THIRD METHOD. 100. Take any distance, on the tape or chain, place one extremity at D, (Fig. 14) and fasten the other, on either side of AD, as at E. Plant a staff at E, take the extremity D of the chain, and carry it round, till it comes on the line DA at A. Place a staff at A, and carry the extremity D of the chain around, until it falls at F, on the line AE produced: then ADF is the right angle sought, being an angle in a semicircle (128)*. a PROBLEM. 101. To ascertain the horizontal distance from a given point to an inaccessible object. FIRST METHOD. a Let A (Pl. I. Fig. 15) be an inaccessible object, and B the point from which the distance is to be estimated. Measure directly from the object A, the horizontal line BE, of any convenient length; then lay off at E a right angle AED, and measure in the direction ED, any convenient distance to D. Place a staff at D; then, at the point B, lay off the right angle ABC, and measure along the line BC till a staff placed at C falls on the line DA. Now, DF being equal to DE-CB, is known: Hence, by similar triangles, DF:FC :: DE: EA: in which proportion all the terms are known except the fourth, which may, therefore, be regarded as found; hence BA is known. a a SECOND METHOD. 102. At the point B, (Fig. 18) lay off BE perpendicular to the line BA, and measure along it any convenient distance BE. At E lay off the right angle BED, and measure any distance in the direction ED. Having placed a staff at the extremity D, place a second one at C, on the line BE, and also on the line DA, and measure either the distance BC or CE. The triangles CDE and CAB being similar, we have CE:ED:: CB: AB, in which all the terms are known except the fourth, THIRD METHOD. 103. Measure any horizontal base line, as BC (Fig. 16). Then, having placed staves at B and C, measure any convenient distances BD and CE, such, that the points D, B, and A, shall be in the same right line, as also, the points E, C, and A; then measure the diagonal lines DC and EB Now, in the triangle BEC, the three sides are known, therefore, the angle ECB can be found (62). In the triangle CDB, the three sides are also known, and the angle CBD may be determined (62). These angles being respectively subtracted from 180°, the two angles ACB and ABC become known; and hence, in the triangle ABC, we have two angles and the included side, to find the side BA. PROBLEM. 104. To find the altitude of an object, when the distance to the vertical line passing through the top of it is known. Let CD (Fig. 17) be the altitude required, and AC the known distance. From A, measure on the line AC, any convenient distance AB, and place a staff vertically at B. Then placing the eye at A, sight to the object D, and let the point, at which the line AD cuts the staff BE, be marked. Measure the distance BE on the staff; then say, As AB: BE :: AC: CD, then, CD becomes known. If the line AC cannot be measured, on account of intervening objects, it may be ascertained by calculation, as in the last problem, and then, having found the horizontal distance, the vertical line is readily determined as before. CHAPTER III. OF THE CONTENT OF GROUND. а 105. It being one of the objects of surveying to ascertain the content or area of given pieces of land, it is important to have a distinct idea of the manner in which this area is to be estimated; and then, we can understand clearly the methods to be pursued, to find the data necessary for the computation. The surface of the ground, being in general so broken and uneven, as to render it impossible, without great trouble and expense, to ascertain its true content, the method of referring it to a horizontal plane has been generally adopted; and now, in estimating the content, we consider the bounding lines as horizontal, and the whole surface reduced to a horizontal plane. This manner of estimating land being established, the sum of the areas of all the parts, into which a tract of land may be divided, is equal to the area, estimating it as an entire piece; which would not be the case, if the areas of the parts had reference to the actual surface, and the area of the whole were calculated from its bounding lines. 106. In surveying, there are two kinds of measurement, linear, and superficial; and each of them has its appropriate unit of measure. The unit of measure of a quantity is an ascertained magnitude of the same kind with the quantity measured: thus, for a line, the unit of measure is a line of a given length, a foot, a yard, a rod, &c.: and the length of a line is known, when we know its unit of measure, and the number of times which the line contains it. For superficies, or areas, the unit of measure is an area of known dimensions, a a 7 du ed linear measure, is generally used. If, therefore, the linear dimensions of ground be estimated in feet, yards, rods, or chains, the superficial measure will be easiest found in square feet, square yards, square rods, or square chains, and when so determined, the number expressing the area will be nothing else, than the number of times which the unit of superficial measure is contained in the land measured. 307. It has already been observed (81) that Gunter's chain of four rods, or 66 feet in length, and which is divided into 100 links, is the chain in general use among surveyors. We shall, therefore, take the length of this chain for the unit of linear measure, and consequently, the square, of which it is the side, for the unit of superficial measure: this unit is generally called a square chain. 108. An acre is a surface equal in extent to ten square chains, or equal to a rectangle, of which one of the sides is 10 chains in length, and the other, one. A rood is one quarter of an acre. A chain being four rods in length, it follows that a square chain contains 16 square rods; and therefore, an acre, or 10 square chains, contains 160 square rods, and a rood, 40. The square rods are sometimes called perches. Land is generally computed in acres, roods, and perches. 109. The dimensions of a survey being taken in chains and links, or in chains and decimals of a chain,(81) it becomes necessary to show, how the area is to be found in acres, roods, and perches, or square rods. Now, 1 square chain=10,000 square links; and consequently, 10 square chains, or one acre,=100,000 square links. If, therefore, there be a rectangle, the sides of which are links, the area is found in acres and decimals of an acre, by multiplying together the sides, which gives the number of square links; and then, dividing the product by 100,000, or, what is the same thing, pointing off five decimal places from the right hand. If the decimal part be then multiplied by 4, and the same number of places pointed off in the product, the figures on the left of the decimal point, will show the roods; and, if the decimal of figures pointed off from the right hand in the product, the result will indicate the perches and decimals of a perch. If one of the dimensions be in links and the other in chains, the chains may be regarded as links, by multiplying them by 100, or annexing two ciphers; or, if the multiplication be made without annexing the ciphers, the product is reduced to acres and decimals of an acre, by pointing off three places of decimals in the product. If both the dimensions be in chains, the product is reduced to acres by dividing by 10, or pointing off one decimal place. From all which, we conclude, 1. That, if links be multiplied by links, the product is reduced to acres by pointing off five decimal places from the right hand. 2. That, if chains be multiplied by links, the product is reduced to acres by pointing off three decimal places from the right hand. 3. That, if chains be multiplied by chains, the product is reduced to acres by pointing off one decimal place from the right hand. 110. As there are 16.5 feet in a rod, a square rod is equal to 16.5 x 16.5=272.25 square feet. If this number be multiplied by 160, we shall have 272.25 x 160=43560.00, equal to the number of square feet in an acre. As there are 9 in a square yard, if the number 43560.00 be divided by 9, the quotient, 4840, is the number of square yards in an acre. square feet PROBLEM. 111. To find the area of a square or rectangular piece of ground. Multiply the two sides together ; the product is the required area (173)* Remark. If the sides be estimated in rods, their product will express the area in square rods or perches: then, dividing the product by 160, and the remainder again by 40, will give the area in acres, roods, and perches. If the sides be estimated in chains and links, the manner of reducing their product to |