\fdO'^ - Old Forge Coal Mines

246 SURVEYING.
table of Radii and Deflections we find the radius of a 3° 15'
14° 12'
curve is 1,763.18 ft.; ^ 1= = 7° OG'; tan 7° 06' =
.12456. Substituting these values in the above formula, we
have T= 1,763 X .12456 = 219.62 ft. Ans.
(669)
See Art. 1252.
(670) The angle of intersection 30° 45', reduced to
decimal form, is 30.75°. The degree of curve 5° 15', reduced
to decimal form, is 5.25°. Dividing the intersection angle
30.75° by the degree of curve 5.25 (see Art. 1252), the
quotient is the required length of the curve in stations of
100 ft. each.
30 75°
^_^ = 5. 8571 full stations equal to 585. 71 ft.
(671) In order to determine the P. C. of the curve, we
must know the tangent distance which, subtracted from the
number of the station of the intersection point, will give us
the P. C. We find the tangent distance T by applying
formula 91, r= ^ tan I /. (See Art. 1251.) From the
table of Radii-and Deflections we find the radius of a 5° curve
is 1,146.28 ft. ; | /= ^IJ^ = 16° 33' ; tan 16° 33' = .29716.
P.C.5°R. Substituting these val-
A ^^ 't^340^3'^J^'^0-^37.8 ^ ues in formula 91, we
J6+97.17]lenSi^'^^O^J!33W have T = 1,146.28 X
'
-29716 = 340.63 ft. In
I
. ^3+59JZ^^'
^^%>\z>
Fig. 67, let A B and
/ ^ C D \>Q. the tangents
/ which intersect in the
/ point E^ forming an
FIG- er.
angle D E F=^^° 06'.
The line of survey is being run in the direction A B, and
the line is measured in regular order up to the intersection
point E, the station of which is 20 + 37.8. Subtracting the
tangent distance, B E= 340.63 ft. from Sta. 20+ 37.8, we
have 16 + 97.17, the station of the P. C. at B. The inter-
section angle 33° 06' in decimal form is 33.1°. Diviling