lines meeting at D, and from D let fall perpendiculars on EB, EA, and GF. Then, comparing triangles GBD, GCD, the angles at G are equal by construction; the angles at B and C are right angles, the side GD common. Hence the triangles are equal in all their parts: BG GC, and BD = DC. By like reasoning, it appears that CF FA, and DA = DC. The point D being equidistant from the right lines E B, EA, which limit angle E, a line bisecting that angle will strike point D. = E OF 6. It may be remarked, therefore, that lines bisecting the vertical angle and the exterior angles contained between the base and the prolongation of the sides of any triangle, will meet in a point equidistant from the base and the said prolongations. We thus have in the figure all the conditions for fitness of the curves. It remains only to solve the triangle G F E, seeing that from its angles the required central angles can be obtained. Example. BA, a 1o, BC, a 6o curve, located: to complete the Y with The longer segment, therefore, is 4,502; the shorter, 511. the longer segment divided by E G = 4,502 ÷ 4,775 = log. 3.653405 Cos. E = - 3.678973 9.974432 cos. 19° 28' angle E. Cos. GFE the shorter segment divided by G F =511÷ log. 3.223236 9.485185 =cos. 720 1,672 = log. 2.708421 12/= angle G F E. = 72° 12' 19° The central angle, BG C, of the 6o curve, is equal to 180 FGE = the sum of the angles at E and F 28' 91° 40', making the arc BC= 1,528 feet. equivalent to 19° 28′ of a 1° curve, = The arc BA, 1,947 feet. Points C and 107° 48', equivalent, on = A being thus ascertained, curve AC may be located. It will consume an angle = 180° 72° 12/ an 80 curve, to 1,347.5 feet. XXXV. TO LOCATE A TANGENT TO A CURVE FROM AN OUTSIDE FIXED POINT. 1. If the ground is open, and the curve can be seen from the fixed point, it may be marked by stakes or poles at short intervals, and the tangent laid off without more ado. 2. Suppose, however, that on cumbered ground a trial tangent, A B, has been run out, intersecting the curve at B: it is required then to find the angle BAE, in order that the true Example. = A B 1,500 feet; DH B, a 4° curve; angle FBD = 20° 13'. First, the angle FBD, between a tangent and a chord, is equal to half the central angle subtended by the same chord. Angle D C B, therefore, 40° 26'. By Table XVI., the chord of 40° 26', for a 1° curve, 3,960.2 feet; for a 4° curve, it is, say, 990 feet DB; and DI IB = 495 feet. The mid. ord. HI is, in like manner, found to be 88.25 feet. from the radius of the 4° curve, we have IC = = Deducting this 1,344.4 feet. Then ICIA tan. IAC; i.e., 1,344.4 ÷ (495 + 1,500) = = 0.674 = =tan. 33° 59′ = angle I A C. HE B = Next, by geometry, the proposed tangent A E=√ADXAB =√2,490 × 1,500 =1,932.6; and E C÷A E=tan. E A C= 1,432.691,932.6=0.7413=tan. 36° 33′ = angle E A C. Then EAC-IAC = 36° 33′ — 33° 59′ = 2° 34′ = angle B A E, the angle required, which can accordingly be laid off from the fixed point A, and the tangent located. XXXVI. TO SUBSTITUTE A CURVE OF GIVEN RADIUS FOR A TANGENT CONNECTING TWO CURVES. Example. 1. A B, a 4° curve; BC = 774 feet; CD, a 6° curve: to put in the 1° curve, EF. Sketch the figure as in margin, H K being parallel and equal In the triangle G III we have then the sides given; namely say, 910 feet, HI= 5,730 - 955 5,730 1,433=4,297 feet: to GH = = - find the angles. Under Case 3, Trigonometry (III.), IH IG+ GH:: IG -GHILLH; i.c., 4,775 : 5,207: 3,387 3,693, the difference of the segments into which the base III is divided by a perpendicular from G. Adding half the difference of the seg 4,775 feet, and GI B K PH = ments thus found to half their sum, the longer segment, I L, is found to be 4,234 feet; subtracting half the difference from half the sum, the shorter segment, LH, is found to be 541 feet. Then HL HG = 541910 0.5945 cos. 53° 31' angle GIII. In like manner, dividing IL by IG, we find the angle GIH to be 9° 49'. The sum of these angles angle EGH 63° 20′, for the reason that each is equal to 180 — HIGI. Finally, EGH KGH = 63° 20' - 58° 19' . = = 5° 01' angle EG B, equivalent to a distance from B of 125 feet around the 4° curve to the P. C. C. at E; and GIH-EG B = 9° 49' 5° 01' = 4° 48': angle CHF, equivalent to a distance from C of 80 feet around the 6° curve to the P. C. C. at F. XXXVII. TO RUN A TANGENT TO TWO CURVES ALREADY LOCATED. 1. If one curve be visible from the other, or if both be visible from some inter-M mediate point, mark them on the ground with stakes at short intervals. The points M or L in the range of the required tangent may then be fixed by one or two trial settings of the transit, and 2. Should obstacles prohibit this plan, measure any convenient line, FG or B CD, from one to the other curve, and, completing the traverse AFGE or ABCDE, determine thence the bearings and distances asunder of the centres A and E. The right triangle A E K, in which EK = the sum of the radii, may then be solved, and the points H and I ascertained as in the following example: = Total northing, 3,514 feet; total easting, 4,391 feet. = = 1.2496 = = tan. 51° 20' bearing = distance A E. A E; and 4,391 Also, EKAE = (1,433955) ÷ 5,573 sin. 25° 22: = 90° 00' 25° 22' 64° 38'. = angle EAK; and angle A EK Hence the bearing of AK or HI is N. 76° 42′ E., and that of AH or IE, N. 13° 18′ W. = Since AB bears N. 20° E., the angle H A B 20° 00'+13° 18'33° 18', equivalent to a distance of 833 feet from B around the 4° curve to the required P. T. at H; and, since DE bears N. 45° 00′ W., the angle IED = = 45° 00' 13° 18' 31° 42', equivalent to a distance of 528 feet from D around the 6° curve to the required P. C. at I. = 3. Should the curves turn in the same direction, the side EK of the triangle A EK is equal to the difference of the radii instead of their sum. In other respects, the method exemplified will apply to that case also. 4. The preceding solution may be useful as an exercise. But the problem is one of rare occurrence, and the conditions must be extraordinary which prevent a close approximation, at least, to the true line in the field. The better way in actual practice, then, is to run out a trial tangent as nearly right as possible. If it errs by passing outside the objective curve, close with a compound (XXIX.); if that error be inadmissible, or if it errs by cutting the objective curve, measure the miss, |