4. In what climate is the north of Spitzbergen? 5. In what climate is Cape Horn?

6. In what climate is Botany Bay, and what other places are situated in the same climate?

PROBLEM XXXIX.

To find the breadths of the several climates between the equator and the polar circles.

Rule. For the northern climates. Elevate the north pole 231° above the northern point of the horizon, bring the sign Cancer to the meridian, and set the index to twelve; turn the globe eastward on its axis till the index has passed over a quarter of an hour; observe that particular point of the meridian passing through Libra, which is cut by the horizon, and at the point of intersection make a mark with a pencil: continue the motion of the globe eastward till the index has passed over another quarter of an hour, and make a second mark; proceed thus till the meridian passing through Libra* will no longer cut the horizon; the several marks brought to the brass meridian will point out the latitude where each climate ends.t

Examples. 1. What is the breadth of the ninth north climate, and what places are situated within it?

Answer. The breadth of the 9th climate is 2° 57', it begins in latitude 49° 2' north, and ends in latitude 51° 59′ north, and all places situated within this space are in the same climate The places will be nearly the same as those enumerated in the first example to the preceding problem.

2. What is the breadth of the second climate, and in what latitude does it begin and end?

3. Required the beginning, end, and breadth of the fifth climate.

4 What is the breadth of the seventh climate north of the equator? in what latitude does it begin and end? and what places are situated within it?

On Adam's globes, the meridian passing through Libra is divided into degrees, in the same manner as the brass meridian is divided; the horizon will, therefore, cut this meridian in the seve ral degrees answering to the end of each climate, without the trouble of bringing it to the brass meridian, or marking the globe. † See a Table of the climates, with the method of constructing it, at pages 16 and 17.

PROBLEM XL.

To find that part of the equation of time which depends on the obliquity of the ecliptic.

Rule. Find the sun's place in the ecliptic, and bring it to the brass meridian; count the number of degrees from Aries to the brass meridian, on the equator and on the ecliptic; the difference reckoning four minutes of time to a degree, is the equa

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tion of time. It the number of Sun faster than the clock in degrees on the ecliptic exceed those on the equator, the sun is faster than the clock; but, if the number of degrees on the equator exceed those on the ecliptic, the sun is slower than the clock.

Note. The equation of time, or diffe ence between the time shown by a well regulated clock, and a true su diai, depends upon two causes; viz. the obliquity of the ecliptic, and the unequal motion of the earth in its orbit. The former of these causes may be explained by the above problem. If two suns were to set off at the same time fr n the point Aries, and move over equal spaces in equal time, the one on the ecliptic, the other on the equator, it is evident they would never come to the meridian together, except at the time of the equinoxes, and on the longest and shortest days. The annexed Table shows how much the sun is faster or slower than the clock ought to be, so far as the variation depends on the obliquity of the ecliptic only The signs of the first and third quadrants of the ecliptic are at the top of the table, and the degrees in these signs on the left hand; in any of these signs, the sun is faster than the clock. The signs of the second and third quadrants are at the bottom of the table, and the degees in these signs at the right hand; in any of these signs, the sun is slower

than the clock.

l'hus, when the sun is in 20 degrees of 8 or m, it is 9 minutes 50 seconds faster than the clock; and when the sun is in 18 degrees of or V, it is 6 minutes 2 seconds slower than the clock.

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M. S. M. S. M. S. 0 0 8 248 46 30 10 208 358 36 29 2 0 408 458 25 28 3 1 08 548 14 27 4 1 199 38 1 26 5 1 399 117 49 25 6 1 99 187 35 24 7 2 189 247 21 23 8 2 379 317 6 22 9 2 569 366 51 21 10 3 169 416 35 20 11 3 349 456 19 19 12 3 539 496 2 18 134 119 515 45 17 14 4 299 535 27 16 154 479 545 9 15 165 49 554 50 14 17 5 219 554 31 13 18 5 389 544 12 12 19 5 549 523 52 11 20 6 109 503 32 10 21 6 269 47 3 12 9 22 6 419 432 1 23 6 559 382 30 24 7 99 332 9 25 7 239 271 48 26 7 369 201 27 27 7 499 131 5 28 8 18 50 43 29 8 138 560 22 30 8 248 460 0

Deg

Sun slower than the clock in

Examples. 1. What is the equation of time on the 17th of July?

Answer The degrees o1. the equator exceed the degrees on the ecliptic by two: hence the sun is eight minutes slower than the clock.*

2. On what four days of the year is the equation of time nothing?

3. What is the equation of time dependant on the obliquity of the ecliptic on the 27th of October?

4. When the sun is 18° of Aries, what is the equation of time?

PROBLEM XLI.

To find the sun's meridian altitude at any time of the year at any given place.

t

Rule. Find the sun's declination, and elevate the pole to that declination; bring the given place to the brass meridian, and count the number of degrees between it and the horizon; these degrees will show the sun's meridian altitude.t

OR,

Elevate the pole so many degrees above the horizon as are equal to the latitude of the place; find the sun's place in the ecliptic, and bring it to that part of the brass meridian which is numbered from the equator towards the poles; count the number of degrees contained on the brass meridian between the sun's place and the horizon, and they will show the altitude.‡

OR, BY THE ANALEMMA.

Elevate the pole so many degrees above the horizon as are equal to the latitude of the place; find the day of

* The learner will observe, that the equation of time here determined, is not the true equation, as noted on the 7th circle on the horizon of Bardin's globes: the equation of time there given cannot be determined by the globe.

† See Problem XXI.

See Problem XXII.

the month on the analemma, and bring it to that part of the brass meridian wnich is numbered from the equator towards the poles; count the number of degrees contained on the brass meridian between the given day of the month and the horizon, and they will show the altitude.

Examples. 1. What is the sun's meridian altitude at London on the 21st of June?

Answer. 62 degrees.

2. What is the sun's meridian altitude at London on the 21st of March?

3. What is the sun's least meridian altitude at London?

4. What is the sun's greatest meridian altitude at Cape Horn?

5. What is the sun's meridian altitude at Madras on the 20th of June?

6. What is the sun's meridian altitude at Bencoolen on the 15th of January

PROBLEM XLII.

When it is midnight at any place in the temperate_or torrid zones, to find the sun's altitude at any place (on the same meridian) in the north frigid zone, where the sun does not descend below the horizon.

Rule. Find the sun's declination for the given day, and elevate the pole to that declination; bring the place (in the frigid zone) to that part of the brass meridian which is numbered from the north pole towards the equator, and the number of degrees between it and the horizon will be the sun's altitude.

OR,

Elevate the north pole so many degrees above the horizon as are equal to the latitude of the place in the frigid zone; bring the sun's place in the ecliptic to the brass meridian, and set the index of the hour circle to twelve; turn the globe on its axis till the index points to the other twelve; and the number of degrees between the sun's place and the horizon, counted on the brass meridian toward that part of the horizon marked north, will be the sun's altitude.

Examples. 1. What is the sun's altitude at the North Cape in Lapland, when it is midnight at Alexandria in Egypt on the 21st of June?

Answer. 5 degrees.

2. When it is midnight to the inhabitants of the island of Sicily on the 22d of May, what is the sun's altitude at the north of Spitzbergen, in latitude 80° north?

3. What is the sun's altitude at the north east of Nova Zembla, when it is midnight at Tobolsk, on the 15th of July?

4. What is the sun's altitude at the north of Baffin's Bay, when it is midnight at Buenos Ayres, on the 28th of May ?

PROBLEM XLIII.

To find the sun's amplitude at any place.

Rule. Elevate the pole so many degrees above the horizon as are equal to the latitude of the given place; find the sun's place in the ecliptic, and bring it to the eastern semi-circle of the horizon; the number of degrees from the sun's place to the east point of the horizon will be the rising amplitude: bring the sun's place to the western semi-circle of the horizon, and the number of degrees from the sun's place to the west point of the horizon will be the setting amplitude.

OR, BY THE ANALEMMA.

Elevate the pole so many degrees above the horizon as are equal to the latitude of the place; bring the day of the month on the analemma to the eastern semi-circle of the horizon: the number of degrees from the day of the month to the east point of the horizon will be the rising amplitude: bring the day of the month to the western semi-circle of the horizon, and the number of degrees from the day of the month to the west point of the horizon will be the setting amplitude.

Examples. 1. What is the sun's amplitude at London on the 21st of June?

Answer. 39° 48′ to the north of the east, and 39° 48′ to the Borth of the west

2. On what point of the compass does the sun riṣe and set at London on the 17th of May?

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