PROBLEM XXXVII. To find the duration of twilight at the north pole. Rule. Elevate the north pole so that the equator may coincide with the horizon; observe what point of the ecliptic, nearest to Libra, passes under 18° below the horizon, reckoned on the brass meridian, and find the day of the month correspondent thereto; the time elapsed from the 23d of September to this time will be the duration of evening twilight. Secondly, observe what point of the ecliptic nearest to Aries; passes under 18° below the horizon, reckoned on the brass meridian, and find the day of the month correspondent thereto; the time elapsed from that day to the 21st of March will be the duration of morning twilight. Example. What is the duration of twilight at the north pole, and what is the duration of dark night there? Answer. The point of the ecliptic nearest to Libra which passes under 18 degrees below the horizon, is 22 degrees in m, answering to the 18th of November; hence the evening twilight continues from the 23d of September (the end of the longest day) to the 13th of November (the beginning of dark night) being 51 days. The point of the ecliptic nearest to Aries which passed under 18 degrees below the horizon is 9 degrees in answering to the 29th of January; hence, the morning twilight continues from the 29th of Jauuary to the 21st of March (the beginning of the longest day) being fifty-one days. From the 23d of September to the 21st of March is 179 days, from which deduct 102 (=51 X2,) the remainder is 77 days, the duration of total darkness at the north pole; but even during this short period, the moon and the Aurora Borealis shine with uncommon splendour. PROBLEM XXXVIII. To find in what climate any given place on the globe is situated. Rule. 1. If the place be not in the frigid zones, find the length of the longest day at that place (by problem XXVIII.) and subtract twelve hours therefrom; the number of half hours in the remainder will show the cli mate. 2. If the place be in the frigid zone,* find the length of the longest day at that place (by Problem XXX,) and if that be less than thirty days, the place is in the twenty-fifth climate, or the first within the polar circle; if more than thirty and less than sixty, it is in the twenty-sixth climate, or the second within the polar circle; if more than sixty and less than ninety, it is in the twenty-seventh climate, or the third within the polar circle, &c. Examples. 1. In what climate is London, and what other remarkable places are situated in the same climate? Answer. The longest day at London is 16 hours, if we deduct 12 therefrom, the remainder will be 44 hours, or nine half hours; hence, London is the 9th climate north of the equator: and, as all places in or near the same latitude are in the same climate, we shall find Amster dam, Dresden, Warsaw, Irkoutsk, the southern part of the peninsula of Kamtschatka, Nootka Sound, the south of Hudson's Bay, the north of Newfoundland, &c. to be the same climate as London. The learner is requested to turn to the note to Definition 69th, page 15. 2. In what climate is the North Cape in the island of Maggeroe, latitude 71° 30′ north? Answer. The length of the longest day is 77 days; these days divided by 30, give two months for the quotient, and a remainder of 17 days; hence, the place is in the third climate within the polar circle, or the 27th climate reckoning from the equator. The southern part of Nova Zembla, the northern part of Siberia, James' island, Baffin's Bay, the northern part of Greenland, &c. are in the same climate. 3. In what climate is Edinburgh, and what other places are situated in the same climate? 4. In what climate is the north of Spitzbergen? 5. In what climate is Cape Horn? * The climates between the polar circles and the poles were unknown to the ancient geographers; they reckoned only seven climates north of the equator. The middle of the first northern climate they made to pass through Meroe, a city of Ethiopia, built by Cambyses, on an island in the Nile, nearly under the tropic of Cancer; the second through Syene, a city of Thebais in upper Egypt, near the cataracts of the Nile; the third through Alexandria; the fourth through Rhodes; the fifth through Rome or the Hellespont; the sixth through the mouth of the Borysthenes or Dnieper, and the seventh through the Riphæan mountains, supposed to be situated near the source of the Tanais or Don river. The southern parts of the earth being in a great measure unknown, the climates received their names from the northern ones, and not from particular towns or places. Thus the climate which was supposed to be at the same distance from the equator southward, as Meroe was northward, was called Antidiameroes, or the opposite climate to Meroe; Antidiasyenes was the opposite climate to Syenes, &c. 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 2310 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. † 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 Adams' globes, the meridian passing through Libra is divided into degrees, in the same manner as the brass meridian is divided; the hori zon will, therefore, cut this meridian in the several degree 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 upon 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 equation of time. If the number Note. The equation of time, or differénce beween the time shown by a well regulated clock, and a true sun-dial depends upon two bauses; 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 from the point Aries, and move over equal places 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 degrees in these signs at the right hand: in any of these signs the sun is slower than the clock. Thus, 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 it is 6 minutes 2 seconds slower than the clock. 9 2 569 366 51 21 10 3 169 41 6 35 20 247 99 33 2 9 29 8 189 560 22 9 3 Sun slower than the clock in 1 of 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. Examples. 1. What is the equation of time on the 17th of July? Answer. The degrees on 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 dependent on the obliquity of the ecliptic on the 27th of October? 4. When the sun is 18° of Aries, what is the equa tion of time? PROBLEM XLI. To find the sun's meridian altitude at any time of the year at any given place. 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 on the brass meridian (the nearest way) to the horizon; these degrees will show the sun's meridian altitude.t Note. The sun's altitude may be found at any particular hour, in the following manner. Find the sun's declination, and elevate the pole to that declination; bring the given place to the brass meridian and set the index to 12; then, if the given time be before noon, turn the globe westward as many hours as the time wants of noon; if the given time be past noon, turn the globe eastward as many hours as the time is past noon. Keep the globe fixed in this position, and screw the quadrant of altitude on the brass meridian over the sun's declination; bring the graduated edge of the quadrant to coincide with the given place, and the number of degrees between that place and the horizon will show the sun's altitude. OR, Elevate the pole so many degrees above the horizon as are equal to the latitude of the place; find the sun's * 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. |