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A USEFUL PROBLEM.

TO FIND THE TRUE AREA OF A FIELD WHICH HAS BEEN MEASURED BY A CHAIN TOO LONG OR TOO SHORT.

Calculate the area as if the chain was of a true length, then institute the following proportion. As the

square of the length of the true chain;
Is to the area, as found by the chain made use of;
So is the square of the length of that chain;
To the true area of the field.

EXAMPLE

Suppose a field, measured by a two rod chain 3 inches too long, is found to contain 41 acres 1 rood and 33 rods, what is the true area.

As the square of 33 feet, the true length of a two rod chain; is to 41 acres 1 rood and 33 rods; so is the square of 33 feet 3 inches, the length of the chain used in the survey; to 42 acres and 13 rods. 33 feet=396 inches. 396X396=156816 square inches.

41 acres 1 rood 33 rods=6633 rods.

33 feet 3 inches=399 inches. 399X399=159201 square inches.

159201X6633-156316=6733 rods.
6733+160=42 acres 13 rods, the true area.

PART II.

LAYING OUT LAND.

PROBLEM I. To lay out any number of acres in the form of a square.

Annex 5 ciphers to the number of acres, which will turn them into square links, the square root of which will be the side of the square in links.

EXAMPLE I. It is required to lay out 810 acres in the form of a square.

Answer. Each side of the square must be 9000 links, or

PROBLEM II. TO LAY OUT ANY NUMBER OF ACRES IN THE FORM OF A PARALLELOGRAM, WHEREOF ONE SIDE IS GIVEN.

Divide the number of acres when turned into square links, by the given side; the quotient will be the side required.

EXAMPLE. What must be the longest side of a rectangle, which is to contain 25 aeres, when the shortest side is 5 chains and 50 links?

Answer. 2500000-550=4545 links for the longest side.

PROBLEM III. TO LAY OUT ANY NUMBER OF ACRES IN A FIELD, 3, 4, 5, 6, &c. TIMES AS LONG AS IT IS BROAD.

Divide the acres when turned into square links, by the ratio between the length and breadth ; the square root of the quotient will be the shortest side.

EXAMPLE. It is required to lay out 100 acres 5 times as long as it is broad.

Answer. 10000000-552000000, the square root of which is 1414 links for the shortest side, and the longest will be 7070 links.

PROBLEM IV. TO MAKE A TRIANGLE WHICH SHALL CONTAIN A GIVEN NUMBER OF ACRES, BEING CONFINED TO A CERTAIN BASE.

Double the given number of acres, to which, annex 5 ciphers, and divide by the base; the quotient will be the perpendicular height in links.

EXAMPLE. Upon a base of 40 chains, to lay out 100 acres in a triangular form.

Answer. 5000 links, or 50 chains, will be the length of the perpendicular

The perpendicular may be erected from any part of the base: thus, the triangle A B C, see Fig. 64, is the same as A B E, each containing 100 acres.

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As different fields are so variously, and many of them irregularly shaped, and as they are required to be divided in many different proportions, it is difficult to give rules which will apply to particular cases. The business of dividing land must therefore be left, in a great measure, to the skill and judgment of the surveyor; who, if he is well acquainted with trigonometry, and with measuring land, will not find it difficult, after a little practice, to divide a field in such a manner as shall be desired. If he has before him a plot of the field, and knows the number of parts into which it is to be divided, and the proportion which each part is to bear to the others, he will readily find out where the dividing lines are to be drawn.

A few Rules and EXAMPLES will be given for the general instruction of the learner.

PROBLEM I. TO CUT OFF ANY NUMBER OF ACRES FROM 4 SQUARE OR RECTANGLE.

Say, as the whole number of acres in the field; is to the length of the square or length or breadth of the rectangle; so is the number of acres proposed to be cut off; to their proportion of the length or breadth.

PROBLEM II. TO CUT OFF ANY NUMBER OF ACRES BY A LINE FROM ANY ANGLE OF A TRIANGLE.

the dividing line is to be drawn; then say, as the number of acres in the whole triangle; is to the whole base; so is the given number of acres; to their part of the base.

Fig. 66.

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EXAMPLE. See Fig. 66.

А.

B

D In the triangle A B C, which contains 48 acres, it is required to cut off 18 acres, by a line proceeding from C to the base A B, which is 40 chaing.

As 48: 40: : 18 : 15 Lay 15 chains on the base from B to D, and draw the line C D. The triangle will then be divided as was proposed; BC D, containing 18 acres.

PROBLEM III. TO TAKE OFF ANY GIVEN NUMBER OF ACRES FROM A MULTANGULAR FIELD.

Fig. 67.

EXAMPLE I. See Fig. 67. B

с

E

IG

А

Let A B C D, &c. be the plot of a field containing 11 acres, from which it is required to cut off 5 acres.

Join two opposite corners of the field, as D and G, with the line D G, (which you may judge to be near the partition line,) and find the area of the part DEFG, which, suppose, may want 140 rods of the quantity proposed to be cut off. Measure the line DG, which, suppose to be 70 rods; divide 140 by 35, the half of DG, and the quotient 4 will be the length of the perpendicular of a triangle, whose, base is 70 and the area 140. Lay off 4 rods from G to I, and draw the line D I, which will be the dividing line. *

*This explanation supposes D G and AG at right angles to each other. When they are not at right angles, the height G I must not

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Rods.

Rods. AB. N. 19° 0' E. 108 FG. West

70.9 BC. S. 77 0 E. 91 GH. N. 36° 0 W. 47 CD. S. 27 0 E. 115 HI. North

64.3 DE. S. 52

58 I A. N. 62 15 W. 59 S 15 30 E.

Acres Rood Rods. Whole area 152-1-25

Find the area of the part I ABC, according to SECTION III. page 64, as follows: set the latitude and departure of the three first sides, I A, A B, and B C, in their proper columns, in a traverse table; and place as much southing, viz. 109.1, equal to the line CK, and as much westing, viz. 71.7, equal to the line K I, as will balance the columns. This southing and westing will be the latitude and departure made by the line C I. The area of I ABC will be found to be 8722 rods, which is less than half the area of the whole field by 3470 rods, the quantity to be contained in the triangle ICN.

Find the bearing and distance of C I by RIGHT ANGLED TRIGONOMETRY, CASE IV. as follows :*

As C K, the southing of C I, 109, nearly 2.037426
: radius

10 000000
:: KI, the westing of CI, 71.7

1.855519

11.855519
2.037426

: tangent course S. 33° 20' W.

9.818093

* The mode given above is undoubtedly the most correct, but the use of the traverse table will save many figures. From that table the

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