How many miles is the Cape of Good Hope South and East of Philadelphia ? In the middle latitude 37°, a degree is 48 miles, then 60: 48: 5619: 4495 miles. Or, 98°×48=4485 G. ms. Ans. 4440 G. miles, S. and 4495 G. miles E. The square root of the sum of the squares of the difference of latitude and departure will give the distance between them. The ship Columbus was spoken in lat. 37° 57′ N. and long. 67° 38′ W. bound to a port in lat. 42° 17' N. and long. 72° 38' W. What is the distance to the port? Ans. 347. To find the bearing of one place from another by Arithmetic. RULE. To the distance, as found in the last case, add half the greater of the other two numbers, whether it be latitude or departure, and say,-As the sum is to the less number, so is 86 to the degrees, or angle required, if the difference of latitude be more than the departure. But if the departure be greater, the fourth number, found as above, must be taken from 90° for the answer, EXAMPLE. The city of A. is north and west of the town of B. The difference of latitude between them is 260 miles, the departure 230, and the distance 347 miles. Required the bearing of A. from B. 347 130 477: 230 :: 86 86 1380 1840 477)19780(411908 Taking or of large numbers, will be sufficiently exact for geographical purposes thus, 34 : 13 47: 23:: 86 23 258 Suppose the difference of latitude between two places is 1970 miles; the difference of longitude, reduced to departure, is 955 miles, and the distance deduced thence 2139: Required the bearing? 21 9 Supposing 1970 miles to be departure, the answer would be 90-26-64°. 46 60 64 Bearing. Ans. 36° 17 27 54 73 - 33 When the difference of latitude and departure are alike, or nearly so, the answer is 45°, and needs no calculation to find it; as in the example from Philadelphia to the Cape of Good Hope, which in that case is 45°, or S. E., as the Cape is South and East of that city. When there is little or no departure, the bearing is North or South; and when it is so with the latitude, the bearing is East or West. A TABLE OF ANGLES, which every Point and Quarter Point of the Compass makes with the Meridian. To reduce degrees of longitude into time. RULE. Multiply the degrees by 4 for minutes of time, and the minutes or miles of longitude by 4 for seconds, and divide the products by 60 for time. EXAMPLE. Required the time answering to 42° 43′ 45′′ of longitude. hrs. ms. sec. 42X4 168÷6C=2 48 0 0 3 Ans. 2 50 55 To reduce time into degrees of longitude. RULE. Multiply the hours by 15 for degrees, and divide the minutes and seconds respectively by 4. EXAMPLES. 1. Reduce 8 hrs. 37 ms. 38 sec. to degrees of longitude. 8X15=120° 0' 37÷4= 9 15 384 0 9 0' 0 30 Ans. 129 24 30 2. What is the difference of longitude and time between Philadelphia, in longitude 75° 19 W. and Alexandria in Egypt in 30° 16' East? Philadelphia 75° 19′ W. Alexandria 30. 16 E. Diff. long. 105 35 105 X4420' 0" 60)422 20 105° 35' hrs. 7 2 20 Hence, when it is noon at Philadelphia, it is 7 hrs. 2 ms. 20 sec. in the afternoon at Alexandria; and when it is noon at Alexandria, it is only 4 hrs. 57 ms. 40 sec. in the morning at Philadelphia-thus, 12 0 0 7 2 20 hrs. 4 57 40 NOTE. When the place required is east of the one given, the time required will be as much later in the day, as the difference of longitude, reduced to time, gives hours and minutes; and when west of the given place, it will be so much earlier. When it is 20 minutes past 2 o'clock in the afternoon at Marseilles, in longitude 5° 22′ E. what time is it at Boston light-house, in longitude 70° 58' W. Ans. 9 hrs. 14 ms. 40 sec. in the morning. 14 To find the part of the Globe opposite to a given place. RULE. Take the same latitude of an opposite name for the place required, and subtract the longitude of the given place from 180°, the remainder is the longitude of the place required, but of an opposite name. EXAMPLES. 1. Required the opposite part of the globe to London, in latitude 51° 31' N. and near the first meridian. London lat. 51° 31' N. long. 0° 00′ 2. Required the opposite part of 180 00 the globe to Boston. long. 108 57 E. 3. Suppose a ship in any of the following placés, what place on the globe would be opposite to it? To estimate the distance of objects at sea. Multiply the square root of the height in feet by 114, and point off the unit's figure in the product for decimal parts, the remainder is the miles required-thus, Height. Sq. Root. Miles. At 16 feet X 111 4,6 the distance. |