bearing cannot be taken. Indeed, any two omissions may always be supplied by calculation. It is far better, however, if possible, to take all the notes on the field. For, when any of them are supplied by calculation, there are no test by which the accuracy of the work can be ascertained, and all the errors of the notes affect also the parts which are supplied. 1. In a survey we have the following notes. What is the bearing and distance from station 3 to 4. Ans. Bearing, S 36° E. 2. In a survey we have the following notes: What is the bearing and distance from 3 to 4? , 2910 , 2.21 ch To determine the angle included between any two courses, when their bearings are known. AC is N 26° W AB is N 46° E AC is N 26° W CAD 180°-92o—88° = AC is N 26° W CAF 180°-40°=1400 When the meridional letters are alike, and those of departure also alike, the difference of the bearings will be the angle between the courses. When the meridional letters are alike, and those of departure unlike, the sum of the bearings will be the angle between the courses. When the meridional letters are unlike, and those of departure alike the angle between the courses will be equal to 180° minus the sum of the bearings. When the meridional letters are unlike, and those of departure also unlike, the angle between the courses will be equal to the difference of the bearings taken from 180o. REMARK. The above rules are determined, under the supposition that the two courses are both run from the angular point. Hence, if it be required to apply the rules to two courses run in the ordinary way, as we go around the field, the bearing of one of them must be reversed before the calculation for the angle is made. 1. The bearings of two courses, from the same point, are N 37° E, and S 85° W: what is the angle included between them? 2. The bearings of two adjacent courses, in going round a piece of land, are N 39° W, and S 48° W: what is the angle included between them? Ans. 87°. 3. The bearings of two adjacent courses, in going round a piece of land, are S 85° W, and N 69° W : what is the angle included between them? Ans. 154°. 4. The bearings of two adjacent courses, in going round a piece of land, are N 55° 30′ E, and S 69° 20' E: what is the angle included between them? น PROBLEM. Ans. 124° 50'. To run a line from a given point in the boundary of a piece of land, so as to cut off on either side of it a given portion of the field. 150. Make a complete survey of the field, by the rules already given. Let us take, as an example, the field whose area is computed at page 106. That field contains 104A 1R 16P, and the following is a plot of it. Let it now be required to run a line from station A, in such a manner as to cut off on the left any part of the field; 26A 2R 31P. say, It is seen, by examining the field, that the division line will probably terminate on the course CD. Therefore, draw a line from A to C, which we will call the first closing line. The bearings and lengths of the courses AB, BC, are table on page 106: hence, the bearing and distance from C to A, can be calculated by the last problem: they are in this example, Bear. S 9° 28' E: Course 22.8 ch. Having calculated the bearing and length of the closing line, find, by the general method, the area which it cuts off: that area, in the present case, is 13A 3R 3P. It is now evident that the division line must fall on the right of the closing line AC, and must cut off an area ACH, equal to the difference between that already cut off, and the given area : that is, an area equal to 26A 2R 31P given area. 13A 3R 3P area already cut off. 12A 3R 28P. Since the bearing of the next course CD, and the bearing of the closing line AC are known, the angle ACD which they form with each other, can be calculated, and is in this example 80° 32°. Hence, knowing the hypothenuse AC, and the angle ACG at the base, the length AG of the perpendicular let fall on the course CD, can be found, and is 22.49 chains. Since the area of a triangle is equal to its base multiplied by half its altitude, it follows, that the base is equal to the area divided by half the altitude. Therefore, if the area 12A 3R 28P be reduced to square chains, and divided by 11.244 chains, which is half the perpendicular AG, the quotient, which is 11.58 chains, will be the base CH. Hence, if we lay off from C, on CD, a distance CH, equal to 11.5 chains, and then run the line AH, it will cut off from the land the required area. REMARK I. If the part cut off by the first closing line, should exceed the given area, the division line will fall on the left of AC. REMARK II. If the difference between the given area and the first area cut off, divided by half the perpendicular AG, line from A to D, and consider it as the first closing line, and let fall a perpendicular on DE. REMARK III. When the point from which the division line is to be drawn, falls between the extremities of a course, dividing the course into two parts, consider one of the parts as an entire course, and the other as forming a new course, having the same bearing. The manner of making the calculation will then be the same as before. Method of determining the area of a Survey by means of the Table of Natural Sines and Cosines. If, in a circle of which the radius is 1, we calculate the sine and cosine for every minute of the quadrant, they form what is called a Table of Natural Sines and Cosines. The natural sine is the perpendicular, and the natural cosine the base of a right angled triangle of which the hypothenuse, or radius of the circle, is 1. Since either leg of a right angled triangle is less than the hypothenuse, it follows that the natural sine or cosine of every arc of the quadrant is less than 1. These sines and cosines are expressed in decimals of the radius 1, and although the decimal point is not written in the table, yet it must always be prefixed to the number before using it. Thus in page 69, the sine of 5° 30′ Sine of 40° 25' (page 73) Cosine of 40° 25' is .09585. 5o 30' .99540. .64834. .76135. When the angle exceeds 45°, the degrees are found at the bottom of the page, and the minutes are counted upwards in the right hand column of the page, as in the table of logarithmic sines. Thus, sine of 84° 20' (page 66) is The cosine of 84° 20′ .99511. 99 .09874. .98362. |