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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

13.A 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

26A 2R 31P given area.

13A 3R 3P area already cut off. to .

12A 3R 28 P. 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

12.A 3R 28P be reduced to square chains, and divided by 11.24chains, 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' is

.09585. The cosine of

5° 30'

.99540. Sine of 40° 25' (page 73)

.64834. Cosine of 40° 25'

.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

.99511. The cosine of 84° 20'

.09874. Sine of 79° 37' (page 67)

.98362. Cosine of 79° 37'

.18023. Sine of 69° 25'

.93616. Cosine of 69° 25'

.35157. Sine of 57° 59'

.84789.

If the Surveying Compass has a vernier which enables you to read the bearings to smaller parts of a degree than 15', greater accuracy may be attained by using the table of natural sines, instead of the Traverse Table, for computing

the area.

We shall now show the method of calculating the latitude and departure of any course, from the table of natural sines.

D Let AD, for example be any course, DAE the bearing, and AC=1 the radius of the table of natural sines.

Having formed the right angled
triangles ACB, ADE, we have DAE=bearing,

AE=dif. of latitude and ED=departure,
AB=cosine of bearing and BC=sine of bearing.
From similar triangles, we have,

AC AB AD : AE; that is,
1 : cosine of bear. :: course : dif. of lat. ; hence,
dif. of latitude=course x cosine of bearing ; that is ;

The difference of latitude is equal to the length of the course multiplied by the cosine of the bearing. Again,

AC : CB :: AD : DE; that is,

1 : sine of bearing :: course : departure; hence,

departure=course x sine of bearing, that is, The departure is equal to the length of the course multiplied by the sine of the bearing.

Ex. 1. Find, froin the Table of natural sines, the latitude and departure of the course 49 yards and bearing 35° 18' Natural cosine of 35° 18'

.81614 Length of the course

49 Product, which is the dif. of latitude

39.99086.

O

.57786

49

Natural sine of 35° 18'
Length of the course
Product, which is the departure

28.31514,

2. The bearing is 65° 39', the course 69.41 chains: what is the latitude, and what the departure ? Natural cosine of 65° 39'

.41231 Length of the course

69.41 Product, which is the Dif. of Latitude

28.6184371, Natural sine of 65° 39'

.91104 Length of the course

69.41 Product, which is the Departure

63.2352864. 3. The bearing is 75° 47', the course 89.75 chains : what is the latitude, and what the departure ? Natural cosine of 75° 47'

.24559 Length of course

89.75 Product, which is the Dif. of Latitude 22.0417025. Natural sine of 75° 47'

.96937 Length of course

89.75 Product, which is the Departure

87.0009575.

4. Find the area of a piece of land from the following notes.

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Calculating the latitude and departure of each course by the rules already given, we have

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8 s 74° 35' W 21.

5.58264
20.24442

5.61385

20.24442 1113.37662 112.043091 135.91501|135.93221|112.70986)112.709851135.92361|135.92361 1112.043091

135.915011

Error in Southing
Half Error

1.33353
0.66676

0.01720 Error in Easting.
0.00860 Half Error.

Instead of balancing by the method explained in Art. 138, we divide each error by two. Now if we subtract half the error in southing from the column of northings and at the same time add it to the column of southings, these two columns will exactly balance. In like manner, if we subtract half the error in easting from the column of westings and at the same time add it to the column of eastings, these columns will also balance.

The errors should be distributed in proportion to the lengths of the courses, but this may be done with sufficient accuracy without making the proportions. If any of the courses have been run over rough ground, the probability is that the errors belong to these courses and they should be distributed among them.

In this example we separate the half error in southing into the three parts .40700, .21302 and .04674, and subtract them respectively from the northings of courses 2, 1 and 3, and then place the northings in the balanced columns. For the southings, we separate the error into the four parts .40772, .20031, .03121, and .02752, and add them respectively to the southings of the courses 4, 5, 8 and 7. We then enter the southings in the balanced columns. As the error in easting is so small we add half of it to the easting of course 3, and subtract half

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