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ter this bearing in the field notes opposite station 1. Then measure the distance from A to B, which we will suppose to be 10 ch., and insert that distance opposite station 1, in the column of distances.

We next take the bearing from B to C, N. 623 E., and then measure the distance BC=9 ch. 25 l., both of which we insert in the notes opposite station 2.

At station C we take the bearing to D, S. 36° E., and then measure the distance CD=7 ch. 60 1., and place them in the notes opposite station 3.

At D we take the bearing to A, S. 451° W., and measure the distance DA = 10 ch. 40 1. We shall then have made all the measurements on the field which are necessary to determine the contents of the ground.

15. REMARK I. The reverse bearing or back sight, from B to A, is the angle ABH; and since the meridians NS and HG are parallel, this angle is equal to the bearing NAB. The reverse bearing is, therefore, S. 31° E.

The reverse bearing from C, is S. 623° W.; that is, it is the angle ICB= GBC.

And generally, a reverse bearing, or back sight, is always equal to the forward bearing, and differs from it in both of the letters by which it is designated.

16. REMARK II. In taking the bearings with the compass, there are two sources of error. 1st. The inaccuracy of the observations: 2d. Local attractions, or the derangement which the needle experiences when brought into the vicinity of iron-ore beds, or any ferruginous substances.

To guard against these sources of error, the reverse bearing should be taken at every station: if this and the forward bearing are of the same value, the work is probably right; but if they differ considerably, they should both be taken again.

17. REMARK III. If the forward and back sights at the end of any course of the survey agree, it may be safely assumed, that no local attraction disturbs the needle at these points; and hence, that the next foresight is also free

local attraction, when it arises, will first show itself in the difference between a true foresight and an erroneous back sight.

When this difference appears, subtract the back sight from the foresight, and call the difference the correction for the next foresight. The correction will be positive when the foresight is the larger, and negative when it is less.

Add this correction, with its proper sign, to the foresight of the next course, when the meridional and longitudinal letters of that course are both the same, or both different from the foresight of the previous course, and subtract it when one of the letters is the same and the other different the result will be the true bearing. The true bearing of any other course may be found by the same process.

True Foresights.

EXAMPLE.

Back Sights.

Foresights of next Foresights
Course.
Corrected.

N 17° 25′ W

1. S 85° 10' WN 85° 05′ ES 16° 15' WS 16° 20′ W 2. N 16° 20' ES 18° 20' WN 15° 25′ W 3. N 17° 25' WS 16° 10' E N 28° 16' E 4. S. 47° 18' EN 48° 10' W

N 27° 01' E

S 49° 15' WN 50° 07' W

NOTE. If there be no course in the survey in which the foreward and back sights agree, take the one in which they agree the nearest, and add half the difference of the bearings to the least, and treat the result as the true bearing.

18. REMARK IV. In passing over the course AB, the northing is found to be HB, and the departure, which is west, is represented by AH. Of the course BC, the northing is expressed by BG, and the departure, which is east, by GC. Of the course CD, the southing is expressed by CI, and the

G

B

W

N

PC

K

E

ing is expressed by KA, and the departure, which is west, by DK. It is seen from the figure, that the sum of the northings is equal to HB+BG=HG; and that the sum of the southings is equal to CI+KA=PA=HG: hence, the sum of the northings is equal to the sum of the southings.

=

If we consider the departures, it is apparent that the sum of the eastings is equal to GC+ CF GF; and that the sum of the westings is equal to AH+DK= GF; hence also, the sum of the eastings is equal to the sum of the westings. We therefore conclude, that when any survey is correctly made, the sum of the northings will be equal to the sum of the southings, and the sum of the eastings to the sum of the westings.

It would indeed appear plain, even without a rigorous demonstration, that after having gone entirely round a piece of land, the distance passed over in the direction due north, must be equal to that passed over in the direction 'due south; and the distance passed over in the direction due east, equal to that passed over in the direction due

west.

Having now explained the necessary operations on the field, we shall proceed to show the manner of computing the contents of the ground. We shall first explain,

THE TRAVERSE TABLE AND, ITS USES.

19. This table shows the latitude and departure corresponding to bearings that are expressed in degrees and quarters of a degree from 0 to 90°, and for every course from 1 to 100, computed to two places of decimals.

The following is the method of deducing the formulas for computing a traverse table; by means of these formulas and a table of natural sines, the latitude and departure of a course may be computed to any desirable degree

Let AD represent any course, and NAD ACB, expressed in degrees and minutes, be its bearing.

unit of measure of the the radius of the table

(Bk. I., Sec. III., Art. 14).

CB parallel to

Let AC be the course, and also of natural sines Draw DE and

NS, and AE perpen

dicular to AS. Then will DE be the

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latitude, and AE the departure of the course, and CB the cosine, and AB the sine of the bearing.

From similar triangles we have these proportions,

1

:

AC : CB :: AD: DE, or

cos of the bearing :: course : latitude, AC: AB :: AD : AE, or : sin of the bearing :: course : departure. Whence,

1

lat. = course X cos of the bearing,

dep. =course X sin of the bearing.

We have then the following practical rule for computing the latitude and departure of any course.

Look in a table of natural sines for the cosine and sine of the bearing. Multiply each by the length of the course, and the first product will be the latitude, and the second will be the departure of the given course.

EXAMPLES.

1. The bearing is 65° 39', the course 69.41 chains: what

is the latitude, and what the departure?

Natural cosine of 65° 39' .

Length of the course

.41231

69.41

Product, which is the Dif. of Latitude, 28.6184371.

Natural sine of 65° 39'

.91104

Length of the course

69.41

2. The bearing is 75° 47', the course 89.75 chains: what is the latitude, and what the departure?

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Product, which is the Dif. of Latitude, 22.0417025.

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20. In this manner the traverse table given at the end of the book has been computed. When the bearing is given in degrees and quarters of a degree, and the difference of latitude and departure are required to only two places of decimals, they may be taken directly from the traverse table.

If the bearing is less than 45°, the angle will be found at the top of the page; if greater, at the bottom. Then, if the distance is. less than 50, it will be found in the column "distance," on the left hand page; if greater than 50, in the corresponding column of the right hand page.

The latitudes or departures of courses of different lengths, but which have the same bearing, are proportional to the lengths of the courses. Thus, in the W figure, the latitudes AG, AC, or the departures GF, CB, are to each other as the courses AF, AB.

N

G

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H

B

E

A

S

Therefore, when the distance is greater than 100, it may be divided by any number which will give an exact quotient, less than 100: then the latitude and departure of the quotient being found and multiplied by the divisor, the products will be the latitude and departure of the whole course. It is also plain, that the latitude or departure of two or more courses, having the same bearing, is equal to the sum of the latitudes or departures of the courses taken separately.

Hence, if we have any number greater than 100, as 614, we have only to recollect that, 610+4= 614; and

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