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 has less effect, in facilitating the calculation, it is not necessary to pay any regard to $A,$ but we have at once

$$\tan \frac{1}{8} w=\tan \frac{1}{3} v \sqrt{\frac{1+C}{\gamma\left(1+\frac{4}{3} T\right)}},$$

and hence the time $t,$ by multiplying the mean notion corresponding to the true anomaly, $w,$  in the Barkerian table, by $\frac{B}{\alpha}.$

42.
We have constructed with sufficient fulness a table, such as we have just described, and have added it to this work, (Table I.). Only the first part pertains to the ellipse; we will explain, further on, the other part, which includes the hyperbolic motion. The argument of the table, which is the quantity $A,$ proceeds by single thousandths from 0 to 0.300 ; the $\log B$  and $C$  follow, which quantities it must be understood are given in ten millionths, or to seven places of decimals, the ciphers preceding the significant figures being suppressed; lastly, the fourth column gives the quantity $T$  computed first to five, then to six figures, which degree of accuracy is quite sufficient, since this column is only needed to get the values of $\log B$  and $C$  corresponding to the argument $T,$  whenever $t$  is to be determined from $v$  by the precept of the preceding article. As the inverse problem which is much more frequently employed, that is, the determination of $v$ and $r$  from $t,$  is solved altogether without the help of $T,$  we have preferred the quantity $A$  for the argument of our table rather than $T,$  which would otherwise have been an almost equally suitable argument, and would even have facilitated a little the construction of the table. It will not be unnecessary to mention, that all the numbers of the table have been calculated from the beginning to ten places, and that, therefore, the seven places of figures which we give can be safely relied upon; but we cannot dwell here upon the analytical methods used for this work, by a full explanation of which we should be too much diverted from our plan. Finally, the extent of the table is abundantly sufficient for all cases in which it is advantageous to pursue the method just explained, since beyond the limit $A=0.3,$ to which answers $T=0.392374,$  or $E=64^{\circ} 7^{\prime},$  we may, as has been shown before, conveniently dispense with artificial methods.


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 == 43. ==

We add, for the better illustration of the preceding investigations, an example of the complete calculation for the true anomaly and radius vector from the time, for which purpose we will resume the numbers in article 38. We put then $e=$ $0.9674567, \log q=9.7656500, t=63.54400,$  whence, we first derive the constants $\log \alpha=0.03052357, \log \beta=8.2217364, \log \gamma=0.0028755.$

Hence we have $\log \alpha t=2.1083102,$ to which corresponds in Barker’s table the approximate value of $v=99^{\circ} 6^{\prime}$  whence is obtained $A=0.022926,$  and from our table $\log B=0.0000040.$  Hence, the correct argument with which Barker’s table must be entered, becomes $\log \frac{\alpha t}{B}=2.1083062,$  to which answers $w=99^{\circ} 6^{\prime}$  $13^{\prime \prime} .14;$  after this, the subsequent calculation is as follows:-

$$\begin{aligned} & \log \tan ^{2} \frac{1}{2} w \quad \text {. . } 0.1385934 \quad \log \tan \frac{1}{2} w \quad \text {. . . . . } 0.0692967 \\ & \log \beta \text {. . . . . } 8.2217364 \quad \log \gamma \text {. . . . . . . } 0.0028755 \\ & \log A \cdot. .8 .3603298 \quad \frac{1}{2} \text { Comp. } \log \left(1-\frac{4}{6} A+C\right) .0 .0040143 \\ & A=\text {. . . . } 0.02292608 \quad \overline{\log \tan \frac{1}{2} v} \text {. . . . . } 0.0761865 \\ & \text { hence } \log B \text { in the same manner as before; } \quad \frac{1}{2} v=\text {. . . . } 50^{\circ} 0^{\prime} 0^{\prime \prime} \\ & C=.0 .0000242 \quad v=. . . . \quad 10000 \\ & 1-\frac{4}{5} A+C=.0 .9816833 \quad \log q \text {. . . . . . } 9.7656500 \\ & 1+\frac{1}{5} A+C=.1 .0046094 \quad \text { 2 Comp. } \log \cos \frac{1}{2} v \text {. . } 0.3838650 \\ & \log \left(1-\frac{4}{6} A+C\right) \text {. . . } 9.9919714 \\ & \text { C. } \log \left(1+\frac{1}{6} A+C\right) \cdot. \quad 9.9980028 \\ & \log r \text {. . . . . . . . } 0.1394892 \end{aligned}$$

If the factor $B$ had been wholly neglected in this calculation, the true anomaly would have come out affected with a very slight error (in excess) of $0^{\prime \prime} .1$  only.

44.
It will be in our power to despatch the hyperbolic motion the more briefly, because it is to be treated in a manner precisely analogous to that which we have thus far expounded for the clliptic motion.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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 We present the equation between the time $t$ and the auxiliary quantity $u$  in. the following form :-

$$(e-1)\left(\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u\right)+\left(\frac{1}{10}+\frac{9}{10} e\right)\left(\frac{1}{2}\left(u-\frac{1}{u}\right)-\log u\right)=k t\left(\frac{e-1}{q}\right)^{\frac{3}{2}}$$

in which the logarithms are hyperbolic, and

$$\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u$$

is a quantity of the first order,

$$\frac{1}{3}\left(u-\frac{1}{u}\right)-\log u$$

a quantity of the third order, when $\log u$ may be considered as a small quantity of the first order. Putting, therefore,

$$\frac{6\left(\frac{1}{2}\left(u-\frac{1}{u}\right)-\log u\right)}{\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u}=4 A, \frac{\frac{1}{20}\left(u-\frac{1}{u}\right)+\frac{9}{10} \log u}{2 \sqrt{ } A}=B,$$

$A$ will be a quantity of the second order, but $B$  will differ from unity by a difference of the fourth order. Our equation will then assume the following form:-

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

which is entirely analogous to equation [1] of article 37. Putting moreover,

$$\left(\frac{u-1}{u+1}\right)^{2}=T$$

$T$ will be a quantity of the second order, and by the method of infinite series will be found

$\frac{A}{T} = 1 + \frac{4}{5} A + \ frac{8}{175} A^2 - \frac{8}{525} A^3 + \frac{1896}{336875} A^4 - \frac{28744}{13138125} A^5 +$ etc.

Wherefore, putting

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

$C$ will be a quantity of the fourth order, and

$$A=\frac{(1+C) T}{1-\frac{1}{8} T}$$

Finally, for the radius vector, there readily follows from equation VII., article 21,

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


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 45.

The latter part of the table annexed to this work belongs, as we have remarked above, to the hyperbolic motion, and gives for the argument $A$ (common to both parts of the table), the logarithm of $B$  and the quantity $C$  to seven places of decimals, (the preceding ciphers being omitted), and the quantity $T$  to five and afterwards to six figures. The latter part is extended in the same manner as the former to $A=0.300,$ corresponding to which is $T=0.241207, u=2.930,$  or $=0.341, F= \pm 52^{\circ} 19^{\prime};$  to extend it further would have been superfluous, (article 36 ).

The following is the arrangement of the calculation, not only for the determination of the time from the true anomaly, but for the determination of the true anomaly from the time. In the former problem, $T$ will be got by means of the formula

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

with $T$ our table will give $\log B$  and $C,$  whence will follow

$$A=\frac{(1+C) T}{1-\frac{C}{5} T} ;$$

finally $t$ is then found from the formula [2] of the preceding article. In the last problem, will first be computed, the logarithms of the constants

$$\begin{aligned} & \alpha=\frac{75 k \sqrt{ }\left(\frac{1}{5}+\frac{9}{5} e\right)}{2 q^{\frac{3}{2}}} \\ & \beta=\frac{5 e-5}{\mathrm{I}+9 e} \\ & \gamma=\sqrt{\frac{5 e+5}{1+9 e}} \end{aligned}$$

$A$ will then be determined from $t$  exactly in the same manner as in the elliptic motion, so that in fact the true anomaly $w$  may correspond in Barker’s table to the mean motion $\frac{\alpha t}{B},$  and that we may have

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

the approximate value of $A$ will be of course first obtained, the factor $B$  being


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 either neglected, or, if the means are at hand, being estimated; our table will then furnish the approximate value of $B,$ with which the work will be repeated; the new value of $B$  resulting in this manner will scarcely ever suffer sensible correction, and thus a second repetition of the calculation will not be necessary. $C$ will be taken from the table with the corrected value of $A,$  which being done we shall have,

$$\tan \frac{1}{2} v=\frac{\gamma \tan \frac{1}{2} w}{\sqrt{\left(1+\frac{4}{3} A+O\right)}}, r=\frac{\left(1+\frac{4}{5} A+C\right) q}{\left(1-\frac{1}{8} A+C\right) \cos ^{2} \frac{1}{2} v}$$

From this it is evident, that no difference can be perceived between the formulas for elliptic and hyperbolic motions, provided that we consider $\beta, A,$ and $T,$  in the hyperbolic motion as negative quantities.

46.
It will not be unprofitable to elucidate the hyperbolic motion also by some examples, for which purpose we will resume the numbers in articles $23,26.$

I. The data are $e=1.2618820, \log q=0.0201657, v=18^{\circ} 51^{\prime} 0^{\prime \prime}: t$ is required. We have

$$\begin{aligned} & 2 \log \tan \frac{1}{2} v \quad \text {. . . . } 8.4402018 \quad && \log T \text {. . . . . } 7.5038375 \\ & \log \frac{e-1}{e+1} \text {. . . . . . } 9.0636357 \quad && \log (1+C) \text {. . } 0.0000002 \\ & \log T \text {. . . . . }7.503875 && \mathrm{C}. \log \left (1 - \frac{4}{5} T \right ) \\ & \log B=\text {. . . . . } 0.0000001 && \log A \text {. . . . . } 7.5049476 \\ & C=\text {. . . . } 0.0000005 \\ & \log \frac{2 B q^{\frac{3}{2}}}{k \sqrt{ }(e-1)} \cdot .2 .3866444 \log \frac{2 B(1+9 e)}{15 k}\left(\frac{q}{e-1}\right)^{\frac{9}{2}} \cdot. \cdot 2.8843582 \\ & \log A^{\frac{1}{2}} \text {. . . . . } 8.7524738 . \log A^{\frac{3}{2}} \text {. . . . . . . . } 6.2574214 \\ & \log 13.77584=. .1 .1391182 \log 0.138605=. . . . .9 .1417796 \\ & 0.13861 \\ & 13.91445=t. \end{aligned}$$

II. $e$ and $q$  remaining as before, there is given $t=65.41236 ; v$  and $r$  are required. We find the logarithms of the constants,


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 $$\begin{aligned} & \log \alpha=9.9758345 \\ & \log \beta=9.0251649 \\ & \log \gamma=9.9807646 \end{aligned}$$

Next we have $\log \alpha t=1.7914943,$ whence by Barker’s table the approximate value of $w=70^{\circ} 31^{\prime} 44^{\prime \prime},$  and hence $A=0.052983.$  To this $A$  in our table answers $\log B=0.0000207;$  from which, $\log \frac{\alpha t}{B}=1.7914736,$  and the corrected value of $w=70^{\circ} 31^{\prime} 36^{\prime \prime} .86.$  The remaining operations of the calculation are as follows:-

$$2 \log \tan \frac{1}{2} w \quad \text {. . } 9.6989398 \quad \log \tan \frac{1}{3} w \quad \text {. . . . } 9.8494699$$

$\log \beta.$. . . . $9.0251649 \log \gamma.$. . . . . 9.9807646

$\log A.$. . 8.7241047 C $\log \left(1+\frac{4}{5} A+C\right) \cdot 9.9909602$

$A=.$. . . . $0.05297911 \log \tan \frac{1}{2} v.$. . . 9.8211947

$\log B$ as before, $\quad \frac{1}{3} v=.$. $33^{\circ} 31^{\prime} 30^{\prime \prime} .02$

$$\begin{aligned} & C=. .0 .0001252 \quad v=\quad. \quad .6730 .04 \\ & 1+\frac{4}{5} A+C=\text {. } 1.0425085 \quad \log q \text {. . . . . . } 0.0201657 \\ & 1-\frac{1}{5} A+C=. .0 .9895294 \quad 2 \text { C. } \log \cos \frac{1}{2} v \quad. . .0 .1580378 \\ & \log \left(1+\frac{4}{5} A+C\right) \quad. \quad .0 .0180796 \\ & \text { C. } \log \left(1-\frac{1}{5} A+C\right) \cdot \cdot 0.0045713 \\ & \log r \text {. . . . . . . } 0.2008544 \end{aligned}$$

Those which we found above (article 26), $v=67^{\circ} 2^{\prime} 59^{\prime \prime} .78, \log r=0.2008541,$ are less exact; and $v$  should properly have resulted $=67^{\circ} 3^{\prime} 0^{\prime \prime} .00,$  with which assumed value, the value of $t$  had been computed by means of the larger tables.


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 “The women of Micamaca call their Emperor the handsomest of men.”

“Remember this, my son,—to raisers of hogs, there is nothing more beautiful than a fat, dull hog.”

There are neither hogs nor raisers of hogs in Micamaca, Your Majesty. The only animals these people eat are lambs as white as snow, and pigeons with feet of cardinals.”

“This only proves that you have been more profoundly enchanted and more successfully fattened than myself. I am but half-dazed. I tell you, my son, if we do not hurry out of this place, we shall soon walk on all fours, and grunt and seek muddy troughs,—like this woman approaching us now.”

“This woman? Dulcinea—Your Majesty—Dulcinea, Empress of Micamaca, whose voice is more beautiful than the song of nightingales, whose face is more radiant than the stars.” Saying this, the Governor-General made many efforts at deep bowing, which he could not accomplish on account of his great body protruding far beyond him, shining with decorations and wide golden bands.

“My Lord, your slave bids you good-day,” spoke or rather sang Dulcinea, Empress of Micamaca, to Don Quijote, Emperor of Micamaca. And she knelt upon her dazzling knees, hidden beneath gorgeous silk. “Take her away, Sancho, my son,—beat her with a stick. Let her not grunt at my feet!


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 , the transverse flute playing the lower partand in Tamerlane (iii. 5) two flutes-à-bec and two transverse flutes, one of each kind playing together in unison. Occasionally parts are marked for "flauto ou traverso," to be played on either kind of flute.

The famous obligato to "Oh, ruddier than the Cherry," in Acis and Galatea, has a rather interesting history. In the first Italian version of the Serenata (1708) the flute-à-bec only is used, but this song does not occur in it, being first introduced in the English version published about 1720, where it is marked "flauto"i.e., flute-à-bec. In the second Italian version (1732) we find the transverse flute used throughout, , Acis and Galatea, "Oh, ruddier than the Cherry." but again this song is omitted. When Mozart re-scored the work, he gave this obligato to a transverse flute, adding some beautiful and characteristic passages, whilst Mendelssohn allotted it to two flutes. At the Ancient Concerts (London) it used to be performed on a flageolet, but nowadays it is usually played on a piccolo; possibly by way of a joking reply to the monster Polyphemus' demand for a pipe for his


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 Drive the sow off! Her paws are muddy. She smells like a litter of hogs.”

The Empress stunned, remained kneeling. The Governor-General opened and closed his mouth,—the broken black teeth of the upper jaw seeking in vain to touch the broken black teeth of the lower jaw. The Emperor walked slowly to the window, sighing deep, circular sighs like mournful winds.

“Arise, Your Majesty,” at last the Governor-General managed to say, trying in vain to bend to lift her.

“It does not matter, dear friend,” the Empress answered, “if my Lord thinks me a sow. I shall be his devoted, his faithful sow.” She arose, and walked off, two tears like two great pearls, like two small stars, piercing her magnificent eyes.

Sancho-Panza, Governor-General of Micamaca, watched the Emperor press his nose against the window-pane.

“My master is mad!” he thought. “But that should not hinder me from embracing the beautiful Micamaciennes who give themselves up like delicious fruit, nor from eating the lambs as white as milk, and the pigeons with feet of cardinals.”

“Sancho, my friend, come here, and watch this great ugly wind-mill turn.”

The Governor-General walked slowly, heavily to the window, his feet clanking with the silver spurs, his studded scimitar beating lightly against


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 his side. He presedpressed [sic] his heavy nose against the window-pane, and looked out.

“I see no wind-mill, Your Majesty.”

“Yonder—there—beyond the trees.”

“That is not a wind-mill, Your Majesty. It is your body-guard, composed of the handsomest officers, drilling.”

“Sancho, my friend, you are lost. They have enchanted you completely, alas!”

The Governor-General had the impulse to kneel before His Majesty, and beg him to come to his senses, but his body protruding far beyond him forbade him from doing it. Instead he took his master’s hand, and kissed it.

“Your Majesty, I beg of you,—be reasonable! See things as they are. When we used to encounter wind-mills in our humble days, you used to think them great armies, and now the flower of the finest army in the world, you consider a wind-mill!”

“O, Sancho, Sancho, faithful shield-bearer to the once great Don Quijote, most valiant knight of Spain—you shall soon walk on all fours and grunt.”

“How should I walk on all fours, when I cannot even bend? A moment ago, I wanted to kneel before you, but could not.”

“Sancho, my friend, I am still enough of the ‘Knight of the Sorrowful Visage’ to speak figuratively. When I spoke of you as walking on all fours and grunting, I meant spiritually. To all the