XXXIII.

HAVING LOCATED A TANGENT, A B, INTERSECTING A CURVE, C D, FROM THE CONVEX SIDE, TO FIND THE POINT E ON SAID CURVE AT WHICH TO BEGIN A CURVE OF GIVEN RADIUS WHICH SHALL MERGE IN THE LOCATED TANGENT.

2. DC, a 1° curve; angle a = 64° 32': to connect with a 49 curve. Here cos. x = (5,730 × 0.43) — 1,433 ÷ (5,730 +1,433) = 0.1439 = cos. 81° 43'; and x — a = 17° 11', equivalent to a distance from B around the 1° curve of 1,718 feet to E, the point at which to start the 4° curve.

son that each is equal to DE.

ter of a semicircle.

Said semicircle touches tangent BA at D,

its middle point; and D E being perpendicular to G F, we have by geometry GE DE DE EF; i.e., GE EF, or RX , DE2. Hence DE BD=DA=√R × r = -R tan. x,

and we are thus enabled to fix the points E and A.

3. In the two foregoing problems, the angle consumed by curve E A is: 180°- -X.

Example.

BE, a 210 curve located; BA, a tangent: to complete the Y with a 6° curve, E A.

log. 3.511482

=

1,3373,247

1.614649, which corresponds to log. cos. 9.614649, or to the decimal number 0.4118, indicating in either case the angle 65° 41' :

=x.

DE=BD=DA = R tan. x = = 2,292 × 0.6455 = €1,479.4. DE may be found also by reference to Table XVI., where the tangent of a 1° curve for 65° 41' is seen to be 3,698.6. Dividing this number by 23, we have 1,479.4, as above.

curves will pass through the points of intersection of the tangents to those curves severally. But lines so drawn in this case bisect also the angles of a triangle, and, demonstrably by geometry, meet in one point equidistant from the three sides of the triangle. That point, therefore, must be a common P. I. for all the curves, and that equidistance the "tangent" length common to them all.

Example.

Given BA, a 3o, and BC, a 4° curve: to complete the Y with a 5° curve, CA.

Adding half the difference to half the sum of the segments of the base EG, we shall have the greater of them; i.e., (3,343 +804) ÷ 2 = 2,073.5, which is the cos. E, E F being radius. Hence 2,073.5 ÷ 3,056

=

log. 3.316704

log.

3.485153 9.831551 cos. 47° 16' = E. By Table XVI., the tangent of a 1° curve corresponding to this angle is 2,507.3: that of a 3° curve, therefore, is 835.8 = the common tangent BD or DA. Multiplying the common tangent by 4, we shall find opposite the product in Table XVI. the central angle of the 4° curve to be 60° 32'; multiplying it by 5, we find, in like manner, the central angle of the 5° curve to be 72° 12'. Arc BA, 47° 16', is equivalent to 1,575 feet on the 3° curve; arc BC, 60° 32', is equivalent to 1,513 feet on the 4° curve. Points being thus fixed at A and C, curve CA can be laid on the ground.

5. If curve BA is concave to the Y, the radii being given, construct the figure as follows:

First draw the triangle GFE, the sides of which are obviously derived from the given radii. Prolong the sides E G and

lines meeting at D, and from D let fall perpendiculars on EB, EA, and GF. Then, comparing triangles GBD, GCD, the angles at G are equal by construction; the angles at B and C are right angles, the side GD common. Hence the triangles are equal in all their parts: BG = GC, and BD = DC. By like reasoning, it appears that CF FA, and DA DC. The point D being equidistant from the right lines E B, EA, which limit angle E, a line bisecting that angle will strike point D.

=

6. It may be remarked, therefore, that lines bisecting the vertical angle and the exterior angles contained between the base and the prolongation of the sides of any triangle, will meet in a point equidistant from the base and the said prolongations. We thus have in the figure all the conditions for fitness of the curves. It remains only to solve the triangle G F E. seeing that from its angles the required central angles can be obtained.

Example.

BA, a 1o, BC, a 6° curve, located: to complete the Y with

=

log.

3.809358

log.

.

3.491782

log.

3.601042

.

The longer segment, therefore, is 4,502; the shorter, 511. Cos. E the longer segment divided by E G = 4,502 ÷ 4,775 — log. 3.653405 - 3.678973 = 9.974432 =cos. 19° 28' = angle E. the shorter segment divided by GF=511 ÷

1,672 =

Cos. GFE = = log. 2.708421 — log. 3.223236 = 9.485185 12= angle G F E.

=cos. 72°

The central angle, B G C, of the 6° curve, is equal to 180 FGE = the sum of the angles at E and F 28' 91° 40', making the arc BC= 1,528 feet.

=

equivalent to 19° 28′ of a 1° curve,

A being thus ascertained, curve AC may be located. It will

=

XXXV.

TO LOCATE A TANGENT TO A CURVE FROM AN OUTSIDE FIXED POINT.

1. If the ground is open, and the curve can be seen from the fixed point, it may be marked by stakes or poles at short intervals, and the tangent laid off without more ado.

2. Suppose, however, that on cumbered ground a trial tangent, A B, has been run out, intersecting the curve at B: it is required then to find the angle BA E, in order that the true