PROBLEM III. Given the three Sides of a Triangle; to find its Area, RULE. Add the three sides together, and take half their sum; subtract each side severally from that half sum, noting the remainders: then, To the logarithm of the half sum add the logarithms of the three remainders; now, the sum of these four logarithms, being divided by 2, will give the area of the triangle. Example. Let the three sides of a triangle be 433, 312, and 205 yards respectively; required its area? Given the Diameter of a Circle; to find its Circumference, and conversely. RULE. To the logarithm of the diameter add the constant logarithm 0. 497150, and the sum will be the logarithm of the circumference. And, to the logarithm of the circumference add the constant logarithm 9.502850, and the sum will be the logarithm of the diameter. Example 1. The earth's diameter is 7917.5 miles; required its circumference? Example 2. If the circumference of the earth be 25000 miles, what is its diameter? Note. The circumference of a circle whose diameter is unity or 1, is 3.14159265; and, since the circumferences of circles are to each other as their diameters, or radii,-therefore, as the diameter 1, is to its circumference 3. 14159265; so is the diameter of any other circle, to its circumference: and hence the above Rule. The constant logarithm is expressed by the logarithm of 3.14159265, or log. arithmetical complement of that number. PROBLEM V. Given the Diameter, or the Circumference of the Earth; to find the whole Area of its Surface. RULE. Since the surface of a sphere or globe whose diameter is unity or 1, is 3.14159265, the logarithm of which is 0.497150, and its arithmetical complement 9. 502850; and since the surfaces of spheres are to each other as the squares of their diameters; therefore, To twice the logarithm of the earth's diameter add the constant logarithm 0. 497150, and the sum will be the logarithm of the area of the earth's surface, in square miles. Or, To twice the logarithm of the earth's circumference add the constant logarithm 9.502850, and the sum will be the logarithm of the earth's surface, in square miles. Example 1. Required the area or superficial measure, in square miles, of the whole of the earth's surface, allowing its diameter to be 7917.5 English miles? Example 2. Required the area or superficial measure of the whole of the earth's surface, in square miles, allowing its circumference to be 24873 English miles? Circumference of the earth = 24873 Twice its log. To the logarithm of the degrees in the given arc, considered as a natural number, add the logarithm of the radius of that arc, and the constant logarithm 8.241877; the sum will be the logarithm of the length of the arc. Example. Required the length of an arc of 45 degrees, the radius being 9 inches? Note. The circumference of a circle whose diameter is unity or 1, is 3.14159265. And since the circumference of a circle contains 360 degrees; and that the measure of the longest arc thereof is that which corresponds to the semi-circle :—therefore, as 180° : 3.14159265::1°, to 0.0174533; the logarithm of which is 8.241877; which is the true constant logarithm for determining the length of the arc of a circle corresponding to any given number of degrees, and parts of a degree. PROBLEM VII. Given the Length of an Arc of a Circle, and the number of Degrees contained therein; to find its Radius. RULE. To the logarithm arithmetical complement of the number of degrees contained in the given arc, esteemed as a common number, add the logarithm of its length, in inches, and the constant logarithm 1.758123; the sum (abating 10 in the index) will be the length of the radius in inches. Note. The constant logarithm made use of in this Rule is the arithmetical complement of that which is used in the preceding Rule. Example. Let the length of the arc of a circle be 10.82 inches, and the number of degrees contained therein 40; required the length of the radius that would sweep that arc? Given number of degrees = Constant logarithm = 40 Log. arith. comp. 8. 397940 Log. = 1.034270 Length of the radius, in inches = 15.5 Logarithm = PROBLEM VIII. Given the length of an Arc of a Circle, and its Radius; to find the number of Degrees contained therein. RULE. To the logarithm arithmetical complement of the length of the radius of the given arc, add the logarithm of the length of the arc, and the constant logarithm 1.758123; the sum, abating 10 in the index, will be the logarithm of the number of degrees contained in the given arc. Note.--The constant logarithm made use of in this Rule is the arithmetical complement of that which is made use of in Problem VI. Example. Let the length of the arc of a circle be 10.82 inches, and its radius 15.5 inches; required the number of degrees contained therein? Number of degrees in the arc =40. . Logarithm = 1.602061 -- . Remark. The two last Problems will be found useful in adapting Pendulums to given spaces, on board a man-of-war, for showing the heel. actual number of degrees that a ship may down. PROBLEM IX. To find the Length of a Pendulum for Vibrating Seconds at the Equator, and also at the Poles of the World. RULE. Since the circumference of a circle whose diameter is unity or 1, is found by computation to be 3. 14159265; and since a pendulum vibrates in the arc of a circle, or cycloid, the radius of which is equal to the length of the pendulum from the centre of oscillation; therefore, if twice the space passed through by a falling body in one second of time, be divided by the square of the above circumference, the quotient will be the length of the pendulum for vibrating seconds. Now, it has been found, taking the mean of many experiments, that a heavy body let fall at the equator will descend, by the force of gravity, 16.04436 feet in one second of time :-hence, by logarithms, 16.04436 ft.=192.53232 inches x 2=385.06464 Logarithm=2.5855336 Circum. of a circle to diameter 1-3.14159265 Tw. its log.=0.9942998 Length of the pendulum, in inches 39.0152, Logarithm=1.5912338; which, therefore, is the correct length of the pendulum for vibrating seconds on all points of the equator. Now, since the ratio of the earth's equatorial semidiameter to its polar semi-axis (taking the mean of many results) is as 305 to 304. 212 ;* and, since gravity decreases in proportion to the square of the distance, and conversely; therefore it will be,-As the square of the polar semiaxis is to the square of equatorial radius, so is the length of the pendulum at the equator, to the length of the pendulum for vibrating seconds at the poles of the world.-Hence, by logarithms, *This is given in round or whole numbers, page 337, as 305 to 304. |