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EXAMPLE.

The transit of the star ẞ Ursa Minoris, whose declination was 74° 46′ was observed as follows:

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March 8th, 1850. The star 8, Canis Majoris, was observed on the Vth wire only. Time of transit over that wire

equat. interv.

6' 54" 3.8

34.42 log. 1.53681

star's dec. (N. Alm.) 28° 46′ 16" 89 log.cos 9.94277

int. on par. of dec. 39*27

log.

1.59404
Correction

---

-39.27

Transit imaginary middle wire 6 53′′ 24•53

To compute the effect of error of level upon the time of meridian transit.— It will be first necessary to determine the inclination of the supporting axis to the horizon. For this purpose place the striding level on the pivots, and take the readings at both ends; suppose as in the diagram the

* III. here and below stand for the imaginary middle wire.

*

e 40 30 20 10

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20 10

30

40

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west end reads 40 and the east 20. It is evident that each end of the bubble stood at 30 when the level was horizontal, and that each has moved 10 divisions, which is obtained by taking the dif ference between the readings of the east and west end, and dividing by 2. So that if i denote the inclination of the supporting axis ab to a horizontal, and i' the inclination of the level to ab, i + i' will denote the inclination of the level to the horizon, and we have (w denoting the west reading and e the east).

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Reversing the level, supposing i' to be greater than i, the east end will

now be the highest, but the incli- e

nation to the horizon will no

longer be the sum, but the differ

ence of the inclinations

hence

and i',

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subtracting (2) from (1), and dividing by 2 we have

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If i be greater than ', after reversing, the west end would still be the highest, and we should have instead of (2)

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Adding (1) and (3), there results after dividing by 2

(3)

The glass tube is a portion of a circle of large radius, so that the movement of the bubble indicates the angular movement of the level. To find the angular value of a division of the level scale, place the level on the telescope of some instrument to which a large vertical circle is attached. Turn the circle till the bubble passes over a number of divisions of the scale, which will be equal to the degrees, minutes, &c., through which the circle has moved; divide this number of degrees and fractions of a degree by the number of divisions of the level scale passed over, and the quotient will be the value of one division of the scale.

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If the east end of the supporting axis be the highest instead of the west end, as we have supposed above, a similar course of reasoning would produce the formula

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which is the same as the above, changing e and e' for w and w'. Both formulas may be expressed in one, thus

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which is the formula always to be employed for obtaining the inclination of the supporting axis.

N. B. If the sum of the east readings exceed the sum of the west readings the east end of the axis is too high, and vice versâ.

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As a verification that the level readings have been correctly noted it may be observed that e+w should be the same in both positions of the level, being the length of the bubble which may be supposed not to change from the effects of temperature during an observation. Thus in the above example 50 + 40 = 24 + 66.

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To compute now the effect of the inclination of the axis as determined above, upon the time of transit, let нzo be the vertical circle in which the telescope would play, when the supporting axis was horizontal, нzo the circle in which it plays when the supporting axis is inclined; then zoz measured by the arc zz is the inclination of the planes of these two circles, and equal

H

to i the inclination of the supporting axis to the horizon. The length of the arc s described by a star at s, in passing from the vertical to the inclined circle in which the telescope plays is expressed by z sin os. (See Spher. Geom., Prop. II., Cor. 6), that is

a being the altitude of the star. similar to s is expressed by s

S = i sin a

But the length of an arc of the equator + cos d, & being the star's declination. (See note on p. 153, and Spher. Geom., Prop. II. Cor. 6.) This arc of the equator is converted into time by dividing by 15. Hence

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is the difference of time between the passage of the star over the middle wire of the telescope when the supporting axis is inclined, and its passage over a vertical circle. N. B. 1. The altitude of the star, designated by a in the above formula, is obtained from its declination given by catalogue, and the latitude of the place as at p. 150. 2. If the east end of the supporting axis is too high, the telescope is thrown to the west of its proper position, and the star, moving as it does from east to west by the diurnal motion, passes the wires too late; the correction therefore found as above is then subtractive. If the west end be too high, the star passes the wires too soon, and the correction is additive.

To compute the azimuth error of the instrument and its effect upon the time of transit.

This error arises from the deviation from the meridian of the vertical circle which the line of collimation describes, or to which the circle that it describes is reduced as above.

The most ready way of determining the amount of deviation and its effect on the time of transit is by "the method of high and low stars," as it is termed, that is by the transit of a star near the zenith, and one remote from it or near the horizon, whose right ascension and consequent time of transit does not differ much from the former.

To understand this let it first be supposed that one star passes the meridian exactly at the zenith, and that the other passes near the horizon, and let it be supposed also that the stars have the same right ascension. If the instrument were exactly adjusted to the meridian so that the optical axis of the telescope moved in the plane of that circle, then the two stars would be on the middle wire at the same instant, if the telescope could be brought down instantaneously from the high to the low star. But if the

circle in which the telescope plays make a small angle (called the angle of deviation or azimuth error) with the meridian, then the supporting axis being supposed horizontal, when the telescope is vertical it will point. to the zenith, in which all vertical circles as well as the meridian intersect, and the zenith star will be on the middle wire at the same instant as before, but the lower down the other star is, the wider will its time of transit over the middle wire differ from that of its transit over the meridian, because the farther will the vertical circle which the telescope describes be from the meridian, the farther we go from the zenith where they intersect. The greater, therefore, the difference between the time of transit of the zenith star and the low star over the wire of the instrument as compared with the difference of their times of transit over the meridian, which in the case supposed is zero, the greater the deviation of the vertical circle described by the instrument from the meridian. Now it is not necessary that the high star should pass exactly at the zenith, but only near it. If the difference of the observed time of transit of the high and low star be equal to the difference of their right ascensions, that is, of their times of passing the meridian, the optical axis of the telescope moves in the plane of the meridian; if not, this axis describes a vertical circle which deviates from the meridian, and the amount of this deviation and the consequent error in the time of meridian transit, we proceed now to show how to determine.

Let the full circle in the diagram represent the horizon, z the zenith, P the pole, and consequently PZ the meridian. Let zs represent the vertical circle described by the telescope, s the place of a star where it crosses it, and appears on the middle wire, and Ps, the declination circle, passing through the star. In the spherical triangle Pzs, in which the hour angle P

P

D

S

represents the time that has elapsed since the star s passed the meridian, PZ, before it reached the wire of the telescope in the vertical zs, we have by the sine proportion

sin P sin z: sin zs sin PS

From which, taking P in place of sin P, since it is a very small angle, and z the small angle of deviation of the vertical zs from the meridian in place of the sine of its supplement rzs, which is also its own sine, and representing zs by and the complement of PS or the declination of the

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