224. To verify the Accuracy of the Cross. Place it at a given station: range with one of the cuts to a welldefined object, and place a stake in the perpendicular; then turn the cross one-quarter round, and if the stake is in the perpendicular, the cross is correct, but if not, the instrument is in error by half the observed deviation.

This will be apparent by reference to Fig. 89. If the angle ACD is acute, the stake will be placed to the left of the true position, as at F. By turning the block one-fourth round, the acute angle will be found at BCE, and the stake will be posited

A

Fig. 89.

DE

G

at G, as far to the right as it was before to the left.

B

225. The Optical Square. The optical square is a much more convenient instrument for drawing perpendiculars than the cross. It consists of a circular box, having a fine vertical slit cut in one side, and directly opposite a circular or oval opening with a vertical line, such as a horsehair stretched across it. The box contains a piece of lookingglass set across it, so as to make an angle of 45° with the line of sight. From the upper half of this glass the silvering must be removed. Half-way between the two openings mentioned is another, to allow the rays coming from an object in the perpendicular to fall on the mirror and be reflected to the eye.

and, looking at the other end of the line, through the openings A and B, directs his assistant, who is seen by reflection through C, to place his rod in such a position that its image shall coincide with the hair across the opening B. HG is then perpendicular to AF.

To find the point in which the perpendicular from a distant point will intersect AF, walk along the line, keeping the line of sight AB directed to the end of the line. When the image of a pole standing at the point from which the perpendicular is to be drawn appears at H, the proper position has been attained.

Set a

226. To test the Accuracy of the Square. Erect a perpendicular with it, as above directed. Then sight along the perpendicular, and if the original line appears perpendicular, the instrument is correct; if it does not, the deviation will equal twice the error of the instrument. pole in the true perpendicular, which will be found as in Art. 224, and alter the position of the glass until the reflected image appears in the proper position. One end of the glass should be movable by screws or by little wedges, so as to allow of its position being rectified.

C.-PARALLELS.

Problem 1.-Through a given point to run a parallel to a

given accessible line.

227. This may be done by Arts. 97, 98, or 99, or thus:

Let AB (Fig. 91) be the line, and C the point. From C to any point D in AB, run out the line CD. From E, any point in CD, run a

line cutting AB in F. Then make EG a fourth proportional

to DE, EF, and EC, or EG

=

EF.EC
ED

and GC will be paral

lel to AB.

Problem 2.-To draw a parallel to an inaccessible line, two points of which are visible.

Problem 3.-To draw a parallel to a given line through an

tersecting DE in E. Measure AB and DE. Run through any point in AB the line BFG, intersecting DE in F.

and CG will be parallel to AB.

DE.BF

Make FG:

=

AB-DE

to AB.

AB: DE:: BC: EC;

BG: FG:: BC: EC, and CG is parallel to EF, or

SECTION III.

OBSTACLES IN RUNNING AND MEASURING LINES.*

A.-OBSTACLES IN RUNNING LINES.

230. IN ranging out lines by the method described in Art. 204, obstacles are frequently met with which prevent the operation being directly carried on. In such cases some contrivance is necessary in order that the line may be prolonged beyond such obstacle. Various methods have been devised for this purpose. The following are among the most simple :—

pendiculars CD and BE long enough to pass the obstacle. Through D and E run the line DG; and at two points F and G beyond the obstacle, set off perpendiculars FH

* In Gillespie's "Land Surveying" may be found a still greater variety of methods for these objects.

and GI equal to CD. Then HIK will be the prolongation of AB.

equilateral triangle FEG, and prolong EG till EH = EB. Describe the equilateral triangle HKI, and KH will be the prolongation of AB.

Fig. 96.,

I

B

C

H

G

233. Instead of making BEH an equilateral triangle, which would sometimes require the point E to be inconveniently remote, run BE (Fig. 96) as before. Set out the perpendicular EG = 1.732 × BE. Describe the equilateral triangle GFI. Bisect FI in H. Then HG will be the prolongation of BC.

F

B.-OBSTACLES IN MEASURING LINES.

234. When, owing to any obstructions, the distance of a line cannot be directly measured, resort should be had to trigonometrical methods. In the absence, however, of the proper instruments, it may be necessary to determine such distances. The following are a few of the many methods that may be employed in such cases :

1. To measure a line when both ends are accessible.

235. Arts. 231, 232, 233, furnish means of determining the distance in this case. By the method Art. 231, BH=