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Also, in triangle D FI, -
DF: DI:: sin. I : sin. k... DF X sin. k=DI X sin. I.

Adding these equations, we have

EC X sin. n +DF X sin. k= (CI+DI) X sin. I.

F

K

H

But EC and D F are each equal to R; sin. n = cos. N; sin. k=cos. K; and CI+DI=2 R. Hence the equation becomes,

RX (cos. N + cos. K)=2R X sin. I.

.. sin. I= (cos. N + cos. K) • 2.

The foregoing elegant solution is abridged from Henck. 2. Angle A = 180 — (n + I); angle B = 180 — (K + I).

To find radius, draw FH parallel, and E H perpendicular, to CD.

Then EH=EF X sin. I. But EH EG + GH; EG=R X sin. A; and GH=R X sin. B.

.. EF X sin. I=RX (sin. A + sin. B).

Example. EF=1,400, N= 30°, K= 200.

Sin. I=(cos. N+cos. K) ; 2.
N= 30°, nat. cos.

0.86603
K= 20°, nat. cos.

0.93969

1.80572 1.80572 • 2= 0.90286 - : nat. sin. 64° 32'.

...I= 64° 32'.

A= 1800 (n+1)= 1800 (600 + 64° 32') = 55° 28'. B - 1800 (k +1)= 180° (700 + 64° 32') = 45° 28'. R=EF X sin. I = (sin. A + sin. B).

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3. The young student should bear in mind that the addition or subtraction of the logarithms of two natural numbers gives a logarithm representing, not the sum or difference, but the product or quotient, of such numbers. When, therefore, as in the two foregoing cases, the sum or difference of two or more trigonometric functions — sines, tangents, and the like - is sought, the logarithm of the sum of the natural functions, and not the sum of their logarithms, is to be used. If, for example, sin. A X sin. B is required, the log. sin. A + log. sin. B= the logarithm of the product of the sines designated; but, if sin. A + sin. B is sought, the natural sines of those angles must be added together, and the logarithm of the sum of these natural functions must be used in making logarithmic calculations.

XLIII.

TO CONNECT TWO DIVERGENT TANGENTS BY A

REVERSED CURVE.

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1. ADVANCING TOWARDS THE INTERSECTION OF TANGENTS.

Given the angle of divergence, N, the initial P. C. at G, the distance GH, and the radii R, r, to find the central angles A and B.

GK=CG X cos. N= R cos. N.
GL=GH X sin. N.
GK - GL=LK or EF, C F being drawn parallel to L E.
Cos. B=DF= DC=(r + EF) • (R+r).
Angle GCK= 90° — N; angle DCF= 90° — B.

Angle A =GCK-DCF= (90° — N) — (90° B) = B - N.

Example.
N= 24° 30', G H=854, R= 1,440, r = 1,146.

GK=Rcos. N.

.

R= 1,440
Cos. N, 24° 30'

log. 3.158362
log. 9.959023

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LK or EF=GK - GL=1,310 — 354 = 956.

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B= 35° 38'.
A =B - N= 35° 38' — 24° 30' = 11° 08'.

JE

2. RECEDING FROM THE INTERSECTION OF TANGENTS.

Given the angle of divergence, N, the initial P. C. at G, the distance GH, and the radii R, r, to find the central angles A and B.

GK=GH X tan. N. KC=GC-GK=R-GK. LC or EF=KC X cos. N, the line C F being drawn parallel to LE.

Cos. B=DF=CD= (r + EF) • (R+r).

Example.
N=18° 30', G H = 920, R=955, r=819.

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KC=R-GK= 955 — 307.8= 647.2.

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B= 36° 08
A=B+N= 36° 08' + 18° 30' = 54° 38'.

XLIV.

TO SHIFT A P. R. C. SO THAT THE TERMINAL TANGENT SHALL MERGE IN A GIVEN TANGENT PARALLEL THERETO.

Given the reversed curve E F G, ending in tangent GV: to find the angle of retreat, A, on the first branch, and the angle C of the second branch, ending in tangent HT, parallel to GV.

Measure the error TG=D, perpendicular to the terminal

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