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To reduce two-pole chains and links to four-pole chains
1. If the number of chains be even, divide them by 2, and to the quotient annex the given number of links. Thus, in 16 two-pole chains and 37 links, there are 8 four-pole chains and 37 links. Or because each link is the hundredth part of a four-pole chain, the four-pole chains and links may be written thus 8.37 four-pole chains.
2. If the number of chains be odd, divide by 2 as before, and for the 1 that is to carry, add 50 to the given number of links. Thus in 17 two-pole chains and 42 links, there are 8 four-pole chains and 92 links, or 8.92 four-pole chains.
To reduce two-pole chains and links, to perches and
decimals of a perch.
Multiply the links by 4 and the chains by 2. If the links when multiplied by 4, exceed a hundred set down the excess and carry 1 to the chains. Thus 17 two-pole chains and 21 links = 34.81 perches; also 15 two-pole chains and 38 links = 31.52 perches.
To reduce four-pole chains and links, to perches and
decimals of a perch.
Multiply the chains and links by 4. Thus 13.64 four To reduce square four-pole chains to acres.
Divide by 10, and the quotient will be acres. If there be decimals in the quotient, multiply by 4 and by 40 to obtain the roods and perches.
EXAMPLE. In 523.2791 square chains, how many acres ?
Ans. 52A. IR. 12P.
Observation on Chaining. All slant or inclined surfaces, as the sides of a hill, should be measured horizontally and not on the plane or surface of the hill. To effect this the hind end of the chain, in ascending a hill, should be raised from the ground, till it is on a level with the fore end, and by means of a plummet and line, should be held perpendicularly above the termination of the preceding chain. In descending a hill the fore end of the chain should be raised in the same manner, and the plummet being suspended from it, will shew the commencement of the succeeding chain.
The bearing or course of a line is its situation in reA line running due north and south is.called a meri. dian line.
The bearing of a line is expressed by the angle contained between it, and a meridian line passing through one of its ends, and is said to be north so many degrees east or west, or south so many degrees east or west, according as the line runs between the north and east or north and west, or between the south and east or south and west.
The bearings of lines are generally taken with an instrument called a Circumferentor, or more commonly a Surveyor's Compass. A description of this instrument or of the method of using it, is deemed unnecessary, as it will be better understood from a few minutes inspection of the instrument itself, and an explanation from a person acquainted with the manner of using it, than from a detailed description in writing.
The bearing of a line taken at one end, is the reverse of the bearing of the same line taken at the other end :* thus if the bearing of a line AB taken at the end A, be north 350 east, its bearing taken at the end B, will be south 35' west.
When the bearings of two lines, running from the same point, are given, the angle contained between them may be found by the following rules.
RULE 1. When the bearings of both lines are between the north and east or north and west, or between the south and east or south and west, subtract the less bear
* Note. This is not strictly true, but the difference is too small to be observed in practice. In the latitude of 40° the greatest difference between the bearing and the reverse bearing of a line a mile in length, is only 44'1.
ing from the greater ; the remainder will be the angle contained between them : thus if AB bear N. 34. E. and AD, N. 58° E. the angle BAD will be = 21°. Fig. 67.
Rule 2. When the bearing of one of the lines is between the north and east and the other between the north and west, or when one is between the south and east and the other between the south and west, add them together; the sum will be the angle contained between them: thus if BA bear S. 340 W. and BC, S. 35° E., the angle ABC will be = 69o. Fig. 67.
Rule 3. When the bearing of one of the lines is between the north and east and the other between the south and east, or when one is between the north and west and the other between the south and west, add them together and subtract the sum from 180°; the remainder will be the angle contained between them : thus if CB bear N. 35o W. and CD, S. 87° W. the angle BCD will be = 58°. Fig. 67.
RULE 4. When the bearing of one of the lines is between the north and east, and the other between the south and west, or when one is between the north and west and the other between the south and east, add 180 to the less bearing, and from the sum subtract the greater; the remainder will be the angle contained between them : thus if DC bear N. 87° E. and DA, S. 58° W. the angle ADC will be = 151°. Fig. 67.
Containing rules for finding the areas of triangles, qua
drilaterals, circles, and ellipses ; also the method of protracting a survey and finding its area by dividing it into triangles and trapeziums.
To find the Area of a Parallelogram, whether it be a
Square, a Rectangle, a Rhombus, or a Rhomboides.
Multiply the length by the height or perpendicular breadth, and the product will be the area.*
Note. Because the length of a square is equal to its height, its area will be found by multiplying the side by itself.
* DEMONSTRATION. Let ABCD (Fig. 68) be a rectangle; and let its length AB and CD, and its breadth AD and BC, be each divided into as many equal parts, as are expressed by the number of times they contain the lineal measuring unit; and let all the opposite points of division be connected by right lines. Then, it is evident that these lines divide the rectangle into a number of squares, each equal to the superficial measuring unit; and that the number of these squares is equal to the number of lineal measuring units in the length, as often repeated as there are lineal measuring units in the breadth, or height; that is, equal to the length drawn into the breadth. But the area is equal to the number of squares or superficial measuring units; and therefore the area of a rectangle is equal to the product of the length and breadth.
Again, a rectangle is equal to any oblique parallelogram of an equal length and perpendicular height (36.1.); therefore the area of every parallelogram is