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objects, and for the purpose of fixing the table in the same relative position in different stations.

There is also an index-rule of brass, IR, fitted with a telescope or sights, one edge of which, called the fiducial edge, is in the same plane with the sights, and by which lines are drawn on the paper to represent the direction of any object observed through the sights. This rule is graduated to serve as a scale of equal parts.

31. PROBLEM V. To survey with the plane table from a station within the field.

Place the table at the station O (fig. to 25); adjust it so that the magnetic needle shall point to north on the compass card, or else observe the bearing of the needle, and fix on some point in the paper on the table for this station; bring the nearer end of the fiducial edge of the index-rule to this point, and direct the sights to the corner A, and draw an obscure line with the pencil or a point along this edge to represent the direction OA; measure OA, and from the scale lay down its length on the obscure line, and then the point A is determined. Draw on the table the lines OB, OC, OD, and OE, exactly in the same way. The points A, B, C, D, E, being now joined, the plan of the field is finished, and its contents may be computed as explained in article 23, by measurements taken on the plan.

The angles at O, subtended by the sides of the field, can also be measured at the same time by placing the frame with that side uppermost which contains the angular divisions, and then the contents of the field can be calculated independently of the plan.

32. PROBLEM VI.-To survey with the plane table, by taking stations at all the corners of the field but one, and measuring all its sides.

Let ABCDE (fig. to 26) be the field; place the table at some corner, as A, and mark a point in the paper where most convenient to represent that station; adjust the instrument, as to the direction of the magnetic needle, as in last problem. Apply the nearer end of the index-rule to this station point, and direct the sights to the station E, and

draw an obscure line as before to denote the direction AE; then, in a similar manner, through the station point, draw a line for the direction AB; measure AE and AB, and with the scale lay off these measures on the obscure lines denoting AE and AB. Remove the instrument now to the second station B, and place it so that the needle shall rest at the same point of the compass card as before. If the index-rule is now laid along the direction of AB, the first station A will coincide with the sights, if the table is properly placed. With the index-rule draw an obscure line through the second station point to represent BC; measure BC, and lay the distance off on the line BC on the paper; remove the table to the third station C, adjust its position as before, and draw a line to represent the direction CD; measure CD, and, by the scale, lay this length off on CD on the paper; and, lastly, place it at D as before, and draw a line to represent DE; this line will meet AE in E, if the work has been correctly performed. The plan of the field is now completed.

33. PROBLEM VII.-To survey a field from two stations.

Let O and Q (fig. to 27) be the two stations. Fix the table at O, and adjust it as formerly; assume a convenient point on the paper for the first station O; draw an obscure line to represent OQ as before; measure OQ, and, with the the scale, lay this length off on OQ on the paper, and the point for the station Q is determined. Then draw obscure lines to represent the lines OC, OB, &c., drawn from O to the angles of the field, without measuring these lines, as in article 31; and having placed the table at Q, and adjusted it, draw obscure lines from Q to represent the lines drawn from Q to the corners of the field; and the intersections of these lines, with the former lines from O, will determine the corners C, B, A, &c., and the plan will be completed.

34. PROBLEM VIII.-To survey more than one field with the plane table.

Having surveyed one of the fields according to any of the methods in the three preceding problems, fix on a station in this field, whose position is known on the paper, and

take some station in the adjoining field at a sufficient distance; then, from the former station, draw an obscure line in the direction of the latter, measure the distance between them, and lay it off from a scale on the paper; and thus the new station in the adjoining field is determined on the plan. Place the table in this station, adjust it, and if it is correctly placed, and the index-rule placed on the line joining the two last stations, the sights will coincide with the station in the first field. Proceed to the planning of this second field; then, in a similar manner, plan the next, and so on till the whole survey is finished, and then measure it as before by means of the plan (23.)

When a new sheet of paper is required in consequence of that on the table being filled, some line must be drawn on the latter at the most advanced part of the work, and the edge of the former being applied to it, the station lines must be produced on this sheet. Before drawing the line, the latter sheet must be held in such a position as is most convenient for continuing the next part of the work upon it. The first sheet being removed from the table, and this one, previously moistened, fixed in by means of the frame, the work may be continued after the paper has got dry. When this sheet is filled, another is similarly fixed on the table; and when the survey is completed, the sheets can all be accurately joined by means of the connecting lines.

At the beginning of the work, the position of some conspicuous object or mark may be laid down on the paper, and at any stage of the subsequent operation, its position may be ascertained, and if it coincide with the first position, it is a proof that the work is correct; if not, some error must have been committed, which must be rectified before proceeding further; with this check, the greatest accuracy may be secured in the survey.


From a station within a hexagonal field the distances of each of its corners were measured, and also their bearings ; required its plan and area, the measurements being as below. Ans. 12 ac. 3 ro. 6.45 pls.

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This exercise is to be solved like that under problem II.


35. It frequently becomes a problem in land-surveying to cut off a certain portion from a field. When the field is of a regular form, this process may be frequently effected by a direct method; but in the case of irregular fields, it can be accomplished only by indirect or tentative methods.

36. PROBLEM IX.-To cut off a portion from a rectangular field by a line parallel to its ends.

Find the area of the field, then as its area is to that of the part to be cut off, so is the length of the field to the length of the part.

Let AD be the field, AF the part

to be cut off, then

'Divide the area of the required part by the breadth of the field, and

the quotient will be the length."






Let a area of AF, and b = the breadth AC, and 7 =

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Or, if A = area of the field AD, and L = its length AB,

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The area of a rectangular field is 10 acres 3 roods 20 poles, its length is 1500 links, and breadth 725 links; it is required to cut off a part from it of the contents of 2 acres 28 poles by a line parallel to its side.

A 10 ac. 3 ro. 20 pls. = 1087500 links
= 217500 ...

a= 2 ac.

0 ro. 28 pls.

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When any aliquot part is to be cut off from the field, find the same part of the base, and it will be the length of the required part.


A rectangular field is 1250 links long and 320 broad; it is required to cut off a part of it, to contain 1 acre 2 roods 16 poles, by a line parallel to one of its ends; what is the length of this part? Ans. 500 links. 37. PROBLEM X.-To cut off any portion from a triangle by a line drawn from its vertex.

Let ABC be the triangle, and APC the part to be cut off, then

Divide the area of the required part by the altitude of the triangle, and the quotient will be half the length of its base.'




If a = area of the required part APC, 1= AP, and


h = altitude of the triangle, then /= or /=


2 a

Or, if A area of the given triangle ABC, L= its base

AB, then A: a=L:l, and l=





The length of one side of a triangular field is 2500 links, and the perpendicular upon it from the opposite corner is 1240 links; it is required to cut off a triangular portion from it, by a line drawn from the same angle to this side, so that its contents shall be 5 acres 16 poles.

a = 5 ac. 0 ro. 16 pls. = 510000 links

2a 1020000

Hence, /= =

h 1240

= 822.6 links = AP.


Cut off from a triangular field, as in the preceding exercise, a part containing 2 acres 1 rood 24 poles, the length of one

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