PROBLEM II. To lay out a Railway Curve by the common Method. Let A B, C D be two straight or tangental portions of a railway, the points B, C being required to be joined by a circular curve B C, to which A B, C D shall be tangents at the points B and C ; and let BO be the radius of the curve, which is supposed to be deter mined by one or other of the cases in Problem I. accordingly as the map is found to be accurate or inaccurate. Put the radius B0 = r; measure on the tangent A B prolonged the distance B p1 = 1 chain, 1 as is usual in practice; set off Pi qi = *at right angles to B P1; 2 r then is the first point in the curve. Through B q1 measure the right line B 91 P2 = twice B p1 =2 chains, and set off P2 92 = 14 = twice p1 9, at right angles to B q1 P2; then q2 is the second point in the curve, repeating the last operation till the curve reach the point C. Lastly, 93 C P5 being measured 2 chains, the last offset p5 D P5 * Demonstration. Complete the semicircle B C Q, prolong B O to Q, and join q, Q, B 42. Then because B P1, BP, B q2 are always, in practice, so very small compared with B Q that they nearly coincide with the curve, and consequently Bp, is approximately Bq2;.. by the nature of the = B91 = = will be found the first offset p11, or half the preceding one p4 C, if the work be right. See the following notes and example. NOTE 1. Only a small part of the operation for a curve of great length is shown in the figure; but as the whole of the work, except the first and last offsets, is alike, to show more would be unnecessary. 1 NOTE 2. The values of for all radii, from 15 chains to 540, are given in 2 r Table No. 4. at the end of the work. Ex. Let B0 = r = 80 chains = 1 mile; then which being multiplied by 792, the number of inches in 1 chain, gives 4.95 inches, the first offset p1 91; whence 4.95 × 2 = 9.9 1 × 792 160 inches, the second offset p2 92. Or, by Table No. 4. opposite 80, in the column marked Radius, &c. stands 4.95 inches, in the column marked Offsets, &c. which is the as above. value of p1919 NOTE 3. By this method the greater part of British, as well as foreign, railway curves, have been laid out, having been invented by the author about 25 years ago, when the Stockton and Darlington railway was laid out. It was eagerly adopted by railway surveyors, as it involves very little calculation, and does not require the use of an angular instrument. It is, however, defective in practice, on account of its requiring the coupling together of so very many short lines, as small errors will unavoidably creep in and multiply, and more especially so if the ground be rough, so that the curve has frequently to be retraced several times before it can be got right, especially if it be a long one. This defect in the above method induced the author to prepare three other methods, which are given in the following Problems, and which he would recommend to engineers according to circumstances. See Observations on the invention of the methods of laying out railway curves at the end of these Problems. NOTE 4. When the curve shall have been laid out with sufficient accuracy, the rods or quills that mark the ends of the offsets, 91, 92, &c. must be taken out, and their places supplied by strong wooden stumps, about 16 or 18 inches in length, and 11⁄2 inch square, each end of the curve being marked with three stumps, or one large one with a cross or some other conspicuous mark on it. The straight portions of the line must also be marked with stumps at the end of every chain. PROBLEM III. To lay out a Railway Curve by Ordinates or Offsets from its Tangents, no material Obstructions being supposed to exist on the convex Side of the Curve, to prevent the use of the Chain. CASE I. When the length of the Curve is less than of its Radius. Let B C be the curve, A B, C D tangents thereto, which are ranged till they meet at T, and B O the radius of the curve, which must be correctly determined by one or other of the cases in Prob. I. according to circumstances. Measure on B T the usual distance Вpi 1 The successive offsets, at the end of every chain being &c.*, or what amounts to the same thing, the = * Demonstration. Draw the radius q, 0, and 3 a 1 to OB. Put O B = O q3=r, and 9, a=p, B=8; then Oa√72-82, and Ba= = P3 93 = O B −Oa =r-√ r2 82. Now, if 8 be taken successively = 12, 22, 32, &c. chains, the values of P1 91, P2 92, P3 93, &c. will be respectively r- √r2—12, r— √r2 —22, r — √√/r2 — 32, &c. But as 12, 22, 32, &c. are very small compared with r2, within the limitation assigned to Case I. of this Problem, the values of √/r2—12, √r2—22, &c. may be taken respectively r 12 2r' 3 22 = exceeds of its radius, only 5 or 6 of the offsets ought to be taken in this manner, and the remainder, say, commencing with p, 97, ought to be taken r- r2-72, &c. Although in a curve of 80 chains' radius, the 10th offset by the contracted method exceeds the same offset by the correct method by only of a link out of 621 links. This difference, however, becomes gradually greater as the distance on the tangent approaches the 20th chain. 2d, 3rd, 4th, &c. offsets, are respectively 22, 32, 42, &c. times the first offset, or 4, 9, 16, &c. times the first one. Having laid out the offsets in this manner, till the last one, which suppose to be p4 94 is either within, or very little more than, a chain from T, make p’4 T P4 T, and lay out the same offsets in an inverted order on T C, as were laid out on B T, that is, beginning with the greatest first and ending with the least. Ex. Let the radius of the curve be 80 chains, then p1 91 P3 93 = = 4.95 inches; whence P2 2 = 4 × 4.95 = 19.8 inches, 44.55 inches = 3 ft. 8.55 in. &c., or the value of the first be taken from the table at the end of the book. 9 × 4.95 offset may To make the Distances of the Stumps equal. As it can rarely happen in practice that the distance 94 4 will be =1 chain, when this is the case, it will be better to set out the curve from C as was done from B: then, if the distance 94 7'4 be less than 1 chain, suppose m links less, lay the chain from q'4 to q′3, and at the m (1—m) 2 r distance q'4 t = m, lay off the offset ts= then s will be the point for the stump, instead of 4, thus making q ́4 s = 1 chain. This operation must be repeated between 93 and 92, 92 and 91, &c., the last point to be changed necessarily falling on the tangent C D. If the distance 94 94 be greater than 1 chain by m links, lay the chain from q' in the direction of 94, and, at the distance of q ́4 v=m links, (1+m) m lay off the offset v u= then u will be the place for the stump; " 2 r next lay the chain from q, in the direction of И, and repeat the previous operation: this must be done to the end of the curve, the last mark to be changed falling, in this case, within the curve. The stumps may now be put down as pointed out in Note 4. Prob. II. 4 Ex. 1. When the distance 94 9'4 is less than 1 chain by 40 links, (1-m) m the radius of the curve being 80 chains. Here t s = 2 r (1-.40) x .40 160 ch.= .60 x.40 x 792 = 1.188 in. = 1 inch nearly. CASE II. When the Curve is any required Length. As it is inconvenient in practice to have the offsets greatly to exceed two chains in length; if, therefore, the curve be a long one, the offsets may be confined within proper limits by dividing the curve into two or more parts, and introducing one or more additional tangents. In the annexed figure the curve B C is divided into two parts in C', at which point the tangent T C' T' is drawn, meeting the tangents BT, TC in T and T', the tangents B T, T C' being each taken = BP3 + P3 93*, wherein B p. is taken the nearest whole number of chains to B O, and P3 93 is the offset corresponding to B p. By thus taking the length of B T, T C', the last offset on BT and the first one on T C will meet at 93, the middle point of B C', and therefore the distances of the ends of the offsets, that form the curve, will all be sufficiently near to equality, i. e. to one chain. Also by trigonometry, BT: BO:: rad. tan. BTO = /BTC'; whence the BT C' becomes known, and consequently the direction of the new tangent, T C' T' is also known. Having, therefore, determined BT and BTC', the offsets may be laid out to P3 93 ; and having made TP3 = T p'3, the same offsets may be laid out, in an inverted order on T C'; the order of the offsets being a second * Demonstration.-By the similar triangles O B T, 93 P3 T, OB: BP3 + P3 T 1 n (BT) :: P3 93: PT; but since Bp3- O B, or the nearest whole number to it, and since p,T must always be very small compared with Bp3, it may be rejected; 1 n ..OB: OB ( = 1 very nearly to B T) :: P3 93 : P3T=¿P3 93, wherein n may be taken any whole number, 8 being the most convenient, as in the example. |