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189. Having calculated the principal lines of the survey, it is, in general, best to plot the work before computing the area, because the division of it into triangles, rectangles, trapezoids, &c., can be made to greater advantage when the whole is at once presented to the eye.

We shall first plot the part surveyed with the theodolite.

Place the centre of the circular protractor at V, near the centre of the paper, and draw the line A'B, passing through the points 0 and 180°. Then lay off from the 0 point, an arc of 38° 13' the direction from A to C (see notes), and draw A'C'. Lay off also the arc of 281° 37', the direction from A to E, and draw A’E. Proceed in the same manner to lay off the directions from B, viz.: BA=180", which gives B A'; BE=206 42', which gives BE; BC=116 16, which gives B C'; and BD= 146° 39, which gives B'D'. The directions from C being laid down, give C'B, and CA, before determined, as also C'D', the direction from C to D, being 172 37'. Laying off in like manner the directions from E, gives the lines E B, E'A'.

This being done, let any line, as AB, be drawn on the paper parallel to AB, to represent the base line. Now, as the line 0 and 180°, on the limb of the theodolite, was parallel at all the stations to the base line AB, the 0 point being in the direction towards A, it follows that it was constantly parallel to the line A'B' of the protractor : and hence the lines A'C', AE, &c. are respectively parallel to their corresponding lines on the ground; and thus the direction of every line is fixed without removing

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It is proposed to make the plot on a scale of 19 rods to the inch. Take a distance of one inch in the dividers, and extend the arms of the sector until the dividers will just reach across the angle, one foot standing at division 19 on one arm, and the other foot at division 19 on the other: the sector is then said to be set. Lay the sector carefully on the table; then take

. the distance, from 97 on one arm to 97 on the other, and apply it on the paper from the point A, marked as one extremity of the base line, to B, which will be the other, and AB will represent the base line. Through B draw BC parallel to B C'; take the distance BC

; from the scale, and apply it on the paper; this determines the place of station C. Through C draw CD parallel to C'D', and lay off CD; then join A and D, and B and D. Lastly, draw AE parallel to A'E', and lay off AE; this fixes the point E; then let the lines AE and BE be drawn. If the semicircular protractor were used, it would perhaps be best to lay off the angles at the several station points.

190. To plot the work done with the compass. Form the following table from the compass notes, and write the differences of latitude and departure in their proper columns.

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The scale used in the plot is one of 19 rods, or 314.5 feet to the inch. If one inch be taken in the dividers, and the arms of the sector extended until the dividers will just reach from 31.45 on one arm, to 31.45 on the other, then each division of

a

correspond to 10 feet. The sector having been thus opened to the proper angle, take from it the difference of latitude of the line Al; this is found, very nearly, by extending the dividers across from 41 to 41; and since AD is a meridian, lay off this distance from A to q. At q erect the perpendicular ql, and lay off on it a distance equal to the departure; join A and l; Al is the first line traced with the compass. At A erect the perpendicular AT, and make it equal to 280 feet (see notes) ; this gives the point T at the shore of the river. Lay off 100 feet from A to a, erect the perpendicular ab, and lay off 220 feet to the river on one side, and 100 feet to the artillery on the other; and the same for the offsets at c, d, and 1. At 1, draw a parallel to AD, to represent a meridian line, and lay down the line 12, and its offsets in the same manner, and similarly for all the other lines.

When there are oblique offsets, as at 3 and 5, they are readily plotted, as their lengths and the angles which they make with the meridian are known. In the present example, they serve to verify the work.

191. To join the work done with the plain table.

If the same scale was used with the plain table as that already used in the plot, we have only to transfer the work from one paper to the other. But if the scales are different, measure the angle contained by the lines BF, BG, on the paper of the table, either with a protractor or a scale of chords; and then extend the dividers from B to G, on the paper

of the table, and refer the distance to the scale used with the table ; this will show the true length of the line BG. Then, on the plot, make the angle FBG equal to the measured angle, and the distance BG equal to the measured distance; the point G will then be determined. In the same manner, the other lines may

be | laid down. Or the work may be transferred by letting fall

perpendiculars on the paper of the table, from all the angular points G, H, I, K, &c. on the line BE, and then laying off on BE the distances from B to the foot of each perpendicular, as also the lengths of the respective perpendiculars.

It is best to plot the whole in pencil before making any of the lines in ink, as the plot will generally not unite exactly at the

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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,

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Therefore, the area BDA=3109.28 square rods

, =
area BCD=1808.87

square

rods area BEA=2144.405 square

66

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rods

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 follow

ing table.

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