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to 15 degrees. This magnificent instrument is situated at Burr Castle, Ireland. It was constructed by the Earl of Rosse, at an expense of $60,000.

Speaking of this telescope, a late writer says: "By the energy and ingenuity of that eminent person (Lord Rosse), an eye is directed to the heavens, having a pupil six feet in diameter, with the most complete optical structure, and the power of ranging about for its objects over a great extent of sky; and thus the quantity of light which the eye secures from any point of the heavens is augmented, it may be, fifty thousand times. The rising Moon is seen from the Observatory in Ireland with the same increase of size and light, as if her solid globe, two thousand miles in diameter, retaining all its illumination, really rested upon the summit of the Alps, to be gazed at by the naked eye. An object which appears to the naked eye a single star may, by this telescope, so far as its power of seeing is concerned, be resolved into fifty thousand stars, each of the same brightness as the obvious star.*

710. A Transit Instrument is a telescope used for observing the transit of celestial objects across the meridian, for the purpose of determining differences of right ascension, or obtaining correct time. They are usually from six to ten feet long, and are mounted upon a horizontal axis, between two abutments of mason-work; so that the instrument, when horizontal, will point exactly to the south. It will then take objects in the heavens, when they are exactly on the meridian.

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Let A D in the out represent the telescope, and E and W the east and west abutments, between which it is placed. On the left is seen, attached to the mason work, a graduated circle; and on the eastern end of the axis of the telescope is seen an arm, n, extending to the circle, as an index. Now, suppose the index n to be at o, in the upper part of the circle, when the telescope is horizontal; then if the meridian alth tude of the object to be taken is 10°, the index must be moved

10° from o, as the

circle and the altitude of the object will correspond

motion?

Plurality of Worlds, American edition, p. 187.

Pemarks upon its

Where located? Cost? Description by a late writer? powers? 710. What is a transit instrument? Size? How mounted? Describe parts as shown in the cut. How set the instrument for the meridian altitude of a star?

711. An Astronomical Clock is a clock adapted to keep exact sidereal time. Taking the vernal equinox in the heavens as the zero point, and reckoning 24 hours eastward to the same point again, the time-reckoning 15° to an hour-when an object crosses the meridian, will always represent the right ascension of the object. Hence right ascension is usually given in hours, minutes, and seconds; or in time by the astronomical clock, set by the vernal equinox.

712. A Mural Circle is a large graduated circle, with a telescope erossing its center, used for the measurement of the altitudes and zenith distances of the heavenly bodies, at the instant of their crossing the B meridian. They are usually fixed upon a horizontal axis, that turns in a socket firmly fixed in a north and south wall. The degrees, minutes, and seconds on the circle are

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read by means of microscopes, and indicate the altitude of the object.

In the cut, A is a reading microscope firmly attached to the wall, and B C D E the wall to which the circle is attached. The telescope would denote an altitude of about 40°, which would leave 50° as the zenith distance.

711. An astronomical clock? How set? How indicate right ascension of objects? 712. Describe a mural circle? Its uses? How mounted? How ascertain altitude and Zenith distances by it?

CHAPTER XX.

PROBLEMS AND TABLES.

PROBLEM I.

TO CONVERT DEGREES, ETC., INTO TIME.

RULE I. Divide the degrees by 15, for hours; and multiply the remainder, if any, by 4, for minutes.

2. Divide the odd minutes and seconds in the same manner by 15 for minutes, seconds, &c., and multiply each remainder by 4, for the next lower denomination.

EXAMPLE 1.-Convert 32° 34′ 45′′ into time.

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Ans. 32° 34′ 45′′-2h. 10′ 19′′

EXAMPLE 2.-If it is 12 o'clock at this place, what is the time 20° east of us?

Thus, fifteen in 20°, once, and five over; the once is 1 hour, and the 5 multiplied by 4, gives 20 minutes; the time is then 1 hour and 20 minutes past 12.

EXAMPLE 3.-The longitude of Hartford is 72° 50′ west of Greenwich; what time is it at Greenwich when it is 12 o'clock at Hartford? Ans. 4h. 51 min. 20 sec.

EXAMPLE 4.-When it is 12 o'clock at Greenwich, what is the time at Hartford? Ans. 7h. 8m. 40s.

NOTE.-Table VIII. is designed to facilitate calculations of this kind. The degrees being placed in one column, and the corresponding time in another, it needs no explanation, except to observe that degrees in the left-hand columns may be considered as so many minutes, instead of degrees; in which case, the corresponding time in the adjoining column, must be read as minutes and seconds, instead of hours and minutes. In like manner, the degrees in the left-hand column may be read as seconds, and the cor. responding time, as seconds and thirds.

EXAMPLE.-Find, by the table, the time correspoding to 82° 84' 45".

Thus

Against 82° is 2h. 8m.

66 84'"

2

16s.

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45. "

8

2h. 10m. 198.

PROBLEM II.

TO CONVERT TIME INTO DEGREES, ETC.

RULE.-Multiply the hours by 15, and to the product add onefourth of the minutes, seconds, &c., observing that every minute of time makes 1°, and every second of time '.

EXAMPLE 1.-In 2 hours, 10 minutes, and 19 seconds; how many degrees?

Thus :

2h. 10m. 19s.

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This problem is readily solved by means of Table IX., without the labor of calculation.

Thus:

2 hours 30°

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Ex. 2. When it is 12 o'clock at Hartford, it is 4 hours, 51 minutes, and 20 seconds past noon at Greenwich; how many degrees is Hartford west of Greenwich?

Thus 15 times 4 is 60-added to of 51, is 72° 45′′, and this increased by of 20, is 72° 50′. Ans.

Ex. 3.-A Liverpool packet, after sailing several days from New York, finds the time by the Sun 2 hours and 40 minutes later than by the ship's chronometer : how far has the ship progressed on her way?

Ex. 4.-A vessel leaves Boston, and having been tossed about in foul weather for some days, finds, that when it is 12 o'clock by the Sun, it is only 11 o'clock and 50 minutes by the watch; is the vessel east or west of Boston; and how many degrees?

Ex. 5.-The moment of greatest darkness, during the annular eclipse of 1831, took place at New Haven, 10 minutes after 1 o'clock. A gentleman reports that it happened precisely at 1, where he observed it; and another, that it was 5 minutes after 1 where he saw it: Quere. How far east or west were these gentlemen from each other, and how many degrees from Ne" Haven?

PROBLEM III.

TO FIND WHAT STARS ARE ON THE MERIDIAN AT NINE O'CLOCK IN THE EVENING OF ANY GIVEN DAY.

RULE.-Look for the given day of the month, at the bottom of the maps, and all the stars having the same degree of right ascension will be on the meridian at that time.

EXAMPLE 1.-What stars will be on the meridian at 9 o'clock, the 19th of January?

Solution.-On Map III. I find that the principal stars standing over against the 19th of January, are Rigel and Capella. Ex. 2.-What stars are on the meridian the 20th of Decem ber? Ans. Menkar and Algol.

PROBLEM IV.

ANY STAR BEING GIVEN, TO FIND WHEN IT CULMINATES.

RULE. Find the star's right ascension in the table, or by the map (on the equinoctial), and the day of the month at the top or bottom of the map will be the day on which it culminates at 9 o'clock.

EXAMPLE 1.-At what time is the bright star Sirius on the meridian?

Solution. I find by the table, and by the map, that the right ascension of Sirius is 6 hours and about 38 minutes; and the time corresponding to this, at the bottom of the map, is the 11th of February.

Ex. 2.—At what time is Alpheratz, in the head of Andromeda, on the meridian? Ans. The 9th of November.

PROBLEM V.

THE RIGHT ASCENSION AND DECLINATION OF A PLANET BEING GIVEN, TO FIND ITS PLACE ON THE MAP.

RULE. Find the right ascension and declination of the planet on the map, and that will be its place for the given day.

EXAMPLE 1.-Venus's right ascension on the 1st of January, 1833, was 21 hours, 30 minutes, and her declination 16° south; required her situation on the map?

Solution. On the right hand of the Plate II. I count off

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