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any convenient length, make the angle A D C the measured angle, lay off the distance DC from the scale, at the point C make the angle D C B = angle AD C, erect the perpendiculars C O, and DO, and from the centre O, with o C or o D as radius, describe the arc D S C, which will be the required curve. Now divide the line D C into any number of equal parts, suppose 100 feet each; at the end of each division erect the perpendicular ordinates aa', bb', cc', &c.; and find the length of each, off the same scale to which you plot the figure.
Now proceed to the point D, on the land, and measure the distance Da; at a, measure the ordinate aa', and placing a picket at a' measure a', at right angles to aa', and = a b; having arrived at 1, erect the perpendicular 1 b,=the difference between aa' and bb'; having arrived at b, measure the line b, at right angles to bb' and equal bc, and erect the perpendicular 'c the difference between bb' and cc'; proceed in this manner up to the crown S. The other half of the arc is traced in a similar manner, the only difference is, that the difference between every two adjacent ordinates is laid off, down towards the line D C.
These points, D, a', b', c', e' and c, which are marked on the ground, may be joined by a regular curve. This method is easy, practical and accurate, when due care is taken of the plot and that to a large scale.
The length of all the ordinates are found by the following calculation, (Fig. 25.) The angles D A C and A C B, A C are given. Through c, b, a, draw the perpendicular ordinates; and draw m 1, s 2, t 3, &c., parallel to A C: these parallels are given, being = o c, o b, o a, &c. Join P M.
Take 180 from the sum of the angles D A C and A C B, and you get the angles P A C and PC A, ... you have the angle P. Now to find the radius; we have A o, and the angles of the triangle POA to find A P. To find PO we
have PA A O' PO'. To find the angle n P m: we have m 1 and P m given, to find the angle; say as Pm: R :: m 1: s, angle m P1. We can also find P 1. And Pn – P1 = 1n. In a similar manner find the angles S P 2, P 2, which taken from the radius will give 2 n, the next ordinate proceed in this manner to find each ordinate.
Having the ordinates and their distance apart given, the curve is traced on the ground as before shown.
2. Method. The following method was communicated by Mr. Bowins, C. E,, to the Institution of Civil Engineers, London.
Let A B (Fig. 24,) be a straight line, it is required to lay out the curve K nm, so as to pass it from the straight line: O is the centre of the curve, and K the point at which the line B K is to touch it ; let S K and K n be each one chain -produce S K so as to make K D = S K; and join D n: by producing B K, it is manifest that D A An. In the similar triangles K Dn and Kno, we have Ko: Kn:: Kn: Dn. Therefore Dn Kn; that is, the set is = the chord squared, divided by the radius of the curve. Now let us suppose the radius of the curve to be 100 chains, the chord 1 chain, all reduced to inches, we have 100 X 792: 792 :: 792: 7.92 inches 7828781300 782300 7.92. Now An = 7.922 3. 96 inches, and Kn 792, .. we get < (792) — (3.96)' >= 791.99 inches, nearly one chain.
Construction on the ground. Take K A = one chain, less the hundredth part of an inch, and lay off the perpendicular An 3.96 inches, by a staff for that purpose; then will E be a point on the curve. Again, produce off Kn to H and make n H one chain, and hold the end of the chain fast at n, move the other end towards m, till the distance Hm become 7.92 inches, the off set calculated. In like manner proceed from one chain to another to the end of the curve. When required to pass from curve m
nk 'to the line A B; produce off n K to Q, making KQ 'one chain, then lay off QB in the proper direction, and to half QS; then join K B and produce it at pleasure. 3. Method. Let A B and D C, (Fig. 25,) be the straight lines you are to connect by a curve, having suppose a radius of 60 chains. Set up two theodolites at the points B and C, and from which range a line at right angles to A B and DC, and at the intersection (Q) which will be the centre of the circle, put up a signal to be seen from any point between B and C. Now produce off AB and DC to meet in the point R, then drive stakes into the ground at the end of each, from B to R, and C to R. Having the radius and the distance between each given, to find the set off in direction of the centre. Reduce them all to inches, then R2 x 1 chain squared is the square of the distance from the centre to the end of the first chain; from the square root of this take the radius and you get the first set in inches, by reducing the radius and the chain to inches. For the second set, R2 x 2 chains squared = the distance from the centre to the end of the second chain, the square root less the radius will be the second set off. If you proceed in manner from B to R, you get the set off at the end of each chain; the same sets will answer for the side R C.
But it is much better in all cases where the sets become long to divide the curve into equal parts, and to make use of two tangent lines for each, as Ho and HC for the half Co, and the same for Bo. In the last method we must find the point H, in the field. This is very easy; you have the angle Q R C given, also the radius of the curve and the angle CQO; therefore you have only to make an angle with QC at the point Q = the angle found, which will give the position of Qo, chain in that direction a line = the radius of the curve which will determine the point o in the curve. Then with the theodolite set off the perpen
dicular o H to meet DC in H. Divide o H and HC as before into any number of parts as chains, and find the sets towards the centre from these points, as d', e', f', &c., up to H; these sets will answer for the other side. Now having obtained the several sets, nothing remains, but to go on the ground and set off these in the direction of the object in the centre, and the course of the curve will be marked out as required.
4. Method. To find the sets at right angles to CH and HO, (see last fig.) It is evident h 1, the first set is = (QC' - Ch3) = hi, and gi, the second set = or r(QCC g') = gi, &c.
} (Demonstration of both methods, 47.)
5. Method. This method was described by Mr. Rankine, in a communication to the Institution of Civil Engineers, London. (See the last figure.) Let B be the point where the curve is to start from, allow B a, a b, bc, &c., each one chain: Having the radius given, which may be found if the position of two straight portions of the line be given, and if not, you may assume the radius, and B a one chain, to find the angle B Qa, and its half the angle, a BR; Say, as 1 : 3.1416 :: 2 Q B: the circumference, and again, as the circumference: 360 :: 792 (1 chain :) the degrees in the arc Ba which is the measure of the angle PQa, and its half the angle a BR.
Or 1 chain + radius, divided by the diameter of the curve, will be equal the sine of the angle a BR nearly.
The method of proceeding in the field: place the theodolite at the point B; then lay off from the line B R, the angle a B R, calculated for, and in the direction of the cross hairs set off one chain, which will give the first point a, in the curve; again, lay off from B R, double the angle a BR, and form a set off another chain, and the point where a x is cuts will give the point b, in the curve; proceed after this manner to complete the whole required curve.
Note. If the point B, and the position of A B and D H are fixed, in this case the diameter of the curve is fixed, or if the points B and R are fixed it will limit the diameter, as it must fix the point C. But if the point B be fixed and not R, nor the position of DC, then will the radius depend upon the point you assume in the line BR, and the angle made there; but if you fix the point B, and assume the radius and the angle, then it will fix both a point in BR and the position of CD. In passing from one curve to
other, in the same direction, but the radius of one greater than that of the other, or into a curve of a contrary flexure; it is evident you have no more to do than to set off the tangent to the curve at that point, and work as before.
Required to connect two parallel directions, D A and B C, (Fig. 26,) be a curve of a contrary flexure. Set up the theodolite at the point A, and find the angle DAB; and measure the line AB. Now plot the lines D A, A B, B C ; bisect A B in o, and erect the perpendicular A Q to DA, and at o make the angle AoQ = angle o AQ; from the centre Q with Qo as radius, describe the curve A n o. Again, erect the perpendicular BP to BC, and make the angle BoP angle o BP: from the centre P with PB. radius, describe the arc B mo. Now having the figure constructed, measure off any number of equidistant ordinates from the same scale to which the figure was laid down; or calculate the ordinates to both curves, by the method shown before. Then go upon the ground with the theodolite and trace out the curve as before directed.
Computation of Earth Work.
The excavations and embankments on a railway or canal are resolvable into prismoids, pyramids, and wedges.
To find the solidity of a prismoid.
Rule. To the sum of the areas of the two ends add four