2. A gentleman, knowing that the area of a circle is greater than that of any other figure of equal perimeter, walls in a circular deerpark of 100 perches diameter, in which he makes an elliptical fishpond 10 perches long by 5 wide; required the length of his wall, content of his park, and area of his pond?

Answer. The wall 314.16 perches, enclosing 49.4. 14P. of which 391 perches, or of an acre nearly, is appropriated to the pond.

PROBLEM XIII.

The area of a circle given to find its diameter.

RULE.

To the logarithm of the area add 0.104909, and half the sum will be the logarithm of the diameter. Or, divide the area by .7854, and the square-root of the quotient will be the diameter.

EXAMPLE.

A horse in the midst of a meadow suppose,
Made fast to a stake by a line from his nose;
How long must this line be, that feeding all round,
Permits him to graze just an acre of ground?

Area in perches 160 log. 2.204120

0.104909

2)2.309029

Diameter

2)14.2733 log. 1.154514

Answer, 7.13665 per 117F. 9In.

PROBLEM XIV.

Allowance for roads.

It is customary to deduct 6 acres out of 106 for roads; the land before the deduction is made may be termed the gross, and that remaining after such deduction, the neat.

RULE.

The gross div.

The neat mul.by 1.06,

quotes the neat.

prod: the gross.

EXAMPLES.

1. How much land must I enclose to have 850A. 2R. 20P.

neat?

850.625 x 1.06=901.6625=901 2 26 the Answer.

2. How much neat land is there in a tract of 901A. 2R. 26P.

gross?

40 26

4 2.65

Acres. A. R. P.

1.06)901.6625(850.625=850 2 20 the Answer.

848

&c.

Note. These two operations prove each other.

PROBLEM XV.

To find the area of a piece of ground, be it ever so irregular, by dividing it into triangles camd trapezia.

PL. 7. fig. 4.

We here admit the survey to be taken and protracted; by having therefore the map, and knowing the scale by which it was laid down, the content may be thus obtained.

Dispose the given map into triangles, by fine pencilled lines, such as are here represented in the scheme, and number the triangles with 1, 2, 3, 4, &c. Your map being thus prepared, rule a table with four columns; the first of which is for the number of the triangle, the second for the base of it, the third for the perpendicular, and the fourth for the content in perches.

Then proceed to measure the base of number 1, from the scale of perches the map was laid down, and place that in the second column of the table, under the word base; and from the angle opposite to the base, open your compasses so, as when one foot is in the angular point, the other being moved backwards and forwards, may just touch the base line, and neither go the least above or beneath it; that distance in the compasses, measured from the same scale, is the length of that perpendicular, which place in the third column, under the word perpendicular.

If the perpendiculars of two triangles fall on one and the same base, it is unnecessary to put down the base twice, but insert the second perpendicular opposite to the number of the triangle in the table, and join it with the other perpendicular by a brace, as No. 1 & 2, 4 & 5, 6 & 7, 9 & 10, &c.

Proceed after this manner, till you have measured all the triangles; and then by prob. 6, find the content in perches of each respective triangle, which severally place in the table opposite to the number of the triangle, in the fourth column, under the word content.

But where two perpendiculars are joined together in the table by a brace, having both one and the same base; find the content of each (being a trapezium) in perches, by prob. 11, which place opposite the middle of those perpendiculars, in the fourth column, under the word content.

Having thus obtained the content of each respective triangle and trapezium, which the map contains, add them all together, and their sum will be the content of the map in perches; which being divided by 160, gives the content in acres. Thus, for

This being divided by 160, will give 25A. 3R. 22P. the content of the map.

Let your map be laid down by the largest scale your paper will admit, for then the bases and perpendiculars can be measured with greater accuracy than when laid down by a smaller scale, and if possible measure from scales divided diagonally.

If the bases and perpendiculars were measured by four-pole chains, the content of every triangle and trapezium, may be had as before, in problems 6 and 11, and consequently the whole content of the map.

If any part of your map has short or crooked bounds, as those represented in plate 7, fig. 5, then by the straight edge of a transparent horn, draw a fine pencilled line, as AB, to balance the parts taken and left out, as also another, BC: these parts when small, may be balanced very nearly by the eye, or they may be more accurately balanced by method the third. Join the points A and C by a line, so will the content of the triangle ABC, be equal to that contained between the line AC, and the crooked boundary from Å to B, and to C: by this method the number of triangles will be greatly lessened, and the content become more certain; for the fewer operations you have, the less subject will you be to err: and if an error be committed, the sooner it may be discovered.

The lines of the map should be drawn small, and neat, as well as the bases; the compasses neatly pointed, and scale accurately divided; without all which you may err greatly. The multiplications should be run over twice at least, as also the addition of the column content.

From what has been said, it will be easy to survey a field, by reducing it into triangles, and measuring the bases and perpendiculars by the chain. To ascertain the content only, it is not material to know at what part of the base the perpendicular was taken: since it has been shown (in cor. to theo. 13, geom.) that triangles on the same base, and between the same parallels, are equal; but if you would draw a map from the bases and perpendiculars, it is evident that you must know at what part of the base the perpendicular was taken, in order to set it off in its due position; and hence the map is easily constructed.

To determine the area of a piece of ground, having the map given, by reducing it to one triangle equal thereto, and thence finding its content.

Let ABCDEFGH be a map of ground, which you would reduce to one triangle equal thereto.

Produce any line of the map, as AH, both ways, lay the edge of a parallel ruler from A to C, having B above it, hold the other side of the ruler, or that next you, fast; open till the same edge touches B, and by it, with a protracting pin, mark the point b, on the produced line, lay the edge of the ruler from b to D. having C above it, hold the other side fast, open till the same edge touches C, and by it mark the point c, on the produced line. A line drawn from c to D will take in as much as it leaves out of the map.

Again lay the edge of the ruler from H to F, having G above it, keep the other side fast, open till the same edge touches G, and by it mark the point g, on the produced line; lay the edge of the ruler from g to E, having F above it, keep the other side fast, open till the same edge touches F, and by it mark the point f, on the produced line. Lay the edge of the ruler from f to D, having E above it, keep the other side fast, open till the same edge touches E, and by it mark the point e, on the produced line. A line drawn from D to e, will take in as much as it leaves out. Thus have you the triangle c D e, equal to the irregular polygon A B C D E F G H.*

If, when the ruler's edge be applied to the points A and C, the point B falls under the ruler, hold that side next the said points fast, and draw back the other to any convenient distance; then hold this last side fast, and draw back the former edge to B, and by it mark b, on the produced line; and thus a parallel may be drawn to any point under the ruler, as well as if it were above it. It is best to keep the point of your protracting pin in the last point in the extended line, till you lay the edge of the ruler from it to the next station, or you may mistake one point for another.

This may also be performed with a scale, or ruler, which has a thin sloped edge, called a fiducial edge; and a fine pointed pair of compasses. Thus,

Lay that edge on the points A and C, take the distance from the point B to the edge of the scale, so that it may only touch it, in the same manner as you take the perpendicular of a triangle; carry

*The demonstration of this is evident from Prob. 19, Geo. page 61, of this Book.