TABLES OF OFFSETS FOR RAILWAY CURVES, AND CORRECTION 223
OF LEVELS FOR CURVATURE, ETC.

Radius

No. 1.

Offsets at the end of the first chain from tangent point of railway curves

of curve

224

TABLES OF OFFSETS FOR RAILWAY CURVES, ETC.

No. 2.-Difference between apparent and true level for distances in chains. Correction in decimals of feet.

No. 3.-Difference between apparent and true level for distances in miles. Correction in feet and decimals.

ADDENDA TO THE METHODS OF LAYING OUT

RAILWAY CURVES.

PROBLEM I.

To join the straight portions A B, C D of a railway by a serpentine curve FOIN M, meeting A B at F between A and B, and CD at M between Cand D; the distance BC and the s ABC, BCD being given, to find the common radius G I = I H.

Take the cotangents of half the angles A B C, BCD to rad. = 1; then as the sum of these two cotangents is to the cotangent of half A B C to rad. = 1, so is BC to BI; hence BC BIIC.

In the triangle BCI; as sin. BGI is to sin. ▲ G BI, so is B I to GI = I H, the common radius.

EXAMPLE.

Let the angle ABC = 71° 40', the angle B C D = 129° 15', and the distance BC 95 chains: required the length of the radius G I = IH of the serpentine curve for uniting A B to CD. Here the cotangents of half the angles A B C, B C D to rad. 1 are respectively 1.3848 and 4743, their sum being 1.8591, Then as 18591: 1·3848 :: 95: 70-76 chains = BI. Whence BC BI= 95 — 70·76=24.24 chains I C.

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Again, as sin. BGI is to sin. G BI, so is 70.63 to 51.09 chains GI IH, which is the common radius of the required serpentine curve.

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PROBLEM II.

To lay out a railway curve by means of tangental angles.

This method, already noticed in the note to Problem IV., p. 169, is preferred by many eminent engineers, where the ground on which the curve is to be laid out is level and free from numerous obstructions, except such as do not fall on the curve itself; I shall therefore give it more in detail than in the note referred to.

If from any point B, in a straight line AD, we lay off any number of equal angles, as DBS, SBT, TBU, &c., making the chords B S, ST, TU, &c., equal to each other; then the points B, S, T, U, &c., will be situated in the circum

ference of a circle, which is tangental to the line AD at the point B.

Fix the theodolite at B, and lay off the tangental angle DBS, the chain being extended from B to S to meet the visual line B S; in the same manner lay off the other angles SBT, TBU, &c., till the fourth point V is reached. If any obstruction, as H, should prevent our seeing from B further than to V, the curve may be continued by removing the instrument to U, the point preceding V; thence sighting first on V, continue to lay off additional tangental angles V UW, WUX, &c., as before. Otherwise, moving the instrument to V instead of U, sight back to U, and lay off first the exterior angle P V W equal to double the tangental angle (P being in U V prolonged), and afterwards continue the tangental angles W V X, X V Y, &c., as before, to the end of the curve.

Finally, in order to pass to the end of the curve at Y to a tangent Y Z, place the instrument at Y, and sighting back to X, lay off the tangental angle X YO; then O Y continued towards Z will be the required tangent to the curve.

The method of finding the quantity of the tangental angles is given in the note, p. 169.

ON THE DIVISION OF LAND.

Three straight fences W X, XZ, ZY are given in position; there is a well at W, and it is required to substitute a new straight fence W Y instead of X Z, that the ground on both sides of the divisional fence may have the use of the well; the value of the land on the side WXZ is to that on the side X Z Y as m to n per acre : required the position of the new fence W Y, that the proprietors on each side of it may neither gain nor lose by the change of boundary.

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Let the required straight fence WY cut XZ in P; draw W Q parallel to Y Z, meeting X Z in Q; then Q is a given point, and the area W QX is also given, which put A, also put Z Q a, QW= b, QP = x, and the sine of the angle Z sine of the angle ZQ W 0; then from the nature of the problem we readily obtain the following quadratic equation:

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From this equation the value of x = QP may be readily found, which determines the position of the point P, through which the new fence W Y must pass.

The quantity A must be regarded as positive or negative, accordingly as the point Q falls in the fence X Z, or in its prolongation to the other side of X W.

Corollary 1.-When the land on both sides of X Z is of equal value; then m=n, and the second power of x in the preceding equation will vanish, and

Corollary 2.-When Y Z is parallel to X W, Q will coincide with X, and A will vanish; whence

QP = =&= = { a = X Z.

NOTE. Changes of bonndaries of this kind are frequently required, not only for the advantages of wells or watering places, but also for ready access to more convenient roads, &c. This important article was inadvertently omitted in the previous editions of this work.