62. LEVELLING. Levelling is the art of determining the difference between the true and the apparent level of two places, or of drawing a line a tangent to the earth's surface. The true level is a curve parallel to the surface of water in a state of rest; and since the earth is very nearly round, it may be considered to be the arc of a great circle. The apparent level is a straight line, which is a tangent to the true level at the point where the observation is made. The circumference of the circle (see ¶ 22, page 72) represents the true level, and the tangent, cl, the apparent level, c being the place of observation; and al is the difference between the true and apparent in reference to the station c, at the distance ca. Various instruments are used for the purpose of levelling; the more common of which are the square, the plumb-line, and the spirit-level. Timber may usually be levelled with sufficient accuracy by the square; and if a plumb-line be attached to a good square, so that it may hang parallel with the shorter arm, when the longer arm is brought to a horizontal line, (or parallel to the apparent level,) it makes a very good instrument for levelling timber; and it may be used for many other purposes when great accuracy is not required. The spirit-level consists of a glass tube nearly filled with spirit of wine, and enclosed in a brass tube, except the upper part. It is so formed, that when the air-bubble is at the middle point, the instrument indicates the apparent level; and by setting up rods, divided into feet and parts of a foot, perpendicular to the horizon at different stations and at equal distances from the spirit-level, and observing where the line of level or horizontal line strikes the rods, the difference of level between the different stations may be obtained; and also the true level. The true mean diameter of the earth is 7912.03 miles; 5280 7920 and if we assume the mean diameter of the earth to be 7920 miles, (which is near the truth,) then the difference between the true and apparent level for one mile will be expressed by 7920 of a mile, or 30 of a foot, which being reduced, gives of a foot, or 8 inches, for the difference, or variation between the true and apparent level at the distance of one mile. And since said difference increases nearly as the square of the distance, we may find the difference between the true and apparent level, nearly, by the following rule:— Multiply the square of the distance in miles by 8, and the product will be the required VARIATION in inches; or, multiply the square of the distance in miles by, and the product will be the variation in feet. And when the variation is given, to find the distance, we may reverse the rule thus : Divide the height, or variation, expressed in feet, by, or .6666, and extract the square root of the quotient for the distance in miles: EXAMPLES. 1. At a certain spot in a village an observation was made for the purpose of determining whether water could be brought into the village from a spring 2 miles distant; and it was found that the spring and place of observation were apparently on a level; the descent from the spring to the place of observation is required. Ans. 4 feet 2 inchesa descent sufficient to cause water to run freely. To find the descent by the sliding rule :— Thus, over 2.5 you will find and over 6 miles you will find Place .666 on C over 1 on D, (or 6 on C over 3 on D,) and over the distance in miles found on D, will be found the required variation in feet on the line C. 4.17, or 4 feet 2 inches, nearly; 24 feet for the variation; and over 10 miles, 663; over 15 miles, 150 feet; and over 30 miles, 600 feet; and if you look for the variation in feet on the line C, under the variation will be found the distance in miles on the line D. Thus, under 1 on C, at the middle of the slider, (calling the 1 one foot,) we find 1.225 miles; under 2 feet, 1.73; under 6 feet, 3 miles; under 10 feet, 3.87 miles; under 20 feet, 5.45 miles; under 100 feet, 12.25 miles; and under 1000 feet, 38.7 miles. 2. The top of a lighthouse is seen at sea from the surface of the water at the distance of 23 miles; what is its height? Ans. 352 feet. The results obtained by the above rules, though sufficiently near the truth for short distances, when great accuracy is not required, nevertheless give results which require several corrections to render them strictly accurate. The principal correction, and the only one necessary for most practical purposes, is for refraction. The rays of light in moving through the atmosphere are bent downwards, so that a height just visible at a given distance when there is refraction, would be lower than one just visible at the same distance when there is no refraction. Thus, in consequence of refraction, the distance at which an object may be seen is increased one-eleventh in ordinary states of the atmosphere; and the height of an object, or the variation, is diminished nearly one-sixth. Therefore, in finding the distance, increase the result obtained by the above rule one-eleventh; and in finding the variation, from the given distance subtract one-twelfth of itself, and then proceed as directed above, and the result will be the true height, or variation, very nearly. Or, To find the distance:-Extract the square root of nine-fifths of the height. And to find the height:-Take five-ninths of the square of the distance. To gauge the distance and height by the sliding rule, allowing for refraction:-Place 1 on C over 1.336 on D, (or 8 on C over 12 on D;) then under the height or variation in feet found on C, will be found the distance in miles on D; and over the distance in miles found on D, will be found the height in feet on C. In the following examples an allowance is made for refrac tion. 3. At what distance can an object, 24 feet high, be seen at sea from the surface of the water? Ans. 6.55, or 6.6 miles, nearly. 4. From what height will the horizon be 12 miles distant? Ans. 80, or 80.6 feet. 5. At what distance can an object, 54 feet high, be seen at sea from the surface of the water? Ans. 9.8 miles. 6. At what distance can the top of a lighthouse, 216 feet high, be seen from the surface of the ocean? Ans. 19.7 miles. 7. Required the distance of the visible horizon from the top of a hill 803 feet high. Ans. 37.9 miles. 8. From what height will the horizon be 36 miles distant? Ans. 726 feet. 9. From a ship's mast, at the height of 120 feet, the top of a lighthouse, 240 feet above the level of the sea, was just visible; required their distance from each other. Ans. 35.3, or 35.6 miles. In the last example the distance for each of the heights must be calculated separately, and their sum will be the whole distance. By the sliding rule, having set the slider, under 120 feet will be found 14.8, and under 240 feet 20.8 miles on D, and 20.8 added to 14.8=35.6 miles. 10. If from the summit of a mountain, 11310 feet high, the distance of the visible horizon is 142 miles; required the earth's diameter. Ans. 7911 miles. 11. What is the difference between the true and apparent level at the distance of of a mile? of? of? of? of 14? and of 3 miles ? Answers in order,-0.414; 1.65; 3.3; 0.103; 10.34; and 81 inches. 12. The difference of level between four stations is taken as follows, to find the difference between the extreme stations; let A be 10 feet above B, B 8 feet below C, and C 12 feet above D; the difference of level between A and D is required. Ans. 14 feet. 13. Let A be 12 feet above B, B 8 feet 3 inches below C, C 10 feet 11 inches above D, and D 3 feet 2 inches below E; what is the difference of level between A and E? Ans. A is 11 feet 6 inches above E. 63. MENSURATION OF HEIGHTS AND DISTANCES. If it be required to find the distance between two objects, one of which is inaccessible, it may be done in the following Let GN represent the bank of a river, and B an object on the opposite bank, it being required to find the width of the stream at B, viz. AB. From the point A, measure the length of a line AE, at right angles to AB, of any convenient length, say 20 yards, and at E set up a staff. Then from A, at right angles to AE, measure the line AC of any convenient length, say 62 yards. Then from C, at right angles to AC, measure a |