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of which 112.7 feet, tan. A E for the 4o curve. Move to E, deflect go R; range out the line E F, made up of EB - A E
112.7 feet, and BF any convenient distance, say 90 feet. This 90 feet is the assumed tangent of some unknown angle on the 6o curve. To find the angle, multiply 90 by 6, and seek the product, 540, in the tan. column of Table XVI., where it is found opposite 10° 46'. By moving then to F, deflecting 100 46' R, and measuring FC = 90 feet, the point C is fixed on the second curve.
3. Should unexpected obstacles be met in carrying out either of these plans, the triangles AGC or EGF may be solved, and the point C fixed by ineans of the lines A G, G C.
4. The application of the foregoing methods to urning obstacles on simple curves needs no special instance.
TO SHIFT A P. C. SO THAT THE CURVE SHALL
TERMINATE IN A GIVEN TANGENT.
1. Suppose a 3o curve AB to have been located, containing an angle of 44° 26', and ending in tangent B E: required, that it shall end in tangent DF, parallel to B E. It is plain, from the
diagram, that if the curve and its initial tangent be moved forward, like the blade of a skate, until the terminal tangent merges in D F, the P. T. will have traversed the line BD, equal and parallel to AC. If, therefore, on the ground at B, the angle E B D, equal to the whole angle consumed by the curve, in this case 44° 26', be laid off to the right, and the distance B D to the range of the proposed terminal tangent be measured, the equal distance A C, from the original to the required P. C., is thus directly ascertained.
Should such direct measurement be impracticable, range out the tangent BE, and, at any convenient point, measure the distance from it square across to the proposed terminal tangent DF, say 56 feet. Then in the right triangle B ED, makthe sine ED= 56 feet. Hence, by trigonometry, ED: nat. sin. 41° 26', or 56 = 0.7, = BD= 80 feet, = distance A C along the initial tangent, from the erroneous to the correct P. C.
2. This problem occurs more frequently than any other in the field; and the young engineer should have it by heart, that the distance square across between terminal tangents, divided by the natural sine of the total angle turned, will give him the distance he is to advance or recede with his P. C. to make a fit.
3. Excepting on precarious rocky steeps, city streets, or like exact confines, to strike within two feet of any point designated in the project, may be considered striking the mark. Astronomical nicety, whether with transit or level, in an ordinary railroad location, is mere waste of time.
4. The observant reader will not fail to perceive that the foregoing rule applies to systems of curves, or to compound lines also, the angle E B D being the angle included between the initial and terminal tangents, let what flexures or indirections soever have been interposed; and that, if the angle referred to be either 189° or 360°, adjustment by shift of P. C. is impracticable. In those cases, a change of radius becomes necessary.
TO SUBSTITUTE FOR A CURVE ALREADY LOCATED, ONE OF DIFFERENT RADIUS, BEGINNING AT THE SAME POINT, CONTAINING THE SAME ANGLE, AND ENDING IN A FIXED TERMINAL TANGENT.
1. Suppose the 4o curve AB, containing an angle of 32° 20', to have been located, and that it is required to substitute for it another curve A C, which shall end in a parallel tangent C F, 60 feet to the right.
FIRST METHOD. — Find the length of the long chord AC, AB + BC. Referring to Table XVI., the chord of a 1o curve
fore, 797.7 feet, say 798 feet, A B. To find BC, solve the triangle B DC, observing that the angle DBC = BAI= one-half of the central angle 32° 20', 16° 10', and that DC
60 feet. Then DC : nat. sin. 16° 10' = 60 • .278 = say 216 feet, BC. Hence AC AB + BC 798 + 216 1,014 feet.
Having thus found the length of chori A C, the radius and rate of curvature may be deduced as in X.
Or, dividing the tabular chord of 32° 20' by chord AC 1,014, the degree of the required curve is ascertained directly to be 3.15, equivalent to 3o 09'.
2. SECOND METHOD. — Find the apex distance AH, = AI +IH. The tabular tangent of 32° 20' divided by 4 gives A I
415 feet. In the triangle KDC, the sile DC • nat. sin. K 60 ; nat. sin. 32° 20', 112 feet = KC=IH. Then АН AI + IH 415 + 112 527 feet; anıl the tabular tangent 1,661 - 527 gives 3.15, equivalent to 3° 09', the degree of the required curve A C, as before.
HAVING LOCATED A CURVE A B C, TO FIND THE POINT B AT WHICH TO COMPOUND INTO ANOTHER CURVE OF GIVEN RADIUS, WHICH SHALL END IN TANGENT EF, PARALLEL TO THE TERMINAL TANGENT OF THE ORIGINAL CURVE, AND A GIVEN DISTANCE FROM IT.
1. To find B, the angle BIC must be found. Call the given uistance between tangents D; the larger radius, R; the smaller one, ge; the required angle, a. Then, referring to the figure, observe that in the triangle IHK, IH being radius, IK is the cosine a; i.e., IK ; III
nat. cosine a. But I HER — r; IK
= IC – KC, and KC= KF or HE + FC,=r+D; i.e., I K= R - 7 – D. Hence nat. cosine
The same reasoning would apply if A B E were the curve first located, and a terminal curve of larger radius required to be put in.
2. We have, then, the following general rule for such cases: Divide the perpendicular distance between terminal tangents by the difference of the radii, and subtract the quotient from unity; the remainder is the natural cosine of the angle of retreat along the located curve to the required P. C. C.
Example. 3. A 3o curve on the ground, to find the P. C. C. of a 5o curve striking 27 feet to the right. Here D = 27; R. = 1,910 — 1,146, = 764; D • R - 27 · 754, .03534; and 1 .03534 .96466 nat. cosine 15° 17'. We must go back, therefore, 509 feet on the 3o curve, to compound into the 5°
Had the 5o curve been located first, we must have gone back 306 feet to begin the 3o curve which should strike 27 feet to the left. In either case, time might be saved by moving directly from E to C, or the reverse, and spotting in the curve backwards. To do this, we have in the right triangle FEC, the angle E= half of 15° 19', = 7° 381', and the side FC=27 feet. Then EC=27 • nat. sin. 7° 38}',=203 feet; and if E were stake 54.20 on the 5o curve, B would fall at stake 54.20
51.14; and C, the P. T. of the 3o curve, at 51.14 + 5.09, stake 56.23.
TO SHIFT A P. C. C., SO THAT THE TERMINAL
BRANCH OF THE CURVE SHALL END IN A GIVEN TANGENT.
FIRST CASE: the terminal branch having the shorter radius.
1. Suppose the compound curve ACN located, and that it is required to fix a new P. C. C. at B, from which
H the terminal branch BM shall merge in tangent ML, a given distance from NO. To fix B, the central angle B HM of the new terminal branch must be found, and substituted for
tance asunder of the terminal tangents, D; the central angle, CIN, =IEK, of the located terminal branch, 6; and the central angle, B II M, = HEF, to be substituted for it, a.
In the right triangle, EIK, EK=E I cos. IEK= (R — r)
In the right triangle II FE, EF=EH cos. HEF=(R — 9)
Also, FK=LO=D, since each is equal to r — - KL.
Then E F=EK-F K; i.e., (R — r) cos. a= (R — r) cos. - D.
Hence nat. cosine a=nat. cosine b—ID: (R — r)].
Were the curve B M located, and the curve C N to be substituted for it, – that is to say, were a given and required, we should have, by transposition, nat. cos. b=nat. cos. at ID :-(R-r)]
Example. A 3o, compounding into a 5o curve at C, which consumes an angle CIN, = 30° 22', and ends in a tangent, N O, which is found, by measurement of LO, to be 34 feet too far to the left. Here, D 34, R
1,910, r 1,146, 8 -30° 22'; and, by the solution, nat. cos. a=nat. cos. 300 22 - · (34 ; (1,910 – 1,146]) = 0.8628 — (31 = 764).
Then 0.8628 0.0445 0.8183 = cos. 35° 05',
angle a; a-b= BIM – CIN=BEC
the angle of retreat from the crroneous P. C. C. 350 05 300 22 = 4° 43', equivalent to 157 feet, on the 3o curve, from C to B.
2. SECOND CASE: the terminal branch having the longer radius.
Let BN represent the terminal branch located with central angle IKO=b, and suppose it required
to determine the new arc CM, with central angle I EF= a. Call the longer radius R, the