Second Method.-Run A C and make < A C D

Otherwise. Bisect A C in I. In BI produced, make ID = BI. Join C D, which will be parallel to A B.

PROPOSITION 9.

To determine a point C in a right line with A B, the point B being a steeple, or other conspicuous object, which is visible from A and C, and inaccessible from both.

Measure the angles of deflection B A D d D E, e EF, DE and F being suitable points for measuring the angles of deflection. The station F being selected as near C as can be judged upon the ground.

The angles of deflection at D and E being to the left, their sum must be diminished by the

angle at A: the remain

der is the deviation from parallelism of the lines A B and E F. This remainder subtracted from 90°, or a right angle gives the angle of deflection f F G at F, to make F G at right angles with A B prolonged. Run F C G, measuring the distance

F G. Observe the an

gles B F G and F G B, the complements of which are F B C and G B C. The sum of these complements is FB G.

As the angle FB G, in degrees or minutes,
Is to the angle F B C in the same measure,
So is F G

To F C,
Or, as

FBG: GBC: GF: CG.

From either of these proportions the point C is determined, which will be in A B produced, F G

is the complementary course of A B. It may be

run out where there is no local disturbance of the needle by setting the instrument at F, and run F G by this complementary course. The base F G of the triangle B F G should be of such a length, that the angle F B G may be only a few degrees; if it should be too great, the distances F C and C G will not increase in the same ratio with the angles F BC and CB G.

N. B.-If F B G is greater than 10°, the segments F C and G C should be found by the usual rules of Trigonometry.

The following proportion is in substance the same as the preceding. From the external angle BG H, take the BF G; then say,

As this remainder is to the difference between the angle B F G and a right one; so is F G to F C as before.

It may sometimes be convenient to take several stations in deflecting from A to F; but in all cases the angles of deflection to the right hand must be added together, and also those on the left of the lines deflected; the difference of these sums will be

the deviation, from parallelism, of the last line deflected from the line A B.

If the deflected angles on the right exceed those to the left, the difference must be laid off to the left, and vice versa: the telescope will then be parallel to A B.

If at every station we arrive at, we set the vernier to the same degrees as at the last station, reversing the telescope on its axis or in its wyes, and bringing it to bear on the last station point; then, having clamped the lower plate to the tripod, bring the telescope in its direct position on the next station, the vernier will perform the additions and subtractions of the angles of deflection; consequently, when we arrive at any station when the instrument is adjusted by the last station point, bring the vernier to zero, the telescope will be parallel to A B-but if set to 90°, it will be at right angles with it. We may make any angle whatever with A B, by setting the vernier by means of the tangent screw to that angle.

Given < F B C 2° 5' ; G B C 2° 55′ and

74

A MANUAL FOR SURVEYORS.

F G 12 ft., to find F C (the points F and G being determined as above).

As F B G (300′) : F B C (125′) : : F G (12 ft.) :
FC 5 ft.

F B G (300′) : C B G (175′) : : F G (12 ft.) :
G C = 7 ft.

Case 2d. When C falls without the triangle FBG.

Find, as before, the angle F B G, the angle G B C, which is equal to the difference between B G C and a right angle, and the distance F G.

As FBG: < GBC::FG: G C, which, being laid off on F G produced, determines C, a point in A B prolonged. The contents of this proposition and Case 1st of Prop. 7th, are believed to be new;

D

G

B

Ꮯ

nothing of a similar character is to be found in any publication with which I am acquainted.

Given FBC= 50'; GB C-10' and F G 8 ft.