PROBLEM LII.

Given two observed altitudes of the sun, the time elapsed 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 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 equa tor 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 de clination, 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. Suppose on the 4th of June, 1813, in north latitude, the sun's altitude, at 29 minutes past 10 in the forenoon, to be 65° 24', and at 31 minutes past 12, 74° 8'; required the latitude.

* 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 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.

Answer. The sun's declination is 22° 27′ north, the elapsed time two hours two min. answering to 30° 30'; the complement of the first altitude 24° 56′, the complement of the second altitude 15° 52′, and the Latitude sought, 36° 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. 51° 32′ north.*

3. When the sun's declination was 22° 40′ north, his altitude at 10 h. 54 m. 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. 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

A great variety of examples, accurately calculated by a general rule, without an assumed latitude, may be seen in Keith's Trigonometry, second edition, page 292, &c.

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.

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

* 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 19 70° 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.

the sun's declination the moon will be vertically eclipsed.

Example. On the 26th of January, 1804, at 58 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 whole of Europe, Africa, and the continent of Asia. For more examples, see the following Table of Eclip

ses.

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. Pingré, a member of the Academy of Sciences, and published in L'Art des verifier les Dates. In olassing these tables, the arrangement of Mr. Ferguson has been followed; see page 267 of his Astronomy, where a catalogue of the visible eclipses is given from 1700 to 1600, 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. Pingré, are not perfectly accurate, and were only designed to show nearly the time when an eclipse may be expected to happen. The limits where these eclipses are visible are generally from the tropic of Cancer in Africa, to the northern extremity of Lapland, and from the 5th degree of north latitude in Asia, to the north polar circle; though some few of them are visible beyond the pole. In longitude, the limits are the 5th and 155th meri. diane, supposing the 20th to pass through Paris; hence, it appears that they are calculated for the meridian of Ferro; which will make their limits from London to be from 12° 46' west long. to 137° 14' east. M. Pingré says, that an eclipse of the sun is visible from 32° to 64° north, and as far south of the place where it is central. In the following table the moon is represented by M, the sun by S, t stands for total, p for partial, M for morning, and A for afternoon; the rest is obvious.