In this example, the whole perimeter of the field is 100 chains, the whole error in latitude .34, the whole error in departure .42, and the length of the first side 18. To find the corresponding errors,

.34 .06 the error in latitude,

100 18:42.08 the error in departure.

The error in latitude is to be added to 10.26 making it 10.32, as in the column of corrected northings; and the error in departure is to be added to 14.79 making it 14.87, as in the column of corrected eastings. After the corrections are made for each of the courses, the remaining part of the calculation is the same as in the preceding examples.

122. If the length and direction of each of the sides of a field except one be given, the remaining side may be easily found by calculation. For the difference between the sum of the northings and the sum of the southings of the given sides, is evidently equal to the northing or southing of the remaining side; and the difference between the sum of the eastings and the sum of the westings of the given sides, is equal to the easting or westing of the remaining side. Having then the difference of latitude and departure for the side required, its length and direction may be found, in the same manner as in the sixth case of plane sailing. (Art. 49.)

Example V.

What is the area of a field of six sides, of which five are given, viz.

123. Plotting by departure and difference of latitude.— A survey may be easily plotted from the northings and southings, eastings and westings. For this purpose, the column of Meridian Distances is used. It will be convenient to add also another column, containing the distance of each station from a given parallel of latitude, and formed by adding the northings and subtracting the southings, or adding the southings and subtracting the northings.

Let AT (Fig. 33.) be a parallel of latitude passing through the first station of the field. Then the southing TB or LM is the distance of B, the second station, from the given parallel. To this adding the southing BH, we have LO the distance of CO from LT. Proceeding in this manner for each of the sides of the field, and copying the 7th column in the table, p. 65, we have the following differences of latitude and meridian distances.

To plot the field, draw the meridian NS, and perpendicular to this, the parallel of latitude LT. From L set off the differences of latitude LM, LO, LR, and LP. Through L, M, O, R, and P, draw lines parallel to LT; and set off the meridian distances AL, BM, CO, DR, and EP. The points A, B, C, D, and E, will then be given.

124. When a field is a regular figure, as a parallelogram, triangle, circle, &c. the contents may be found by the rules in Mensuration, Sec. I, and II.

125. The area of a field which has been plotted, is sometimes found by reducing the whole to a TRIANGLE of the same area. This is done by changing the figure in such a manner as, at each step, to make the number of sides one less, till they are reduced to three.

Let the side AB (Fig. 35.) be extended indefinitely both ways. To reduce the two sides BC and CD to one, draw a line from D to B, and another parallel to this from C, to intersect AB continued. Draw also a line from D to the point of intersection G. Then the triangles DBC and DBG are

equal. (Euc. 37. 1.) Taking from each the common part DBH, there remains BGH equal to DCH. If then the triangle DCH be thrown out of the plot, and BGH be added, we shall have the five-sided figure AGDEF equal to the sixsided figure ABCDEF.

In the same manner, the line EL may be substituted for the two sides AF and EF; and then DM, for EL and ED. This will reduce the whole to the triangle MGD, which is equal to the original figure. The area of the triangle may then be found by multiplying its base into half its height; and this will be the contents of the field.

In practice, it will not be necessary actually to draw the parallel lines BD, GC, &c. It will be sufficient to lay the edge of a rule on C, so as to be parallel to a line supposed to pass through B and D, and to mark the point of intersection G.

126. If after a field has been surveyed, and the area computed, the chain is found to be too long or too short; the true contents may be found, upon the principle that similar figures are to each other as the squares of their homologous sides. (Euc. 20. 6.) The proportion may be stated thus;

As the square of the true chain, to the square of that by which the survey was made;

So is the computed area of the field, to the true area.

3

Ex. If the area of a field measured by a chain 66.4 feet long, be computed to be 32.6036 acres; what is the area as measured by the true chain 66 feet long?

Ans. 33 acres.

127. A plot of a field may be changed to a different scale, that is, it may be enlarged or diminished in any given ratio, by drawing lines parallel to each of the sides of the original plan.

To enlarge the perimeter of the figure ABCDE (Fig. 36.) in the ratio of aG to AG; draw lines from G through each of the angular points. Then beginning at a, draw ab parallel to AB, bc parallel to BC, &c.

It is evident that the angles are the same in the enlarged figure, as in the original one. And by similar triangles,

AG aG: BG: bG:: CG; cG :: &c.

And

AG: aG:: AB: ab :: BC: bc:: &c. Therefore ABCDE and abcde are similar figures. (Enc. Def. 1. 6.)

In the same manner, the smaller figure a'b'c'd'e' may be drawn, so as to have its perimeter proportioned to ABCDE as a'G to AG.

SECTION II.

METHODS OF SURVEYING IN PARTICULAR CASES.

ART. 128. MEASURING round a field, in the manner explained in the preceding section, is by far the most common method of surveying. The following problems are sometimes useful. They may serve to verify or correct the surveys which are made by the usual method.

PROBLEM I.

To survey a field from TWO STATIONS.

129. FIND THE DISTANCE OF THE TWO STATIONS, AND THEIR BEARINGS FROM EACH OTHER; THEN TAKE THE BEARINGS OF THE SEVERAL CORNERS OF THE FIELD FROM EACH OF THE STATIONS.

In the field ABCDE, (Fig. 37.) let the distance of the two stations S and T be given, and their bearings from each other. By taking the bearing of A from S and T, or the angles AST and ATS, we have the direction of the lines drawn from the two stations to one of the corners of the field. The point A is determined by the intersection of these lines. In the same manner, the point B is determined by the intersection of SB and TB; the point C, by the intersection of SC and TC; &c. &c. The sides of the field are then laid down, by connecting the points ABCD, &c.

The area is obtained, by finding the areas of the several triangles into which the field is divided by lines drawn from one of the stations. Thus the area of ABCDE (Fig. 37.) is equal to

ABT+BCT+CDT+DET+EAT

or to

ABS+BCS+CDS+DES+EAS.

Now we have the base line ST given and the angles, in the triangle AST, to find AS and AT; in the triangle BST, to find BS and BT, &c. After these are found, we have two