Difference northing and southing (4595'07-566-65)

4028-42;

97882 natural tangent R NG 44° 23' course

of N MN. 44° 23' west, and angle SMD = 39.25, or 44°

23'-5°.

To calculate MN make the difference of latitude 402842 = cosine 44° 23′, and the required distance N M = radius. Then we have

3.751025

To N M 5636-7

The triangles ANI and BMI being similar, we have by logarithms (Davies' Legendre, book II., prop. X)—that is, by "composition and division:"

=

As NM 5636'7
Is to R.

So is sum of radii 4605 (2000+2605)

3.751025 10.000000

To cosine ANI=BMI= 35° 13' Having now determined the angle RNI angle A Ñ I = 35° 13′, the angle RNA becomes ence= 9° 10′.

9-912205

44° 23', and the to their differ

Therefore continue the curve from R towards A, 9° 10′ of curvature, and we have the tangent point A required. Again, we have SMI 39° 23', and the angle BMI 35° 13', consequently curve from S to B 4° 10′ of curvature, and we have the tangent point B required.

Now to find the length of tangent A B, multiply the sum of the radii 4605 by the natural tangent of 35° 13', and we have the length required.

Suppose the two curves to be connected by a common tangent, instead of running in opposite directions as in Case 1st, curve the same way, as GHS and CDEL. It is required to find the position of the tangent S D.

Assume the points H and E; from H lay off tangent HI; from E lay off tangent EF; join F and I by a straight line, if convenient, or by a traverse, if there be obstructions. Let A H be an artificial meridian, and, as in Case 1st, calculate the distance AB, also its course = angle H AG; this will give also the angle EBA.

Suppose radius A H= 14325, tangent HI= 500 feet, angle MIF=6°, IF=1000 feet, NFT = 8°, EF = 600 feet, and radius EB 2865 feet. = We will then have the following traverse, by

which to find the course and distance of A B:

DBA

SA G.

=

6° 28'

Now 47° 10' -40° 42' 6° 28' HAS. Then curve from H

162 feet nearly to S. of 40° 42′ + 8° + 6° : = 54° 42'

Now A B makes with B E an angle Hence we must curve from E to D. 377 feet distance. The

7° 32' curvature

54° 42′ — 47° 10′ : points S and D will be the termini of the required tangent. Then difference of radii x natural tangent (DBE = 47° 10′) = 1432.5 x 1.07864 = 1545 15 = : AKSD = length of tangent. Now when the two curves are so situated as to be seen the one from the other, assume two points as near as you can judge to the true termini of common tangent. Cause about a dozen small

straight stakes or pins to be set up end way about twenty feet apart from one of the assumed points or curves. Then set the instrument at the other, and see how tangent from instrument strikes the row of stakes. Note the difference, and move the instrument until tangent therefrom strikes as tangent to the row of stakes. Make a point where it does. Set the instrument over said point, and in like manner see how tangent from instrument strikes the other curve. Thus we dispense with all the previous calcula

tion.

Having located two curves connected by a tangent, as in Case 2d, Prop. IX., it is required to throw out the tangent, and introduce instead a curve with given radius.

=

Let the radius AS 1432.5 feet, BD = 1637 feet, and their common tangent SD = 220 feet. It is required to find on the two

curves two tangent points, X and Y, from which, if the required radius (say 2865 feet) be drawn, it will pass through the points A and B, intersecting in the centre P, equi-distant from X and Y.

Now in the triangle B A K we have given, difference of radii

BK 1637-1432·5 204·5; also, AK

SD = 220, to find the

angle K A B, its complement KBA SAG,* and the distance A B.

=*92954 = natural tangent of 42° 54′ = KAB.

Therefore its complement K BA=SAG = 47° 5. Now BK × secant KB A = 2045 × 1468801 300-37AB; call it 300 feet. Again, in the triangle BAP we have AB 300, A P = 2865 1432.5 14325, BP = 2865-1637 1228. To find the angles ABP, BPA, and BA P, make AP = 1432·5 feet the base, and let Q be the foot of the perpendicular from B. nometry we have: AP: BP

Then by trigo

BA:: BP-BA: PQ-QA, or

1432.5 1228 + 300 :: 1228 - 300: 989-8 PQ

14325+ 989-8

=

PA

1228

QA. Then

1432.5

=73783 nat. cos. of B A P = 42° 27′.

98628 nat. cos. of BPA = 9° 30′.

PQ 1211.15

Now YBP being a straight line, the angle Y B A ≈ 42° 27′ + 9° 30′ = 51° 57′ and X AP being a straight line, the angle X A G = BAP 42° 27'. Now the angle SAG being 47° 51′ the angle SAX will equal 47° 5-42° 27' 4° 38', and Y BD = 51° 57′

We therefore move back from S 4° 38′ of curvature, or 116 feet to X; also from D 4° 51' of curvature, or 139 feet to Y; we then have the points X and Y, which are to be connected with a 2° curve of 2865 feet radius.

PROPOSITION XI. FIG. 11.

Having located a compound curve terminating in a given tangent, it is required to change the p. c. c., also the length of the last radius, so as to pass through the same terminating point with a given difference in the direction of the tangent.

Let the given curve H A be a 2° of 2865 feet radius compounded to A B, a 2° 30' curve 2292 feet radius, 800 feet in length, and containing 20° of curvature; it is required to move the p. c. c. forward from A towards B, curving therefrom with a shorter radius than 2292 feet, passing through the fixed point B on to tangent with 2° 30' additional curvature.

The following method, though not perfectly accurate, will be

* Because the three angles in the triangle KAB = 180°. Also the sum of the angles on one side the line BG 180°. Subtracting from 180° the angle A and the right angle at K, we have left the angle at B. Subtracting from 180° the angle A (as before) and the right angle 8 A K, we have the angle SAG; hence the angle KBA the angle SA G.