To calculate MN make the difference of latitude 4028:42 = cosine 44° 23', and the required distance N M = = radius. Then we have 3.751025 The triangles A NI and BMI being similar, we have by logarithms (Davies' Legendre, book II., prop. X)—that is, by "compo sition and division:" As N M = 5636'7 Is to R. So is sum of radii 4605 BMI 35° 13' To cosine ANI Having now determined the angle RNI angle A NI 35° 13', the angle RNA becomes ence 9° 10'. 9.912205 44° 23', and the to their differ Therefore continue the curve from R towards A, 9° 10' of curvature, and we have the tangent point A required. Again, we have SMI 39° 23', and the angle BMI = 35° 13', consequently curve from S to B 4° 10' of curvature, and we have the tangent point B required. Now to find the length of tangent AB, multiply the sum of the radii 4605 by the natural tangent of 35° 13', and we have the length required. Suppose the two curves to be connected by a common tangent, instead of running in opposite directions as in Case 1st, curve the same way, as GHS and CDEL. It is required to find the position of the tangent S D. Assume the points H and E; from H lay off tangent HI; from E lay off tangent EF; join F and I by a straight line, if convenient, or by a traverse, if there be obstructions. Let A H be an artificial meridian, and, as in Case 1st, calculate the distance A B, also its course angle H AG; this will give also the angle E B A. Suppose radius A H 14325, tangent HI=500 feet, angle MIF 6°, IF 1000 feet, NFT 8°, EF 600 feet, and radius EB 2865 feet. We will then have the following traverse, by which to find the course and distance of A B: = = Now A B makes with B E an angle Hence we must curve from E to D 377 feet distance. The Now 47° 10' -40° 42': = 6° 28' HAS. Then curve from H 6° 28' 162 feet nearly to S. of 40° 42′ + 8° +6° 54° 42' 54° 42'-47° 10' : 7° 32′ curvature points S and D will be the termini of the required tangent. Then difference of radii x natural tangent (D BE = 47° 10′')= 1432.5 × 107864 = 1545 15 = AK = SD = length of tangent. Now when the two curves are so situated as to be seen the one from the other, assume two points as near as you can judge to the true termini of common tangent. Cause about a dozen small Fig 10. straight stakes or pins to be set up endway about twenty feet apart from one of the assumed points or curves. Then set the instrument at the other, and see how tangent from instrument strikes the row of stakes. Note the difference, and move the instrument until tangent therefrom strikes as tangent to the row of stakes. Make a point where it does. Set the instrument over said point, and in like manner see how tangent from instrument strikes the other curve. Thus we dispense with all the previous calcula tion. PROPOSITION X. FIG. 10. Having located two curves connected by a tangent, as in Case 2d, Prop. IX., it is required to throw out the tangent, and introduce instead a curve with given radius. Let the radius AS 1432.5 feet, BD = 1637 feet, and their common tangent SD = 220 feet. It is required to find on the two curves two tangent points, X and Y, from which, if the required radius (say 2865 feet) be drawn, it will pass through the points A and B, intersecting in the centre P, equi-distant from X and Y. Now in the triangle B A K we have given, difference of radii BK 1637 - - 1432·5 2045; also, A K=SD 220, to find the angle K A B, its complement K B A = S A G,* and the distance A B. •92954 natural tangent of 42° 54′ = K A B. 47° 5. 1228. Therefore its complement K BASAG 1432.5 Now BK × call it 300 300, AP = To find the feet the base, Then by trigo AP: BP + BA:: BP-BA: PQ-QA, or 1228 + 300 :: 1228 300: 989-8 = PQ 1432 5989.8 1432.5 -989.8 = PQ = 1211·15, and = QA. Then = QA = 221·35. 2 2 Now YBP being a straight line, the angle Y BA = 42° 27′ + 9° 30′ 51° 57′ and X A P being a straight line, the angle X A G = BAP 42° 27'. Now the angle SAG being 47° 5' the angle SAX will equal 47° 54′-42° 27′ = 4° 38', and Y BD = 51° 57′| — 47° 54′ — 4° 51'. -- We therefore move back from S 4° 38' of curvature, or 116 feet to X; also from D 4° 51' of curvature, or 139 feet to Y; we then have the points X and Y, which are to be connected with a 2° curve of 2865 feet radius. PROPOSITION XI. FIG. 11.. Having located a compound curve terminating in a given tangent, it is required to change the p. c. c., also the length of the last radius, so as to pass through the same terminating point with a given difference in the direction of the tangent. Let the given curve H A be a 2° of 2865 feet radius compounded to A B, a 2° 30' curve 2292 feet radius, 800 feet in length, and containing 20° of curvature; it is required to move the p. c. c. forward from A towards B, curving therefrom with a shorter radius than 2292 feet, passing through the fixed point B on to tangent with 2° 30' additional curvature. The following method, though not perfectly accurate, will be * Because the three angles in the triangle KAB=180°. Also the sum of the angles on one side the line BG 180°. Subtracting from 180° the angle A and the right angle at K, we have left the angle at B. Subtracting from 180° the angle A (as before) and the right angle S AK, we have the angle SAG; hence the angle KBA the angle SA G. found sufficiently so for most practical purposes. Had the 20 curve H A been continued 800 feet farther, to a point C, the variation BC would be equal 28 feet.* Now by compounding to a 2° 30' curve I turn off with the instrument for the chord AB 2° the instrument set at the required point P, with a backsight on A, and a foresight on B, I turn off 20° 2° 30' 11° 15', that is 3° 15' 2 instrumental deflection over and above that required for a con tinuous 2° curve to C; the curve P B will therefore be shorter than A B in the ratio of 3° 15' to 2°; hence the proportion : 32:800 492 length of curve P B. AP then will equal 800-492308 feet of 2° curve; but 308 feet of a 2° curve gives 6° 10' of curvature; hence PB contains 22° 30' -6° 10' 16° 20' of curvature in 492 feet distance; then 16.333... we have 4.92 = 3.3198° = 3o 19', or 1728 feet radius for the curve PB. It will be sufficiently accurate, however, to continue the 2° curve 310 feet to P, and then run 490 feet of a 3° 20' curve. Were HA a tangent by making A P the same length and rate of curvature as above, the curve PB would be the same also. *2 x 1.75 x 8 = 28. |