practised, especially where the fences are high, or the ground hilly, thus preventing the directions of the straight fences being seen to distant chain lines, as in the cases of the fence a b, o c, with respect to the chain line A D.

THE METHOD OF MEASURING HILLY GROUND.

When the ground, over which lines are measured, rises or falls, or both alternately, the horizontal distances are what are required in plotting the survey, as well as for finding the content thereof, and not the actual distances measured along the surface of the ground.

For many ordinary purposes the horizontal measurement may be obtained by holding the end of the chain up so as to keep it horizontal, as nearly as can be judged by the eye, the arrow being placed vertical under the end so held up: but when a large and accurate survey is required, the distances must be measured along the line of ground, and the angles of elevation and of depression of the several inclined parts of the line must be taken, either with a common quadrant, or afterwards with the theodolite (to be hereafter described), and the lengths of the several rises and falls must be noted; from which the corresponding correct horizontal distances may be readily computed. The following table shews the number of links to be subtracted from every chain, or 100 links, for the angles there set down.

TABLE shewing the reduction in links and decimals of a link upon 100 links for every half degree of inclination from 3° to 20° 30'.

By this table the trouble of computation is avoided, only the

distance, measured on each rise or fall, requiring to be multiplied by the reduction in chains corresponding to the angle of each rise or fall, and the product, subtracted from that distance, will give the correct distance, as in the following

EXAMPLES.

1. A line was measured 12.43 chains, on ground having a rise of 8 degrees, required the horizontal length of the line. Here to 810, or 8° 30′ corresponds the reduction 1.10 links, whence

13.6730 links

12.29.327 horizontal distance, in which the decimal, being less than half a link, is rejected; thus making the correct horizontal distance 12.29 chains, or 1229 links.

2. The acclivity of a hill rises 20°, and measures 16.14 chains, its declivity falls 11° and measures 32-28 chains, required the horizontal distance between the extremities of the line thus measured over the hill, it being level at the top 2.80 chains,

49.60 chains, the horizontal distance required.

NOTE. When fences are crossed, stations made, &c., on the acclivity or declivity of a hill, the horizontal distance up to such points must be found. Some surveyors place the arrow forward a distance equal to reduction, due to the angle of acclivity or declivity, at the end of every chain measured, and thus obviate the necessity of reducing the line afterwards, having for the purpose a small pocket quadrant, so graduated that the plumb-line thereof shews, on observing the angle of elevation or depression, the reduction required for each chain. great deal of trouble is thus saved as the theodolite cannot be conveniently carried about for this purpose. Such pocket quadrants are not made by the mathematical instrument makers, being the productions of clock makers or other mechanics, according to the various designs of surveyors.

A

THE USE OF THE PARALLEL RULER,

IN REDUCING CROOKED FENCES TO STRAIGHT ONES, TO FACILITATE THE COMPUTATION OF THE CONTENTS OF FIELDS.

As some surveyors prefer the parallel ruler to the method already given, for reducing crooked fences to straight ones, the method of using that instrument for this purpose is here given. This method is founded on a well-known proposition of Euclid, in which it is shewn that triangles on the same base, and between the same parallels, are equal.

Let A B C, ABD be triangles on the same base AB, and between the same parallels A B, CD; then the triangle ABC is equal

to the triangle A B D. And, if the triangle A E B, which is common to the other two triangles, be taken away, the remaining triangles A E C, BED will also be equal; whence equal areas may be transferred from one side of a line to the other, which is the principle, on which, as already said, the following Problems are founded.

PROBLEM I.

IT IS REQUIRED TO REDUCE THE OFFSET-PIECE A B C DE TO A RIGHT ANGLED TRIANGLE AEC, BY AN EQUALIZING LINE Ec, WITH THE PARALLEL RULER.

to B, which will cut A c at a, where a mark must be made. Lay the ruler from a to D, and the further side thereof being now held fast, bring the near side to C, marking A c at b. Lay the ruler from 6 to E, move it parallel to D, marking Ac Join Ec; then A Ec is a right angled triangle required, and its area may be found by taking half the product of A E and A c.

at c.

THE FOLLOWING IS A GENERAL RULE FOR SOLVING PROBLEMS OF THIS KIND.

Draw a temporary line, as A c, at right angles, or at any other angle to the chain line, as A E, of the offsets.

1. Lay the ruler from the first to the third angle, and move it parallel to the second angle; then make the first mark on the temporary line.

2. Lay the ruler from the first mark on the temporary line to the fourth angle, and move it parallel to the third angle; then make the second mark on the temporary line.

3. Lay the ruler from the last named mark to the fifth angle, and move it parallel to the fourth angle; then make the third mark on the temporary line.

4. Lay the ruler from the last named third mark on the temporary line to the sixth angle, and move it parallel to the fifth angle; then make the fourth mark on the temporary line.

In this manner the work of casting by the parallel ruler may be conducted to any number of angles. Great care must be taken, during the operation, to prevent the ruler slipping, as such an accident will derange the whole of the work, if not discovered and immediately corrected.

PROBLEM II.

TO REDUCE A CURVED OFFSET-PIECE TO A RIGHT-ANGLED TRIANGLE.

5

Let A abcde B be the curved offset-piece. Divide the curve by points a, b, &c., 4 so that the parts A a, a b, &c., may be straight or nearly so; and draw A 5 perpendicular to A A B. Lay the ruler from A

1

to b; move it parallel to a, and mark A 5 at 1. Lay the ruler from 1 to c; move it parallel to b, and mark A 5 at 2. Lay the ruler from 2 to d; move it parallel to c, and mark A5 at 3 Lay the ruler from 3 to e; move it parallel to d, and mark A 5 at 4. Lay the ruler from 4 to B; move it parallel to e, and mark A5 at 5. Draw the line B5; then will A B 5 be a right angled triangle equal in area to the offset-piece A a b c de B, as required.

EXAMPLES FOR PRACTICE ON THE TWO PRECEDING

PROBLEMS.

1. Lay down a right-lined offset-piece, from the fol.cwing notes; reduce it to a triangle by the parallel ruler; and find its content, both by calculation from the offsets and the casting of the ruler.

The area found by calculation from the offsets is Oa. 2r. 17p. and the perpendicular of the triangle, found by the casting of

the parallel ruler, is 161 links. Hence

751 X 161
2

= 0.60455

It hence

square links, or somewhat more than Oa. 2r. 16p. appears that the method of casting by the ruler, gives the content very near the truth. In fact this method is mathematically accurate; but the danger of error, as already said, arises from the accidental slipping of the ruler during the operation.

2. Lay down and find the area of a curve-lined offset-piece, from the following field notes: and find its area both by calculation, and by the casting of the parallel ruler.

By both the methods the area is found to be a. 2r. 151p. nearly, the perpendicular on A B being found 199 links by the casting operation.

PROBLEM III.

TO REDUCE THE IRREGULAR FIELD ABCDEFGHK TO A

TRAPEZIUM BY THE PARALLEL RULER.

Prolong the line A K at pleasure. Lay the ruler from K to G; move it parallel to H, and mark A K prolonged at 1. Lay the ruler from 1 to F; move it parallel to G, and mark A K