=AC, the base of the hill. If a whole chain cannot be carried horizontally, half a chain, or less, may, and the sum of these half-chains, or links, will give the base, as before.

If the inclined side of the hill be the plane surface, the angle of the hill's inclination may be taken, and the slant height may be measured on the surface; and thence (by Case 1 of rightangled trigonometry) the horizontal line answering to the top may be found; and if we have the angle of inclination given on the other side, with those already given, we can find the horizontal distance across the hill, by Case 2 of oblique trigonometry.

All inclined surfaces are considered as horizontal ones; for all trees which grow upon any inclined surface do not grow perpendicular thereto, but to the plane of the horizon: thus, if Ad, ef, gh, &c. were trees on the side of a hill, they grow perpendicular to the horizontal base AC, and not to the surface AB: hence the base will be capable to contain as many trees as are on the surface of the hill, which is manifest from the continuation of them thereto. And this is the reason that the area of the base of a hill is considered to be equal in value to the hill itself.

Besides, the irregularities of the surfaces of hills in general are such, that they would be found impossible to be determined by the most able mathematicians. Certain regular curve surfaces have been investigated, with no small pains, by the most

* The number of chains taken down in the field-book is longer than the lines Ao, op, pq, &c., because the chain, being elevated above the surface of the earth (though stretched with a force at both ends), forms a curve, which approaches a right line, according as the force is more or less applied; but does not coincide with it: as, for example.-Let the chain be stretched from d to e (Pl. 8. fig. 4); it does not coincide with de, but forms a curve line, which must be longer than de or its equal Ao, and so is fg or op shorter than the chain, and in like manner with all the rest. And de, fg, &c. Ao, op, &c. AC; consequently, the number of chains, being greater than de, ef, &c. or Ao, op, &c. is greater than AC; therefore, the horizontal line AC (by surveyors in general) is made too long, therefore a deduction must be made for every chain in the field-book; the sum to be taken from AC may be found by making an experiment on a two-pole chain (when extended above the surface of the earth by a force at both extremities), and measuring the distance from its middle point to the middle of the right line which would join its extremities, which call a, and call the length of the chain b; then b2-ade, or half the right line; therefore 2/b2-a2=de, or the right line; from whence 262√b2—a2= the excess for every chain which is measured or taken down in the field-book: calling the number of chains c, then c.2b-2(b-a”)= the whole excess on the horizontal line AC. From what is here demonstrated, the practitioner will be able to find the sum to be taken from every horizontal line in surveying hills, &c.

eminent; therefore an attempt to determine in general the infinity of irregular surfaces which offer themselves to our view, to any degree of certainty, would be idle and ridiculous, and for this reason also, the horizontal area is only attempted.

Again, if the circumjacent lands of a hill be planned or mapped, it is evident we shall have a plan of the hill's base in the middle but were it possible to put the hill's surface in lieu thereof, it would extend itself into the circumjacent lands, and render the whole a heap of confusion: so that if the surfaces of hills could be determined, no more than the base could be mapped.

Roads are usually measured by a wheel for that purpose, called the perambulator, to which there is fixed a machine, at the end whereof there is a spring, which is struck by a peg in the wheel once in every rotation; by this means the number of rotations is known; if such a wheel were 3 feet 4 inches diameter, one rotation would be 10 feet, which is half a plantation perch; and because 320 perches make a mile, therefore 640 rotations will be a mile also; and the machinery is so contrived that by means of a hand, which is carried round by the work, it points out the miles, quarters, and perches, or sometimes the miles, furlongs, and perches.

Or roads may be measured by a chain more accurately; for 80 four-pole, or 160 two-pole chains, or 320 perches, make a mile as before and if roads are measured by a statute chain, it will give you the miles English, but if by a plantation chain, the miles will be Irish. Hence an English mile contains 1760, and an Irish mile 2240 yards; and because 14 half-yards is an Irish, and 11 half-yards is an English perch, therefore 11 Irish perches, or Irish miles, are equal to 14 English ones.

Since some surveys are taken by a four-pole and others by a two-pole chain, and as ground for houses is measured by feet, we will show how to reduce one to the other in the following problems.

PROBLEM I.

To reduce two-pole chains and links to four-pole ones.

If the number of chains be even, the half of them will be the four-pole ones, to which annex the given links. Thus:

1. In 16ch. 371. of two-pole chains, how many four-pole ones? Answer 8ch. 371. But if the number of chains be odd, take the half of them for

1

chains, and add 50 to the links, and they will be four-pole chains and links.

Thus :

2. In 17ch. 421. of two-pole chains how many four-pole Answer 8ch. 921.

ones ?

PROBLEM II.

To reduce four-pole chains and links to two-pole ones.

Double the chains, to which annex the links if they be less than 50; but if they exceed 50 double the chains, add one to them, and take 50 from the links, and the remainder will be the links. Thus :

ch. l.

1. In 8. 37 of four-pole chains how many two-pole ones?

2.

2

16. 37

ch. l.

In 8. 82 of four-pole chains how many two-pole ones?

2. 50

17. 32, Answer.

PROBLEM III.

To reduce four-pole chains and links to perches and decimals of a perch. The links of a four-pole chain are decimal parts of it, each link being the hundredth part of a chain; therefore if the chain and links be multiplied by 4 (for 4 perches are a chain), the product will be the perches and decimal parts of a perch.

Thus:

ch. l.

How many perches in 13. 64 of four-pole chains?

4

Answer, 54. 56 perches.

PROBLEM IV.

To reduce two-pole chains and links to perches and decimals of a perch.

They may be reduced to four-pole ones (by prob. 1), and thence to perches and decimals (by the last); or,

If the links be multiplied by 4, carrying one to the chains when the links are, or exceed, 25; and the chains by 2, adding 1 if

occasion be; the product will be the perches and decimals of a perch. Thus :

ch. l.

1. In 17. 21 of two-pole chains how many perches? 2. 4

Answer, 34. 84 perches.

ch. l.

2. In 15. 38 of two-pole chains how many perches? 2. 4

Answer, 31. 52 perches.

PROBLEM V.

To reduce perches and decimals of a perch to four-pole chains and links. Divide by 4, so as to have two decimal places in the quotient, and that will be four-pole chains and links. Thus: In 31.52 perches how many four-pole chains and links? ch. l.

4)31.52(7. 88, Answer.

35

32

PROBLEM VI.

To reduce perches and decimals of a perch to two-pole chains and links. The perches may be reduced to four-pole chains (by the last), and from thence to two-pole chains (by prob. 2); or,

Divide the whole number by 2, the quotient will be chains to the remainder annex the given decimals, and divide by 4; the last quotient will be the links. Thus:

In 31.52 perches how many two-pole chains and links?

ch. l.

2)31.52(15. 38, Answer.

11

4)152(38

PROBLEM VII.

To reduce chains and links to feet and decimal parts of a foot.

If they be two-pole chains, reduce them to four-pole ones (by prob. 1): these being multiplied by the feet in a four-pole chain will give the feet and decimals of a foot. Thus:

In 17ch. 217. of two-pole chains how many feet?

ch. l.

8. 71 of four-pole chains.

66 feet 1 chain.

5226

5226 Answer, 574 ft. 10 in.

Feet 574.86

12

Inches 10.32

4

1.28

PROBLEM VIII.

To reduce feet and inches to chains and links.

Reduce the inches to the decimal of a foot, and annex that to the feet; that divided by the feet in a four-pole chain will give the four-pole chains and links in the quotient; these may be reduced to two-pole chains and links, if required, by prob. 2. Thus :

In 217ft. 9in. how many two-pole chains? 12)9.00(.75 the decimal of 9 inches.

60

66)217.75(3.29 of four-pole chains, or 6ch. 291.

197

655

61