CHAPTER II.

OBSTACLES TO ALINEMENT.

A. TO PROLONG A LINE.

(409) The instrument being set at the farther end of a line, and directed back to its beginning, the sights of the Compass, if that be used, will at once give the forward direction of the line. They serve the purpose of the rods described in Art. (169). A distant point being thus obtained, the Compass is taken to it and the process repeated. The use of the Transit or Theodolite, for this purpose, was fully explained in Art. (376).

(410) By perpendiculars. When a tree, or house, obstructing the line, is met with, place the instru

Fig. 278.

and a fourth at E, which last will be in the direction of AB pro

longed. If perpendiculars cannot be conveniently used, let BC and DE make any equal angles with the line AB, so as to make CD parallel to it.

sure a distance CD = BC. Then will D be a point in the line AB prolonged. At D, turn 60° from CD prolonged, and the new direction will be in the line of AB prolonged. This method re quires the measurement of one angle less than the preceding.

Fig. 280.

B

EF

(412) By triangulation. Let AB be the line to be prolonged. A Choose some station C, whence can be seen A, B, and a point beyond the obstacle. Measure AB and the angles A and B, of

C

the triangle ABC, and thence calculate the side AC. Set the instrument at C, and measure the angle ACD, CD being any line which will clear the obstacle. Let E be the desired point in the lines AB and CD prolonged. Then in the triangle ACE, will be known the side AC and its including angles, whence CE can be calculated. Measure the resulting distance on the ground, and its extremity will be the desired point E. Set the instrument at E, sight to C, and turn an angle equal to the supplement of the angle AEC, and you will have the direction, EF, of AB prolonged.

In

(413) When the line to be prolonged is inaccessible. this case, before the preceding method can be applied, it will be necessary to determine the lengths of the lines AB and AC, and the angle A, by the method given in Art. (430).

(414) To prolong a line with only an angular instrument. This may be done when no means of measuring any distance can be obtained. Let AB be the line

Fig. 281.

Also, at

point D where this direction intersects the direction BD. C, deflect 90° from B. Then, at D, deflect 90° from DB. The intersections of these last directions will fix a point E. At E deflect 135° from EC or ED, and a line EF, in the direction of AB will be obtained and may be continued.

B. TO INTERPOLATE POINTS IN A LINE.

(415) The instrument being set at one end of a line and directed to the other, intermediate points can be found as in Art. (177), &c. If a valley intervenes, the sights of the Compass, (if the Compass-plate be very carefully kept level cross-ways), or the telescope of the Transit or Theodolite, answer as substitutes for the plumb-line of Art. (179).

(416) By a random line. When a wood, hill, or other obsta

cle, prevents one end of the line, Z,

from being seen from the other, A, run

a random line AB with the Compass or A

Transit, &c., as nearly in the desired

Fig. 282.

Z

B

direction as can be guessed, till you arrive opposite the point Z. Measure the error, BZ, at right angles to AB, as an offset. Multiply this error by 571%, and divide the product by the distance ABВ. The quotient will be the degrees and decimal parts of a degree, contained in the angle BAZ. Add or subtract this angle to or from the Bearing or reading with which AB was run, according to the side on which the error was, and start from A, with this corrected Bearing or reading, to run another line, which will come out at Z, if no error has been committed.*

Example. A random line was run, by compass, with a Bearing of S. 80° E. At 20 chains' distance a point was reached opposite to the desired point, and 10 links distant from it on its right. Required the correct Bearing.

Ans. By the rule,

10 × 57°.3
2000

= 0°.2865 = 17'. The cor

rect Bearing is therefore S. 80° 17′ E. If the Transit had been used, its reading would have been changed for the new line by the same 17'. A simple diagram of the case will at once shew whether the correction is to be added to the original Bearing or angle, or subtracted from it.

If Trigonometrical Tables are at hand, the correction will be

BZ

more precisely obtained from this equation; Tan. BAZ = AB In this example, AB

BZ 10
2000

=

.005 = tan. 17'.

The 57°.3 rule, as it is sometimes called, may be variously modified. Thus, multiply the error by 86°, and divide by one and a half times the distance; or, to get the correction in minutes, multiply by 3438 and divide by the distance; or, if the error is given in feet and the distance in four-rod chains, multiply the former by 52 and divide by the distance, to get the correction in minutes.

The correct line may be run with the Bearing of the random line, by turning the vernier for the correction, as in Art. (312).

Fig. 283.

(417) By Latitudes and Departures. When a single line, such as AB, cannot be run so as to come opposite to the given point Z, proceed thus, with the Compass. Run any number of zig-zag courses, AB, BC, CD, DZ, in any convenient direction, so as at last to arrive at the desired point. Calculate the Latitude and Departure of each of these courses and take their algebraic sums. The sum of the Latitudes will be equal to AX, and that

XZ

of the Departures to XZ. Then is Tan. ZAX= ;

A

i. e. the algebraic sum of the Departures divided by the algebraic sum of the Latitudes is equal to the tangent of the Bearing.*

(418) When the Transit or Theodolite is used, any line may be taken as a Meridian, i. e. as the line to which the following lines are referred; as in "Traversing," Art. (373), page 254, all the successive lines were referred to the first line. In the figure, on the next page, the same lines as in the preceding figure are repre

sented, but they are referred to the first course, AB, instead of to the Magnetic Meridian as before, and their Latitudes are measured along its produced line, and its Departures perpendicular to it. As before, a right-angled triangle will be formed, and the angle ZAY will be the angle at A between the first line AB and the desired line AZ.

Fig. 284.
Z

B

D

Y

This method of operation has many useful A applications, such as in obtaining data for running Railroad Curves, &c., and the student should master it thoroughly.

The desired angle (and at the same time the distance) can be obtained, approximately, in this and the preceding case, by finding in a Traverse Table, the final Latitude and Departure of the desired line (or a Latitude and Departure having the same ratio) and the Bearing and Distance corresponding to these will be the angle and distance desired.

portion, CD: CA :: EF: EG, will give the distance EG, from

E to a point in the line AB. So for other points.

and the angle ACB, and thence calculate the angle CAB. Then observe any angle ACD, beyond the obstacle. In the triangle ACD, a side and its including angles are known, to find CD. Measure it, and a point, D, in the desired line, will be obtained.