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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 G D common. Hence the triangles are equal in all their parts: BG=GC, and B D=DC. By like reasoning, it appears that CF= FA, and D A 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.

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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 FE, seeing that from its angles the required central angles can be obtained.

Example. BA, a 1°, BC, a 6o curve, located: to complete the Y with

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In triangle G FE, –

EF= 5,730 – 717 = 5,013.
EG= 5,730 — 955 = 4,775.
G F ; 955 + 717 = 1,672.

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The longer segment, therefore, is 4,502; the shorter, 511. Cos. E the longer segment divided by EG=4,502 -- 4,775 = log. 3.653405 — 3.678973 - 9.974432 = cos. 19° 28' angle E.

Cos. GFE the shorter segment divided by GF=511 1,672 log. 2.708421 – log. 3.223236 9.485185

cos. 720 12 = angle G F E.

The central angle, BGC, of the 6o curve, is equal to 180 FGE the sum of the angles at E and F = 72° 12 + 19° 28' 91° 40', making the arc BC= 1,528 feet. The arc B A, equivalent to 19° 28' of a 1° curve, 1,947 feet. Points C and A being thus ascertained, curve AC may be located. It will consume an angle 180° - 720 12 107° 48', equivalent, on an 8° 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 B A E, in order that the true

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Example. AB= 1,500 feet; DHB, a 4o curve; angle FBD= 20° 13'.

First, the angle FBP, 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 4o curve, it is, say, 990 feet

DB; and DI=IB= 495 feet. The versin. H I is, in like manner, found to be 88.25 feet. Deducting this from the radius of the 4o curve, we have I C= 1,314.4 feet.

Then IC:IA tan. I AC; i.e., 1,314.4 · (495 + 1,500) = 0.674 tan. 33° 59' angle I AC.

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B

Next, by geometry, the proposed tangent A E=NAD XAB =V2,490 X 1,500=1,932.6; and E C:A E=tan. E A C= 1,432.69 -- 1,932.6=0.7413=tan. 36° 33' = angle E AC. Then E AC-I AC=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 4o curve; BC 774 feet; CD, a 6o curve: to put in the 1° curve, EF.

Sketch the figure as in margin, H K being parallel and equal to BC. Then KG BG BK or CH 1,433 955 478 feet; KH • GK 774 :- 478 = 1.62 tan. 580 19 angle KGH; and KII ; sin. 58° 19 = 774 ; 0.851 - 909.6

B

H

In the triangle G H I we have then the sides given; namely GH say, 910 feet, HI=5,730 955 = 4,775 feet, and GI

= 5,730 — 1,433=4,297 feet: to find the angles.

Under Case 3, Trigonometry (III.), IH :IG + GH :: IG -GH: IL-LH; i.e., 4,775 : 5,207 :: 3,387 : 3,693, the difference of the segments into which the base IH is divided by a perpendicular from G. Adding half the difference of the seginents 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 sun, the shorter segment, L H, is found to be 541 feet.

Then HL; HG 541 • 910 0.5915 cos, 53° 31' = angle GHI. 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 HGI. Finally, EGH – KGH=63° 20' — 58° 19' = 5° 01'

angle E GB, equivalent to a distance from B of 125 feet around the 4o curve to the P. C. C. at E; and GIH – EGB go 49

5° 01' = 4° 48' = angle CHF, equivalent to a distance from C of 80 feet around the 6o curve to the P. C. C.

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at F.

XXXVII.

TO RUN A TANGENT TO TWO CURVES ALREADY

LOCATED.

E

M--

F

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 C D, from one to the other curve, and, completing the traverse A FGE or A B C D E, determine thence the bearings and distances asunder of the centres A and E. The right triangle A EK, in which EK = the sum of the radii, may then be solved, and the points H and I ascertained as in the following example: –

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Example.
FB, a 4o curve; G D, a 6o curve.

N.
A B, N. 20° E., 1,433 feet 1,316.6
B C, East,

3,570 feet
CD, N. 31° E., 1,800 feet 1,492.2
D E, N. 45° W., 955 feet

675.2

E. 490.0 3,570.0 1,006.2

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675.2

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Total northing, 3,514 feet; total éasting, 4,391 feet.

Then 4,391 · 3,514 1.2496 tan. 50° 20' bearing A E; and 4,391 • sin. 50° 20' : 5,704 feet distance A E. Also, EK : AE (1,433 + 955) 5,704 sin. 24° 45' angle EAK; and angle AEK 900 000 24° 45' = 65° 15'. Hence the bearing of A K or H I is N. 75° 05' E., and that of A H or I E, N. 14° 55' W.

Since AB bears N. 20° E., the angle H AB - 20° 00' + 14° 55' = 34° 55', equivalent to a distance of 873 feet from B around the 4o curve to the required P. T. at H; and, since D E bears N. 45° 00' W., the angle IED : 45° 00' 11° 55' = 30° 05', equivalent to a distance of 501 feet from. D around the 6o 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,

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