Ex. 2. Wanting to know the distance between two inaccessible objects, C and D, I measured a base-line, AB, 28.76 rods, and found the angle CAB=33°, CAD=66°, DBA=59° 45', and DBC=76°. What is the distance from C to D? Ans., 97.696 rods. THE DETERMINATION OF AREAS. (151.) The area or content of a tract of land is the horizontal surface included within its boundaries. When the surface of the ground is broken and uneven, it is very difficult to ascertain exactly its actual surface. Hence it has been agreed to refer every surface to a horizontal plane; and for this reason, in measuring the boundary lines, it is nec essary to reduce them all to horizontal lines. The measuring unit of surfaces chiefly employed by surveyors is the acre, or ten square chains. One quarter of an acre is called a rood. Since a chain is four rods in length, a square chain contains sixteen square rods; and an acre, or ten square chains, contains 160 square rods. Square rods are called perches. The area of a field is usually expressed in acres, roods, and perches, designated by the letters A., R., P. When the lengths of the bounding lines of a field are given in chains and links, the area is obtained in square chains and square links. Now, since a link is of a chain, a square link will be TXT of a square chain; that is, Too of a chain. Hence we have the following 1 100 TABLE. 1 100 1 square chain=10,000 square links. 1 1 acre 10 square chains=100,000 square links. If, then, the linear dimensions are links, the area will be expressed in square links, and may be reduced to square chains by cutting off four places of decimals; if five places be cut off, the remaining figures will be acres. If the decimal part of an acre be multiplied by 4, it will give the roods, and the resulting decimal, multiplied by 40, will give the perches. (152.) The difference of latitude, or the northing or southing of a line, is the distance that one end is further north or south than the other end. Thus, if NS be a meridian passing through the end A of the line AB, and BC be perpendicular to NS, then is AC the difference of latitude, or northing of AB. The departure, or the easting or westing of a B .ine, is the distance that one end is further east or west than the other end. Thus BC is the departure or westing of the line AB. It is evident that the distance, difference of latitude, and departure form a right-angled triangle, of which the distance is the hypothenuse. N A The meridian distance of a point is the perpendicular let fall from the given point on some assumed meridian, and is east or west according as this point lies on the east or west side of the meridian. The meridian distance of a line is the distance of the middle point of that line from some assumed meridian. (153.) When a piece of ground is to be surveyed, we begin at one corner of the field, and go entirely around the field, measuring the length of each of the sides with a chain, and their bearings with a compass. Plotting a Survey. When a field has been surveyed, it is easy to draw a plan of it on paper. For this purpose, draw a line to represent the meridian passing through the first station; then lay off an angle equal to the angle which the first side of the field makes with the meridian, and take the length of the side from a scale of equal parts. Through the extremity of this side draw a second meridian parallel to the first, and proceed in the same manner with the remaining sides. This method will be easily understood from an example. EXAMPLE 1. Draw a plan of a field from the following courses and distances, as given in the field-book. N B Draw NS to represent a meridian line; in NS take any convenient point, as A, for the first station, and lay off an angle, NAB, equal to 45°, the bearing from A to B, which will give the direction from A to B. Then, from the scale of equal parts, make AB equal to 9.30, the length of the first A side; this will give the station B. Through B draw a second meridian parallel to NS; lay off an angle of 60°, and make the line BC equal to 11.85. Proceed in the same E D manner with the other sides. If the survey is correct, and the plotting accurately performed, the end of the last side, EA, will fall on A, the place of beginning. This plot is made on a scale of 10 chains to an inch. (154.) To avoid the inconvenience of drawing a meridian through each angle of the field, the sides may be laid down from the angles which they make with each other, instead of the angles which they make with the meridian. Reverse one of the bearings, if necessary, so that both bearings may run from the same angular point; then the angle which any two contiguous sides make with each other may be determined from the following RULES. 1. If both courses are north or south, and both east or west, subtract the less from the greater. 2. If both are north or south, but one east and the other west, add them together. 3. If one is north and the other south, but both east or west, subtract their sum from 180°. 4. If one is north and the other south, one east and the other west, subtract their difference from 180°. Thus the angle CAB is equal to N cordingly find the angle ABC-105°. BCD=100°. DEA=101°. EAB=104°. CDE=130°. With these angles the field may be plotted without drawing parallels. EXAMPLE 2. The following field notes are given to protract the survey: (155.) The accompanying traverse table shows the difference of latitude and the departure to four decimal places, for distances from 1 to 10, and for bearings from 0° to 90°, at intervals of 15'. If the bearing is less than 45°, the angle will be found on the left margin of one of the pages of the table, and the distance at the top or bottom of the page; the difference of latitude will be found in the column headed Lat. at the top of the page, and the departure in the column headed Dep. If the bearing is more than 45°, the angle will be found on the right margin, and the difference of latitude will be found in the column marked Lat. at the bottom of the page, and the departure in the other column. The latitudes and departures for different distances with the same bearing are proportional to the distances. Therefore the distances may be reckoned as tens, hundreds, or thousands, if the place of the decimal point in each departure and difference of latitude be changed accordingly. Ex. 1. To find the latitude and departure for the course 45° and the distance 93. Under distance 9 on page 141, and opposite 45°, will be found latitude 6.3640 and departure 6.3640. Hence, for distance 90, the latitude is 63.640, and adding the latitude for the distance 3, viz., 2.121, we find the latitude for distance 93 to be 65.761. Ex. 2. To find the latitude and departure for the course 60° and the distance 11.85. The latitude for 10 is 5.0000.| Departure for 10 is 8.6603. Latitude for 11.85 is 5.9250.| Depart. for 11.85 is 10.2624. Ex. 3. To find the latitude and departure for the course 20° and the distance 5.30. Ans. Latitude 4.98, and departure 1.81. The traverse table may be used not only for obtaining departure and difference of latitude, but for finding by inspection the sides and angles of any right-angled triangle; for the latitude and departure form the two legs of a right-angled triangle, of which the distance is the hypothenuse, and the course is one of the acute angles. In this manner we find the latitude and departure for each side of the field given in Example 1, page 105, to be as in the following table: |