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Draw the tangents, radii, and curves, fixing the P. R. C. midway of D.

Draw the chords GI, I E, the line B F perpendicular to GI, and the line E H in prolongation of radius CE to an intersection with B H passed through centre B parallel to tangents.

That I falls midway of D, follows from the necessary symmetry of the figure; and GIE must be a straight line, because the radii BI, IC, perpendicular to a common tangent at the same point, form a straight to which the chords GI, IE, are equally inclined. CH: CB=

= cos. A; but CH=2 R — D, and CB= 2 R.

.. cos. A (2R — D) • 2 R. BH= B C sin. A= 2 R sin. A;

GF R sin. 1 A; GE=4 GF. .:GE=4 R sin. 1 A, and GI or IE=2 R sin. f A. Observe, that, in the right triangles G KE and B G F, the angles at G and B are each equal to 1 A: hence the triangles are similar.

Escample.
R 800 feet, D 24 feet.
To find angle A.

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Cos. A

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(2 R - D) - 2 R=1,576 • 1,600

= 0.985 = nat. cos. 9° 56'.

BH may then be found : 2 R sin. A 1,600 X 0.1725 276 feet, and laid off from the P. C. at G to K, the point E being fixed by a right angle from K. Or GE may be found 4 R sin. } A

3,200 X 0.866 277.1 feet, and laid off from G to E, the point I being fixed 138.5 feet from G, and angle KGE made equal to half of A= 4° 58'.

2. The distances G K and D given, to find R. In triangle GKE, KE= D.

D ; GK

GE; and GE :

tan. } A; D = sin. 1 A

sin. A=4R.

Or, having found G E, we have from the congruity of triangles G KE, BFG,

D: GE :: 4 GE or GF: R.

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TO CONNECT TWO PARALLEL TANGENTS BY A REVERSED CURVE HAVING UNEQUAL RADII.

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1. Given the perpendicular distance, D, between two parallel tangents, and the unequal radii, R and r, of a reversed curve, to find the central angles, A, the chords, and the Cos. A = CH = BC; but CH= (R – r) – D, and

BC=R+r.
.:. Cos. A = (R+p — D) • (R + r).

The straight reach GK=BH= (R + r) sin. A.
The sum of the chords GE=GK; cos. 1 A.

GI=2 R sin. $ A.
IE= 2 r sin. A=GE – GI.

Example D=28, R=955, r=574.

Cos. A

.

= (R+r — D) • (R+r) = 1,501 • 1,529.
1,501

log. 3.176381
1,529

log. 3.184407 Cos. A, 10° 59'

9.991974

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2. The distances GK and D, and one of the unequal radii, R, given, to find the other radius, r, and the central

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Given the angles A and B, and the length A B of a straight line connecting two diverging tangents, to find theradius of a reversed curve to close the angles.

AI=R x tan. { A; BI= R X tan. 1 B.

.. AB=R X (tan. A + tan. 1 B).

Example A = 16°, B= 10°, A B = 840.

B

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A B, 840

log. 2.924279
*A= 80, nat. tan. 0.14054

{ B=5°, nat. tan. 0.08749
Tan. A + tan. I B= 0.22803 log. -1.357992
R=3,600 .

3.566287

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XLII.

A REVERSED CURVE BETWEEN FIXED POINTS.

Given the angles N and K, and the length of the straight line E F connecting two divergent tangents, to find the radius of a reversed curve from E to F, connecting the tangents.

1. Denote the angle EIC or DIF by I; the angle CEI, complement of N, by n; and the angle D FI, complement of K, by k.

Then, in triangle ECI, –

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