or both south, the sun rises before six, and sets after six, by a space of time equal to the ascensional difference; but if the sun's declination and the latitude be of contrary names, viz. the one north and the other south, the sun rises after six, and sets before six.

Examples. 1. Required the sun's right ascension, obtique ascension, oblique descension, ascensional difference, and time of rising and setting at London, on the 15th of April.

An we. The right ascension is 23° 30', the oblique ascension is 9° 45', the ascensional difference (23° 30′-9° 45′ =) 13° 45′ or 55 minutes of time; consequently the sun rises 55 minutes before 6, or 5 min past 5, and sets 55 min. past 6. The oblique descen sion is 37° 15'; consequently the descensional difference is (37 15'- 23° 30′ =) 13° 45′, the same as the ascensional difference

2. What are the sun's right ascension, oblique ascension, and oblique descension, on the 27th of September at London? what is the ascensional difference, and at what time does the sun rise and set?

3. What are the sun's right ascension, declination, oblique ascension, rising amplitude, oblique descension, and setting amplitude at London, on the 1st of May? what is the ascensional difference, and at what time does the sun rise and set?

4. What are the sun's right ascension, declination, oblique ascension, rising amplitude, oblique descension, and setting amplitude at Petersburg, on the 21st of June? what is the ascensional difference, and at what time does the sun rise and set?

5. What are the sun's right ascension, declination, oblique ascension, rising amplitude, oblique descension, and setting amplitude at Alexandria, on the 21st of December? what is the ascensional difference, and at what time does the sun rise and set?

PROBLEM LI.

Given the day of the month, and the sun's amplitude, to find the latitude of the place of observation.

Rule. Find the sun's place in the ecliptic, and bring it to the eastern or western part of the horizon (according as the eastern or western amplitude is given), elevate or depress the pole till the sun's place coincides with the given amplitude on the horizon, then the ele vation of the pole will show the latitude.

OR THUS:

Elevate the north pole to the complement* of the amplitude, and screw the quadrant of altitude upon the brass meridian, over the same degree; bring the equinoctial point Aries to the brass meridian, and move the quadrant of altitude till the sun's declination for the given day (counted on the quadrant) coincides with the equator; the number of degrees between the point Aries and the graduated edge of the quadrant will be the latitude sought.

Examples. 1. The sun's amplitude was observed to be 39° 48' from the east towards the north, on the 21st of June; required the latitude of the place? Answe. 51° 32′ north.t

2. The sun's amplitude was observed to be 15° 30′ from the east towards the north, at the same time his declination was 15° 30'; required the latitude?

Answer. The latitude was nothing.

3. On the 29th of May, when the sun's declination was 21° 30′ north, his rising amplitude was known to to be 22° northward of the east; required the latitude?

Answer. 12° north.

4. When the sun's declination was 20 north, his rising amplitude was 4o north of the east; required the latitude?

Answer. 60° north.

PROBLEM LII.

Given two observed altitudes of the sun, the time elaps ed between them, and the sun's declination, to find the latitude.+

Rule. Find the sun's declination, either by the globe or an ephemeris; take the number of degrees contained

*The complement of the amplitude is found by subtracting the amplitude from 90°. The rule is exactly the same as above; for it is formed from a right angled spherical triangle, the base being the complement of the amplitude, the perpendicular the latitude of the place, and the hypothenuse the complement of the sun's declination.

† See Keith's Trigonometry, page 237.

‡ Dr. Wilson, in his Dissertation on the Rise and Progress of Navigation, prefixed to Robertson's Treatise, says, this problem was first solved by the globe by Mr Robert Hues, and published in 1594; and Dr Mackay, in page 158 of his Complete Navigator, mentions the same circumstance. I have not been able to procure

therein from the equator with a pair of compasses, and apply the same number of degrees upon the meridian passing through Libra* from the equator northward or southward, and mark where they extend to; turn the elapsed time into degrees, and count those degrees upon the equator from the meridian passing through Libra; bring that point of the equator where the reckoning ends to the graduated edge of the brass meridian, and set off the sun's declination from that point along the edge of the meridian, the same way as before; then take the complement of the first altitude from the equator in your compasses, and, with one foot in the sun's declination, and a fine pencil in the other foot, describe an arc; take the complement of the second altitude in a similar manner from the equator, and, with one foot of the compasses fixed in the second point of the sun's declination, cross the former arc: the point of intersection brought to that part of the brass meridian which is numbered from the equator towards the poles, will stand under the degree of latitude sought.

Examples. 1. On the 4th of June 1808, in north latitude, the sun's altitude, at 29 minutes past 10 in the forenoon, was 65° 24', and at 31 minutes past 12, the altitude was 74° 8'; required the latitude?

Answer. The sun's declination was 22° 27′ north, the elapsed time was two hours two min, answering to 30° 30'; the complement of the first altitude was 24° 36', the complement of the second altitude 15° 52', and the latitude sought 56° 57' north.

2. Given the sun's declination 19° 39′ north, his altitude in the forenoon 38° 19', and, at the end of one hour and a half, the same morning, the altitude was 50° 25'; required the latitude of the place, supposing it to be north?

Answer. 519 32' north. #

this book, nor have I ever seen a solution to the problem by the globe.

*

Any meridian will answer the purpose as well as that which passes through Libra; on Adams' globes this meridian is divided like the brass meridian.

† See the method of turning time into degrees, Prob. XIX.

A great variety of examples, accurately calculated by the gen eral rule, without an assumed latitude, may be seen in Keith's Trigonometry, page 289, &c.

3. When the sun's declination was 22° 40′ north, his altitude at 10 h. 54 min. in the forenoon was 53° 29', and at 1 h. 17 m. in the afternoon it was 52° 48′; required the latitude of the place of observation, supposing it to be north?

Answer. 57° 8' north.

4. In north latitude, when the sun's declination was 22° 23′ south, being on the 5th of December, the sun's altitude in the afternoon was observed to be 14° 46′, and, after 1 h. and 22 m. had elapsed, his altitude was 8° 27′; required the latitude?

Answer. 50° 34′ north.

PROBLEM LIII.

The day and hour being given when a solar eclipse will happen, to find where it will be visible.

Rule. Find the sun's declination, and elevate the pole agreeably to that declination; bring the place, at which the hour is given, to that part of the brass meridian which is numbered from the equator towards the poles, and set the index of the hour circle to twelve; then, if the given time be before noon, turn the globe westward till the index has passed over as many hours as the given time wants of noon; if the time be past noon, turn the globe eastward as many hours as it is past noon, and exactly under the degree of the sun's declination on the brass meridian you will find the place on the globe where the sun will be vertically eclipsed: at all places within 70 degrees of this place, the eclipse may* be visible, especially if it be a total eclipse.

Example. On the 11th of February 1804, at 27 min. past ten o'clock in the morning at London, there was an eclipse of the sun; where was it visible, supposing the moon's penumbral shadow to extend northward 70 degrees from the place where the sun was vertically eclipsed?

Answer. London, &c. For more examples consult the Table of Eclipses, following the next problem.

* When the moon is exactly in the node, and when the axes of the moon's shadow and penumbra pass through the centre of the earth, the breadth of the earth's surface under the penumbral shadow is 70o 20'; but the breadth of this shadow is variable; and, if it be not accurately determined by calculation, it is impossible to tell by the globe to what extent an eclipse of the sun will be visible.

PROBLEM LIV.

The day and hour being given when a lunar eclipse will happen, to find where it will be visible.

Rule. Find the sun's declination for the given day, and note whether it be north or south; if it be north, elevate the south pole so many degrees above the horizon as are equal to the declination; if it be south, elevate the north pole in a similar manner; bring the place at which the hour is given to that part of the brass meridian which is numbered from the equator towards the poles, and set the index of the hour circle to twelve; then, if the given time be before noon, turn the globe westward as many hours as it wants of noon; if after noon, turn the globe eastward as many hours as it is past noon; the place exactly under the degree of the sun's declination will be the antipodes of the place where the moon is vertically eclipsed. Set the index of the hour circle again to twelve, and turn the globe on its axis till the index has passed over twelve hours; then to all places above the horizon the eclipse will be visible; to those places along the western edge of the horizon the moon will rise eclipsed; to those along the eastern edge she will set eclipsed; and to that place immediately under the sun's declination the moon will be vertically eclipsed.

Example. On the 26th of January 1804, at 56 min. past seven in the afternoon, at London, there was an eclipse of the moon; where was it visible?

Answer. It was visible to the waole of Europe, Africa, and the continent of Asia For more examples, see the following Table

of Eclipses.

NOTE. The substance of the following Table of Eclipses was extracted from Dr. Hutton's translation of Montucla's edition of Ozanam's Mathematical and Physical Recreations, published by Mr. Kearsley in Fleet-street. These eclipses were originally calculated by M. Pringré, a member of the Academy of Sciences, and published in L'Art des verifier les Dates. In classing these tables the arrangement of Mr. Ferguson has been followed; see page 267 of his Astronomy, wh re a catalogue of the visible eclipses is given from 1700 to 1800, taken from L'Art des verifier les Dates. It may be necessary to inform the learner, that the times of these eclipses, as calculated by M Pringré, are not perfectly accurate, and were only designed to show nearly the time when an eclipse may be ex