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

Draw the chords G I, I E, the line BF perpendicular to GI, and the line E H in prolongation of radius C E 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 line, to which the chords GI, IE, are equally inclined. CH: CB= = cos. A; but CH=2R – D, and C B=2 R.

.. cos. A - (2 R – D) = 2 R. BH - B C sin. A=2 R sin. A;

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

Example.
R= 800 feet, D= 24 feet.
To find angle A.

the

Cos. A

(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.

GE; and GE =

D : GK= tan. 4 A; D = sin. 1 A

sin. } A=4R.

Or, having found GE, 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 — q) – D, and

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

=

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

GI=2 R sin. $ A.
- 2 r sin. Í A=GE – GI.

ΙΕ:

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

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Cos. A= (R+9 — 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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IE=GE -- GI= 292.6 — 182.8= 109.8.

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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A REVERSED CURVE HAVING UNEQUAL ANGLES.

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. 4 B,

.. AB=RX (tan. A + tan. } B).

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A B, 840

log. 2.924279
fA= 80, nat. tan. 0.14054

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

3.566287

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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