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28  :Yet when her name was heard no more,
 * And when the robe her mother gave,
 * And small, light moccasin she wore,
 * Had slowly wasted on her grave,
 * Unmarked of him the dark maids sped
 * Their sunset dance and moonlit play;
 * No other shared his lonely bed,

No other fair young head upon his bosom lay.

 * A lone, stern man. Yet, as sometimes
 * The tempest-smitten tree receives
 * From one small root the sap which climbs
 * Its topmost spray and crowning leaves,
 * So from his child the Sachem drew
 * A life of Love and Hope, and felt
 * His cold and rugged nature through

The softness and the warmth of her young being melt.

 * A laugh which in the woodland rang
 * Bemocking April’s gladdest bird,—
 * A light and graceful form which sprang
 * To meet him when his step was heard,—
 * Eyes by his lodge-fire flashing dark,
 * Small fingers stringing bead and shell
 * Or weaving mats of bright-hued bark,—

With these the household-god had graced his wigwam well.

 * Child of the forest! strong and free,
 * Slight-robed, with loosely flowing hair,
 * She swam the lake or climbed the tree,
 * Or struck the flying bird in air.
 * O’er the heaped drifts of winter’s moon
 * Her snow-shoes tracked the hunter’s way;
 * And dazzling in the summer noon

The blade of her light oar threw off its shower of spray!

 * Unknown to her the rigid rule,
 * The dull restraint, the chiding frown,
 * The weary torture of the school,
 * The taming of wild nature down.
 * Her only lore, the legends told
 * Around the hunter’s fire at night;
 * Stars rose and set, and seasons rolled,

Flowers bloomed and snow-flakes fell, unquestioned in her sight.

 * Unknown to her the subtle skill
 * With which the artist-eye can trace
 * In rock and tree and lake and hill
 * The outlines of divinest grace;
 * Unknown the fine soul’s keen unrest,
 * Which sees, admires, yet yearns alway;
 * Too closely on her mother’s breast

To note her smiles of love the child of Nature lay!

 * It is enough for such to be
 * Of common, natural things a part,
 * To feel, with bird and stream and tree,
 * The pulses of the same great heart;
 * But we, from Nature long exiled,
 * In our cold homes of Art and Thought
 * Grieve like the stranger-tended child,

Which seeks its mother’s arms, and sees but feels them not.

 * The garden rose may richly bloom
 * In cultured soil and genial air,
 * To cloud the light of Fashion’s room
 * Or droop in Beauty’s midnight hair;
 * In lonelier grace, to sun and dew
 * The sweetbrier on the hillside shows
 * Its single leaf and fainter hue,

Untrained and wildly free, yet still a sister rose!

 * Thus o’er the heart of Weetamoo
 * Their mingling shades of joy and ill
 * The instincts of her nature threw;
 * The savage was a woman still.
 * Midst outlines dim of maiden schemes,
 * Heart-colored prophecies of life,
 * Rose on the ground of her young dreams

The light of a new home, the lover and the wife.

Cool and dark fell the autumn night, But the Bashaba’s wigwam glowed with light, For down from its roof, by green withes hung, Flaring and smoking the pine-knots swung.

And along the river great wood-fires Shot into the night their long, red spires, Showing behind the tall, dark wood, Flashing before on the sweeping flood.

In the changeful wind, with shimmer and shade, Now high, now low, that firelight played,


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Page:The story of the flute (IA storyofflute1914fitz).djvu/146

 (which was a favourite with Mendelssohn), is for four strings and flute. The "Polonaise" in this work is for flute and bass only, the former being given a bravura variation; the flute is also very prominent in the bright "Badinerie." Bach has also written a concerto in F, ib. Badinerie. major for flute, oboe, violin, and trumpet, two for flute, violin and clavier, and one for two flutes and clavier all with string accompaniment.

Handel employs the flute much less frequently. Quite a large number of his operas and oratorios have no flute part; The Messiah, for instance (to which, however, Mozart has added a flute and also a piccolo part). In many works he introduces the flute in one or two numbers only; in none is it used anything approaching continuously throughout, as it often is by Bach. Handel's flute parts do not contain a single really difficult passage. Strange to say, his few "flauto piccolo" passages are harder and much more elaborate than those for the flute (he hardly ever uses both instruments together); the piccolo solo in Rinaldo being about the most difficult he ever wrote for any kind of flute. Handel rarely travels beyond the two octaves D′ to D′′′ of the flute (it is to be noticed that the compass of the older flutes-à-bec was two octaves as a rule); occasionally we

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6.

Let us proceed now to the comparison of the motion with the time. Putting, as in Art. 1, the space described about the sun in the time $t=\frac{1}{2} g,$ the mass of the moving body $=\mu,$  that of the sun being taken $=1,$  we have $g=k t \sqrt{p} \sqrt{1+\mu}.$  The differential of the space $=\frac{1}{2} r r d v,$  from which there results $k t \sqrt{p} \sqrt{1+\mu}=\int r r \operatorname{d} v,$  this integral being so taken that it will vanish for $t=0.$  This integration must be treated differently for different kinds of conic sections, on which account, we shall now consider each kind separately, beginning with the ELLIPSE.

Since $r$ is determined from $v$  by means of a fraction, the denominator of which consists of two terms, we will remove this inconvenience by the introduction of a new quantity in the place of $v.$  For this purpose we will put $\tan \frac{1}{2} v \sqrt{\frac{1-e}{1+e}}=\tan \frac{1}{2} E,$  by which the last formula for $r$  in the preceding article gives

$$r=\frac{p \cos ^{2} \frac{1}{2} E}{(1+e) \cos ^{2} \frac{1}{2} v}=p\left(\frac{\cos ^{2} \frac{1}{2} E}{1+e}+\frac{\sin ^{2} \frac{1}{2} E}{1-e}\right)=\frac{p}{1-e e}(1-e \cos E)$$

Moreover we have $\frac{\operatorname{d} E}{\cos ^{2} \frac{1}{2} E}=\frac{\operatorname{d} v}{\cos ^{2} \frac{1}{2} v} \sqrt{\frac{1-e}{1+e}},$ and consequently $\operatorname{d} v=\frac{p \operatorname{d} E}{r \sqrt{1-e}};$  hence

$$r r \operatorname{d} v=\frac{r p \operatorname{d} E}{\sqrt{1-e e}}=\frac{p p}{(1-e e)^{\frac{3}{2}}}(1-e \cos E) \operatorname{d} E$$

and integrating,

$$k t \sqrt{p} \sqrt{1+\mu}=\frac{p p}{(1-e e)^{\frac{3}{2}}}(E-e \sin E)+\text { Constant. }$$

Accordingly, if we place the beginning of the time at the perihelion passage, where $v=0,$ $E=0,$  and thus constant $=0,$  we shall have, by reason of $\frac{p}{1-e e}=a,$

$$E-e \sin E=\frac{k t \sqrt{1+\mu}}{a^{\frac{3}{2}}} .$$

In this equation the auxiliary angle $E,$ which is called the eccentric anomaly, must be expressed in parts of the radius. This angle, however, may be retained in degrees, etc., if $e \sin E$ and $\frac{k t \sqrt{1+\mu}}{a^{\frac{3}{2}}}$  are also expressed in the same manner; these quantities will be expressed in seconds of are if they are multiplied by the


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 number $206264.81$. We can dispense with the multiplication by the last quantity, if we employ directly the quantity $k$ expressed in seconds, and thus put, instead of the value before given, $k=3548^{\prime \prime}{.}18761,$  of which the logarithm $=3.5500065746$. The quantity $\frac{k t \sqrt{1+\mu}}{a^{\frac{3}{2}}}$ expressed in this manner is called the mean anomaly, which therefore increases in the ratio of the time, and indeed every day by the increment $\frac{k \sqrt{1+\mu}}{a^{\frac{3}{2}}},$  called the mean daily motion. We shall denote the mean anomaly by $M.$

7.

Thus, then, at the perihelion, the true anomaly, the eccentric anomaly, and the mean anomaly are $=0;$ after that, the true anomaly increasing, the eccentric and mean are augmented also, but in such a way that the eccentric continues to be less than the true, and the mean less than the eccentric up to the aphelion, where all three become at the same time $=180^{\circ};$  but from this point to the perihelion, the eccentric is always greater than the true, and the mean greater than the eccentric, until in the perihelion all three become $=360^{\circ},$  or, which amounts to the same thing, all are again $=0.$  And, in general, it is evident that if the eccentric $E$  and the mean $M$  answer to the true anomaly $v,$  then the eccentric $360^{\circ}-E$  and the mean $360^{\circ}-M$  correspond to the true $360^{\circ}-v.$  The difference between the true and mean anomalies, $v-M,$  is called the equation of the centre, which, consequently, is positive from the perihelion to the aphelion, is negative from the aphelion to the perihelion, and at the perihelion and aphelion vanishes. Since, therefore, $v$ and $M$  run through an entire circle from $0$  to $360^{\circ}$  in the same time, the time of a single revolution, which is also called the periodic time, is obtained, expressed in days, by dividing $360^{\circ}$  by the mean daily motion $\frac{k \sqrt{1+\mu}}{a^{\frac{3}{2}}},$  from which it is apparent, that for different bodies revolving about the sun, the squares of the periodic times are proportional to the cubes of the mean distances, so far as the masses of the bodies, or rather the inequality of their masses, can be neglected.


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 8.

Let us now collect together those relations between the anomalies and the radius vector which deserve particular attention, the derivation of which will present no difficulties to any one moderately skilled in trigonometrical analysis. Greater elegance is attained in most of these formulas by introducing in the place of $e$ the angle the sine of which $=e.$  This angle being denoted by $\varphi,$  we have

$$\begin{aligned} & \sqrt{1-ee}=\cos \varphi, \quad \sqrt{1+e}=\cos(45^{\circ}-\tfrac{1}{2} \varphi) \sqrt{2}, \\ & \sqrt{1-e}=\cos (45^{\circ}+\tfrac{1}{2} \varphi) \sqrt{2}, \quad \sqrt{\frac{1-e}{1+e}}=\tan (45^{\circ}-\tfrac{1}{2} \varphi), \\ & \sqrt{1+e}+\sqrt{1-e}=2 \cos \tfrac{1}{2} \varphi, \quad \sqrt{1+e}-\sqrt{1-e}=2 \sin \tfrac{1}{2} \varphi. \end{aligned}$$

The following are the principal relations between $a,$ $p,$  $r,$  $e,$  $\varphi,$  $v,$  $E,$  $M.$

1. $p=a \cos ^{2} \varphi$

2. $r=\frac{p}{1+e \cos v}$

3. $r=a(1-e \cos E)$

4. $\cos E=\frac{\cos v+e}{1+e \cos v},$ or $\cos v=\frac{\cos E-e}{1-e \cos E}$

5. $\sin \frac{1}{2} E =\sqrt{\frac{1}{2}(1-\cos E)}=\sin \frac{1}{2} v \sqrt{\frac{1-e}{1+e \cos v}}$ $\phantom{\sin \frac{1}{2} E}=\sin \frac{1}{2} v \sqrt{\frac{r(1-e)}{p}}=\sin \frac{1}{2} v \sqrt{\frac{r}{a(1+e)}}$ | $\cos \frac{1}{2} E=\sqrt{\frac{1}{2}(1+\cos E)}=\cos \frac{1}{2} v \sqrt{\frac{1+e}{1+e \cos v}}$ $\phantom{\cos \frac{1}{2} E}=\cos \frac{1}{2} v \sqrt{\frac{r(1+e)}{p}}=\cos \frac{1}{2} v \sqrt{\frac{r}{a(1-e)}}$ | $\tan \frac{1}{2} E=\tan \frac{1}{2} v \tan \left(45^{\circ}-\frac{1}{2} \varphi\right)$ | $\sin E=\frac{r \sin v \cos \varphi}{p}=\frac{r \sin v}{a \cos \varphi}$ | $r \cos v=a(\cos E-e)=2 a \cos \left(\frac{1}{2} E+\frac{1}{2} \varphi+45^{\circ}\right) \cos \left(\frac{1}{2} E-\frac{1}{2} \varphi-45^{\circ}\right)$ | $\sin \frac{1}{2}(v-E)=\sin \frac{1}{2} \varphi \sin v \sqrt{\frac{r}{p}}=\sin \frac{1}{2} \varphi \sin E \sqrt{\frac{a}{r}}$ | $\sin \frac{1}{2}(v+E)=\cos \frac{1}{2} \varphi \sin v \sqrt{\frac{r}{p}}=\cos \frac{1}{2} \varphi \sin E \sqrt{\frac{a}{r}}$ | $M=E-e \sin E.$ |undefined


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 9.

If a perpendicular let fall from any point whatever of the ellipse upon the line of apsides is extended in the opposite direction until it meets the circle described with the radius $a$ about the centre of the ellipse, then the inclination to the line of apsides of that radius which corresponds to the point of intersection (understood in the same way as above, in the case of the true anomaly), will be equal to the eccentric anomaly, as is inferred without difficulty from equation IX. of the preceding article. Further, it is evident that $r \sin v$ is the distance of any point of the ellipse from the line of apsides, which, since by equation VIII. it $=a \cos \varphi \sin E,$ will be greatest for $E=90^{\circ},$  that is in the centre of the ellipse. This greatest distance, which $=a \cos \varphi=\frac{p}{\cos \varphi}=\sqrt{a p},$ is called the semi-axis minor. In the focus of the ellipse, that is for $v=90^{\circ},$ this distance is evidently $=p,$  or equal the semi-parameter.

10.

The equations of article 8 comprise all that is requisite for the computation of the eccentric and mean anomalies from the true, or of the eccentric and true from the mean. Formula VII. is commonly employed for deriving the eccentric from the true; nevertheless it is for the most part preferable to make use of equation X. for this purpose, especially when the eccentricity is not too great, in which case $E$ can be computed with greater accuracy by means of X. than of VII. Moreover, if X. is employed, the logarithm of sine $E$ required in XII. is had immediately by means of VIII.: if VII. were used, it would be necessary to take it out from the tables; if, therefore, this logarithm is also taken from the tables in the latter method, a proof is at once obtained that the calculation has been correctly made. Tests and proofs of this sort are always to be highly valued, and therefore it will be an object of constant attention with us to provide for them in all the methods delivered in this work, where indeed it can be conveniently done. We annex an example completely calculated as a more perfect illustration.


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 Given $v=310^{\circ} 55^{\prime} 29^{\prime \prime} .64$, $\varphi=14^{\circ} 12^{\prime} 1^{\prime \prime} .87$ , $\log r=0.3307640$ ; $p,$ $a,$  $E,$  $M,$  are required. [The letter $n$ affixed to a logarithm signifies that the number corresponding to it is negative.]

hence $\tfrac{1}{2}(v-E) = -4^{\circ}58^{\prime}22^{\prime\prime}.94;$ $v-E=-9^{\circ}56^{\prime}45^{\prime\prime};$  $E=320^{\circ}52^{\prime}15^{\prime\prime}.52.$

Further, we have

hence $e\sin E$ in seconds $=31932^{\prime\prime}.14=8^{\circ}52^{\prime}12^{\prime \prime} .14;$  and $M=329^{\circ} 44^{\prime} 27^{\prime \prime} .66.$

The computation of $E$ by formula VII. would be as follows:

$$ \begin{array}{l} \frac{1}{2} v=155^{\circ} 27^{\prime} 44^{\prime \prime} .82 \\ 45^{\circ}-\tfrac{1}{2} \varphi=37^{\circ} 53^{\prime} 59^{\prime \prime} .065 \\ \vphantom{\frac{1}{2}} \end{array} \qquad \begin{array}{ll} \log \tan \frac{1}{2} v \quad.\quad.\quad. & 9.6594579 n \\ \log \tan (45^{\circ}-\tfrac{1}{2} \varphi) \;\;. & 9.8912427 \\ \hline \log \tan \frac{1}{2} E \;\;\;.\quad.\quad. & 9.5507006 n \end{array}$$

whence $\frac{1}{2} E=160^{\circ} 26^{\prime} 7^{\prime \prime} .76,$ and $E=320^{\circ} 52^{\prime} 15^{\prime \prime} .52,$  as above.


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 11.

The inverse problem, celebrated under the title of Kepler’s problem, that of finding the true anomaly and the radius vector from the mean anomaly, is much more frequently used. Astronomers are in the habit of putting the equation of the centre in the form of an infinite series proceeding according to the sines of the angles $M,$ $2 M,$  $3 M,$  etc., each one of the coefficients of these sines being a series extending to infinity according to the powers of the eccentricity. We have considered it the less necessary to dwell upon this formula for the equation of the centre, which several authors have developed, because, in our opinion, it is by no means so well suited to practical use, especially should the eccentricity not be very small, as the indirect method, which, therefore, we will explain somewhat more at length in that form which appears to us most convenient.

Equation XII., $E=M+e \sin E,$ which is to be referred to the class of transcendental equations, and admits of no solution by means of direct and complete methods, must be solved by trial, beginning with any approximate value of $E,$  which is corrected by suitable methods repeated often enough to satisfy the preceding equation, that is, either with all the accuracy the tables of sines admit, or at least with sufficient accuracy for the end in view. If now, these corrections are introduced, not at random, but according to a safe and established rule, there is scarcely any essential distinction between such an indirect method and the solution by series, except that in the former the first value of the unknown quantity is in a measure arbitrary, which is rather to be considered an advantage since a value suitably chosen allows the corrections to be made with remarkable rapidity. Let us suppose $\varepsilon$ to be an approximate value of $E,$  and $x$  expressed in seconds the correction to be added to it, of such a value as will satisfy our equation $E=\varepsilon+x.$  Let $e \sin \varepsilon,$  in seconds, be computed by logarithms, and when this is done, let the change of the $\log \sin \varepsilon$  for the change of $1^{\prime \prime}$  in $\varepsilon$  itself be taken from the tables; and also the variation of $\log e \sin \varepsilon$  for the change of a unit in the number $e \sin \varepsilon;$  let these changes, without regard to signs, be respectively $\lambda,$  $\mu,$  in which it is hardly necessary to remark that both logarithms are presumed to contain an equal number of decimals. Now, if $\varepsilon$ approaches so near the correct value of $E$


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 that the changes of the logarithm of the sine from $\varepsilon$ to $\varepsilon+x,$  and the changes of the logarithm of the number from $e \sin \varepsilon$  to $e \sin (\varepsilon+x),$  can be regarded as uniform, we may evidently put

$$e \sin (\varepsilon+x)=e \sin \varepsilon \pm \frac{\lambda x}{\mu},$$

the upper sign belonging to the first and fourth quadrants, and the lower to the second and third. Whence, since

$\displaystyle \varepsilon+x=M+e \sin (\varepsilon+x)$, we have $\displaystyle x=\frac{\mu}{\mu+\lambda}(M+e \sin \varepsilon-\varepsilon),$

and the correct value of $E,$ or

$$\varepsilon+x=M+e \sin \varepsilon \pm \frac{\lambda}{\mu+\lambda}(M+e \sin \varepsilon-\varepsilon)$$

the signs being determined by the above-mentioned condition.

Finally, it is readily perceived that we have, without regard to the signs, $\mu: \lambda=1: e \cos \varepsilon,$ and therefore always $\mu>\lambda,$  whence we infer that in the first and last quadrant $M+e \sin \varepsilon$  lies between $\varepsilon$  and $\varepsilon+x,$  and in the second and third, $\varepsilon+x$  between $\varepsilon$  and $M+e \sin \varepsilon,$  which rule dispenses with paying attention to the signs. If the assumed value $\varepsilon$ differs too much from the truth to render the foregoing considerations admissible, at least a much more suitable value will be found by this method, with which the same operation can be repeated, once, or several times if it should appear necessary. It is very apparent, that if the difference of the first value $\varepsilon$ from the truth is regarded as a quantity of the first order, the error of the new value would be referred to the second order, and if the operation were further repeated, it would be reduced to the fourth order, the eighth order, etc. Moreover, the less the eccentricity, the more rapidly will the successive corrections converge.

12.

The approximate value of $E,$ with which to begin the calculation, will, in most cases, be obvious enough, particularly where the problem is to be solved for several values of $M$  of which some have been already found. In the absence of other helps, it is at least evident that $E$ must fall between $M$  and $M \pm e,$  (the eccentricity $e$  being expressed in seconds, and the upper sign being used in the


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 first and second quadrants, the lower in the third and fourth), wherefore, either $M,$ or its value increased or diminished by any estimate whatever, can be taken for the first value of $E.$  It is hardly necessary to observe, that the first calculation, when it is commenced with a value having no pretension to accuracy, does not require to be strictly exact, and that the smaller tables are abundantly sufficient. Moreover, for the sake of convenience, the values selected for $\varepsilon$  should be such that their sines can be taken from the tables without interpolation; as, for example, values to minutes or exact tens of seconds, according as the tables used proceed by differences of minutes or tens of seconds. Every one will be able to determine without assistance the modifications these precepts undergo if the angles are expressed according to the new decimal division.

13.

Example. Let the eccentricity be the same as in article 10. $M=332^{\circ} 28^{\prime} 54^{\prime \prime}.77.$ There the $\log e$  in seconds is $4.7041513,$  therefore $e=50600^{\prime \prime}=14^{\circ} 3^{\prime} 20^{\prime \prime}.$  Now since $E$  here must be less than $M,$  let us in the first calculation put $\varepsilon=326^{\circ},$  then we have by the smaller tables

$$ \begin{array}{ll} \log \sin \varepsilon .\quad.\quad.\quad.\quad. &9.74756 n, \\ \log e \text { in seconds } \quad.\quad.& 4.70415 \\ \hline & 4.45171 n; \end{array} \begin{array}{c} \scriptscriptstyle \text { Change for } 1^{\prime} \ldots 19 \text {, whence } \lambda=0.32.\\ \\ \\ \end{array} $$

$$\frac{0.32}{1.28} \times 4960^{\prime \prime}=1240^{\prime \prime}=20^{\prime} 40^{\prime \prime}.$$

Wherefore, the corrected value of $E$ becomes $324^{\circ} 37^{\prime} 20^{\prime \prime}-20^{\prime} 40^{\prime \prime}=324^{\circ} 16^{\prime} 40^{\prime \prime},$  with which we repeat the calculation, making use of larger tables.

$$ \begin{array}{ll} \log \sin \varepsilon \quad.\quad.\quad.\quad. &9.7663058n\\ \log e \quad.\quad.\quad.\quad.\quad. &4.7041513\\ \hline &4.4704571n \end{array} \begin{array}{l} \qquad \lambda=29.25\\ \\ \qquad \mu=147 \end{array} $$


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 $$\begin{aligned} & e \sin \varepsilon=-29543^{\prime \prime} .18=-8^{\circ} 12^{\prime} 23^{\prime \prime} .18 \\ & M+e \sin \varepsilon \quad \text {. . . . } 3241631.59 \\ & \text { differing from } \varepsilon \text {. . . } 8.41 \text {. } \end{aligned}$$

This difference being multiplied by $\frac{\lambda}{\mu-\lambda}=\frac{29.25}{117.75}$ gives $2^{\prime \prime} .09,$  whence, finally, the corrected value of $E=324^{\circ} 16^{\prime} 31^{\prime \prime} .59-2^{\prime \prime} .09=324^{\circ} 16^{\prime} 29^{\prime \prime} .50,$  which is exact within $0^{\prime \prime} .01.$

14.
The equations of article 8 furnish several methods for deriving the true anomaly and the radius vector from the eccentric anomaly, the best of which we will explain.

I. By the common method $v$ is determined by equation VII., and afterwards $r$  by equation II.; the example of the preceding article treated in this way is as follows, retaining for $p$  the value given in article 10.

$$\begin{aligned} & \frac{1}{2} E=162^{\circ} 8^{\prime} 14^{\prime \prime} .75 \quad \log e \quad \text {. . . . } 9.3897262 \\ & \log \tan \frac{1}{2} E \text {. . . } 9.5082198 n \quad \log \cos v \text {. . . . } 9.8496597 \\ & \log \tan \left(45^{\circ}-\frac{1}{2} \varphi\right) \cdot 9.8912427 \quad 9.2393859 \\ & \log \tan \frac{1}{2} v. . . .9 .6169771 n \quad e \cos v \quad=0.1735345 \\ & \frac{1}{2} v=157^{\circ} 30^{\prime} 41^{\prime \prime} .50 \quad \log p \text {. . . . } 0.3954837 \\ & v=315 \quad 123.00 \quad \log (1+e \cos v) \cdot .0 .0694959 \\ & \log r \text {. . . . } 0.3259878 \text {. } \end{aligned}$$

II. The following method is shorter if several places are to be computed, for which the constant logarithms of the quantities $\sqrt{a(1+e)}, \sqrt{a(1-e)}$ should be computed once for all. By equations V. and VI. we have

$$\begin{aligned} & \sin \frac{1}{2} v \sqrt{ } r=\sin \frac{1}{2} E \sqrt{a(1+e)} \\ & \cos \frac{1}{2} v \sqrt{ } r=\cos \frac{1}{2} E \sqrt{a(1-e)} \end{aligned}$$

from which $\frac{1}{2} v$ and $\log \sqrt{ } r$  are easily determined. It is true in general that if we have $P \sin Q=A, P \cos Q=B, Q$ is obtained by means of the formula tan $Q=\frac{A}{B},$  and then $P$  by this, $P=\frac{A}{\sin Q},$  or by $P=\frac{B}{\cos Q}:$  it is preferable to use


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 the former when $\sin Q$ is greater than $\cos Q;$  the latter when $\cos Q$  is greater than $\sin Q.$  Commonly, the problems in which equations of this kind occur (such as present themselves most frequently in this work), involve the condition that $P$  should be a positive quantity; in this case, the doubt whether $Q$  should be taken between 0 and $180^{\circ},$  or between $180^{\circ}$  and $360^{\circ},$  is at once removed. But if such a condition does not éxist, this decision is left to our judgment.

We have in our example $e=0.2453162.$

$$\begin{aligned} & \log \sin \frac{1}{2} E && 9.486732 && \log \cos \frac{1}{2} E && 9.486732 n \\ & \log \sqrt{a \left ( 1+e \right ) } && 0.2588593 && \log \sqrt{a \left ( 1+e \right ) } &&  0.1501020 \\ \end{aligned} $$

Hence $$\begin{aligned}

& \log \sin \frac{1}{2} v\sqrt{r} && 9.7456225 && \text {whence} && \log \tan\frac{1}{2} v = 9.169771 n \\ & \ log \cos \frac{1}{2} v\sqrt{r} && 0.1286454 n && && \frac{1}{2}v = 157^\circ 30' 41''.50 \\ & \ log \cos \frac{1}{2} v && 9.9656515 n && && v = 315^\circ 1' 23''.00 \\ & \log \sqrt{r} && 0.1629939 \\ & \log r && 0.3259878 \\ \end{aligned} $$

III. To these methods we add a third which is almost equally easy and expeditious, and is much to be preferred to the former if the greatest accuracy should be required. Thus, $r$ is first determined by means of equation III, and after that, $v$  by $\mathrm{X}.$  Below is our example treated in this manner.

Formula VIII, or XI, is very convenient for verifying the calculation, particularly if $v$ and $r$  have been determined by the third method. Thus;


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15.
Since, as we have seen, the mean anomaly $\dot{M}$ is completely determined by means of $v$  and $\varphi,$  in the same manner as $v$  by $M$  and $\varphi,$  it is evident, that if all these quantities are regarded as variable together, an equation of condition ought to exist between their differential variations, the investigation of which will not be superfluous. By differentiating first, equation VII., article 8, we obtain

$$\frac{d E}{\sin E}=\frac{d v}{\sin v}-\frac{d \varphi}{\cos \varphi}$$

by differentiating likewise equation XII., it becomes

$$\mathrm{d} M=(1-e \cos E) \mathrm{d} E-\sin E \cos \varphi \mathrm{d} \varphi .$$

If we eliminate $\mathrm{d} E$ from these differential equations we have

$$\mathrm{d} M=\frac{\sin E(1-e \cos E)}{\sin v} \mathrm{~d} v-\left(\sin E \cos \varphi+\frac{\sin E(1-e \cos E)}{\cos \varphi}\right) \mathrm{d} \varphi$$

or by substituting for $\sin E, 1-e \cos E,$ their values from equations VIII., III.,

$$\mathrm{d} M=\frac{r r}{a a \cos \varphi} \mathrm{d} v-\frac{r(r+p) \sin v}{a a \cos ^{2} \varphi} \mathrm{d} \varphi$$

or lastly, if we express both coefficients by means of $v$ and $\varphi$  only,

$$\mathrm{d} M=\frac{\cos ^{3} \varphi}{(1+e \cos v)^{2}} \mathrm{~d} v-\frac{(2+e \cos v) \sin v \cos ^{2} \varphi}{(1+e \cos v)^{2}} \mathrm{~d} \varphi .$$

Inversely, if we consider $v$ as a function of the quantities $M, \varphi,$  the equation has this form:-

$$\mathrm{d} v=\frac{a a \cos \varphi}{r r} \mathrm{~d} M+\frac{(2+e \cos v) \sin v}{\cos \varphi} \mathrm{d} \varphi$$

or by introducing $E$ instead of $v$

$$\mathrm{d} v=\frac{a a \cos \varphi}{r r} \mathrm{~d} M+\frac{a a}{r r}(2-e \cos E-e e) \sin E \mathrm{~d} \varphi .$$


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16.
The radius vector $r$ is not fully determined by $v$  and $\varphi,$  or by $M$  and $\varphi,$  but depends, besides these, upon $p$  or $a;$  its differential, therefore, will consist of three parts. By differentiating equation II. of article 8, we obtain

$$\frac{\mathrm{d} r}{r}=\frac{\mathrm{d} p}{p}+\frac{e \sin v}{1+e \cos v} \mathrm{~d} v-\frac{\cos \varphi \cos v}{1+e \cos v} \mathrm{~d} \varphi .$$

By putting here

$$\frac{\mathrm{d} p}{p}=\frac{\mathrm{d} \alpha}{\alpha}-2 \tan \varphi \mathrm{d} \varphi$$

(which follows from the differentiation of equation I.), and expressing, in conformity with the preceding article, $\mathrm{d} v$ by means of $\mathrm{d} M$  and $\mathrm{d} \varphi,$  we have, after making the proper reductions,

$$\begin{aligned} & \frac{\mathrm{d} r}{r}=\frac{\mathrm{d} a}{a}+\frac{a}{r} \tan \varphi \sin v \mathrm{~d} M-\frac{a}{r} \cos \varphi \cos v \mathrm{~d} \varphi, \\ & \mathrm{d} r=\frac{r}{a} \mathrm{~d} a+a \tan \varphi \sin v \mathrm{~d} M-a \cos \varphi \cos v \mathrm{~d} \varphi. \end{aligned}$$

Finally, these formulas, as well as those which we developed in the preceding article, rest upon the supposition that $v, \varphi,$ and $M,$  or rather $\mathrm{d} v, \mathrm{~d} \varphi,$  and $\mathrm{d} M,$  are expressed in parts of the radius. If, therefore, we choose to express the variations of the angles $v, \varphi,$ and $M,$  in seconds, we must either divide those parts of the formulas which contain $\mathrm{d} v, \mathrm{~d} \varphi,$  or $\mathrm{d} M,$  by 206264.8, or multiply those which contain $\mathrm{d} r, \mathrm{~d} p, \mathrm{~d} a,$  by the same number. Consequently, the formulas of the preceding article, which in this respect are homogeneous, will require no change.

17.
It will be satisfactory to add a few words concerning the investigation of the greatest equation of the centre. In the first place, it is evident in itself that the difference between the eccentric and mean anomaly is a maximum for $E=90^{\circ},$ where it becomes $=e$  (expressed in degrees, etc.); the radius vector at this point $=a,$  whence $v=90^{\circ}+\varphi,$  and thus the whole equation of the centre $=\varphi+c,$


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 which, nevertheless, is not a maximum here, since the difference between $v$ and $E$  may still increase beyond $\varphi.$  This last difference becomes a maximum for $\mathrm{d}(v-E)=0$  or for $\mathrm{d} v=\mathrm{d} E,$  where the eccentricity is clearly to be regarded as constant. With this assumption, since in general

$$\frac{\mathrm{d} v}{\sin v}=\frac{\mathrm{d} E}{\sin E}$$

it is evident that we should have $\sin v=\sin E$ at that point where the difference between $v$  and $E$  is a maximum; whence we have by equations VIII., III.,

$$r=a \cos \varphi, e \cos E=1-\cos \varphi, \text { or } \cos E=+\tan \frac{1}{2} \varphi$$

In like manner $\cos v=-\tan \frac{1}{2} \varphi$ is found, for which reason it will follow * that

hence again

$$v=90^{\circ}+\arcsin \tan \frac{1}{2} \varphi, E=90^{\circ}-\arcsin \tan \frac{1}{3} \varphi$$

$$\sin E=V\left(1-\tan ^{2} \frac{1}{2} \varphi\right)=\frac{V \cos \varphi}{\cos \frac{1}{2} \varphi}$$

so that the whole equation of the centre at this point becomes

$$2 \arcsin \tan \frac{1}{2} \varphi+2 \sin \frac{1}{2} \varphi \sqrt{\cos \varphi},$$

the second term being expressed in degrees, etc. At that point, finally, where the whole equation of the centre is a maximum, we must have $\mathrm{d} v=\mathrm{d} M,$ and so according to article $15, r=a \vee \cos \varphi;$  hence we have

$$\cos v=-\frac{1-\cos ^{\frac{3}{2} \varphi} \varphi}{e}, \cos E=\frac{1-\sqrt{\cos \varphi}}{e}=\frac{1-\cos \varphi}{e(1+\sqrt{\cos \varphi)}}=\frac{\tan \frac{1}{2} \varphi}{1+\sqrt{\cos \varphi}},$$

by which formula $E$ can be determined with the greatest accuracy. $E$ being found, we shall have, by equations X., XII.,

$$\text { equation of the centre }=2 \arcsin \frac{\sin \frac{1}{2} \varphi \sin E}{\sqrt[4]{\cos \varphi}}+e \sin E \text {. }$$

We do not delay here for an expression of the greatest equation of the centre by means of a series proceeding according to the powers of the eccentricities, which several authors have given. As an example, we annex a view of the three maxima which we have been considering, for Juno, of which the eccentricity, according to the latest elements, is assumed $=0.2554996.$ ? =0

??=0


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 $$\begin{array}{|c|c|c|c|c|} \hline \text{Maximum.} & E & E-M & 0-E & 0-M \\ \hline \begin{array}{c} E-M \\ v-E \\ v-M \\ \end{array} & \begin{array}{lll} 90^{\circ} & 0^{\prime} & 0^{\prime \prime} \\ 82 & 32 & 9 \\ 86 & 14 & 40 \\ \end{array} & \includegraphics[max width=\textwidth]{2024_02_19_0baa176114aa2645ceebg-042} & \includegraphics[max width=\textwidth]{2024_02_19_0baa176114aa2645ceebg-042(1)} & \begin{array}{llll} 29^{\circ} & 26^{\prime} & 32^{\prime \prime} .05 & \\ 29 & 26 & 35 & .80 \\ 29 & 30 & 16 & .96 \\ \end{array} \\ \hline \end{array}$$

18.

In the PARABOLA, the eccentric anomaly, the mean anomaly, and the mean motion, become $=0;$ here therefore these ideas cannot aid in the comparison of the motion with the time. In the parabola, however, there is no necessity for an auxiliary angle in integrating $r \mathrm{~d} v;$ for we have

and thus,

$$r r \mathrm{~d} v=\frac{p p \mathrm{~d} v}{4 \cos ^{4} \frac{1}{2} v}=\frac{p p \mathrm{~d} \tan \frac{1}{2} v}{2 \cos ^{2} \frac{1}{2} v}=\frac{1}{2} p p\left(1+\tan ^{2} \frac{1}{2} v\right) \mathrm{d} \tan \frac{1}{2} v$$

$$\int r r d v=\frac{1}{2} p p\left(\tan \frac{1}{2} v+\frac{1}{3} \tan ^{3} \frac{1}{2} v\right)+\text { Constant. }$$

If the time is supposed to commence with the perihelion passage, the Constant $=0;$ therefore we have

$$\tan \frac{1}{2} v+\frac{1}{3} \tan ^{3} \frac{1}{2} v=\frac{2 t k v(1+\mu)}{p^{\frac{3}{2}}}$$

by means of which formula, $t$ may be derived from $v,$  and $v$  from $t,$  when $p$  and $\mu$  are known. In the parabolic elements it is usual, instead of $p,$ to make use of the radius vector at the perihelion, which is $\frac{1}{2} p,$  and to neglect entirely the mass $\mu.$  It will scarcely ever be possible to determine the mass of a body, the orbit of which is computed as a parabola; and indeed all comets appear, according to the best and most recent observations, to have so little density and mass, that the latter can be considered insensible and be safely neglected.

19.

The solution of the problem, from the true anomaly to find the time, and, in a still greater degree, the solution of the inverse problem, can be greatly abbreviated by means of an auxiliary table, such as is found in many astronomical works.


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 If the variations of the anomaly $v$ are wanted in seconds, both parts also of $\mathrm{d} v$  must be expressed in this manner, that is, it is necessary to take for $/ c$  the value $3548^{\prime \prime} .188$  given in article 6. If, moreover, $\frac{1}{2} p=q$ is introduced instead of $p,$  the formula will bave the following form:

$$\mathrm{d} v=\frac{k \sqrt{ } 2 q}{r r} \mathrm{~d} t-\frac{3 k t}{r r \sqrt{ }^{2} q} \mathrm{~d} q$$

in which are to be used the constant logarithms

$$\log k \sqrt{ } 2=3.7005215724, \log 3 k \sqrt{ } \frac{1}{8}=3.8766128315$$

Moreover the differentiation of the equation

furnishes

$$r=\frac{p}{2 \cos ^{2} \frac{1}{2} v}$$

$$\frac{\mathrm{d} r}{r}=\frac{\mathrm{d} p}{p}+\tan \frac{1}{2} v \mathrm{~d} v$$

or by expressing $\mathrm{d} v$ by means of $\mathrm{d} t$  and $\mathrm{d} p,$

$$\frac{\mathrm{d} r}{r}=\left(\frac{1}{p}-\frac{3 k t \tan \frac{1}{2} v}{2 r r \sqrt{ } p}\right) \mathrm{d} p+\frac{k \sqrt{ } p \tan \frac{1}{2} v}{r r} \mathrm{~d} t .$$

By substituting for $t$ its value in $v,$  the coefficient of $\mathrm{d} p$  is changed into

$\frac{1}{p}-\frac{3 p \tan ^{2} \frac{1}{2} v}{4 r r}-\frac{p \tan ^{\frac{1}{2}} \frac{1}{2} v}{4 r r}=\frac{1}{r}\left(\frac{1}{8}+\frac{1}{2} \tan ^{2} \frac{1}{2} v-\frac{3}{2} \sin ^{2} \frac{1}{2} v-\frac{1}{2} \sin ^{2} \frac{1}{2} v \tan ^{2} \frac{1}{2} v\right)=\frac{\cos v}{2 r} ;$

but the coefficient of $\mathrm{d} t$ becomes $\frac{k \sin v}{r \sqrt{ } p}.$  From this there results

$$\mathrm{d} r=\frac{1}{2} \cos v \mathrm{~d} p+\frac{k \sin v}{\sqrt{p}} \mathrm{~d} t$$

or if we introduce $q$ for $p$

\mathrm{d} r=\cos v \mathrm{~d} q+\frac{k \sin v}{\sqrt{2} q} \mathrm{~d} t .</


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 $$r=\frac{p}{2 e \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}$$

For $v=0,$ the factors $\cos \frac{1}{2}(v-\psi),$  and $\cos \frac{1}{2}(v+\psi),$  in the denominator of this fraction become equal, the second vanishes for the greatest positive value of $v,$  and the first for the greatest negative value. Putting, therefore,

$$\frac{\cos \frac{1}{2}(v-\psi)}{\cos \frac{1}{2}(v+\psi)}=u$$

we shall have $u=1$ in perihelion; it will increase to infinity as $v$  approaches its limit $180^{\circ}-\psi;$  on the other hand it will decrease indefinitely as $v$  is supposed to return to its other limit $-\left(180^{\circ}-\psi\right);$  so that reciprocal values of $u,$  or, what amounts to the same thing, values whose logarithms are complementary, correspond to opposite values of $v.$

This quotient $u$ is very conveniently nsed in the hyperbola as an auxiliary quantity; the angle, the tangent of which is

$$\tan \frac{1}{2} v \sqrt{\frac{e-1}{e+1}}$$

can be made to render the same service with almost equal elegance; and in order to preserve the analogy with the ellipse, we will denote this angle by $\frac{1}{3} F.$ In this way the following relations between the quantities $v, r, u, F$  are easily brought together, in which we put $a=-b,$  so that $b$  becomes a positive quantity. I. $b=p \operatorname{cotan}^{2} \psi$

II. $r=\frac{p}{1+e \cos v}=\frac{p \cos \psi}{2 \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}$

III. $\tan \frac{1}{2} F=\tan \frac{1}{8} v \sqrt{\frac{e-1}{e+1}}=\tan \frac{1}{2} v \tan \frac{1}{2} \psi=\frac{u-1}{u+1}$

IV. $u=\frac{\cos \frac{1}{2}(v-\psi)}{\cos \frac{1}{2}(v+\psi)}=\frac{1+\tan \frac{1}{2} F}{1-\tan \frac{1}{2} F^{\prime}}=\tan \left(45^{\circ}+\frac{1}{2} F^{\prime}\right)$ V. $\frac{1}{\cos F}=\frac{1}{2}\left(u+\frac{1}{u}\right)=\frac{1+\cos \psi \cos v}{2 \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}=\frac{e+\cos v}{1+e \cos v}.$

By subtracting 1 from both sides of equation $V.$ we get,

VI. $\sin \frac{1}{2} v \sqrt{ } \frac{\sin \frac{1}{2} F}{} \sqrt{\frac{p}{(e-1) \cos F}}=\sin \frac{1}{2} F \sqrt{\frac{(e+1) b}{\cos F}}$

$$=\frac{1}{2}(u-1) \sqrt{\frac{p}{(e-1) u}}=\frac{1}{2}(u-1) \sqrt{\frac{(e+1) b}{u} .}$$


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 meet an E′′′, very rarely an F′′′, and I have only noticed a single case in which he has written a G′′′ for the flute. He uses the three lowest notes F to D much less frequently than Bach, who introduces them freely. Though he sometimes uses flutes in tutti movements along with oboes, trumpets, bassoons, and occasionally horns, as a rule, like Bach, he reserves them for special effects and to accompany vocal solos or duets.

Although Handel introduced the transverse flute in a work written in 1705, in his earlier operas he almost exclusively uses the flute-à-bec, which instrument he continued to employ throughout his whole career, including his last work, The Triumph of Truth and Time (1757). But this production was really a rechauffé of an earlier work written in 1708, and it is interesting to notice that in some numbers where the flute-à-bec was used in 1708, the transverse flute is substituted for it in the later version. One number in this version in which the flute-à-bec is retained is taken bodily from his early opera Agrippina. In Giustino, i. 4 (1737), he uses a bass flute-à-bec, and in Riccardo (iii. 2) we have the solitary instance of his use of a bass transverse flute. It tranposes from F minor to G minor, and is allotted a melody, extending from G′ to B′′♭, the accompaniment being played on violins and violas only. Sometimes Handel uses both forms of flute in the one work. Thus in La Resurrezione (1708) and in Rodelinda (ii. 5) we have two flutes-à-bec and also a transverse flutein the latter work they are used


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 But the Barkerian is by far the most convenient, and is also annexed to the admirable work of the celebrated OLBERS, (Abhandlung über die leichteste und bequemste Methode die Baln eines Cometen zu berechnen: Weimar, 1797.) It contains, under the title of the mean motion, the value of the expression $75 \tan \frac{1}{2} v+25$ $\tan ^{3} \frac{1}{3} v,$  for all true anomalies for every five minutes from 0 to $180^{\circ}.$  If therefore the time corresponding to the true anomaly $v$  is required, it will be necessary to divide the mean motion, taken from the table with the argument $v,$  by $\frac{150 k}{p^{\frac{3}{2}}},$  which quantity is called the mean daily motion; if on the contrary the true anomaly is to be computed from the time, the latter expressed in days will be multiplied by $\frac{150 k}{p^{\frac{3}{2}}},$  in order to get the mean motion, with which the corresponding anomaly may be taken from the table. It is further evident that the same mean motion and time taken negatively correspond to the negative value of the $v;$ the same table therefore answers equally for negative and positive anomalies. If in the place of $p,$ we prefer to use the perihelion distance $\frac{1}{2} p=q,$  the mean daily motion is expressed by $\frac{k \sqrt{ } 2812.5}{q^{\frac{3}{2}}},$  in which the constant factor $k \sqrt{ } 2812.5=$  0.912279061, and its logarithm is 9.9601277069. The anomaly $v$ being found, the radius vector will be determined by means of the formula already given,

$$r=\frac{q}{\cos ^{2} \frac{1}{2} v}$$

20.
By the differentiation of the equation

$$\tan \frac{1}{2} v+\frac{1}{3} \tan ^{8} \frac{1}{2} v=2 t k p^{-\frac{8}{2}},$$

if all the quantities $v, t, p,$ are regarded as variable, we have

$$ \frac { \mathrm{d} v}{2 \cos ^{4} \frac{1}{2} v}=2 k p^{-\frac{8}{2}} \mathrm{~d} t-3 t k p^{-\frac{8}{2}} \mathrm{~d} p $$

$$ \mathrm{~d} v=\frac{k \sqrt{ } p}{r r} \mathrm{~d} t-\frac{3 t k}{2 r r \sqrt{ } p} \mathrm{~d} p$$


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 In the same manner, by adding 1 to both sides, it becomes

VII. $\cos \frac{1}{2} v \sqrt{ } r=\cos \frac{1}{2} F \sqrt{\frac{p}{(e+1) \cos F}}=\cos \frac{1}{3} F \sqrt{\frac{(e-1) b}{\cos F}}$

$$=\frac{1}{2}(u+1) \sqrt{\frac{p}{(e+1) u}}=\frac{1}{2}(u+1) \sqrt{\frac{(e-1) b}{u} .}$$

By dividing VI. by VII. we should reproduce III.: the multiplication produces VIII. $r \sin v=p \operatorname{cotan} \psi \tan F=b \tan \psi \tan F$

$$=\frac{1}{2} p \operatorname{cotan} \psi\left(u-\frac{1}{u}\right)=\frac{1}{2} b \tan \psi\left(u-\frac{1}{u}\right) \text {. }$$

From the combination of the equations II. V. are easily derived

IX. $r \cos v=b\left(e-\frac{1}{\cos F}\right)=\frac{1}{2} b\left(2 e-u-\frac{1}{u}\right),$ X. $r=b\left(\frac{e}{\cos F}-1\right)=\frac{1}{2} b\left(e\left(u+\frac{1}{u}\right)-2\right).$

22.
By the differentiation of the formula IV. (regarding $\psi$ as a constant quantity) we get

$$\frac{\mathrm{d} u}{u}=\frac{1}{2}\left(\tan \frac{1}{2}(v+\psi)-\tan \frac{1}{2}(v-\psi)\right) \mathrm{d} v=\frac{r \tan \psi}{p} \mathrm{~d} v$$

hence,

$$r r \mathrm{~d} v=\frac{p r}{u \tan \psi} \mathrm{d} u$$

or by substituting for $r$ the value taken from $\mathrm{X}.$

$$r r \mathrm{~d} v=b b \tan \psi\left(\frac{1}{2} e\left(1+\frac{1}{u u}\right)-\frac{1}{u}\right) \mathrm{d} u$$

Afterwards by integrating in such a manner that the integral may vanish at the perihelion, it becomes

$$

\int r r \mathrm{d}v = b b a \tan \psi \left ( \frac{1}{2} e \left ( u - \frac{1}{u} \right ) -\log u \right ) = kt \sqrt{p} \sqrt { \left ( 1 + \mu \right ) } = kt \tan \psi \sqrt{b}  \sqrt{ 1 + \mu } $$

The logarithm here is the hyperbolic; if we wish to use the logarithm from Brigg’s system, or in general from the system of which the modulus $=\lambda,$ and


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 the mass $\mu$ (which we can assume to be indeterminable for a body moving in an hyperbola) is neglected, the equation assumes the following form:-

XI. $\frac{1}{2} \lambda e \frac{u u-1}{u}-\log u=\frac{\lambda k t}{b^{\frac{3}{2}}},$

or by introducing $F,$

$$\lambda e \tan F-\log \tan \left(45^{\circ}+\frac{1}{3} F^{\prime}\right)=\frac{\lambda k t}{b^{\frac{3}{2}}} \text {. }$$

Supposing Brigg’s logarithms to be used, we have

$$\log \lambda=9.6377843113, \quad \log \lambda k=7.8733657527$$

but a little greater precision can be attained by the immediate application of the

 * hyperbolic logarithms. The hyperbolic logarithms of the tangents are found in several collections of tables, in those, for example, which Schutze edited, and still more extensively in the Magnus Canon Triangulor. Iogarithmicus of Benjamin Ursin, Cologne, 1624, in which they proceed by tens of seconds.

Finally, formula XI. shows that opposite values of $t$ correspond to reciprocal values of $u,$  or opposite values of $F$  and $v,$  on which account equal parts of the hyperbola, at equal distances from the perihelion on both sides, are described in equal times.

23.
If we should wish to make use of the auxiliary quantity $u$ for finding the time from the true anomaly, its value is most conveniently determined by means of equation IV.; afterwards, formula II. gives directly, without a new calculation, $p$ by means of $r,$  or $r$  by means of $p.$  Having found $u,$  formula XI. will give the quantity $\frac{\lambda k t}{b^{\frac{3}{2}}},$ which is analogous to the mean anomaly in the ellipse and will be denoted by $N,$  from which will follow the elapsed time after the perihelion transit. Since the first term of $N,$ that is $\frac{2 e(u u-1)}{2 u}$  may, by means of formula VIII. be made $=\frac{\lambda r \sin v}{b \sin \psi},$ the double computation of this quantity will answer for testing its accuracy, or, if preferred, $N$  can be expressed without $u,$  as follows: -

XII. $N=\frac{\lambda \cdot \tan \psi \sin v}{2 \cos \frac{1}{2}(v+\psi) \cos \frac{1}{2}(v-\psi)}-\log \frac{\cos \frac{1}{2}(v-\psi)}{\cos \frac{1}{2}(v+\psi)}.$


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 Example. - Let $e=1.2618820,$ or $\psi=37^{\circ} 35^{\prime} 0^{\prime \prime}, v=18^{\circ} 51^{\prime} 0^{\prime \prime}, \log r=$  0.0333585. Then the computation for $u, p, b, N, t,$ is as follows:-

First term of $N=.$ 0.0621069

$\log u=\quad 0.0491129$

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \multirow{4}{*}{}\begin{array}{l} N= \\ \log \lambda k \\ \frac{3}{2} \log b \\ \end{array} & & \therefore & 0.0129940 & \log N &. & & . & 8.1137429 \\ \hline & . . & - . & 7.8733658\} & &  &  &  &  \\ \hline & \cdot &. . & 0.9030900\} & Difference & \includegraphics[max width=\textwidth]{2024_02_19_0baa176114aa2645ceebg-048} & - & & 6.9702758 \\ \hline & &  &  & \begin{array}{l} \log t \\ t= \\ \end{array} & &  &  & \begin{array}{r} 1.1434671 \\ 13.91448 \\ \end{array} \\ \hline \end{array}$$

24.
If it has been decided to carry out the calculation with hyperbolic logarithms, it is best to employ the auxiliary quantity $F,$ which will be determined by equation III., and thence $N$  by XI.; the semi-parameter will be computed from the radius vector, or inversely the latter from the former by formula VIII.; the second part of $N$  can, if desired, be obtained in two ways, namely, by means of the formula hyp. $\log \tan \left(45^{\circ}+\frac{1}{2} F\right),$ and by this, hyp. $\log \cos \frac{1}{2}(v-\psi)-$ hyp. $\log$ $\cos \frac{1}{2}(v+\psi).$  Moreover it is apparent that here where $\lambda=1$  the quantity $N$


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 will come out greater in the ratio $1: \lambda,$ than if Brigg’s logarithms were used. Our example treated according to this method is as follows:-

$\log \tan \frac{1}{2} \psi.$. . . 9.5318179

$\log \tan \frac{1}{2} v.$. . . 9.2201009

$\log \tan \frac{1}{2} F.$. . . 8.7519188

$$F=3^{\circ} 13^{\prime} 58^{\prime \prime} .12$$

$\log e.$. . . . . 0.1010188

$\log \tan F.$. . . 9.0543366

9.1553554

C. hyp. $\log \cos \frac{1}{2}(v-\psi)=0.01342266$

e $\tan F=.$. . . 0.14300638

C. hyp. $\log \cos \frac{1}{2}(v+\psi)=0.12650930$

hyp. $\log \tan \left(45^{\circ}+\frac{1}{2} F\right)=0.11308666$

Difference. . . . $=0.11308664$

$N=.$. . . . 0.02991972

$\log N.$. . . . . 8.4759575

$\log \%.$. . . 8.2355814$\}$

$\left.\frac{8}{2} \log b \cdot. . . . \quad 0.9030900\right\}$

Difference. . . . . 7.3324914

$\log t.$. . . . . 1.1434661

$t=\quad 13.91445$

25.
For the solution of the inverse problem, that of determining the true anomaly and the radius vector from the time, the auxiliary quantity $u$ or $F$  must be first derived from $N=\lambda k b^{-\frac{3}{2}} t$  by means of equation XI. The solution of this transcendental equation will be performed by trial, and can be shortened by devices analogous to those we have described in article 11. But we suffer these to pass without further explanation; for it does not seem worth while to elaborate as carefully the precepts for the hyperbolic motion, very rarely perhaps to be exhibited in celestial space, as for the elliptic motion, and besides, all cases that can possibly occur may be solved by another method to be given below. Afterwards $F$ or $u$  will be found, thence $v$  by formula III., and subsequently $r$  will be determined either by II. or VIII; $v$ and $r$  are still more conveniently obtained by means of formulas VI. and VII.; some one of the remaining formulas can be called into use at pleasure, for verifying the calculation.


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 26.

Example - Retaining for $e$ and $b$  the same values as in the preceding example, let $t=65.41236: v$  and $r$  are required. Using Briggs’s logarithms we have $\log t.$. . . . 1.8156598

$\log \lambda \hbar b^{-\frac{3}{2}}.$. . 6.9702758

$\log N.$. . . . 8.7859356, whence $N=0.06108514.$ From this it is seen that the equation $N=\lambda e \tan F-\log \tan \left(45^{\circ}+\frac{1}{2} F\right)$  is satisfied by $F=25^{\circ} 24^{\prime} 27^{\prime \prime} .66,$  whence we have, by formula III.,

$\log \tan \frac{1}{2} F.$. . 9.3530120

$\log \tan \frac{1}{2} \psi.$. . . 9.5318179

$\log \tan \frac{1}{2} v.$. 9.8211941, and thus $\frac{1}{2} v=33^{\circ} 31^{\prime} 29^{\prime \prime} .89,$ and $v=$  $67^{\circ} 2^{\prime} 59^{\prime \prime} .78.$  Hence, there follows,

C. $\left.\log \cos \frac{1}{2}(v+\psi) \quad \cdot 0.2137476\right\}$

C. $\left.\log \cos \frac{1}{2}(v-\psi) \cdot 0.0145197\right\}$

$\log \frac{p}{2 e}.$. . . 9.9725868

difference. . . . . . . 0.1992279

$\log \tan \left(45^{\circ}+\frac{1}{3} F\right) \quad.$. 0.1992280

$\log r.$. . . 0.2008541.

27.

If equation IV. is differentiated, considering $u, v, \psi,$ as variable at the same time, there results,

$$\frac{\mathrm{d} u}{u}=\frac{\sin \psi \mathrm{d} v+\sin v \mathrm{~d} \psi}{2 \cos \frac{1}{2}(v-\psi) \cos \frac{1}{2}(v+\psi)}=\frac{r \tan \psi}{p} \mathrm{~d} v+\frac{r \sin v}{p \cos \psi} \mathrm{d} \psi .$$

By differentiating in like manner equation XI, the relation between the differential variations of the quantities $u, \psi, N,$ becomes,

$$\frac{\mathrm{d} N}{\lambda}=\left(\frac{1}{2} e\left(1+\frac{1}{u u}\right)-\frac{1}{u}\right) \mathrm{d} u+\frac{(u u-1) \sin \psi}{2 u \cos ^{2} \psi} \mathrm{d} \psi,$$

or

$$\frac{\mathrm{d} N}{\lambda}=\frac{r}{b u} \mathrm{~d} u+\frac{r \sin v}{b \cos \psi} \mathrm{d} \psi.$$


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 Hence, by eliminating $\mathrm{d} u$ by means of the preceding equation we obtain

or

$$\frac{\mathrm{d} N}{\lambda}=\frac{r r}{b b \tan \psi} \mathrm{d} v+\left(1+\frac{r}{p}\right) \frac{r \sin v}{b \cos \psi} \mathrm{d} \psi$$

$$\begin{aligned} \mathrm{d} v & =\frac{b b \tan \psi}{\lambda r r} \mathrm{~d} N-\left(\frac{b}{r}+\frac{b}{p}\right) \frac{\sin v \tan \psi}{\cos \psi} \mathrm{d} \psi \\ & =\frac{b b \tan \psi}{\lambda r r} \mathrm{~d} N-\left(1+\frac{p}{r}\right) \frac{\sin v}{\sin \psi} \mathrm{d} \psi. \end{aligned}$$

28.

By differentiating equation $\mathrm{X},$ all the quantities $r, b, e, u,$  being regarded as variables, by substituting

$$\mathrm{d} e=\frac{\sin \psi}{\cos ^{2} \psi} \mathrm{d} \psi$$

and eliminating $\mathrm{d} u$ with the help of the equation between $\mathrm{d} N, \mathrm{~d} u, \mathrm{~d} \psi,$  given in the preceding article, there results,

$$\mathrm{d} r=\frac{r}{b} \mathrm{~d} b+\frac{b b e(u u-1)}{2 \lambda u r} \mathrm{~d} N+\frac{b}{2 \cos ^{2} \psi}\left\{\left(u+\frac{1}{u}\right) \sin \psi-\left(u-\frac{1}{u}\right) \sin v\right\} \mathrm{d} \psi \text {. }$$

The coefficient of $\mathrm{d} N$ is transformed, by means of equation VIII., into $\frac{b \sin v}{\lambda \sin \psi};$  but the coefficient of $\mathrm{d} \psi,$  by substituting from equation IV.,

$$u(\sin \psi-\sin v)=\sin (\psi-v), \quad \frac{1}{u}(\sin \psi+\sin v)=\sin (\psi+v)$$

is changed into

$$\frac{b \sin \psi \cos v}{\cos ^{2} \psi}=\frac{p \cos v}{\sin \psi}$$

so that we have

$$\mathrm{d} r=\frac{r}{b} \mathrm{~d} b+\frac{b \sin v}{\lambda \sin \psi} \mathrm{d} N+\frac{p \cos v}{\sin \psi} \mathrm{d} \psi$$

So far, moreover, as $N$ is considered a function of $b$  and $t,$  we have

$$\mathrm{d} N=\frac{N}{t} \mathrm{~d} t-\frac{8}{2} \frac{N}{b} \mathrm{~d} b$$

which value being substituted, we shall have $\mathrm{d} r,$ and also $\mathrm{d} v$  in the preceding article, expressed by means of $\mathrm{d} t, \mathrm{~d} b, \mathrm{~d} \psi.$  Finally, we have here to repeat our


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 previous injunction, that, if the variations of the angles $v$ and $\psi$  are conceived to be expressed, not in parts of the radius, but in seconds, either all the terms containing $\mathrm{d} v, \mathrm{~d} \psi,$  must be divided by 206264.8, or all the remaining terms must be multiplied by this number.

29.
Since the auxiliary quantities $\varphi, E, M,$ employed in the ellipse obtain imaginary values in the hyperbola, it will not be out of place to investigate their connection with the real quantities of which we have made use: we add therefore the principal relations, in which we denote by $i$  the imaginary quantity $\sqrt{ }-1.$

$$\begin{aligned} & \sin \varphi=e=\frac{1}{\cos \psi} \\ & \tan \left(45^{\circ}-\frac{1}{2} \varphi\right)=\sqrt{\frac{1-e}{1+e}}=i \sqrt{\frac{e-1}{e+1}}=i \tan \frac{1}{2} \psi \\ & \tan \varphi=\frac{1}{2} \operatorname{cotan}\left(45^{\circ}-\frac{1}{2} \varphi\right)-\frac{1}{2} \tan \left(45^{\circ}-\frac{1}{2} \varphi\right)=-\frac{i}{\sin \psi} \\ & \cos \varphi=i \tan \psi \\ & \varphi=90^{\circ}+i \log (\sin \varphi+i \cos \varphi)=90^{\circ}-i \log \tan \left(45^{\circ}+\frac{1}{2} \psi\right) \\ & \tan \frac{1}{2} E=i \tan \frac{1}{2} F=\frac{i(u-1)}{u+1} \\ & \frac{1}{\sin E}=\frac{1}{2} \operatorname{cotan} \frac{1}{2} E+\frac{1}{2} \tan \frac{1}{2} E=-i \operatorname{cotan} F, \end{aligned}$$

or

$$\begin{aligned} & \sin E=i \tan F=\frac{i(u u-1)}{2 u} \\ & \operatorname{cotan} E=\frac{1}{2} \operatorname{cotan} \frac{1}{2} E-\frac{1}{2} \tan \frac{1}{2} E=-\frac{i}{\sin F} \end{aligned}$$

or

$$\begin{aligned} & \tan E=i \sin F=\frac{i(u u-1)}{u u+1} \\ & \cos E=\frac{1}{\cos F}=\frac{u u+1}{2 u} \\ & i E=\log (\cos E+i \sin E)=\log \frac{1}{u}, \end{aligned}$$

or

$$\begin{aligned} & E=i \log u=i \log \left(45^{\circ}+\frac{1}{2} F\right) \\ & M=E-e \sin E=i \log u-\frac{i e(u u-1)}{2 u}=-\frac{i N}{\lambda} \end{aligned}$$

The logarithms in these formulas are hyperbolic.


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 30.

Since none of the numbers which we take out from logarithmic and trigonometrical tables admit of absolute precision, but are all to a certain extent approximate only, the results of all calculations performed by the aid of these numbers can only be approximately true. In most cases, indeed, the common tables, which are exact to the seventh place of decimals, that is, never deviate from the truth either in excess or defect beyond half of an unit in the seventh figure, furnish more than the requisite accuracy, so that the unavoidable errors are evidently of no consequence: nevertheless it may happen, that in special cases the effect of the errors of the tables is so augmented that we may be obliged to reject a method, otherwise the best, and substitute another in its place. Cases of this kind can occur in those computations which we have just explained; on which account, it will not be foreign to our purpose to introduce here some inquiries concerning the degree of precision allowed in these computations by the common tables. Although this is not the place for a thorough examination of this subject, which is of the greatest importance to the practical computer, yet we will conduct the investigation sufficiently far for our own object, from which point it may be further perfected and extended to other operations by any one requiring it.

31.
Any logarithm, sine, tangent, etc. whatever, (or, in general, any irrational quantity whatever taken from the tables,) is liable to an error which may amount to a half unit in the last figure: we will designate this limit of error by $\omega,$ which therefore is in the common tables $=0.00000005.$  If now, the logarithm, etc., cannot be taken directly from the tables, but must be obtained by means of interpolation, this error may be slightly increased fron two eauses. In the first place, it is usual to take for the proportional part, when (regarding the last figure as unity) it is not an integer, the next greatest or least integer; and in this way, it is easily perceived, this error may be increased to just within twice its actual amount. But


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 we shall pay no attention to this augmentation of the error, since there is no objection to our affixing one more than another decimal figure to the proportional part, and it is very evident that, if the proportional part is exact, the interpolated logarithm is not liable to a greater error than the logarithms given directly in the tables, so far indeed as we are authorized to consider the changes in the latter as uniform. Thence arises another increase of the error, that this last assumption is not rigorously true; but this also we pretermit, because the effect of the second and higher differences (especially where the superior tables computed by TaYtor are used for trigonometrical functions) is evidently of no importance, and may readily be taken into account, if it should happen to turn out a little ton great. In all cases, therefore, we will put the maximum unavoidable error of the tables $=\omega,$ assuming that the argument (that is, the number the logarithm of which, or the angle the sine etc. of which, is sought) is given with strict accuracy. But if the argument itself is only approximately known, and the variation $\omega^{\prime}$ of the logarithm, etc. (which may be defined by the method of differentials) is supposed to correspond to the greatest error to which it is liable, then the maximum error of the logarithm, computed by means of the tables, can amount to $\omega+\omega^{\prime}.$

Inversely, if the argument corresponding to a given logarithm is computed by the help of the tables, the greatest error is equal to that change in the argument which corresponds to the variation $\omega$ in the logarithm, if the latter is correctly given, or to that which corresponds to the variation $\omega+\omega^{\prime}$  in the logarithm, if the logarithm can be erroneous to the extent of $\omega^{\prime}.$  It will hardly be necessary to remark that $\omega$  and $\omega^{\prime}$  must be affected by the same sign.

If several quantities, correct within certain limits only, are added together, the greatest error of the sum will be equal to the sum of tlie greatest individual errors affected by the same sign; wherefore, in the subtraction also of quantities approximately correct, the greatest error of the difference will be equal to the sum of the greatest individual errors. In the multiplication or division of a quantity not strictly correct, the maximum error is increased or diminished in the same ratio as the quantity itself.


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 32.

Let us proceed now to the application of these principles to the most useful of the operations above explained.

I. If $\varphi$ and $E$  are supposed to be exactly given in using the formula VII., article 8, for computing the true anomaly from the eccentric anomaly in the elliptic motion, then in $\log \tan \left(45^{\circ}-\frac{1}{2} \varphi\right)$  and $\log \tan \frac{1}{3} E,$  the error $\omega$  may be committed, and thus in the difference $=\log \tan \frac{1}{2} v,$  the error $2 \omega;$  therefore the greatest error in the determination of the angle $\frac{1}{2} v$  will be

$$\frac{3 \omega \mathrm{d} \frac{1}{2} v}{\mathrm{~d} \log \tan \frac{1}{2} v}=\frac{3 \omega \sin v}{2 \lambda}$$

 denoting the modulus of the logarithms used in this calculation. The error, therefore, to which the true anomaly $v$ is liable, expressed in seconds, becomes

$$\frac{3 \omega \sin v}{\lambda} 206265=0^{\prime \prime} .0712 \sin v,$$

if Brigg’s logarithms to seven places of decimals are employed, so that we may be assured of the value of $v$ within $0^{\prime \prime} .07;$  if smaller tables to five places only, are used, the error may amount to $7^{\prime \prime} .12.$

II. If $e \cos E$ is computed by means of logarithms, an error may be committed. to the extent of

$$\frac{3 \omega e \cos E}{\lambda}$$

therefore the quantity

$$1-e \cos E, \text { or } \frac{r}{a},$$

will be liable to the same error. In computing, accordingly, the logarithm of this quantity, the error may amount to $(1+\delta) \omega,$ denoting by $\delta$  the quantity

$$\frac{3 e \cos E}{1-e \cos E}$$

taken positively: the possible error in $\log r$ goes up to the same limit, $\log a$  being assumed to be correctly given. If the eccentricity is small, the quantity $\delta$ is always confined within narrow limits; but when $e$  differs but little from 1, $1-e \cos E$  remains very small as long as $E$  is small; consequently, $\delta$  may


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 increase to an amount not to be neglected: for this reason formula III, article 8, is less suitable in this case. The quantity $\delta$ may be expressed thus also,

$$\frac{3(a-r)}{r}=\frac{3 e(\cos v+e)}{1-e e}$$

which formula shows still more clearly when the error $(1+\delta) \omega$ may be neglected.

III. In the use of formula X., article 8, for the computation of the true from the mean anomaly, the $\log \sqrt{ } \frac{a}{r}$ is liable to the error $\left(\frac{1}{2}+\frac{1}{2} \delta\right) \omega,$  and so the $\log$  $\sin \frac{1}{2} \varphi \sin E \sqrt{ } \frac{a}{r}$  to that of $\left(\frac{5}{2}+\frac{1}{2} \delta\right) \omega;$  hence the greatest possible error in the determination of the angles $v-E$  or $v$  is

$$\frac{\omega}{\lambda}(7+\delta) \tan \frac{1}{2}(v-E),$$

or expressed in seconds, if seven places of decimals are employed,

$$\left(0^{\prime \prime} .166+0^{\prime \prime} .024 \delta\right) \tan \frac{1}{2}(v-E)$$

When the eccentricity is not great, $\delta$ and $\tan \frac{1}{2}(v-E)$  will be small quantities, on account of which, this method admits of greater accuracy than that which we have considered in I.: the latter, on the other hand, will be preferable when the eccentricity is very great and approaches nearly to unity, where $\delta$  and $\tan \frac{1}{2}(v-E)$  may acquire very considerable values. It will always be easy to decide, by means of our formulas, which of the two methods is to be preferred.

IV. In the determination of the mean anomaly from the eccentric by means of formula XII., article 8, the error of the quantity $e \sin E,$ computed by the help of logarithms, and therefore of the anomaly itself, $M,$  may amount to

$$\frac{3 \omega e \sin E}{\lambda}$$

which limit of error is to be multiplied by $206265^{\prime \prime}$ if wanted expressed in seconds. Hence it is readily inferred, that in the inverse problem where $E$ is to be determined from $M$  by trial, $E$  may be erroneous by the quantity

$$\frac{3 \omega e \sin E}{\lambda} \cdot \frac{\mathrm{d} E}{\mathrm{~d} M} \cdot 206265^{\prime \prime}=\frac{3 \omega e a \sin E}{\lambda r} \cdot 206265^{\prime \prime} \text {, }$$

even if the equation $E-e \sin E=M$ should be satisfied with all the accuracy which the tables admit.


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 The true anomaly therefore computed from the mean may be incorrect in two ways, if we consider the mean as given accurately; first, on account of the error committed in the computation of $v$ from $E,$  which, as we have seen, is of slight importance; second, because the value of the eccentric anomaly itself may be erroneous. The effect of the latter cause will be expressed by the product of the error committed in $E$ into $\frac{\mathrm{d} v}{\mathrm{~d} E},$  which product becomes

$$\frac{3 \omega e \sin E}{\lambda} \cdot \frac{\mathrm{d} v}{\mathrm{~d} M} \cdot 206265^{\prime \prime}=\frac{3 \omega e a \sin v}{\lambda r} \cdot 206265^{\prime \prime}=\left(\frac{e \sin v+\frac{1}{2} e e \sin 2 v}{1-e e}\right) 0^{\prime \prime} .0712$$

if seven places of decimals are used. This error, always small for small values of $e,$ may become very large when $e$  differs but little from unity, as is shown by the following table, which exhibits the maximum value of the preceding expression for certain values of $e.$

V. In the hyperbolic motion, if $v$ is determined by means of formula III., article 21, from $F$  and $\psi$  accurately known, the error may amount to

$$\frac{3 \omega \sin v}{\lambda} \cdot 206265^{\prime \prime}$$

but if it is computed by means of the formula

$$\tan \frac{1}{2} v=\frac{(u-1) \tan \frac{1}{2} \psi}{u+1}$$

$u$ and $\psi$  being known precisely, the limit of the error will be one third greater, that is,

$$\frac{4 \omega \sin v}{\lambda} \cdot 206265^{\prime \prime}=0^{\prime \prime} .09 \sin v$$

for seven places.

VI. If the quantity

$$\frac{\lambda k t}{b^{\frac{3}{2}}}=N$$

is computed by means of formula XI, article 22, with the aid of Briggs’s loga-


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 rithms, assuming $e$ and $u$  or $e$  and $F$  to be known exactly, the first part will be liable to the error

$$\frac{5(u u-1) e \omega}{2 u},$$

if it has been computed in the form

$$\frac{\lambda e(u-1)(u+1)}{2 u}$$

or to the error

$$\frac{3(u u+1) e \omega}{2 u}$$

if computed in the form

$$\frac{1}{2} \lambda e u-\frac{\lambda e}{2 u}$$

or to the error $3 e \omega \tan F$ if computed in the form $\lambda e \tan F,$  provided we neglect the error committed in $\log \lambda$  or $\log \frac{1}{2} \lambda.$  In the first case the error can be expressed also by $5 e \omega \tan F,$  in the second by $\frac{3 e \omega}{\cos F},$  whence it is apparent that the error is the least of all in the third case, but will be greater in the first or second, accurding as $u$  or $\frac{1}{u}>2$  or $<2,$  or according as $\pm F^{\prime}>36^{\circ} 52^{\prime}$  or $<36^{\circ} 52^{\prime}.$  But, in any case, the second part of $N$  will be liable to the error $\omega.$

VII. On the other hand, it is evident that if $u$ or $F$  is derived from $N$  by trial, $u$  would be liable to the error

$$(\omega \pm 5 e \omega \tan F) \frac{\mathrm{d} u}{\mathrm{~d} N}$$

or to

$$\left(\omega+\frac{3 e \omega}{\cos F}\right) \frac{\mathrm{d} u}{\mathrm{~d} N}$$

according as the first term in the value of $N$ is used separated into factors, or into terms; $F,$  however, is liable to the error

$$(\omega \pm 3 e \omega \tan F) \frac{\mathrm{d} F}{\mathrm{~d} N} .$$

The upper signs serve after perihelion, the lower before perihelion. Now if $\frac{\mathrm{d} v}{\mathrm{~d} N}$ is substituted here for $\frac{\mathrm{d} u}{\mathrm{~d} N}$  or for $\frac{\mathrm{d} F}{\mathrm{~d} N},$  the effect of this error appears in the determination of $v,$  which therefore will be


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 $$\frac{b b \tan \psi(1 \pm 3 e \tan F) \omega}{\lambda r r} \text { or } \frac{b b \tan \psi(1+3 e \sec F) \omega}{\lambda r r},$$

if the auxiliary quantity $u$ has been employed; on the other hand, if $F$  has been used, this effect becomes,

$$\frac{b b \tan \psi(1 \pm 3 e \tan F) \omega}{\lambda r r}=\frac{\omega}{\lambda}\left\{\frac{(1+e \cos v)^{2}}{\tan ^{8} \psi} \pm \frac{3 e \sin v(1+e \cos v)}{\tan ^{2} \psi}\right\} .$$

If the error is to be expressed in seconds, it is necessary to apply the factor 206265 ". It is evident that this error can only be considerable when $\psi$ is a small angle, or $e$  a little greater than 1. The following are the greatest values of this third expression, for certain values of $e,$ if seven places of decimals are employed:

To this error arising from the erroneous value of $F$ or $u$  it is necessary to apply the error determined in $\mathrm{V}.$  in order to have the total uncertainty of $v.$

VIII. If the equation XI., article 22, is solved by the use of hyperbolic logarithms, $F$ being employed as an auxiliary quantity, the effect of the possible error in this operation in the determination of $v,$  is found by similar reasoning to be,

$$\frac{(1+e \cos v)^{2} \omega^{\prime}}{\tan ^{8} \psi} \pm \frac{3 e \sin v(1+e \cos v) \omega}{\lambda \tan ^{2} \psi}$$

where by $\omega^{\prime}$ we denote the greatest uncertainty in the tables of hyperbolic logarithms. The second part of this expression is identical with the second part of the expression given in VII.; but the first part in the latter is less than the first in the former, in the ratio $\lambda \omega^{\prime}: \omega,$ that is, in the ratio $1: 23,$  if it be admissible to assume that the table of Ursin is everywhere exact to eight figures, or

$$\omega^{\prime}=0.000000005$$


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 33.

The methods above treated, both for the determination of the true anomaly from the time and for the determination of the time from the true anomaly, ${ }^{*}$ do not admit of all the precision that might be required in those conic sections of which the eccentricity differs but little from unity, that is, in ellipses and hyperbolas which approach very near to the parabola; indeed, unavoidable errors, increasing as the orbit tends to resemble the parabola, may at length exceed all limits. Larger tables, constructed to more than seven figures would undoubtedly diminish this uncertainty, but they would not remove it, nor would they prevent its surpassing all limits as soon as the orbit approached too near the parabola. Moreover, the methods given above become in this case very troublesome, since a part of them require the use of indirect trials frequently repeated, of which the tediousness is even greater if we work with the larger tables. It certainly, therefore, will not be superfluous, to furnish a peculiar method by means of which the uncertainty in this case may be avoided, and sufficient precision may be obtained with the help of the common tables.

34.
The common method, by which it is usual to remedy these inconveniences, rests upon the following principles. In the ellipse or hyperbola of which $e$ is the eccentricity, $p$  the semi-parameter, and therefore the perihelion distance

$$\frac{p}{1+e}=q,$$

let the true anomaly $v$ correspond to the time $t$  after the perihclion; in the parabola of which the semi-parameter $=2 q,$  or the perihelion distance $=q,$  let the true anomaly $w$  correspond to the same time, supposing in each case the mass $\mu$  to be either neglected or equal. It is evident that we then have

 * Since the time contains the factor $a^{\frac{3}{2}}$ or $b^{\frac{3}{2}},$  the greater the values of $a=\frac{p}{1-e e},$  or $b=\frac{p}{e^{2}-1},$  the more the error in $M$  or $N$  will be increased.


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 $$\int \frac{p p \mathrm{~d} v}{(1+e \cos v)^{2}}: \int \frac{4 q q \mathrm{~d} w}{(1+\cos v)^{2}}=\sqrt{ } p: \sqrt{ } 2 q$$

the integrals commencing from $v=0$ and $v=0,$  or

$$\int \frac{(1+e)^{\frac{3}{2}} d v}{(1+e \cos v)^{2} \sqrt{2}}=\int \frac{2 d w}{(1+\cos w)^{2}}$$

Denoting $\frac{i-e}{1+e}$ by $\alpha, \tan \frac{1}{2} v$  by $\theta,$  the former integral is found to be

$$\sqrt{ }(1+\alpha) \cdot\left(\theta+\frac{1}{3} \theta^{3}(1-2 \alpha)-\frac{1}{5} \theta^{5}(2 \alpha-3 \alpha \alpha)+\frac{1}{7} \theta^{7}\left(3 \alpha \alpha-4 \alpha^{3}\right)-\text { etc. }\right)$$

the latter, $\tan \frac{1}{2} w+\frac{1}{3} \tan ^{3} \frac{1}{2} v.$ From this equation it is easy to determine $v$  by $\alpha$  and $v,$  and also $v$  by $\alpha$  and $w$  by means of infinite series: instead of $\alpha$  may be introduced, if preferred,

$$1-e=\frac{2 \alpha}{1+\alpha}=\delta$$

Since evidently for $\alpha=0,$ or $\delta=0,$  we have $v=w,$  these series will have the following form :-

$$\begin{aligned} & v=v+\delta v^{\prime}+\delta \delta v^{\prime \prime}+\delta^{3} v^{\prime \prime \prime}+\text { etc. } \\ & v=v+\delta w^{\prime}+\delta \delta v^{\prime \prime}+\delta^{3} v^{\prime \prime \prime}+\text { etc. } \end{aligned}$$

where $v^{\prime}, v^{\prime \prime}, v^{\prime \prime},$ etc. will be functions of $v,$  and $v^{\prime}, v^{\prime \prime}, w^{\prime \prime \prime},$  functions of $w.$  When $\delta$  is a very small quantity, these series converge rapidly, and few terms suffice for the determination of $v$  from $v,$  or of $v$  from $w. \quad t$ is derived from $v,$  or $w$  from $t,$  by the method we have explained above for the parabolic motion.

35.
Our Bessel has developed the analytical expressions of the three first coeffcients of the second series $w^{\prime}, w^{\prime \prime}, w^{\prime \prime \prime},$ and at the same time has added a table constructed with a single argument $w$  for the numerical values of the two first $v^{\prime}$  and $u^{\prime \prime},$  (Von Zuch Monatliche Correspondena, vol. XII., p. 197). A table for the first coefficient $w^{\prime},$ computed by Simpson, was already in existence, and was annexed to the work of the illustrious OLBERs above commended. By the use of this method, with the help of BeSSEL’s table, it is possible in most cases to determine the true anomaly from the time with sufficient precision; what remains to be desired is reduced to nearly the following particulars:


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FOR SEVERAL months now Don Quijote had been Emperor of Micamaca, and his faithful shield-bearer, Sancho Panza, Governor-Generola. Greater than all modern countries was Micamaca, and more beautiful. An endless garden. A Paradise without a “tree of knowledge,” without a treacherous snake, a silly woman, and an over-bearing lord. A traveller could not tell what struck him most—the deep scarlet of the roses, or the lily-whiteness of the womens’ faces, or the noble bearing of the men. An eternal Summer was Micamaca, where the rain fell only to deepen the lustre of things, and the wind blew only to scatter perfumes of great delight.

Profound were the love and the admiration of the people of Micamaca for their Emperor and their Governor-General. The women like exquisite fruit hanging ripe upon trees implored to be culled and bitten into. The men begged to be dressed as soldiers to wage war against any nation.

But Don Quijote, Emperor, looked in the great silver-mirror, and though he saw himself dazzling


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 I. In the inverse problem, the determination of the time, that is, from the true anomaly, it is requisite to have recourse to a somewhat indirect method, and to derive $w$ from $v$  by trial. In order to meet this inconvenience, the first series should be treated in the same manner as the second: and since it may be readily perceived that $-v^{\prime}$ is the same function of $v$  as $v^{\prime}$  of $v,$  so that the table for $w^{\prime}$  might answer for $v^{\prime}$  the sign only being changed, nothing more is required than a table for $v^{\prime \prime},$  by which either problem may be solved with equal precision.

Sometimes, undoubtedly, cases may occur, where the eccentricity differs but little from unity, such that the general methods above explained may not appear to afford sufficient precision, not enough at least, to allow the effect of the third and higher powers of $\delta$ in the peculiar method just sketched out, to be safely neglected. Cases of this kind are possible in the hyperbolic motion especially, in which, whether the former methods are chosen or the latter one, an error of several seconds is inevitable, if the common tables, constructed to seven places of figures only, are employed. Although, in truth, such cases rarely occur in practice, something might appear to be wanting if it were not possible in all cases to determine the true anomaly within $0^{\prime \prime} .1,$ or at least $0^{\prime \prime} .2,$  without consulting the larger tables, which would require a reference to books of the rarer sort. We hope, therefore, that it will not seem wholly superfluous to proceed to the exposition of a peculiar method, which we have long had in use, and which will also commend itself on this account, that it is not limited to eccentricities differing but little from unity, but in this respect admits of general application.

36.
Before we proceed to explain this method, it will be proper to observe that the uncertainty of the general methods given above, in orbits approaching the form of the parabola, ceases of itself, when $E$ or $F$  increase to considerable magnitude, which indeed can take place only in large distances from the sun. To show which, we give to

$$\frac{3 \omega e a \sin v}{\lambda r} \cdot 206265^{\prime \prime}$$

the greatest possible error in the ellipse, which we find in article $32, I V,$ the following form,


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 $$\frac{3 \omega e V(1-e e) \cdot \sin E}{\lambda(1-e \cos E)^{2}} \cdot 206265^{\prime \prime}$$

from which is evident of itself that the error is always circumscribed within uarrow limits when $E$ acquires considerable value, or when $\cos E$  recedes further from unity, however great the eccentricity may be. This will appear still more distinctly from the following table, in which we have computed the greatest numerical value of that formula for certain given values of $E,$ for seven decimal places.

$$E=\begin{array}{r|r} 10^{\circ} & \text { maximum error }= \\ 20 & 3^{\prime \prime} .04 \\ 30 & 0.76 \\ 40 & 0.34 \\ 50 & 0.19 \\ 60 & 0.12 \\ & 0.08 \end{array}$$

The same thing takes place in the hyperbola, as is immediately apparent, if the expression obtained in article 32, VII., is put into this form,

$$\frac{\omega \cos F(\cos F+3 e \sin F) V(e e-1)}{\lambda\left(e-\cos F^{\prime}\right)^{2}} 206265^{\prime \prime}$$

The following table exhibits the greatest values of this expression for certain given values of $F.$

When, therefore, $E$ or $F$  exceeds $40^{\circ}$  or $50^{\circ}$  (which nevertheless does not easily occur in orbits differing but little from the parabola, because heavenly bodies moving in such orbits at such great distances from the sun are for the most part withdrawn from our sight) there will be no reason for forsaking the general method. For the rest, in such a case even the series which we treated in article


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 34 might converge too slowly; and therefore it is by no means to be regarded as a defect sof the method about to be explained, that it is specially adapted to those cases in which $E$ or $F$  has not yet increased beyond moderate values.

37.
Let us resume in the elliptic motion the equation between the eccentric anomaly and the time,

$$E-e \sin E=\frac{k t \sqrt{ }(1+\mu)}{a^{\frac{3}{2}}}$$

where we suppose $E$ to be expressed in parts of the radius. Henceforth, we shall leave out the factor $V(1+\mu);$ if a case should occur where it is worth while to take it into account, the symbol $t$  would not express the time itself after perihelion, but this time multiplied by $\sqrt{ }(1+\mu).$  We designate in future by $q$  the perihelion distance, and in the place of $E$  and $\sin E,$  we introduce the quantities

$$E-\sin E, \text { and } E-\frac{1}{10}(E-\sin E)=\frac{9}{10} E+\frac{1}{10} \sin E:$$

the careful reader will readily perceive from what follows, our reason for selecting particularly these expressions. In this way our equation assumes the following form:-

$$(1-e)\left(\frac{9}{10} E+\frac{1}{10} \sin E\right)+\left(\frac{1}{10}+\frac{9}{10} e\right)(E-\sin E)=k t\left(\frac{1-e}{q}\right)^{\frac{9}{2}}$$

As long as $E$ is regarded as a quantity of the first order,

$$\frac{9}{10} E+\frac{1}{10} \sin E=E-\frac{1}{60} E^{3}+\frac{1}{1200} E^{5}-\text { etc. }$$

will be a quantity of the first order, while

$$E-\sin E=\frac{1}{6} E^{3}-\frac{1}{120} E^{5}+\frac{1}{5040} E^{7}-\text { etc. },$$

will be a quantity of the third order. Putting, therefore,

$$\begin{aligned} & \frac{6(E-\sin E)}{\frac{1}{10} E+\frac{1}{10} \sin E}=4 A, \frac{\frac{9}{0} E+\frac{1}{10} \sin E}{2 \sqrt{ } A}=B, \\ & 4 A=E^{2}-\frac{1}{30} E^{4}-\frac{1}{60} E^{6}-\text { etc. } \end{aligned}$$

will be a quantity of the second order, and

$$B=1+\frac{3}{2800} E^{4}-\text { etc. }$$

will differ from unity by a quantity of the fourth order. But hence our equation becomes


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 $$B\left(2(1-e) A^{\frac{1}{2}}+\frac{2}{15}(1+9 e) A^{\frac{1}{2}}\right)=k t\left(\frac{1-e}{q}\right)^{\frac{3}{2}} \ldots. \cdot[1]$$

By means of the common trigonometrical tables, $\frac{9}{10} E+\frac{1}{10} \sin E$ may be computed with sufficient accuracy, but not $E-\sin F,$  when $E$  is a small angle; in this way therefore it would not be possible to determine correctly enough the quantities $\dot{A}$  and $B.$  A remedy for this difficulty would be furnished by an appropriate table, from which we could take out with the argument $E,$  either $B$  or the logarithm of $B;$  the means necessary to the construction of such a table will readily present themselves to any one even moderately versed in analysis. By the aid of the equation

$$\frac{9 E+\sin E}{20 B}=\sqrt{ } A$$

$\checkmark A$ can be determined, and hence $t$  by formula [1] with all desirable precision.

The following is a specimen of such a table, which will show the slow increase of $\log B;$ it would be superfluous to take the trouble to extend this table, for further on we are about to describe tables of a much more convenient form.

38.

It will not be useless to illustrate by an example what has been given in the preceding article. Let the proposed true anomaly $=100^{\circ},$ the eccentricity $=0.96764567, \log q=9.7656500.$  The following is the calculation for $E, B,$  $A,$  and $t$  :

UTF8 mj 一

$\log \tan \frac{1}{2} v.$. . . 0.0761865

$\log \vee^{\frac{1-e}{1+e}}.$. . . 9.1079927.

$\log \tan \frac{1}{8} E.$. . 9.1841792, whence $\frac{1}{2} E=8^{\circ} 41^{\prime} 19^{\prime \prime} .32,$ and $E=$


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 $17^{\circ} 22^{\prime} 38^{\prime \prime} .64.$ To this value of $E$  corresponds $\log B=0.0000040;$  next is found in parts of the radius, $E=0.3032928, \sin E=0.2986643,$  whence $\frac{9}{20} E+\frac{1}{20} \sin E$  $=0.1514150,$  the logarithm of which $=9.1801689,$  and so $\log A^{\frac{1}{2}}=9.1801649.$  Thence is derived, by means of formula [1] of the preceding article,

$$\begin{aligned} & \log \frac{2 B q^{\frac{3}{2}}}{k \sqrt{ }(1-e)} \cdot .2 .4589614 \log \frac{2 B(1+9 e)}{15 k}\left(\frac{q}{1-e}\right)^{\frac{3}{2}} \cdot 2.7601038 \\ & \log A^{\frac{1}{2}} \text {. . . . } 9.1801649 \log A^{\frac{3}{2}} \text {. . . . . . . . } 7.5404947 \\ & \log 43.56386=. \quad 1.6391263 \log 19.98014=. . . . .1 .3005985 \\ & 19.98014 \\ & \overline{63.54400}=t \text {. } \end{aligned}$$

If the same example is treated according to the common method, $e \sin E$ in seconds is found $=59610^{\prime \prime} .79=16^{\circ} 33^{\prime} 30^{\prime \prime} .79,$  whence the mean anomaly $=$  $49^{\prime} 7^{\prime \prime} .85=2947^{\prime \prime} .85.$  And hence from

$$\log k\left(\frac{1-e}{q}\right)^{\frac{3}{2}}=1.6664302$$

is derived $t=63.54410.$ The difference, which is here only $\frac{1}{0} \frac{1}{00}$  part of a day, might, by the errors concurring, easily come out three or four times greater. It is further evident, that with the help of such a table for $\log B$ even the inverse problem can be solved with all accuracy, $E$  being determined by repeated trials, so that the value of $t$  calculated from it may agree with the proposed value. But this operation would be very troublesome: on account of which, we will now show how an auxiliary table may be much more conveniently arranged, indefinite trials be altogether avoided, and the whole calculation reduced to a numerical operation in the highest degree neat and expeditious, which seems to leave nothing to be desired.

39.

It is obvious that almost one half the labor which those trials would require, could be saved, if there were a table so arranged that $\log B$ could be immediately taken out with the argument $A.$  Three operations would then remain; the first indirect, namely, the determination of $A$  so as to satisfy the equation


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 [1], article 37 ; the second, the determination of $E$ from $A$  and $B,$  which may be done directly, either by means of the equation

or by this,

$$E=2 B\left(A^{\frac{1}{2}}+\frac{1}{15} A^{\frac{3}{2}}\right)$$

$$\sin E=2 B\left(A^{\frac{1}{2}}-\frac{3}{5} A^{\frac{3}{2}}\right)$$

the third, the determination of $v$ from $E$  by means of equation VII., article 8. The first operation, we will bring to an easy calculation free from vague trials; the second and third, we will really abridge into one, by inserting a new quantity $C$ in our table by which means we shall have no need of $E,$  and at the same time we shall obtain an elegant and convenient formula for the radius vector. Each of these subjects wè will follow out in its proper order.

First, we will change the form of equation [1] so that the Barkerian table may be used in the solution of it: For this purpose we will put

from which comes

$$A^{\frac{1}{2}}=\tan \frac{1}{3} w \sqrt{\frac{5-5 e}{1+9 e}}$$

denoting by $\alpha$ the constant

$$75 \tan \frac{1}{8} v w+25 \tan \frac{1}{2} w^{3}=\frac{75 k t \sqrt{ }\left(\frac{1}{5}+\frac{1}{b} e\right)}{2 B q^{\frac{3}{2}}}=\frac{\alpha t}{B},$$

$$\frac{75 k \sqrt{ }\left(\frac{1}{5}+\frac{9}{5} e\right)}{2 q^{\frac{3}{2}}} .$$

If therefore $B$ should be known, $w$  could be immediately taken from the Barkerian table containing the true anomaly to which answers the mean motion $\frac{\alpha t}{B} ; A$  will be deduced from $w$  by ineans of the formula

$$A=\beta \tan ^{2} \frac{1}{2} w$$

denoting the constant

$$\frac{5-5 e}{1+9 e} \text { by } \beta \text {. }$$

Now, although $B$ may be finally known from $A$  by means of our auxiliary table, nevertheless it can be foreseen, owing to its differing so little from unity, that if the divisor $B$  were wholly neglected from the beginning, $w$  and $A$  would be affected with a slight error only. Therefore, we will first determine roughly $w$ and $A,$  putting $B=1;$  with the approximate value of $A,$  we will find $B$  in our


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 auxiliary table, with which we will repeat more exactly the same calculation; most frequently; precisely the same value of $B$ that had been found from the approximate value of $A$  will correspond to the value of $A$  thus corrected, so that a second repetition of the operation would be superfluous, those cases excepted in which the value of $E$  may have been very considerable.

Finally, it is hardly necessary to observe that, if the approximate value of $B$ should in any other way whatever be known from the beginning, (which may always occur, when of several places to be computed, not very distant from each other, some few are already obtained,) it is better to make use of this at once in the first approximation: in this manner the expert computer will very often not have occasion for even a single repetition. We have arrived at this most rapid approximation from the fact that $B$ differs from unity, only by a difference of the fourth order, and is multiplied by a very small numerical coefficient, which advantage, as will now be perceived, was secured by the introduction of the quantities $E-\sin E, \frac{9}{10} E+\frac{1}{10} \sin E,$  in the place of $E$  and $\sin E.$

40.
Since, for the third operation, that is, the determination of the true anomaly, the angle $E$ is not required, but the $\tan \frac{1}{2} E$  only, or rather the $\log \tan \frac{1}{2} E,$  that operation could be conveniently joined with the second, provided our table supplied directly the logarithm of the quantity

$$\frac{\tan \frac{1}{2} E}{\sqrt{ } A}$$

which differs from unity by a quantity of the second order. We have preferred, however, to arrange our table in a somewhat different manner, by which, notwithstanding the small extension, we have obtained a much more convenient interpolation. By writing, for the sake of brevity, $T$ instead of the $\tan ^{2} \frac{1}{2} E,$  the value of $A,$  given in article 37 ,

$$\frac{15(E-\sin E)}{9 E+\sin E}$$

is easily changed to


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 $$A=\frac{T-\frac{6}{8} T^{2}+\frac{9}{1} T^{8}-\frac{12}{8} T^{4}+1 \frac{5}{1} T^{6}-\text { etc. }}{1-\frac{8}{18} T+\frac{7}{25} T^{2}-\frac{8}{36} T^{18}+\frac{9}{45} T^{4}-\text { etc. }},$$

in which the law of progression is obvious. Hence is deduced, by the inversion of the series,

$$\frac{A}{T}=1-\frac{4}{5} A+\frac{8}{175} A^{2}+\frac{8}{526} A^{3}+\frac{1}{33 \frac{8}{8} \frac{9}{8} \frac{6}{75}} A^{4}+\frac{28744}{1313 \frac{4}{81} \frac{4}{26}} A^{5}+\text { etc. }$$

Putting, therefore,

$$\frac{A}{T}=1-\frac{4}{5} A+C,$$

$C$ will be a quantity of the fourth order, which being included in our table, we can pass directly to $v$  from $A$  by means of the formula,

$$\tan \frac{1}{3} v=\sqrt{\frac{1+e}{1-e}} \sqrt{\frac{A}{1-\frac{c}{6} A+C}}=\frac{\gamma \tan \frac{1}{2} w}{\sqrt{\left(1-\frac{c}{6} A+C\right)}},$$

denoting by $\gamma$ the constant

$$\sqrt{\frac{5+5 e}{1+9 e}} .$$

In this way we gain at the same time a very convenient computation for the radius vector. It becomes, in fact, (article 8, VI.),

$$r=\frac{q \cos ^{2} \frac{1}{2} E}{\cos ^{2} \frac{1}{2} v}=\frac{q}{(1+T) \cos ^{2} \frac{1}{2} v}=\frac{\left(1-\frac{4}{5} A+C\right) q}{\left(1+\frac{1}{6} A+C\right) \cos ^{2} \frac{1}{2} v}$$

41.

Nothing now remains but to reduce the inverse problem also, that is, the determination of the time from the true anomaly, to a more expeditious form of computation: for this purpose we have added to our table a new column for $T.$ $T,$  therefore, will be computed first from $v$  by means of the formula

$$T=\frac{1-e}{1+e} \tan ^{2} \frac{1}{2} v$$

then $A$ and $\log B$  are taken from our table with the argument $T,$  or, (which is more accurate, and even more convenient also), $C$  and $\log B,$  and hence $A$  by the formula

$$A=\frac{(1+O) T}{1+\frac{4}{3} T}$$

finally $t$ is derived from $A$  and $B$  by formula [1], article 37. If it is desired to call into use the Barkerian table here also, which however in this inverse problem


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 with gold and precious stones, turned his face in a vast despair. He uttered sighs, deep and rumbling and circular like sorrowful winds.

“How long, my dear friends, shall we remain chained to this evil dream, this monstrous incantation?”

Sancho Panza, Governor-General, unlocked his arms from the waist of a beautiful Micamacienne, and approached his master. His feet clanked with the silver spurs, and his scimitar, studded with jewels of all colors, beat lightly against his side, while his body, protruding far beyond him, shone with decorations and wide golden bands.

“What evil dreams, Your Majesty?”

“This enchanted hut; this pig-sty.”

Sancho Panza screwed his small eyes, until they vanished completely in their velvet bed of fat.

“This pig-sty?” he muttered at last, dazed.

“Look at yourself, Sancho. How they have fattened you! You roll upon your short legs like a heavy barrel.”

Sancho Panza, Governor-General, was insulted.

“The beautiful women of Micamaca call my bearing elegant, Your Majesty.”

Don Quijote, Emperor, continued without hearing the answer,—“And look at me. Where is the long face of the ‘Knight of the Sorrowful Visage?’ A moon—reddish and full—gazing dully.”