. Engineering and Contracting . e to argue that recent high prices are attributab!* oats. llj lb., at ..... 27!o8 to decreased productivity in America.—Editor.] ^rl^kx^. .^!.. !^^ ^.^ .^^... eiP (157) 618 Eiifjineering and Contracting for June 2, 1920. A New Method for Laying OutCircular Curves by Deflec-tions from the P. I.* By T. F. HICKERSON, Associate Professur Civil EiigineeriiiB, University of North Carolina. The writer hopes that the tables based upon formulas givenbelow will fill the longfelt need of a smple and time-savingmethod for laying out circular curves by deflections fro
. Engineering and Contracting . e to argue that recent high prices are attributab!* oats. llj lb., at ..... 27!o8 to decreased productivity in America.—Editor.] ^rl^kx^. .^!.. !^^ ^.^ .^^... eiP (157) 618 Eiifjineering and Contracting for June 2, 1920. A New Method for Laying OutCircular Curves by Deflec-tions from the P. I.* By T. F. HICKERSON, Associate Professur Civil EiigineeriiiB, University of North Carolina. The writer hopes that the tables based upon formulas givenbelow will fill the longfelt need of a smple and time-savingmethod for laying out circular curves by deflections from thepoint of intersection of the tangents (the P. 1). thus avoidingthe trouble of moving the instrument and resetting the ver-nier. Referring to Fig, 1. P is any point on the circular arc CBand A is the point of intersection of the tangents. Also C isthe point of curve (P. C), and B is the point of tangent(P. T.). Lines from points A and O to point P make angles ofe and oc with the line AO, these angles being plus when meas-. Fig. above AO and minus when below it. PN is drawn per-pendicular to AO. The deflection angle is called A-PN R sin ct Tan e = = = AX E -f R — R cos ocR sin oc R (sec % A — 1) + R — R cos oc sin ocHence, tan 9^- (1). sec % A — cos aFormula (1) shows that for a given value of A. the angle eis independent of the radius of the curve or the length of thecurve. Imagine the curve divided into ten equal parts, then for-mula (1) gives the deflections to these points of division asfollows sin 4/10 A oc = 4/10 A. tan ©i = ~ sec % A — cos 4/10 Asin 3/10 A cc = 3/10 A. tan 9: = — sec % A — cos 3/10 Asin 2/10 A oc = 2/10 A. tan «, = sec \<z A — cos 2/10 Asin 1/10 A oc = 1/10 A. tan 9. = sec 1/4 A — cos 1/10 Aec = 0 e. = occ = — 1/10 —e, oc=—2/10 , = —9,oc =—3/10 A. e, = — 6: cc= —4/10 = —9, cc=—5/10 A. e,„ = —1/2 (180 —A »? Values of 9„ 9:, «,. etc., computed by means of the aboveformulas for diff
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