In the afternoon of the same day, and the same distance of time from twelve o'clock, watch the lengthening shadows as before, making marks on the circles where the shadow falls.

Lastly, find the middle point exactly between two marks on the same circle, and draw a line from the centre through that point, which will be the true meridian line or north point.

Remove the pin, and fix a short-pointed pin in its place, on which place the magnetic needle; when it is at rest, mark the point it cuts on one of the circles at the north end; from that point draw another line through the centre, measure that angle accurately, which will be the number of degrees' variation. To measure the angle, see Problem 14.

DIVISION OF LAND.

The division of land is applied to many cases; as the general inclosure of a parish in which there are sundry claimants, each receiving land in proportion to their claims, the quantity being guided by the value per acre.*

In some cases an exchange of land is made between two adjoining proprietors for the mutual improvement to their estates, taking value for value, or quantity for quantity.

In either case it requires a correct system to arrive at a true balance, therefore a division has to be made according to circumstances, to accomplish which the following examples are given.

In small plots of level ground having straight fences, whether rectangular or triangular, there is no difficulty in laying out a division without a plan, as the dimensions required for the calculation can be made at the same time the division is made. In all other cases an accurate plan, and quantity of the land

* Whatever the value is per acre, in calculating it must be reduced to shillings and decimals, to obtain an accurate result, in the same manner as the quantities are entered. The value per acre is usually marked in private characters, such as letters of any particular town or object, containing the exact number of letters in lieu of figures, as "Altringham," "Mayflower," &c.

to be divided, must be made before any calculation can be made, or the allotments staked out.

To facilitate the calculations required in the divisions, see Tables, Nos. 11 and 13.

Fig. 1, Plate 27.

Problem 36.

Example 1. It is required to cut off 1 acre 3 roods from a parallelogram containing 3 acres, parallel to the side A B, equal to 600 links. See Table 13.

Rule. Divide the square links in the quantity to be cut off, by the number of links in the side; the product will be the length of the other side; thus:

1.75000600 = 292 links

Note.-Explanation to Table 13. The first column shows the number of acres, the second column the number of square links contained in the required number of acres; the roods and perches the same; each part is taken out separately and added together.

Fig. 2, Plate 27.

Problem 37.

Example 2. It is required to cut off 2 acres from a rectangular piece of ground, containing 3 acres 2 roods 16 perches, from a fixed point at a.

Rule. Draw an assumed line, as a b; then find the quantity of a b C D equal to 2.550; that being an excess of .55000 square links, therefore divide that sum by half the length of a b, equal to 620 links; the product will be 88 links (nearly), provided it was to be set out parallel; the figure of the excess, a b c, being a triangle, the perpendicular cd will then be 176 links; then will a c CD be the quantity required.

Fig. 3, Plate 27.

Problem 38.

Example 3. To divide a piece of building land of a rectangular figure, containing 6 acres 3 roods, between A B C, being of equal value per acre; the proportionate share of A equal to 7 parts, B = 5 parts, and C = 3 parts.

Rule. As the sum total of the proportionate shares is to the whole quantity of the estate to be divided, so is each proportionate share to its respective claim; thus:

Example 4. The same estate is to be divided between A B C, the value of the whole being 7507.; A is to be allotted land equal in value to 350l., B equal to 250l., and C equal to 1507.; thus:

To set out the above quantities on the ground.

Rule. Divide each separate quantity by the length of the side DE equal to 675 links; thus:

And so on in like manner for any number of shares.

Fig. 4, Plate 27.

Problem 41.

Example 5. To divide a triangular piece of ground into a given number of parts by right lines drawn from an angle to its opposite side.

Rule. Divide the base of the triangle into the required number of parts, then draw a line from the vertex to the division point; it will be the proportion required.

K

A triangular piece of ground, containing 26 acres 2 roods, is to be divided into three parts, bearing the proportion to the numbers 4, 2, 1, equal to 7; the length A B equal to 28 chains. By construction:

Divide the base A B, equal 28 chains, into 7 equal parts; from the angle C draw a line to the fourth division as Ca, and from C to the sixth division as C b, dividing the triangle as required.

Arithmetically:

Rule. As the sum of the ratios is to the length of the base A B, so is each respective part to the length required. As:

To find the quantity contained in each. As:

Example 6. The triangular field ABC is to be divided between two persons in the proportion of 2 to 5, having an equal right to the pond, or an occupation road to that point.

Geometrically:

Rule. Divide the side of the triangle A B into 7 equal parts; from C draw the line CE; parallel to which from a, the fifth point, draw the line a D; then draw the line D E, the division fence required.

Problem 43.

To lay this out on the ground:

From the plan, scale the length from C to D equal to 300 links, and from D to B 312 links; if correct, fix a stake at the point D.

Fig. 6, Plate 27.

Problem 44.

Example 7. The triangular field A B C is to be divided into three equal parts, reserving to each the right of water at the pond, or as an entrance from the occupation road.

Rule. Divide the line A B into three equal parts, as A a, a b, bB; draw the line CD; then from a, draw the line a c parallel to CD; and from b, draw the line bd parallel to CD; draw the lines c D and d D, which will be the division fences required. To lay out the same on the ground, measure the lengths from the plan as before, and drive stakes at the points c and d.

Fig. 7, Plate 27.

Problem 45.

Example 8. To divide a triangular field A B C, containing 4A. 1R. 8P., or 4.300, into three equal parts, parallel to the side B C, the length of each side being 10 chains.

Rule. Divide the side A B into three equal parts as a b; then find the mean proportion between A B and A a; by multiplying the two together, the square root of their product will be the mean proportion required, as A c; then draw a line from c parallel to B C, and B c C d will be equal to of the triangle.

Thus: A B = 1000 × A a = 666'√666000 = 816 = A c the mean

In the same manner divide the remaining part, A c, into two equal parts as at b; find the mean proportion between A c and Ab as before; mark off the length from A to e; draw the line from e parallel to B c, then cdef will be the second part, and the triangle A ef will be the third part.

Thus: A c 816 × A b = 408 =

Another method:

332928576 A e the mean

Problem 46.

=

Example 9. Rule. Find the content of the whole triangle and subtract from it the quantity to be cut off. As similar triangles are in proportion to the squares of their like sides, the triangle