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Let ABCD represent a rectangular parallelogram, or oblong; let the side AB or DC contain eight equal parts, and the side AD or BC three of such parts; let the line AB be moved in the 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 eight; 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 contains equal parts. After the same manner it will appear 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.

PROBLEM I.

To find the contents of a square piece of ground.

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

Any area, or contents in perches, being divided by 160, will give the contents in acres; the remaining perches, if more than 40, being divided by 40, will give the roods, and the last re mainder, 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 the left: the figures on the left are acres; because 10 square four-pole chains make an acre, and the remaining figures on the right are decimal parts of an acre. Multiply the five figures to the right by four, cutting five figures

from the 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, 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 contain 160 square perches; and 160 perches make an acre; therefore 10 times 16 perches, or 10 square four-pole chains, make an acre.

Note.—The chains given or required, in any of the following problems, are supposed to be 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.
PL. 1. fig. 17.

Let ABCD be a square field, whose side is 14ch. 291. ; required the contents in acres. By problem 4, section 1, part 2, 14ch. 291. are equal to 29.16 perches 29.16

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14ch. 291. are equal to 7ch. 291, of four-pole chains, by prob

lem 1, section 1, part 2.
G 3

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A. R. P. Acres 531441 contents, as before, 5 1 10. 4

Rood 1257.64
40

Perches 1030560

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 twenty small, or two of the large divisions are an inch: make a square whose side is that length (by prob. 9, geom.), and it is done.

PROBLEM II.
To find the side of a square whose contents are given.

Extract the square root of the given contents in perches, and you have the side in perches, and consequently in chains.

ExAMPLE.

It is required to lay out a square piece of ground which shall contain 12A. 3R. 16P. Required the number of chains in each side of the square ; and to lay down a map of it by a scale of 40 perches to an inch.

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To draw the map. From a scale where 4 of the large or 40 of the small divisions are an inch, take 45.34 the perches of the side, of which make a square.

PROBLEM III.
To find the contents of an oblong piece of ground.

Multiply the length by the breadth, for the contents.

EXAMPLE.

PL. 1. fig. 3. Let ABCD be an oblong piece of ground, whose length AB is 14ch. 25l. and breadth 8ch. 37l. Required the contents in acres, and also to lay down a map of it, by a scale of 20 perches to an inch. ch. l. perches.

14.25=29.00
8.37 = 17.48 } By prob. 4, sect. 1, part 2.

15732 3496

A. R. P. 160)506.9200(3 0 27 contents.

26 perches, or near 27.
Or thus:
four-pole ch.
ch. l. ch. l.

14.25-7.25
8.37=4.37 : By prob. 1, sect. 1, part 2.

5075 2175 2900

Acres 3|16825
4

Rood 67.300
- 4

Perches 269200

To draw the map. Make an oblong (by schol. to prob. 9, geom.) whose length, from a scale of 20 to an inch, may be 29 perches, and breadth 17.48 perches. ->

PROBLEM IV. The contents of an oblong piece of ground and one side given, to find the other. Divide the contents in perches by the given side in perches,

the quotient is the side required in perches; and thence it may be easily reduced to chains.

EXAMPLE.

There is a ditch 14ch. 25l. long, by the side of which it is required to lay out an oblong piece of ground which shall contain 3A. 0R. 27.P. What breadth must be laid off at each end of the ditch to enclose the 3A. 0R. 27.P. 2

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