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\fdO'^ - Old Forge Coal Mines

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(656)<br />

(657)<br />

(658)<br />

(659)<br />

(660)<br />

See Art. 1245.<br />

See Art. 1246.<br />

See Art. 1246.<br />

See Art. 1248.<br />

See Art. 1 249.<br />

SURVEYING. 245<br />

(661) A 5° curve is one in which a central angle of<br />

5° will subtend a chord of 100 ft. at its circumference. Its<br />

radius is practically one-fifth of the radius of a 1° curve,<br />

and equal to 5,730 ft. -^ o = 1,146 ft.<br />

(662) The degree of curve is always twice as great as<br />

the deflection angle.<br />

(663) See Art. 1 249 and Fig. 282.<br />

(664) Formula 90, C' = 2 i^ sin Z>. (See Art. 1250.)<br />

(665) Formula 91, T= R tan ^ /. (See Art. 1251.)<br />

(666) The intersection angle C E F^ being external to<br />

the triangle A E C^ ys, equal to the sum of the opposite<br />

interior angles A and C. A =22° 10' and C = 2S° 15'.<br />

Their sum is 45° 25' = C E F. Ans.<br />

The angle A E C^ISO" - (22° 10' + 23° 15') = 134° 35'.<br />

From the principles of trigonometry (see Art. 1243),<br />

we have<br />

sin 134° 35' : sin 23° 15':: 253.4 ft. : sidey^^fi";<br />

whence, side A E = 140.44 ft., nearly. Ans.<br />

Also, sin 134° 35' : sin 22° 10' :: 253.4 ft. : side C E\<br />

whence, side C E = 134.24 ft., nearly. Ans.<br />

(667) We find the tangent distance T by applying<br />

formula 91, r=i^ tan ^ /. (See Art. 1251.) From the<br />

table of Radii and Deflections we find the radius of a 6° 15'<br />

curve = 917.19 ft.; \ /=^^^^i^=17° 35'; tan 17° 35' =<br />

.31G9. Substituting these values in the formula, we have<br />

7^= 917.10 X .3169 = 290.66 ft. Ans.<br />

(668) We find the tangent distance T by applying<br />

formula 91, r= A' tan ^7. (See Art. 1251.) From the

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