 3x^2\}\{2\})\; =\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{b\}\; \backslash cdot\; \backslash pi\; (\backslash frac\{b^2\; -\; 4b^2\; x^2\; +\; 3x^2\}\{2\})\; =\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{2b\}\; \backslash cdot\; \backslash pi\; (b^2\; -\; 4b^2\; x^2\; +\; 3x^2)$. If we now suppose another parallel extremely near to A $a$, as represented by the dotted line, the distance $L\; l$ between them being $x$, we shall have the fluxion of the momentum of the spheroid $m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{2b\}\; \backslash cdot\; \backslash pi\; (b^2\; x\; -\; 4b^2\; x^2\; x\; +\; 3x^2\; x)$, of which the fluent is $m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{2b\}\; \backslash cdot\; \backslash pi\; (b^2\; x\; -\; 4b^2\; \backslash frac\{x^3\}\{3\}\; +\; \backslash frac\{3x^5\}\{5\})$. This expresses the momentum of the zone EA $a\; Q$, contained between the equator and the parallel of latitude A $a$. Now let $x$ become $=\; b$, and we shall obtain the momentum of the hemispheroid $=\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{2b\}\; \backslash cdot\; \backslash pi\; (b^2\; -\; \backslash frac\{4\}\{3\}\; b^2\; +\; \backslash frac\{3\}\{5\}\; b^2)$, and that of the spheroid $=\; m\; n\; f\; \backslash cdot\; \backslash frac\{d\}\{b\}\; \backslash cdot\; \backslash pi\; (b^2\; -\; \backslash frac\{4\}\{3\}\; b^2\; +\; \backslash frac\{3\}\{5\}\; b^2)\; =\; \backslash frac\{4\}\{15\}\; m\; n\; f\; \backslash cdot\; d\; \backslash cdot\; \backslash pi\; b^2$.

This formula does not express any motion, but only a pressure tending to produce motion, and particularly tending to produce a libration by its action on the cohering matter of the earth, which is affected as a number of levers. It is similar to the common mechanical formula to $d$, where $w$ means a weight, and $d$ its distance from the fulcrum of the lever.

It is worthy of remark, that the momentum of this protuberant matter is just one-fifth of what it would be if it were all collected at the point O of the equator: for the matter in the spheroid is to that in the inscribed sphere as $a^3$ to $b^3$, and the contents of the inscribed sphere is $\backslash frac\{1\}\{2\}\; \backslash pi\; b^3$. Therefore $a^3\; :\; a^3\; -\; b^3\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; b^3\; :\; \backslash frac\{1\}\{2\}\; \backslash pi\; b^3$. $\backslash frac\{a^3\; -\; b^3\}\{a^3\}$, which is the quantity of protuberant matter

It is worthy of remark, that the momentum of this protuberant matter is just one-fifth of what it would be if it were all collected at the point O of the equator: for the matter in the spheroid is to that in the inscribed sphere as $a^3$ to $b^3$, and the contents of the inscribed sphere is $\backslash frac\{1\}\{2\}\; \backslash pi\; b^3$. Therefore $a^3\; :\; a^3\; -\; b^3\; =\; \backslash frac\{1\}\{2\}\; \backslash pi\; b^3\; :\; \backslash frac\{1\}\{2\}\; \backslash pi\; b^3$. $\backslash frac\{a^3\; -\; b^3\}\{a^3\}$, which is the quantity of protuberant matter

Precession. We may, without sensible error, suppose $\backslash frac\{a^2\; -\; b^2\}\{a\}\; =\; 2d$; then the protuberant matter will be $\backslash frac\{4\}\{3\}\backslash pi\; b^2\; d$. If all this were placed at O, the momentum would be $\backslash frac\{4\}\{3\}\backslash pi\; a\; b^2\; f\; \backslash cdot\; OH\; \backslash cdot\; HC\; =\; \backslash frac\{4\}\{3\}\; m\; n\; f\; d\; b^2$, because $OH\; \backslash cdot\; HC\; =\; m\; n\; b^2$; now $\backslash frac\{4\}\{3\}$ is 5 times $\backslash frac\{1\}\{15\}$.

Also, because the sum of all the rectangles $OH\; \backslash cdot\; HC$ round the equator is half of as many squares of $OC$, it follows that the momentum of the protuberant matter placed in a ring round the equator of the sphere or spheroid is one half of what it would be if collected in the point O or E; whence it follows that the momentum of the protuberant matter in its natural place is two-fifths of what it would be if it were disposed in an equatorial ring. It was in this manner that Sir Isaac Newton was enabled to compare the effect of the sun's action on the protuberant matter of the earth, with his effect on a rigid ring of moons. The preceding investigation of the momentum is nearly the same with his, and appears to us greatly preferable in point of perspicuity to the fluxionary solutions given by later authors. These indeed have the appearance of greater accuracy, because they do not suppose all the protuberant matter to be condensed on the surface of the inscribed sphere: nor were we under the necessity of doing this, only it would have led to very complicated expressions had we supposed the matter in each line $AQ$ collected in its centre of oscillation or gyration. We made a compensation for the error introduced by this, which may amount to $\backslash frac\{1\}\{15\}$ of the whole, and should not be neglected, by taking $d$ as equal to $\backslash frac\{a^2\; -\; b^2\}\{2a\}$ instead of $\backslash frac\{a^2\; -\; b^2\}\{a\; +\; b\}$.

The consequence is, that our formula is the same with that of the later authors.

Thus far Sir Isaac Newton proceeded with mathematical rigour; but in the application he made two assumptions, or, as he calls them, hypotheses, which have been found to be unwarranted. The first was, that when the ring of protuberant matter is connected with the inscribed sphere, and subjected to the action of the disturbing force, the same quantity of motion is produced in the whole mass as in the ring alone. The second was, that the motion of the nodes of a rigid ring of moons is the same with the mean motion of the nodes of a solitary moon. But we are now able to demonstrate, that it is not the quantity of motion, but of momentum, which remains the same, and that the nodes of a rigid ring move twice as fast as those of a single particle. We proceed therefore to

Prob. 2. To determine the deviation of the axis, and the retrograde motion of the nodes which result from this libratory momentum of the earth's protuberant matter.

But here we must refer our readers to some fundamental propositions of rotatory motions which are demonstrated in the article ROTATION.

If a rigid body is turning round an axis $A$, passing through its centre of gravity with the angular velocity $a$, and receives an impulse which alone would cause it to turn round an axis $B$, also passing through its centre of gravity, with the angular velocity $b$, the body will now turn round a third axis $C$, passing through its centre of gravity, and lying in the plane of the axes $A$ and $B$, and the sine of the inclination of this third axis to the axis $A$ will be to the sine of inclination to the axis $B$ as the velocity $b$ to the velocity $a$.

When a rigid body is made to turn round any axis Precession. by the action of an external force, the quantity of momentum produced (that is, the sum of the products of every particle by its velocity and by its distance from the axis) is equal to the momentum or similar product of the moving force or forces. 32

If an oblate spheroid, whose equatorial diameter is $a$ and polar diameter $b$, be made to librate round an equatorial diameter, and the velocity of that point of the equator which is farthest from the axis of libration be $v$, the momentum of the spheroid is $\backslash frac\{4\}\{15\}\backslash pi\; a^2\; b^2\; v$. 33

The two last are to be found in every elementary book of mechanics.

Let $ANan$ (fig. 4.) be the plane of the earth's equator, cutting the ecliptic $CNKn$ in the line of the nodes or equinoctial points $Nn$. Let $OAS$ be the section of the earth by a meridian passing through the sun, so that the line $OCS$ is in the ecliptic, and $CA$ is an arch of an hour-circle or meridian, measuring the sun's declination. The sun not being in the plane of the equator, there is, by prop. 1. a force tending to produce a libration round an axis $ZOz$ at right angles to the diameter $Aa$ of that meridian in which the sun is situated, and the momentum of all the disturbing forces is $\backslash frac\{4\}\{15\}\; m\; n\; f\; d\; \backslash pi\; b^2\; i$. The product of any force by the moment $i$ of its action expresses the momentary increment of velocity; therefore the momentary velocity, or the velocity of libration generated in the time $i$ is $\backslash frac\{4\}\{15\}\; m\; n\; f\; d\; \backslash pi\; b^2\; i$. This is the absolute velocity of a point at the distance $i$ from the axis, or it is the space which would be uniformly described in the moment $i$, with the velocity which the point has acquired at the end of that moment. It is double the space actually described by the libration during that moment; because this has been an uniformly accelerated motion, in consequence of the continued and uniform action of the momentum during this time. This must be carefully attended to, and the neglect of it has occasioned very faulty solutions of this problem.

Let $v$ be the velocity produced in the point $A$, the most remote from the axis of libration. The momentum excited or produced in the spheroid is $\backslash frac\{4\}\{15\}\backslash pi\; a^2\; b^2\; v$ (as above), and this must be equal to the momentum of the moving force, or to $\backslash frac\{4\}\{15\}\; m\; n\; f\; d\; \backslash pi\; b^2\; i$; therefore we obtain $v\; =\; \backslash frac\{\backslash frac\{4\}\{15\}\; m\; n\; f\; d\; \backslash pi\; b^2\; i\}\{\backslash frac\{4\}\{15\}\backslash pi\; a^2\; b^2\}$, that is $v\; =\; m\; n\; f\; d\; i\; \backslash frac\{b^2\}\{a^2\}$ or very nearly $m\; n\; f\; d\; i$, because $\backslash frac\{b^2\}\{a^2\}\; =\; 1$ very nearly. Also, because the product of the velocity and time gives the space uniformly described in that time, the space described by $A$ in its libration round $ZOz$ is $m\; n\; f\; d\; i^2$, and the angular velocity is $\backslash frac\{m\; n\; f\; d\; i\}\{a\}$.

Let $r$ be the momentary angle of diurnal rotation. The arch $A\; r$, described by the point $A$ of the equator in this moment $i$ will therefore be $a\; r$, that is, $a\; \backslash times\; r$, and the velocity of the point $A$ is $\backslash frac\{a\; r\}\{i\}$, and the angular velocity of rotation is $\backslash frac\{r\}\{i\}$.

Here then is a body (fig. 5.) turning round an axis Fig. 5. OP₂

Precession. OP, perpendicular to the plane of the equator $z\backslash bar\{z\}z$, and therefore situated in the plane $ZP\backslash bar\{z\}$; and it turns round this axis with the angular velocity $\backslash frac\{r\}\{t\}$. It has received an impulse, by which alone it would librate round the axis $Z\backslash bar\{z\}$, with the angular velocity $\backslash frac\{mnfdi\}\{a\}$. It will therefore turn round neither axis ($n^o$ 31.), but round a third axis $OP\text{'}$, passing through $O$, and lying in the plane $ZP\backslash bar\{z\}$, in which the other two are situated, and the sine $P\backslash Pi$ of its inclination to the axis of libration $Z\backslash bar\{z\}$ will be to the sine $P\backslash rho$ of its inclination to the axis $OP$ of rotation as $\backslash frac\{r\}\{t\}$ to $\backslash frac\{mnfdi\}\{a\}$.

Now $A$, in fig. 4. is the summit of the equator both of libration and rotation; $mnfdi^2$ is the space described by its libration in the time $i$; and $ar$ is the space or arch $Ar$ (fig. 4.) described in the same time by its rotation: therefore, taking $Ar$ to $Ac$ (perpendicular to the plane of the equator of rotation, and lying in the equator of libration,) as $ar$ to $mnfdi^2$, and completing the parallelogram $Armc$, $Am$ will be the compound motion of $A$ ($n^o$ 31.), and $ar:mnfdi^2\; =\; t:\; \backslash frac\{mnfdi^2\}\{a\}$, which will be the tangent of the angle $mAr$, or of the change of position of the equator. But the axes of rotation are perpendicular to their equator; and therefore the angle of deviation $w$ is equal to this angle $rAm$. This appears from fig. 5.; for $\backslash Pi\; P\text{'}\text{'}\; =\; P\backslash rho\; =\; O\backslash rho:\; P\backslash rho\; =\; OP:\; \backslash tan.\; POP$; and it is evident that $ar:mnfdi^2\; =\; \backslash frac\{r\}\{t\}\; mnfdi^2$, as is required by the composition of rotations.

In consequence of this change of position, the plane of the equator no longer cuts the plane of the ecliptic in the line $Nn$. The plane of the new equator cuts the former equator in the line $AO$, and the part $AN$ of the former equator lies between the ecliptic and the new equator $AN\text{'}$, while the part $Am$ of the former equator is above the new one $Am\text{'}$; therefore the new node $N\text{'}$, from which the point $A$ was moving, is removed to the westward, or farther from $A$; and the new node $n\text{'}$, to which $A$ is approaching, is also moved westward, or nearer to $A$; and this happens in every position of $A$. The nodes, therefore, or equinoctial points, continually shift to the westward, or in a contrary direction to the rotation of the earth; and the axis of rotation always deviates to the east side of the meridian which passes through the sun.

This account of the motions is extremely different from what a person should naturally expect. If the earth were placed in the summer solstice, with respect to us who inhabit its northern hemisphere, and had no rotation round its axis, the equator would begin to approach the ecliptic, and the axis would become more upright; and this would go on with a motion continually accelerating, till the equator coincided with the ecliptic. It would not stop here, but go as far on the other side, till its motion were extinguished by the opposing forces; and it would return to its former position, and again begin to approach the ecliptic, playing up

and down like the arm of a balance. On this account Precession. this motion is very properly termed libration; but this very slow libration, compounded with the incomparably swifter motion of diurnal rotation, produces a third motion extremely different from both. At first the north pole of the earth inclines forward toward the sun; after a long course of years it will incline to the left hand, as viewed from the sun, and be much more inclined to the ecliptic, and the plane of the equator will pass through the sun. Then the south pole will come into view, and the north pole will begin to decline from the sun; and this will go on (the inclination of the equator diminishing all the while) till, after a course of years, the north pole will be turned quite away from the sun, and the inclination of the equator will be restored to its original quantity. After this the phenomena will have another period similar to the former, but the axis will now deviate to the right hand. And thus, although both the earth and sun should not move from their places, the inhabitants of the earth would have a complete succession of the seasons accomplished in a period of many centuries. This would be prettily illustrated by an iron ring poised very nicely on a cap like the card of a mariner's compass, having its centre of gravity coinciding with the point of the cap, so that it may whirl round in any position. As this is extremely difficult to execute, the cap may be pierced a little deeper, which will cause the ring to maintain a horizontal position with a very small force. When the ring is whirling very steadily, and pretty briskly, in the direction of the hours of a watch-dial, hold a strong magnet above the middle of the nearer semicircle (above the 6 hour point) at the distance of three or four inches. We shall immediately observe the ring rise from the 9 hour point, and sink at the 3 hour point, and gradually acquire a motion of precession and nutation, such as has been described.

If the earth be now put in motion round the sun, or the sun round the earth, motions of libration and deviation will still obtain, and the succession of their different phases, if we may so call them, will be perfectly analogous to the above statement. But the quantity of deviation, and change of inclination, will now be prodigiously diminished, because the rapid change of the sun's position quickly diminishes the disturbing forces, annihilates them by bringing the sun into the plane of the equator, and brings opposite forces into action.

We see in general that the deviation of the axis is always at right angles to the plane passing through the sun, and that the axis, instead of being raised from the ecliptic, or brought nearer to it, as the libration would occasion, deviates sideways; and the equator, instead of being raised or depressed round its east and west points, is twisted sideways round the north and south points; or at least things have this appearance; but we must now attend to this circumstance more minutely.

The composition of rotation shows us that this change of the axis of diurnal rotation is by no means a translation of the former axis (which we may suppose to be the axis of figure) into a new position, in which it again becomes the axis of diurnal motion; nor does the equator of figure, that is, the most prominent section of the terrestrial spheroid, change its position, and in this new position continue to be the equator of rotation. This was indeed supposed by Sir Isaac Newton;

tion; and this supposition naturally resulted from the train of reasoning which he adopted. It was strictly true of a single moon, or of the imaginary orbit attached to it; and therefore Newton supposed that the whole earth did in this manner deviate from its former position, still, however, turning round its axis of figure. In this he has been followed by Walmsley, Simpson, and most of his commentators. D'Alembert was the first who entertained any suspicion that this might not be certain; and both he and Euler at last showed that the new axis of rotation was really a new line in the body of the earth, and that its axis and equator of figure did not remain the axis and equator of rotation. They ascertained the position of the real axis by means of a most intricate analysis, which obscured the connection of the different positions of the axis with each other, and gave us only a kind of momentary information. Father Frisius turned his thoughts to this problem, and fortunately discovered the composition of rotations as a general principle of mechanical philosophy. Few things of this kind have escaped the penetrating eye of Sir Isaac Newton. Even this principle had been glanced at by him. He affirms it in express terms with respect to a body that is perfectly spherical (cor. 22. prop. 66. B. I.). But it was reserved for Frisius to demonstrate it to be true of bodies of any figure, and thus to enrich mechanical science with a principle which gives simple and elegant solutions of the most difficult problems.

But here a very formidable objection naturally offers itself. If the axis of the diurnal motion of the heavens is not the axis of the earth's spherical figure, but an imaginary line in it, round which even the axis of figure must revolve; and if this axis of diurnal rotation has so greatly changed its position, that it now points at a star at least 12 degrees distant from the pole observed by Timochares, how comes it that the equator has the very same situation on the surface of the earth that it had in ancient times? No sensible change has been observed in the latitudes of places.

The answer is very simple and satisfactory: Suppose that in 12 hours the axis of rotation has changed from the position PR (fig. 6.) to $\backslash rho\; r$, so that the north pole, instead of being at P, which we may suppose to be a particular mountain, is now at $\backslash rho$. In this 12 hours the mountain P, by its rotation round $\backslash rho\; r$, has acquired the position $\backslash pi$. At the end of the next 12 hours, the axis of rotation has got the position $\backslash pi\; r$, and the axis of figure has got the position $\backslash rho\; r$, and the mountain P is now at $\backslash rho$. Thus, on the noon of the following day, the axis of figure PR is in the situation which the real axis of rotation occupied at the intervening midnight. This goes on continually, and the axis of figure follows the position of the axis of rotation, and is never further removed from it than the deviation of 12 hours, which does not exceed $\backslash frac\{1\}\{360\}$th part of one second, a quantity altogether imperceptible. Therefore the axis of figure will always sensibly coincide with the axis of rotation, and no change can be produced in the latitudes of places on the surface of the earth.

Application of this reasoning to nutation and precession. We have hitherto considered this problem in the most general manner; let us now apply the knowledge we have gotten of the deviation of the axis or of the momentary action of the disturbing force to the explanation of the phenomena: that is, let us see what precession and

what nutation will be accumulated after any given time of action. Precession.

For this purpose we must ascertain the precise deviation which the disturbing forces are competent to produce. This we can do by comparing the momentum of libration with the gravitation of the earth to the sun, and this with the force which would retain a body on the equator while the earth turns round its axis.

The gravitation of the earth to the sun is in the proportion of the sun's quantity of matter M directly, and to the square of the distance A inversely, and may therefore be expressed by the symbol $\backslash frac\{M\}\{A^2\}$. The disturbing force at the distance 1 from the plane of illumination is to the gravitation of the earth's centre to the sun as 3 to A, (A being measured on the same scale which measures the distance from the plane of illumination). Therefore $\backslash frac\{3M\}\{A^2\}$ will be the disturbing force $f$ of our formula.

Let $\backslash rho$ be the centrifugal force of a particle at the distance 1 from the axis of rotation; and let $t$ and $T$ be the times of rotation and of annual revolution, viz. sidereal day and year. Then $\backslash rho\; =\; \backslash frac\{M\}\{A^2\}\; =\; \backslash frac\{1\}\{t^2\}\; =\; \backslash frac\{A\}\{T^2\}$. Hence

we derive $\backslash frac\{3M\}\{A^2\}\; =\; 3\backslash rho\; \backslash frac\{t^2\}\{T^2\}$. But since $\backslash dot\{r\}$ was the angular velocity of rotation, and consequently $1\; \backslash times\; \backslash dot\{r\}$ the space described, and $\backslash frac\{1\; \backslash times\; \backslash dot\{r\}\}\{\backslash dot\{t\}\}$ the velocity; and since the centrifugal force is as the square of the velocity divided by the radius, (this being the measure of the generated velocity, which is the proper measure of any accelerating force), we have $\backslash rho\; =\; \backslash frac\{1\; \backslash times\; \backslash dot\{r\}^2\}\{1\; \backslash times\; \backslash dot\{t\}^2\}\; =\; \backslash frac\{\backslash dot\{r\}^2\}\{\backslash dot\{t\}^2\}$, and $f\; =\; \backslash frac\{3\; \backslash dot\{r\}^2\}\{\backslash dot\{t\}^2\}$

$\backslash times\; \backslash frac\{t^2\}\{T^2\}$. Now the formula $f\; m\; n\; d\; \backslash frac\{\backslash dot\{t\}^2\}\{a\}$ expressed the sine of the angle. This being extremely small, the sine may be considered as equal to the arc which measures the angle. Now, substitute for it the value now found, viz. $\backslash frac\{3\; \backslash dot\{r\}^2\}\{\backslash dot\{t\}^2\}\; \backslash times\; \backslash frac\{\backslash dot\{t\}^2\}\{T^2\}$, and we obtain the angle of deviation $w\; =\; \backslash dot\{r\}\; \backslash frac\{3\; t^2\}\{T^2\}\; m\; n\; \backslash frac\{d\}\{a\}$, and this is the simplest form in which it can appear. But it is convenient, for other reasons, to express it a little differently: $d$ is nearly equal to $\backslash frac\{a^2\; -\; b^2\}\{2\; a^2\}$, therefore $w\; =\; \backslash dot\{r\}\; \backslash times\; \backslash frac\{3\; t^2\}\{2\; T^2\}\; m\; n\; \backslash frac\{a^2\; -\; b^2\}\{a^2\}$, and this is the form in which we shall now employ it.

The small angle $\backslash dot\{r\}\; \backslash frac\{3\; t^2\}\{2\; T^2\}\; m\; n\; \backslash frac\{a^2\; -\; b^2\}\{a^2\}$ is the angle in which the new equator cuts the former one. It is different at different times, as appears from the variable part $m\; n$, the product of the sine and cosine of the sun's declination. It will be a maximum when the declination is in the solstice, for $m\; n$ increases all the way to $45^\backslash circ$, and the declination never exceeds $23\backslash frac\{1\}\{2\}^\backslash circ$. It increases, therefore, from the equinox to the solstice, and then diminishes.

Let

Precession.
Fig. 7.

Let ESL (fig. 7.) be the ecliptic, EAC the equator, BAD the new position which it acquires by the momentary action of the sun, cutting the former in the angle $BAE\; =\; i\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; mn\; \backslash frac\{a^2\; -\; b^2\}\{2^2\}$. Let S be the sun's place in the ecliptic, and AS the sun's declination, the meridian AS being perpendicular to the equator. Let $\backslash frac\{a^2\; -\; b^2\}\{a^2\}$ be $k$. The angle BAE is then $=\; i\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; mn$. In the spherical triangle BAE we have $\backslash text\{fin.\; \}\; B\; :\; \backslash text\{fin.\; \}\; AE\; =\; \backslash text\{fin.\; \}\; A\; :\; \backslash text\{fin.\; \}\; BE$, or $=\; AB\; :\; BE$, because very small angles and arcs are as their sines. Therefore BE, which is the momentary precession of the equinoctial point E, is equal to $A\; \backslash frac\{\backslash text\{fin.\; \}\; AE\}\{\backslash text\{fin.\; \}\; B\}\; =\; i\; \backslash times\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; mn$, $\backslash frac\{\backslash text\{fin.\; \}\; R.\; \backslash text\{\; ascen.\}\}\{\backslash text\{fin.\; \}\; \backslash text\{obl.\; ecl.\}\}$.

The equator EAC, by taking the position BAD, recedes from the ecliptic in the colure of the solstices CL, and CD is the change of obliquity or the nutation. For let CL be the solstitial colure of BAD, and $c$ the solstitial colure of EAC. Then we have $\backslash text\{fin.\; \}\; B\; :\; \backslash text\{fin.\; \}\; E\; =\; \backslash text\{fin.\; \}\; LD\; :\; \backslash text\{fin.\; \}\; c$, and therefore the difference of the arcs LD and $c$ will be the measure of the difference of the angles B and E. But when BE is indefinitely small, CD may be taken for the difference of LD and $c$, they being ultimately in the ratio of equality. Therefore CD measures the change of the obliquity of the ecliptic, or the nutation of the axis with respect to the ecliptic.

The real deviation of the axis is the same with the change in the position of the equator, Pp being the measure of the angle EAB. But this not being always made in a plane perpendicular to the ecliptic, the change of obliquity generally differs from the change in the position of the axis. Thus when the sun is in the solstice, the momentary change of the position of the equator is the greatest possible; but being made at right angles to the plane in which the obliquity of the ecliptic is computed, it makes no change whatever in the obliquity, but the greatest possible change in the precession.

In order to find CD the change of obliquity, observe that in the triangle CAD, $R\; :\; \backslash text\{fin.\; \}\; AC$, or $R\; :\; \backslash text\{cos.\; \}\; AE\; =\; \backslash text\{fin.\; \}\; A\; :\; \backslash text\{fin.\; \}\; CD$, $=\; A\; :\; CD$ (because A and CD are exceedingly small). Therefore the change of obliquity (which is the thing commonly meant by nutation) $CD\; =\; A\; \backslash times\; \backslash text\{cos.\; \}\; AE\; =\; i\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; mn$, $\backslash text\{cos.\; \}\; AE\; =\; i\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; \backslash times\; \backslash text\{fin.\; declin.\}\; \backslash times\; \backslash text\{cos.\; declin.\}\; \backslash times\; \backslash text\{cos.\; \}\; R.\; \backslash text\{\; ascen.\}$

But it is more convenient for the purposes of astronomical computation to make use of the sun's longitude SE. Therefore make

In the spherical triangle EAS, right-angled at A (because AS is the sun's declination perpendicular to the equator), we have $R\; :\; \backslash text\{fin.\; \}\; ES\; =\; \backslash text\{fin.\; \}\; E\; :\; \backslash text\{fin.\; \}\; AS$, and $\backslash text\{fin.\; \}\; AS\; =\; p\; x$. Also $R\; :\; \backslash text\{cos.\; \}\; AS\; =\; \backslash text\{cos.\; \}\; AE\; :\; \backslash text\{cos.\}$

ES, and $\backslash text\{cos.\; \}\; ES$ or $y\; =\; \backslash text\{cos.\; \}\; AS\; \backslash times\; \backslash text\{cos.\; \}\; AE$. Therefore $p\; x\; y\; =\; \backslash text\{fin.\; \}\; AS\; \backslash times\; \backslash text\{cos.\; \}\; AS\; \backslash times\; \backslash text\{cos.\; \}\; AE\; =\; mn\; \backslash times\; \backslash text\{cos.\; \}\; AE$.

Therefore the momentary nutation $CD\; =\; i\; \backslash times\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; p\; x\; y$.

We must recollect that this angle is a certain fraction of the momentary diurnal rotation. It is more convenient to consider it as a fraction of the sun's annual motion, that so we may directly compare his motion on the ecliptic with the precession and nutation corresponding to his situation in the heavens. This change is easily made, by augmenting the fraction in the ratio of the sun's angular motion to the motion of rotation, or multiplying the fraction by $\backslash frac\{T\}\{t\}$; therefore

the momentary nutation will be $i\; \backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; k\; p\; x\; y$. In this va-

lue $\backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}$ is a constant quantity, and the momentary nutation is proportional to $x\; y$, or to the product of the sine and cosine of the sun's longitude, or to the sine of twice the sun's longitude; for $x\; y$ is equal to half the sine of twice $x$.

If therefore we multiply this fraction by the sun's momentary angular motion, which we may suppose, with abundant accuracy, proportional to $x$, we obtain the fluxion of the nutation, the fluent of which will express the whole nutation while the sun describes the arch $x$ of the ecliptic, beginning at the vernal equinox. Therefore in place of $y$ put $\backslash sqrt\{1-x^2\}$, and in place

of $x$ put $\backslash frac\{x\}\{\backslash sqrt\{1-x^2\}\}$, and we have the fluxion of the nutation for the moment when the sun's longitude is $x$, and the fluent will be the whole nutation. The fluxion resulting from this process is $\backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; x\; x$, of which the

fluent is $\backslash frac\{3\}\{4\}\; \backslash frac\{t^2\}\{T^2\}\; x^2$. This is the whole change produced on the obliquity of the ecliptic while the sun moves along the arch $x$ ecliptic, reckoned from the vernal equinox. When this arch is $90^\backslash circ$, $x^2$ is 1, and therefore $\backslash frac\{3\}\{4\}\; \backslash frac\{t^2\}\{T^2\}$ is the nutation produced while the sun moves from the equinox to the solstice.

The momentary change of the axis and plane of the equator (which is the measure of the changing force) is $\backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; m\; n$.

The momentary change of the obliquity of the ecliptic is $\backslash frac\{3\}\{2\}\; \backslash frac\{t^2\}\{T^2\}\; p\; x\; x$.

The whole change of obliquity is $\backslash frac\{3\}\{4\}\; \backslash frac\{t^2\}\{T^2\}\; x^2$.

Hence we see that the force and the real momentary change of position are greatest at the solstices, and diminish to nothing in the equinoxes.

The momentary change of obliquity is greatest at the solstices, being proportional to $x\; x$ or to $x\; y$.

The whole accumulated change of obliquity is greatest at the solstices, the obliquity itself being then smallest.

We must in like manner find the accumulated quantity.

Precession. tivity of the precession after a given time, that is, the arch BE for a finite time.

42 Quantity of precession in a given time.
We have $ER\; :\; CD\; =\; \backslash sin.\; EA\; :\; \backslash sin.\; CA$ (or $\backslash cos.\; EA$) $=\; \backslash tan.\; EA\; :\; 1$, and $EB\; :\; ER\; =\; 1\; :\; \backslash sin.\; B$. Therefore $EB\; :\; CD\; =\; \backslash tan.\; EA\; :\; \backslash sin.\; B$. But $\backslash tan.\; EA\; =\; \backslash cos.\; E\; \backslash times\; \backslash tan.\; ES$, $=\; \backslash cos.\; E\; \backslash times\; \backslash frac\{\backslash sin.\; \backslash text\{long.\}\}\{\backslash cos.\; \backslash text\{long.\}\}\; =\; \backslash frac\{q\; \backslash times\}\{\backslash sqrt\{1-x^2\}\}$.

Therefore $EB\; :\; CD\; =\; \backslash frac\{q\; \backslash times\}\{\backslash sqrt\{1-x^2\}\}\; p$, and $CD\; =\; EB\; :\; \backslash frac\{\backslash sin.\; \backslash text\{obliq.\; eclipt.\}\}\{\backslash tan.\; \backslash text\{long.\; \}\; \backslash odot\}$. If we now substitute for CD its value found in N^o 40. viz. $\backslash frac\{3tkp\}\{2T\}\; x$, we obtain $EB\; =$

$\backslash frac\{3t\}\{2T\}\; \backslash times\; \backslash frac\{kq\; \backslash times^3\}\{\backslash sqrt\{1-x^2\}\}$, the fluxion of the precession of the equinoxes occasioned by the action of the sun. The

fluent of the variable part $\backslash frac\{x^3\; \backslash dot\{x\}\}\{\backslash sqrt\{1-x^2\}\}\; =\; x\; \backslash dot\{y\}$, of which the fluent is evidently a segment of a circle whose arch is $x$ and sine $x$, that is, $=\; \backslash frac\{x-x\backslash sqrt\{1-x^2\}\}\{2\}$, and the whole precession, while the sun describes the arch $x$, is $\backslash frac\{3t\}\{2T\}\; \backslash times\; \backslash frac\{kq\}\{2\}\; (2-x\backslash sqrt\{1-x^2\})$. This is the precession of the equinoxes while the sun moves from the vernal equinox along the arch $x$ of the ecliptic.

In this expression, which consists of two parts, $\backslash frac\{3tkq\}\{4T\}\; x$, and $\backslash frac\{3tkq\}\{4T\}\; (-x\backslash sqrt\{1-x^2\})$, the first is incomparably greater than the second, which never exceeds 1", and is always compensated in the succeeding quadrant. The precession occasioned by the sun will be $\backslash frac\{3tkq\}\{4T\}\; x$, and from this expression we see that the precession increases uniformly, or at least increases at the same rate with the sun's longitude $x$, because the quantity $\backslash frac\{3tkq\}\{4T\}$ is constant.

43 Mode of using the formulae.
In order to make use of these formulae, which are now reduced to very great simplicity, it is necessary to determine the values of the two constant quantities $\backslash frac\{3tkp\}\{4T\}$, $\backslash frac\{3tkq\}\{4T\}$, which we shall call N and P, as factors of the mutation and precession. Now $t$ is one sidereal day, and $T$ is 366$\backslash frac\{1\}\{2\}$. $k$ is $\backslash frac\{a^2-b^2\}\{a^2\}$, which according to Sir Isaac Newton is $\backslash frac\{231^2-230^2\}\{231^2\}\; =\; \backslash frac\{1\}\{115\}$; $p$ and $q$ are the sine and cosine of 23$^\{\backslash circ\}$ 28', viz. 0.39822 and 0.91729.

These data give $N\; =\; \backslash frac\{1\}\{141030\}$ and $P\; =\; \backslash frac\{1\}\{61224\}$ of which the logarithms are 4.85069 and 5.21308, viz. the arithmetical complements of 5.14931 and 4.78692.

44 example of the utility of the investigation.
Let us, for an example of the use of this investigation, compute the precession of the equinoxes when the sun has moved from the vernal equinox to the summer solstice, so that $x$ is 90$^\{\backslash circ\}$, or 324000".

VOL. XVII. Part I.

Log 324000" = $x$ - - - - - 5.51055
Log P - - - - - 5.21308
Log 5".292 - - - - - 0.72363

The precession therefore in a quarter of a year is 5.292 seconds; and, since it increases uniformly, it is 21".168 annually.

We must now recollect the assumptions on which this computation proceeds. The earth is supposed to be homogeneous, and the ratio of its equatorial diameter to its polar axis is supposed to be that of 231 to 230. If the earth be more or less protuberant at the equator, the precession will be greater or less in the ratio of this protuberance. The measures which have been taken of the degrees of the meridian are very inconsistent among themselves; and although a comparison of them all indicates a smaller protuberance, nearly $\backslash frac\{1\}\{100\}$ instead of $\backslash frac\{1\}\{1000\}$, their differences are too great to leave much confidence in this method. But if this figure be thought more probable, the precession will be reduced to about 17" annually. But even though the figure of the earth were accurately determined, we have no authority to say that it is homogeneous. If it be denser towards the centre, the momentum of the protuberant matter will not be so great as if it were equally dense with the inferior parts, and the precession will be diminished on this account. Did we know the proportion of the matter in the moon to that in the sun, we could easily determine the proportion of the whole observed annual precession of 50$\backslash frac\{1\}\{2\}$" which is produced by the sun's action. But we have no unexceptionable data for determining this; and we are rather obliged to infer it from the effect which the produces in disturbing the regularity of the precession, as will be considered immediately. So far, therefore, as we have yet proceeded in this investigation, the result is very uncertain. We have only ascertained unquestionably the law which is observed in the solar precession. It is probable, however, that this precession is not very different from 20" annually; for the phenomena of the tides show the disturbing force of the sun to be very nearly $\backslash frac\{2\}\{3\}$ of the disturbing force of the moon. Now 20" is $\backslash frac\{1\}\{3\}$ of 50".

But let us now proceed to consider the effect of the moon's action on the protuberant matter of the earth; and as we are ignorant of her quantity of matter, and consequently of her influence in similar circumstances with the sun, we shall suppose that the disturbing force of the moon is to that of the sun as $m$ to 1. Then earth (ceteris paribus) the precession will be to the solar precession $\backslash pi$ in the ratio of the force and of the time of its action jointly. Let $t$ and $T$ therefore represent a periodical month and year, and the lunar precession will be $=\; \backslash frac\{m\backslash pi\; t\}\{T\}$. This precession must be reckoned on the plane of the lunar orbit, in the same manner as the solar precession is reckoned on the ecliptic. We must also observe, that $\backslash frac\{m\backslash pi\; t\}\{T\}$ represents the lunar precession only on the supposition that the earth's equator is inclined to the lunar orbit in an angle of 23$\backslash frac\{1\}\{2\}$ degrees. This is indeed the mean inclination; but it is sometimes increased to above 28$^\{\backslash circ\}$, and sometimes reduced to 18$^\{\backslash circ\}$. Now in the value of the solar precession the cosine of the obliquity was employed. Therefore whatever is the

M m

Precession. the angle E contained between the equator and the lunar orbit, the precession will be $=\; \backslash frac\{m\; \backslash pi\; t\}\{T\}\; \backslash cdot\; \backslash frac\{\backslash text\{Cos.\; \}\; E\}\{\backslash text\{Cos.\; \}\; 23\backslash frac\{1\}\{2\}^\backslash circ\}$ and it must be reckoned on the lunar orbit.

47 Fig. 8. Now let $\backslash psi\; B$ (fig. 8.) be the immoveable plane of the ecliptic, $\backslash psi\; ED\; \backslash perp\; \backslash psi\; F$ the equator in its first situation, before it has been deranged by the action of the moon, $AGRDBH$ the equator in its new position, after the momentary action of the moon. Let $EGNFH$ be the moon's orbit, of which $N$ is the ascending node, and the angle $N\; =\; 5^\backslash circ\; 8\text{'}\; 46\text{'}\text{'}$.

43 Lunar precession in a month reduced to the ecliptic. In order to reduce the lunar precession to the ecliptic, we must recollect that the equator will have the same inclination at the end of every half revolution of the sun or of the moon, that is, when they pass through the equator, because the sum of all the momentary changes of its position begins again each revolution. Therefore if we neglect the motion of the node during one month, which is only $1\backslash frac\{1\}\{2\}$ degrees, and can produce but an insensible change, it is plain that the moon produces, in one half revolution, that is, while she moves from $H$ to $G$, the greatest difference that she can in the position of the equator. The point $D$, therefore, half-way from $G$ to $H$, is that in which the moveable equator cuts the primitive equator, and $DE$ and $DF$ are each $90^\backslash circ$. But $S$ being the solstitial point, $\backslash psi\; S$ is also $90^\backslash circ$. Therefore $DS\; =\; \backslash psi\; E$. Therefore, in the triangle $DGE$, we have $\backslash sin.\; ED\; :\; \backslash sin.\; G\; =\; \backslash sin.\; EG\; :\; \backslash sin.\; D\; =\; EG\; :\; D$. Therefore $D\; =\; EG\; \backslash times\; \backslash sin.\; G\; =\; EG\; \backslash times\; \backslash sin.\; E$ nearly. Again, in the triangle $\backslash psi\; DA$ we have $\backslash sin.\; A\; :\; \backslash sin.\; \backslash psi\; D$ (or $\backslash cos.\; \backslash psi\; E$) $=\; \backslash sin.\; D\; :\; \backslash sin.\; \backslash psi\; A\; =\; D\; :\; \backslash psi\; A$. Therefore

42 This is the lunar precession produced in the course of one month, estimated on the ecliptic, not constant like the solar precession, but varying with the inclination or the angle $E$ or $F$, which varies both by a change in the angle $N$, and also by a change in the position of $N$ on the ecliptic.

50 Notation in the same time. We must find in like manner the nutation $SR$ produced in the same time, reckoned on the colure of the solstices $RL$. We have $R\; :\; \backslash sin.\; DS\; =\; D\; :\; RS$, and $RS\; =\; D\; \backslash cdot\; \backslash sin.\; DS\; =\; D\; \backslash cdot\; \backslash sin.\; \backslash psi\; E$. But $D\; =\; EG\; \backslash cdot\; \backslash sin.\; E$. Therefore $RS\; =\; EG\; \backslash cdot\; \backslash sin.\; E\; \backslash cdot\; \backslash sin.\; \backslash psi\; E\; =\; \backslash frac\{m\; \backslash pi\; t\; \backslash cos.\; E\}\{T\; \backslash cdot\; \backslash cos.\; \backslash psi\}\; \backslash times\; \backslash sin.\; E\; \backslash times\; \backslash sin.\; \backslash psi\; E$. In this expression we must substitute

the angle $N$, which may be considered as constant during the month, and the longitude $\backslash psi\; N$, which is also nearly constant, by observing that $\backslash sin.\; E\; :\; \backslash sin.\; \backslash psi\; N\; =\; \backslash sin.\; N\; :\; \backslash sin.\; \backslash psi\; E$. Therefore $RS\; =\; \backslash frac\{m\; \backslash pi\; t\}\{T\}\; \backslash times\; \backslash frac\{\backslash sin.\; N\; \backslash cdot\; \backslash sin.\; \backslash psi\; N\; \backslash cdot\; \backslash cos.\; E\}\{\backslash cos.\; \backslash psi\}$. But we must exterminate the angle $E$, because it changes by the change of the position of $N$. Now, in the triangle $\backslash psi\; EN$ we have $\backslash cos.\; E\; =\; \backslash cos.\; \backslash psi\; N\; \backslash cdot\; \backslash sin.\; N\; \backslash cdot\; \backslash sin.\; \backslash psi\; -\; \backslash cos.\; N\; \backslash cdot\; \backslash cos.\; \backslash psi\; =\; yca\; -\; db$. And because the angle $E$ is necessarily obtuse, the perpendicular will fall without the triangle, the cosine of $E$ will be negative, and we shall have $\backslash cos.\; E\; =\; b\; d\; -\; a\; c\; y$. Therefore the nutation for one month will be $=\; \backslash frac\{m\; \backslash pi\; t\}\{T\}\; \backslash times\; \backslash frac\{c\; x\; (b\; d\; -\; a\; c\; y)\}\{b\}$, the node being supposed all the while in $N$.

These two expressions of the monthly precession and may be considered as momentary parts of the moon's action, corresponding to a certain position of the node and inclination of the equator, or as the fluxions of the whole variable precession and nutation, while the node continually changes its place, and in the space of 18 years makes a complete tour of the heavens.

We must, therefore, take the motion of the node as the fluent of comparison, or we must compare the fluxions of the node's motion with the fluxions of the precession and nutation; therefore, let the longitude of the node be $x$, and its monthly change $=\; \backslash dot\{x\}$; we shall then have

$\backslash left(\; \backslash frac\{d\; b\; x\; \backslash dot\{x\}\}\{\backslash sqrt\{1-x^2\}\}\; -\; a\; c\; x\; \backslash dot\{x\}\; \backslash right)$. The fluent of this is $m\; \backslash pi\; n\; \backslash frac\{c\}\{e\; b\}\; \backslash left(\; -d\; b\; \backslash sqrt\{1-x^2\}\; -\; \backslash frac\{a\; c\; x^2\}\{2\}\; \backslash right)$. (Vide Simpson's Fluxions, § 77.) But when $x\; =\; 0$, the nutation must be $=\; 0$, because it is from the position in the equinoctial points that all our deviations are reckoned, and it is from this point that the period of the lunar action recommences. But if we make $x\; =\; 0$ in this expression, the term $-\backslash frac\{a\; c\; x^2\}\{2\}$ vanishes, and the term $-d\; b\; \backslash sqrt\{1-x^2\}$ becomes $=\; -d\; b$; therefore our fluent has a constant part $+d\; b$; and the complete fluent is $m\; \backslash pi\; n\; \backslash frac\{c\}\{e\; b\}\; \backslash left(\; d\; b\; -\; d\; b\; \backslash sqrt\{1-x^2\}\; -\; \backslash frac\{a\; c\; x^2\}\{2\}\; \backslash right)$. Now this is equal to $m\; \backslash pi\; n\; \backslash frac\{c\}\{e\; b\}\; (d\; b\; \backslash times\; \backslash text\{versed\; sine\; \}\; x\; -\; \backslash frac\{1\}\{2\}\; a\; c\; \backslash times\; \backslash text\{versed\; sine\; \}\; 2x)$: For the versed sine of $x$ is equal to $(1\; -\; \backslash cos.\; x)$; and the square of the sine of an arch is $\backslash frac\{1\}\{2\}$ the versed sine of twice that arch.

This, then, is the whole nutation while the moon's ascending node moves from the vernal equinox to the longitude $\backslash psi\; N\; =\; x$. It is the expression of a certain number of seconds, because $\backslash pi$, one of its factors, is the solar precession in seconds; and all the other factors are numbers, or fractions of the radius 1; even $e$ is expressed in terms of the radius 1.

The fluxion of the precession, or the monthly precession, 53 54

Precession. son, is to that of the nutation as the cotangent of $\backslash psi\; E$ is to the sine of $\backslash psi$. This also appears by considering fig. 7. $P\backslash phi$ measures the angle $A$, or change of position of the equator; but the precession itself, reckoned on the ecliptic, is measured by $P\backslash phi$, and the nutation by $p\backslash phi$; and the fluxion of the precession is equal to the fluxion of

55 Let us now express this in numbers: When the node has made a half revolution, we have $x\; =\; 180^\backslash circ$, whose versed sine is 2, and the versed sine of $2x$, or $360^\backslash circ$, is $\backslash equiv\; 0$; therefore, after half a revolution of the node, the nutation ($n^\backslash circ\; 52$) becomes $\backslash frac\{m\; \backslash pi\; n\; c\}\{e\; b\}\; 2\; b\; d$. If, in this expression, we suppose $m\; =\; 2\backslash frac\{1\}\{2\}$, and $n\; =\; 14\backslash frac\{1\}\{2\}\text{'}\text{'}$, we shall find the nutation to be $19\backslash frac\{1\}\{2\}\text{'}\text{'}$.

56 Now the observed nutation is about $18\text{'}\text{'}$. This requires $m$ to be $2\backslash frac\{1\}\{2\}$, and $n\; =\; 16\backslash frac\{1\}\{2\}\text{'}\text{'}$. But it is evident that no astronomer can pretend to warrant the accuracy of his observations of the nutation within $1\text{'}\text{'}$.

To find the lunar precession during half a revolution of the node, observe, that then $x$ becomes $\backslash frac\{e\}\{2\}$, and the sine of $x$ and of $2x$ vanish, $d^2$ becomes $1\; -\; e^2$, and the precession becomes $\backslash frac\{m\; \backslash pi\; n\}\{2\}\; (d^2\; -\; \backslash frac\{1\}\{2\}\; c^2)\; =\; \backslash frac\{m\; \backslash pi\; n\}\{2\}\; (1\; -\; \backslash frac\{1\}\{2\}\; c^2)$, and the precession in 18 years is $m\; \backslash pi\; n\; 1\; -\; \backslash frac\{1\}\{2\}\; c^2$.

57 We see, by comparing the nutation and precession for nine years, that they are as $\backslash frac\{4\; c\; d\}\{e\}$ to $1\; -\; \backslash frac\{1\}\{2\}\; c^2$ nearly as 1 to $17\backslash frac\{1\}\{2\}$. This gives $313\text{'}\text{'}$ of precession, corresponding to $18\text{'}\text{'}$, the observed nutation, which is about $35\text{'}\text{'}$ of precession annually produced by the moon.

And thus we see, that the inequality produced by the moon in the precession of the equinoxes, and, more particularly, the nutation occasioned by the variable obliquity of her orbit, enables us to judge of her share in the whole phenomenon; and therefore informs us of her disturbing force, and therefore of her quantity of matter. This phenomenon, and those of the tides, are the only facts which enable us to judge of this matter: and this is one of the circumstances which has caused this problem to occupy so much attention. Dr Bradley, by a nice comparison of his observations with the mathematical theory, as it is called, furnished him by Mr Machin, found that the equation of precession computed by that theory was too great, and that the theory

would agree better with the observations, if an ellipse Precession, Precise. were substituted for Mr Machin's little circle. He thought that the shorter axis of this ellipse, lying in the colure of the solstices, should not exceed $16\text{'}\text{'}$. Nothing can more clearly show the astonishing accuracy of Bradley's observations than this remark: for it results from the theory, that the pole must really describe an ellipse, having its shorter axis in the solstitial colure, and the ratio of the axes must be that of 18 to 16.8; for the mean precession during a half revolution of the node is $\backslash frac\{m\; \backslash pi\; n\}\{2\}\; (d^2\; -\; \backslash frac\{1\}\{2\}\; c^2)$; and therefore, for the longitude $x$, it will be $\backslash frac\{m\; \backslash pi\; n\}\{e\}\; (d^2\; -\; \backslash frac\{1\}\{2\}\; c^2)$; when this is taken from the true precession for that longitude ($n^\backslash circ\; 54$), it leaves the equation of precession $\backslash frac\{m\; \backslash pi\; n\}\{a\; b\; e\}\; \backslash left(\; (b^2\; -\; a^2)\; d\; c\; \backslash right.$ sine $x\; -\; \backslash frac\{1\}\{2\}\; a\; b\; c$ sine $2x$); therefore, when the node is in the solstice, and the equation greatest, we have it $\backslash equiv\; \backslash frac\{m\; \backslash pi\; n\; c\; d\}\{a\; b\; e\}\; (b^2\; -\; a^2)$. We here neglect the second term as insignificant.

This greatest equation of precession is to $\backslash frac\{2\; m\; \backslash pi\; n\; c\; d\}\{c\}$ Greatest equation of precession. the nutation of $18\text{'}\text{'}$, as $b^2\; -\; a^2$ to $2\; a\; b$; that is as radius to the tangent of twice the obliquity of the ecliptic. This gives the greatest equation of precession $16\text{'}\text{'}.8$, not differing half a second from Bradley's observations.

Thus have we attempted to give some account of this curious and important phenomenon. It is curious, because it affects the whole celestial motions in a very intricate manner, and received no explanation from the more obvious application of mechanical principles, which so happily accounted for all the other appearances. It is one of the most illustrious proofs of Sir Isaac Newton's sagacity and penetration, which caught at a very remote analogy between this phenomenon and the libration of the moon's orbit. It is highly important to the progress of practical and useful astronomy, because it has enabled us to compute tables of such accuracy, that they can be used with confidence for determining the longitude of a ship at sea. This alone fixes its importance: but it is still more important to the philosopher, affording the most incontestable proof of the universal and mutual gravitation of all matter to all matter. It left nothing in the solar system unexplained from the theory of gravity but the acceleration of the moon's mean motion; and this has at last been added to the list of our acquisitions by Mr de la Place.

Quæ toties animos veterum torfere Sophorum,
Queque scholas frustra rauco certamine vexant,
Obvia conspicimus, nube pellente Mathesi,
Jam dubios nulla caligine prægravat error
Quæ superum penetrare domos, atque ardua cœli
Scandere sublimis genii concessit acumen.
Nec fas est proprius mortali attingere divos.

Halley.