. ithms: As sine 8° = H C LIs to 4- C D = 300So is kToa-r= OD = 2855-0 9-143555 2-477121 10-000000 TS3356G * Because nCL -| COD. 396 FoRMULiB FOR RuNNING LiNES, When the ground will not admit of this method, ascertain bymeasurement or calculation the distance from B to D. 2 (B F) X —ti? - B L. Now angle L B D = -^ . The tri-angle will then have the two sides B D and B L, and the includedangle B, to find the angle L D B =: CD Kow in the triangle B C D we have the angle B C D = to the sup-B 4- T)piement of , also the angle at D, co
. ithms: As sine 8° = H C LIs to 4- C D = 300So is kToa-r= OD = 2855-0 9-143555 2-477121 10-000000 TS3356G * Because nCL -| COD. 396 FoRMULiB FOR RuNNING LiNES, When the ground will not admit of this method, ascertain bymeasurement or calculation the distance from B to D. 2 (B F) X —ti? - B L. Now angle L B D = -^ . The tri-angle will then have the two sides B D and B L, and the includedangle B, to find the angle L D B =: CD Kow in the triangle B C D we have the angle B C D = to the sup-B 4- T)piement of , also the angle at D, consequently the angle at B. These angles, together with base B D, determine the chord C D;from which, with the angle HCL, calculate R as before. HCLbecomes known from the fact, that C B D gives C B M = G C B, B 4- T)C B M being B — C B D. This taken from —~— will give the angle HCL required. PROPOSITION XVII. Fig. tivo tangents to locate a curve passing through a given point. \Suppose A B and G D to be the tangents permanently fixed with0. B 4- D * Because N B L (isosceles) = ^ exterior angles at N and M = —x—, B being = NBD,andB-?^=?^= = (CLB = —T—) — L B D which will make all the angles known. I Locating Side Tracks, Etc. 39Y reference to some agreement between individuals; and let F be thegiven point at which it is necessarj^ to keep a given distance fromsome building or other object. JSuppose AB and CD produced tomeet in E. The angle O E D, and consequently its half E B D, areknown. The distance I E is also known. Let the angle O E D = 60°, let I F r= 17*5 feet. It is required tofind the point B, so that the angle FBI shall = 30°. By natural sines: ——-— = 35 rzr FB =H D. sme 30 Now 4/(35 + 17-5) X (35 —17-5)* == Vbl-o x 17-5 = 30*3 == I E measures 462 feet. Then B E will equal 462 + 30*3= similar triangles F B : B E :: BI : B K, or 35 : 492 3 :: 30*3 : 426-2 = B K = D B D = 852-4 and B H == 852-4—3
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Keywords: ., bookcentury1800, bookdecade1850, booksubjectenginee, bookyear1856
