17. This mode of running curves secures a record of each step in the proceeding; so that, if any error occurs, it can readily be detected. At each turning-point, the number in the "tangent" column must correspond with the central angle due to the length of curve to that point; and at the P. T. that number must correspond with the total central angle. The work can thus be checked with facility during its progress, and checks itself at the end.

18. The young transitman is recommended to rule blanks after the pattern given, and exercise himself thoroughly in computing the parts, and recording the field-notes of various curves assumed at will: drawings are not necessary.

XXII.

TURNING OBSTACLES TO VISION IN TANGENT.

1. Draw CF parallel to A B. Let lines BC, CE, FG, cut these parallels at equal

Suppose B to be a stake 24.50 on the tangent A B, and that a deflection left of 10° be made there for 200 feet to a point C. Set transit at C, vernier reading 10° left. BS to B, and deflect 20° right. Vernier will now read 10° right, and telescope will be in line CE. Make CE= 200 feet. Move to E. See that vernier still reads 10° right. BS to C, and turn 10° left. Vernier will now read zero, and telescope will be in line EG, or tangent A B prolonged.

=

Distance BE = 2BC cos. I
394 feet. Then E = 24.50 + 394,

If a parallel line CF were run, a deflection of 10° right would

=

be made at each of the points C and F. If CF were 250 feet, then BG would be: = 250 + 394 644 feet, and the point G would fall at stake 30.94 on tangent A B prolonged. 2. If angle I = 60°, the other conditions of above method being observed, triangle B HE will be equilateral, and BE

A

B

H

E

G

=

BH=HE. If the parallel D C or DF be run, BE = BD +

=

DC, and BG BD + DF. For field work see last example. 3. In turning obstacles by either of these methods, the

angles should be measured with extreme niceness. Handle the instrument lightly, to avoid jarring the vernier; and, if possible, observe well-defined distant objects in the several short ranges, that the lines of foresight and backsight may accurately coincide.

In locating, the following method is preferable to those given above, and should always be used on long tangents.

4. Having established points A and B on the centre line, the farther apart the better within limits of distinct vision, set off the equal rectangular distances AE, BF,

ranging clear of

the obstacle.

F

Place the transit at E or F, fix points G and H on the forward range, and, rectangularly to these points, establish others on the forward range of the centre line at C and D. The offset distances should be measured very carefully with the rod, or with a steel tape if they exceed in length the pocket rule which every engineer should have about him.

XXIII.

TURNING OBSTACLES TO MEASUREMENT IN

TANGENT.

1. Fix a point on tangent A B prolonged at E. Lay off at B a perpendicular of any convenient length. Move the instrument to D, make the angle BDA

= BDE, and mark the point of intersection A. By reason of symmetry in the triangles ADB, BDE, AB = BE, and may be measured on the ground.

2. Or, fix the point E, and lay

A

C

B

F

off the perpendicular BD as before. Move to D, direct the telescope to E, turn a right angle E D C, mark the point of intersection C, and measure CB. Then, by reason of similarity in the triangles C B D, DBE, CB:BD::BD: BE, .. BE=BD2 BC.

Example.

Suppose BD to be 60 feet, and BC 40 feet. Then BD2 ÷ BC 360040: = 90 feet = = BE.

=

3. Or, with the instrument at D, measure the angle BDE. Then BEBD tan. BDE.

Example.

BD= 120 feet, angle BDE = 54° 40'. BD tan. BDE= 120 X 1.41: = 169.2 feet = BE.

4. Or, without an instrument, lay off any convenient lines B F

or CH. Mark the middle point D. Line out H G, parallel to A B. Mark on it the point G in range with D and E. Then GF BE, or GH = CE.

5. Should the use of a right angle be inconvenient, turn any angle EBD = x, measure BD

about equal by estimation to BE, if the ground permits, and

Then the angle B E D, or 2,

=

180 (x + y), and, by trigonom

etry, sin. z sin. y :: BD: BE, ... BE=BD sin. y÷sin. z.

Example.

Let x=44° 02', y = 71° 48', BD = 300 feet.

=

(x + y) = 180°. (44° 02'71° 48') BE = BD sin. y ÷ sin. z = 300 300 X.95.90316.6 feet.

Then z 180° 180° · 115° 50' = = 64° 10'. sin. 71° 48'sin. 64° 10' :

=

The calculation by logarithms would be as follows:

If E is invisible from B, extend the line DB towards C, until a line CE clears the obstacle. The point E must then be established by intersection of the sides CE, DE, in triangle CDE. Supposing the extension BC to have been 120 feet, the side CD will be 420 feet, the angle y 71° 48'; and, by a calculation similar to the above, the side DE, opposite angle x in the lesser triangle, identical with DE in the larger one, will be found to be 231.7 feet. The sum of the angles at the base CE of the triangle CDE = 180° — y = = 180° - 71° 48' = = 108° 12. By trigonometry, two sides and the included angle being known in any plane triangle, the sum of the known sides is to their difference as the tangent of the half sum of angles at base is to the tangent of half their difference. In triangle CDE, therefore, CD + DE or 651.7 CD-DE or 188.3 :: tan. 108.122 or tan. 54° 06′ tan. 54° 06′ x 188.3651.7 = .399 tan. 21° 45', = half the difference of the angles at the

base.

Log. 188.3..

=

=

2.274850

The angle at C, being evidently the lesser of the two angles at the base, is equal to the half sum of these angles decreased by their half difference, =54° 06′ - 21° 45' 32° 21'.

Set the transit, then, at C, foresight to D, deflect 32° 21' left, and fix in that range two points F and G, between which a cord may be stretched, and as nearly as can be judged on opposite sides of E. Move to D, foresight to C, deflect 71° 48' right, and establish a point E at intersection with FG. Cross to E, BS to D, and deflect the angle z = 64° 10′ into the line of the tangent A B prolonged.