point where the work ends; in which case, if the error be small, it may be divided among the nearer lines; if it be large, it shows a mistake somewhere, which must be rectified.

192. We are now to compute the area, and will begin with that portion of the survey that was made with the theodolite.

The area ADCBEA

=7062.555 square rods,

which being divided by 160 (108), gives for the entire area

OF SURVEYING IN GENERAL.

115

We are next to ascertain the area of that portion of the land surveyed with the compass. Of this area, a part lies without the lines traced with the compass, and a part within them. The latter portion will be first computed. There are two methods by which this area may be calculated; first, by dividing it into triangles, trapezoids, &c.; and secondly, by the general method of computing surveys made with the compass. If the latter be chosen, and it is considered preferable, the bearings of the lines BC, CD, and DA, must be found; for which there are already sufficient data.

The bearing from A to D is due north; the angle DAB is 90° 22′ 04′′; therefore, the bearing from A to B is S. 89° 38′ W. nearly; or from B to A, N. 89° 38′ E. (168). But the angle ABC=63° 44'; therefore, the bearing of C from B is N. 25° 54' E., and the distance BC=61.337 rods, or 1012 feet. The bearing of B from C is S. 25° 54′ W., and the angle BCD being equal to 123° 39′, the bearing of D from C is N. 82° 15' E., and the distance CD=70.854 rods, equal to 1169 feet. The bearing from A to D being north, the bearing from D to A is south, the distance is 1057.8 feet.

The bearings and distances being known, we form the following table.

We are next to compute the area which lies without the lines traced with the compass.

Beginning at station A, and regarding the lines which join the extremities of the offsets as right, which we may do without any sensible error, the area of the trapezoids

To compute the area of the quadrilateral legh.

Join 1 and g. Then, in the right-angled triangle 1gh, there are known 1h=140, and hg=40: the remaining parts can therefore be found; they are, the angle glh=15° 56′ 43′′, the hypothenuse 1g=145.62, and its area =2800.

But the bearing from A to 1 is N. 27° E., consequently, from 1 to A, S. 27° W. (168); and since le is perpendicular to Al, its bearing is known, viz. S. 63 E. The bearing of the line 1 2 is N. 11 E.; hence the angle hle is known, it being equal to 180°- (63°+11° 30′)=105° 30'; and therefore the angle gle is known, being equal to 105° 30′-15° 56′ 43′′= 89° 33′ 17′′. Therefore, the area of the triangle gle can be found, two sides and the contained angle being known; it is equal to 8007.9 square feet.

The areas of the quadrilaterals 2ikl, at station 2, and 3nmp at station 3, are computed in a similar manner by dividing them into the triangles 2ik, 2kl, and 3nm, 3mp; and the same for the quadrilaterals at stations 4, 5, 6, and 7.

The areas of the trapezoids and quadrilaterals will be arranged with reference to the lines to which they correspond, and for the sake of convenient reference, will be numbered

193. We are yet to find the area of the ground surveyed with the plain table. This is done by dividing it into triangles, and measuring their bases and perpendiculars by means of the scale of equal parts. Thus, in the triangle BFM (the line BGM being nearly right), by applying the base BM to the scale of equal parts, it is found to be equal to 720 feet, and by apply. ing the perpendicular Ff, it is found equal to 220 feet; and similarly, MQ=860, QW=250, BE, before found, =1623.65, EO=385, RN=470, OS=418, RS=310, TR=320, SS'= 130, and AA"=210.

The area of the triangle BMF= 79200 square feet.