around the square previously sold. Required the width of this strip. Once more, in 1839, 42 acres more were sold, to be set off around the preceding piece. Required the dimensions of this third portion. The answer can be proved by calculating if the dimensions of the remaining rectangle will give the content which it should have, viz. 250—(70 +29 +42) = 109 Acres. The figure is on a scale of 40 chains to 1 inch =1:31680.

(499) New countries. The operations of laying out land for the purposes of settlers, are required on a large scale in new countries, in combination with their survey. There is great difficulty in uniting the necessary precision, rapidity and cheapness. "Triangular Surveying" will ensure the first of these qualities, but is deficient in the last two, and leaves the laying out of lots to be subsequently executed. Compass Surveying" possesses the last two qualities, but not the first. The United States system for surveying and laying out the Public Lands admirably combines an accurate determination of standard lines (Meridians and Parallels) with a cheap and rapid subdivision by compass. The subject is so mportant and extensive that it will be explained by itself in Part XII.

CHAPTER II.

PARTING OFF LAND.

(500) It is often required to part off from a field, or from an indefinite space, a certain number of acres by a fence or other boundary line, which is also required to run in a particular direction, to start from a certain point, or to fulfil some other condition. The various cases most likely to occur will be here arranged according to these conditions. Both graphical and numerical methods will generally be given.*

The given lines will be represented by fine full lines; the lines of construction by broken lines, and the lines of the result by heavy full lines.

The given content is always supposed to be reduced to square chains and decimal parts, and the lines to be in chains and decimals.

A. BY A LINE PARALLEL TO A SIDE.

(501) To part off a rectangle. If the sides of the field adjacent to the given side make right angles with it, the figure parted off by a parallel to the given side will be a rectangle, and its breadth will equal the required content divided by that side, as in Art. (493).

If the field be bounded by a curved or zigzag line outside of the given side, find the content between these irregular lines and the given straight side, by the method of offsets, subtract it from the content required to be parted off, and proceed with the remainder as above. The same directions apply to the subsequent problems.

Fig. 344.

D

(502) To part off a parallelogram. If the sides adjacent to the given side be parallel, the figure parted off will be a parallelogram, and its perpendicular width, CE, will be obtained as above. The length of one of the parallel

(503) To part off a trapezoid. When the sides of the field adjacent to the given side are not parallel, the figure parted off

will be a trapezoid.

When the field or figure is given on the ground, or on a plat,

the content of ABDC. Divide the difference of it and the required

content by CD. Set off the quotient perpendicular to CD, (in this figure, outside of it,) and it will give a new line, GH, a still nearer approximation to that desired. The operation may be repeated, if found necessary.

(504) When the field is given by Bearings, deduce from them, as in Art. (243), the angles at A and B. The required sides will then be given by these formulas:

B

Fig. 346.

D

When the sides AD and BC diverge, instead of converging, as in the figure, the negative term, in the expression for CD, becomes positive; and in the expressions for both AD and BC, the first factor becomes (CD — AB).

or

The perpendicular breadth of the trapezoid = AD. sin. A; = BC. sin. B.

=

Example. Let AB run North, six chains; AD, N. 80° E.; BC, S. 60° E. Let it be required to part off one acre by a fence parallel to AB. Here AB=6.00, ABCD=10 square chains, A = 80°, B = 60°. Ans. CD=4.57, AD=1.92, BC=2.18, and the breadth = 1.89.

The figure is on a scale of 4 chains to 1 inch 1:3168.

B. BY A LINE PERPENDICULAR TO A SIDE.

(505) To part off a triangle. Let FG be the required line.

DEB. Then the required distance BF, from the angular point

to the foot of the desired perpendicular,

= BD

'BFG

Example. Let BD = 30 chains; ED=12 chains; and the Then BF 35.22 chains.

desired area = 24.8 acres.

=

The scale of the figure is 30 chains to 1 inch = 1:23760.

(506) When the field is given by Bearings, find the angle B from the Bearings; then is 2 × BFG

BF = √(tang. B

Example. Let BA bear S. 75° E., and BC

N. 60° E., and let five acres be required to be

B

Fig. 348.

parted off from the field by a perpendicular to BA. Here the angle B = 45°, and BF 10.00 chains.

The scale of the figure is 20 chains to 1 inch =1:15840.

(507) To part off a quadrilateral. Produce the converging sides to meet at B. Calculate the

Fig. 369.

content of the triangle HKB, whe

ther on the ground or plat, or from Bearings. Add it to the content

of the quadrilateral required to be B

H

A

parted off, and it will give that of the triangle FGB, and the me thod of the preceding case can then be applied.

(598) To part off any figure. If the field be very irregularly shaped, find by trial any line which will part off a little less than the required area. This trial line will represent HK in the preceding figure, and the problem is reduced to parting off, according to the required condition, a quadrilateral, comprised between the trial line, two sides of the field, and the required line, and containing the difference between the required content and that parted off by the trial-line.

C. BY A LINE RUNNING IN ANY GIVEN DIRECTION.

(509) To part off a triangle. By construction, on the ground or the plat, proceed nearly as in Art. (505), setting out a line in the required direction, calculating the triangle thus formed, and obtaining BF by the same formula as in that Article.

(510) If the field be given by Bearings, find from them the angles CBA and GFB; then is A (2 × BFG sin (B+ F)).

BF = √(2

sin. B. sin. F

Example. Let BA bear S. 30° E.; BC, N. 80° E.; and a fence be required to run, from some point in BA, a due North course, and to part off one acre. Required the distance from B to the point F, whence it must start. Ans. The angle B 70°, and F = 30°. Then BF= 6.47.

B

Fig. 350.

EG

The scale of Fig. 350 is 6 chains to 1 inch = 1:4752.

Fig. 351.

(511) To part off a quadrilateral. Let it be required to part off, by a line running in a given direction, a quadrilateral from a field in which

C

of the triangle ABE. Measure the side AE. From ABE subtract the area to be cut off, and the remainder will be the content of the triangle CDE. From A set out a line AF parallel to the given direction. Find the content of ABF. Take it from ABE, and thus obtain AFE. Then this formula, ED=AEDE, will fix the point D, since AD AE-ED.

=

(512) When the field and the dividing line are given by Bear ings, produce the sides as in the last article. Find all the angles from the Bearings. Calculate the content of the triangle ABE, by the formula for one side and its including angles. Take the