Draw a blank line with a ruler of a length greater than the diameter of the protractor; pitch upon any convenient point therein, to which apply the centrehole of your protractor with your pin, turning the arc upwards if the angle be less than 180, and downwards if more; and observe to keep the upper edge of the scale, or 180 and 0 degrees upon the line; then prick off the number of degrees contained in the given angle and draw a line from the first point through the point at the degrees; upon which lay the stationary distance. Let this line be lengthened forwards and backwards, keeping your first station to the right, and second to the left; and lay the centre of your protractor over the second station, with your pin, turning the arc upwards, if the angle be less than 180, and downwards, if more; and keeping the 180, and 0 degrees on the line, prick off the number of degrees contained in the given angle, and thro' that point and the last station draw a line, on which lay the stationary distance and in like manner proceed through the whole.

:

In all protractions, if the end of the last station falls exactly in the point you began at, the field-work and protraction are truly taken, and performed; if not, an error must have been committed in one of them; in such case make a second protraction: if this agrees with the former, and neither meet or close, the fault is in the field-work, and not in the protraction; and

193

SECTION IV.

Containing two Methods by which the Areas of rightlined figures may be determined.

DEFINITION.

THE area or content of any plane surface in perches, is the number of square perches that surface contains.

Let ABCD represent a rectangular parallelogram or oblong: let the side AB, or DC, contain 8 equal parts; and the side AD, or BC, three of such parts; let the line AB be moved in direction of AD, till it has come to EF; where AE, or BF (the distance of it from its first situation) may be equal to one of the equal parts. Here it is evident, that the generated oblong ABEF, will contain as many squares as the side AB contains equal parts, which are 8; each square having for its side one of the equal parts, into which AB, or AD, is divided. Again let AB move on till it comes to GH, so as GE, or HF, may be equal to AE, or BF; then it is plain that the oblong AGHB, will contain twice as many squares as the side AB con

tain's equal parts. After the same manner it will ap pear, that the oblong ADCB will contain three times as many squares as the side AB contains equal parts; and in general, that every rectangular parallelogram, whether square or oblong, contains as many squares as the product of the number of equal parts in the base, multiplied into the number of the same equal parts, in the height, contains units, each square having for its side one of the equal parts.

Hence arises the solution of the following problems.

PROB. I.

To find the content of a square piece of ground.

1. Multiply the base in perches, into the perpendicular in perches (or spuare the base) the product will be the content in perches; and because 160 perches make an acre, it must thence follow, that

Any area, or content in perches, being divided by 160, will give the content in acres; the remaining perches, if more than 40, being divided by 40, will give the roods, and the last remainder, if any, will be perches.

Or, thus ;

2. Square the side in four-pole chains and links, and the product will be square four-pole chains and links; divide this by 10, or cut off one more than the decimals, which are five in all from the right towards

10 square four-pole chains make an acre, and the remaining figures are decimal parts of an acre. Multiply the figures to the right by 4, cutting 5 figures from the first product, and if any figure be to the left of them, it is a rood, or roods; multiply the last cut off figures by 40, cutting off five or (which is the same thing) by 4, cutting off four; and the remaining figures to the left, if any, are perches.

1. The first part is plain, from considering that a piece of ground in a square form, whose side is a perch, must contain a perch of ground; and that 40 such perches make a rood, or stang, and four roods an acre; or which is the same thing, that 160 square perches make an acre as before.

2. A square four-pole chain (that is a piece of ground four-poles or perches every way) must con tain 16 square perches; and since 160 perches make an acre, therefore 10 times 16 perches, or 10 square four-pole chains make an acre.

Note, That the chains given or required, in any of the following problems, are supposed two-pole chains, that chain being most commonly used; but they must be reduced to four-pole chains or perches for calculation, because the links will not operate with them as decimals.

EXAMPLES,

fig. 17. D.

B

Let ABCD be a square field, whose side is 14.20; I demand the content in acres.

Ch. L.

By problem 4. section 3. 14.29 are equal to 29.16 perches; then

A. R. P.

29. 16 x 29.16-850.3056-5 1 10

Or, thus:

Ch. L.

Ch. L.

14.26 are equal to 7. 29 of four-pole chains, by prob. 1. sect. 3. and 7. 29+7. 29=53. 1441 chains, which divided by 10 gives the content as before.

It is required to lay down a map of this piece of ground, by a scale of twenty perches to an inch.

Take 29.16 the perches of the given side, from the small diagonal on the common surveying scale, where 20 small, or two of the large divisions are an inch; make a square whose side is that length (by prob. 9,