The Field Engineer a Handy book of Practice in the Survey Location And Track

Cover The Field Engineer a Handy book of Practice in the Survey Location And Track
The Field Engineer a Handy book of Practice in the Survey Location And Track
William Findlay Shunk
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fore, = 797. 7 feet, say 798 feet, = A B. To find B C, solve the triangle BDC, observing that the angle DBC = B AI = one-half of the central angle 32 20', = 16 10', and that t> C = 00 feet. Then D C -f- nat. *in. 16 10' = 60 -=- . 278 = say 216 feet, = B C. Hence AC = AB-fBC = 798 -f 216 = 1, 014 feet.
Having thus found the length of chord A C, the radius and rate of curvature may be deduced as in X.
Or, dividing the tabular chord of 32 20' by chord A C = 1, 014, the degree of the required curv
...e is ascertained directly to be 3. 15, equivalent to 3 09'.
2. SECOND METHOD. Find the apex distance AH, = AI + I H. The tabular tangent of 32 20' divided by 4 gives A I = 415 feet. In the triangle KDC, the side DC -i- nat. Sin. K = 60 -f- nat. Sin. 32 20V = 112 feet = KG = I H. Then AH = A I -f- IH = 415 + 112 = 527 feet; and the tabular tangent 1, 661 -f- 527 gives 3. 15, equivalent to 3 09', the degree of the required curve A C, as before.
XXX.
HAVING LOCATED A CURVE A B C, TO FIND THE POINT B AT WHICH TO COMPOUND INTO ANOTHER CURVE OF GIVEN RADIUS, WHICH SHALL END IN TANGENT E F, PARALLEL TO THE TERMINAL TANGENT OF THE ORIGINAL CURVE, AND A GIVEN DISTANCE FROM IT.


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