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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 B F perpendicular to G I, 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 line, to which the chords GI, IE, are equally inclined.

CH CB =

.. GE

=cos. A; but CH2 RD, and C B = 2 R. ... cos. A = (2 R— D) ÷ 2 R.

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GFR sin. † A; GE=4 GF.

4 R sin. A, and GI or IE = 2 R sin. A.

Observe, that, in the right triangles G KE and BG F, angles at G and B are each equal to

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the

A: hence the triangles

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1,600 X 0.1725

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BH may then be found 2 R sin. A

276 feet, and laid off from the P. C. at G to K, the point E being fixed by a right angle from K.

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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 KG E made equal to half of A = 4° 58'.

2. The distances G K and D given, to find R.

In triangle GKE, KE = D.

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

D: GE 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

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Cos. A CH BC; but CH (R − r) — D, and BC=R+r.

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The straight reach G K = BH = (R+ r) sin. A.

The sum of the chords G E = GK÷cos.

GI= 2 R sin. A.

IE=2r sin. A = GE — GI.

Example.

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

A.

Cos. A(R+ r−D) ÷ (R+ r) = 1,501 1,529.

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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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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=RX tan. A; BI

=

RX tan. B.

.. AB=RX (tan. § A + tan. § B).

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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 DFI, complement of K, by k.

Then, in triangle E CI,

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