PROBLEM III. To take off any given number of Acres from a multangular Field.

EXAMLPE I. See PLATE III. Fig. 65.

Let ABCD, &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 DG (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 a Perpendicular whose Base is 70 and Area 140. Lay off 4 Rods from G to I, and draw the Line DI, which will be the dividing Line.

EXAMPLE II. See PLATE III. Fig. 60.

Let ABCD, &c. be a Tract of Land, to be divided into two equal parts, by a Line from I to the opposite Side CD: To find Arithmetically on what part of the Line CD the dividing Line IN will fall; or to find the Distance CN.

Find the Area of the part IABCI, according to SECTION III. Page 59, as follows: Set the Latitude and Departure of the three first Sides IA, AB and BC 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 KI, as will balance the Columns. This Southing and Westing will be the Latitude and Departure made by the Line CI. The Area of IABCI 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

Find the Bearing and Distance of CI by RIGHT ANGLED TRIGONOMETRY, CASE IV. as follows:

As CK, the Southing of CI, 109

2.03743 10.00000

1.85552

11.85552

2.03743

Note. In this way the Course and Distance may be found from one Angle of a Field to another.

Having found the Line CI divide 3470, the number of Rods to be contained in the Triangle ICN, by one half the Line CI, viz. 65, the Quotient will be the length of the Perpendicular PN, viz. 53.4.

Now, by the Bearings of CI and CD it appears that they form an Angle of 60° 20′; wherefore in the Triangle CPN are given the side PN 53.4 and the Angle at C 60° 20′, to find the Hypothenuse CN.

As Sine PCN 60° 20′

: Side PN 53.4

:: Radius

9.93898

1.72754

10.00000

11.72754

9.93898

: Hyp. CN 61.5

1.78856

Thus the dividing Line must go from I to a Point on the Line CD, which is 61.5 Rods from C. The Bearing and Distance of this Line may be found by the directions given above for finding the Bearing and Distance of the Line CI. Or, they may be found by Oblique Trigonometry CASE III.

K

4

Another method of finding the Distance CN. Having ascertained the Latitude and Departure of the Line CI, set them down in a Traverse Table; find the Latitude and Departure of the Line CD, and place them in the Table; the Difference between the Northing of the Line IC and the Southing of the Line CD will be the Southing of the Line DI. viz. 6.6; and the Sum of the Eastings of those Lines, as they are both Easterly will be the Westing of the Line DI, viz. 123.9. Proceed to calculate the Area of the Triangle ICD, which will be found to be 6522 Rods, nearest.

Note. As in this Triangle two Sides and their contained Angle are given, the Area may be found by PROB. IX. Rule 4. Page 39.

Having found the Area of this Triangle, proceed to find CN according to PROB. II. Page 71, as follows:

As the Area of the Triangle; Is to CD the Base; So is the quantity to be contained in the Triangle ICN; To CN its proportion of the Base.

As 6522: 115:: 3470: 61.2

A third method of finding the Distance CN.

To the Logarithm of double the Area to be contained within the Triangle ICN add Radius; from this Sum subtract the Logarithmic Sine of the angle at C; and from the Remainder subtract the Logarithm of the Side IC; the last Remainder will be the Logarithm of the Side CN.

The double Area of the Triangle ICN is 6940; the Angle at C is 60° 20′; the Side IC is 130.

Double Area 6940

3.84136

Note. Radius may be added by placing a Unit before the Index of the Logarithm for the double Area, without the trouble of setting down the Cyphers.

By Natural Sines.

Divide the double Area by the Natural Sine of the given Angle, and that Qotient by the given Side; the last Qotient will be the Side CN.

Nat. Sine of the Angle at C 60° 20′ 0.86892 6940-0.86892=7986.92

7986.92 130=61.43

From the above the following general Rule may be. drawn.

To find the Side of a Triangle when the Area is given, with one of the Sides and the Angle contained between the given Side and the Side Required.

To the Logarithm of double the Area add Radius; from this Sum subtract the Logarithmic Sine of the given Angle, and from the Remainder subtract the Logarithm of the given Side; the last Remainder will be the Logarithm of the Side required.

Or, By Natural Sines: Divide the double Area by the Nat. Sine of the given Angle, and that Quotient by the given Side; the last Quotient will be the Side required.

CONCLUDING REMARKS.

Other methods of surveying Fields are taught by some authors on this Subject. The preceding, however, will be found most useful in actual practice. Other instruments besides those mentioned in this Book are also sometimes used; such as the Plain Table, Semicircle, Perambulator, Theodolite, &c. But of these instruments very little use is made in NewEngland; and they are not often to be met with. For general practice none will be found more useful than a common Chain, and a Compass upon Rittenhouse's construction. A Surveyor should also provide himself with an Offset Staff, ten Links in length, and accurately divided into Links. This should be made of firm, hard wood, and will be found very convenient in taking Offsets, and also in measuring the Chain; which should be often done, as from a variety of causes a Chain is liable to become inaccurate.

It will be observed that in this Work there are no descriptions of Mathematical and Surveying instruments. The Compiler omitted such descriptions froma belief that nothing which can be written on the Subject will enable a person to understand them without an actual inspection of the instruments themselves, and some instruction from those acquainted with them.

The general principles here taught may be applied to the surveying of Townships, Roads, Rivers, Harbors, &c.