common tangents at the same points, the radii of the three portions of the curve being respectively O B = OC, O'C = O'C' and 0′′ C′ = 0′′ C′′. This kind of curve is adopted where the line is required to pass through given points, as C and C', to avoid obstructions, or where a principal station or terminus is at or near C"; in the latter case the radius O"C" may, if required, be less than 80 chains.

The compound curve may consist of two, three, or more portions of different arcs; thus the curve bc c' consists of two portions, bc, c c'.

PROBLEM VI.

1. To find the several radii of the Compound Curve mechanically. Let A B, D C" be the tangental portions of the line, which are required to be joined by a curve passing through the points C, C', the point C" not being given. Select a curve-ruler such that being applied to touch A B at B, it may also pass through C: if this curve do not pass through C', but through some other point E, another curve-ruler of less radius, in this case, must be selected, and such that it may touch the arc B C at C without cutting it or its prolongation towards E, and also pass through the point C': if this curve cut the prolongation C" T' of the tangent D C", another curve-ruler of less radius than the last one must be selected, and such that it may touch curve C C' at C' and the tangent D C" at C": thus completing the curve B C C' C". The radii O C, O' C', O' C" may be determined from the curve-rulers, as in Case I. Prob. I.

2. To find the radius c' o' of the Compound Curve b c c' geometrically, the starting point b and the radius bo being given.

From the given point b in the tangent a b, draw the given radius bo to a b; and draw the curve to some point c, where it is found convenient to change the radius: draw the radius o c, and thereto draw ct, meeting the tangent d t in ť' ; make ť c′ = ť'c, and from e' draw co' to te' meeting co prolonged, if necessary, in o'; then o' is the centre of the arc c c' of the curve, conformable to the nature of tangents.

The method of constructing the curve, when it consists of three or more parts, is sufficiently obvious.

3. One of the two radii of the Compound Curve, and its starting and closing points being given, to find the other radius.

Let a b, c'd be the tangents, b and c' the starting and closing points of the curve. Draw the perpendiculars boc'h = given radius to the tangents; join o h, and bisect it in f; draw ƒ o' to o h, meeting c'h prolonged in o'; join o' o and prolong it till o c = c'h : then the points o, o', are the centres of the arcs bc, c c', which constitute the compound curve, o'co' c' being the radius required.*

NOTE. - In the compound curve BCC' C", where the radius C" O", which is to be found, is less than the preceding radius C O, the C" H is made CO; HO is joined and bisected in F; and F O" drawn to H O', meeting C" H in O", which is the centre of the arc C' C", &c.

Definition of the Serpentine or S Curve.

The serpentine curve B G C is used in railways, when obstructions or some other cause render its adoption preferable; it consists of two circular arcs of different or the same radii, having their convex sides turned in opposite directions, like the letter S, whence it is sometimes called the S Curve; the two portions B G, G C of the curve have a common normal O G O' at their point of junction G, and therefore a common tangent at the same point. This curve affords the most easy means of joining two parallel, or nearly parallel, portions of a line of railway.

PROBLEM VII.

1. When one radius and its tangental point are given, to find the other radius and tangental point of the serpentine curve geometrically.

From the given tangental point C draw the given radius COL to the tangent C D, and draw the curve C G to some point G,

=

* Demonstration. Since of fh, and fo' is to oh, oooh, also bo was made choc; ..oc = o'c', and the normal o'oc is common to the arcs of the curve. Q. E. D.

=

where it is found convenient that it should have its point of contrary flexure: through O G draw the normal O G ; from

G draw GT

to O GO to meet the tangent A T; make TB = T G ; and draw B OL to A T, meeting O G O' in O'; then O' is the centre, and O' B = 0 G is the radius of the curve B G, as is evident from the nature of tangents.

NOTE.

The radius O' B, and tangental point B may be found mechanically, i. e. by the curve-rulers, as in Problems I. and VI.

2. When the tangental points and one of the radii of the Serpentine Curve are given, to find the other radius geometrically.

From the given tangental points C and B draw C O, B H, repectively to the tangents C D and B A, and equal to the given radius; join O H, and bisect it in F; draw FO' to O H, meeting HB prolonged in O', and join O, O'; making O'GO' B; then O' is the centre, and O' B = 0′ G is the radius of the portion B G of the curve, as required. *

NOTE 1.

case.

NOTE 2.

· The radius B O' may be found mechanically as in the preceding

The radius O' B may be found from the following formula, wherein
LTBC and a = 4 T' BC.

8 BC, r = given radius O C, a =

=

Ex. Let 8 200 chains, r = 110 chains, a = 45°, and a′ = 20°; then sin.

* The demonstration in this case is similar to the one given to Prob. VI.

It may thus be readily ascertained whether the required radius O' B is greater, equal to, or less than 80 chains; if less, the given radius r ought either to be diminished, or the distance BC of the tangental points ought to be increased according to circumstances, in order that O'B may be of the required length, assuming that the portion B G of the curve is not near a terminus or principal station.

PROBLEM VIII.

1. When the two portions of the Serpentine Curve have the same radius, to determine that radius geometrically, the tangental points and their distance being given.

Let AB, DC, be the tangents, B and C the given tangental points, and BC the given distance. Draw Bo Co' respectively to

A B, DC, and of any convenient length; through o, parallel to B C, draw o q indefinitely; with the compasses apply o'o" 2 Co' = 2 Bo; through C, o" draw C o′′ O, meeting B O prolonged in O; and through O, parallel to o'o', draw O O', meeting Co' prolonged in O'; then O and O' are the centres, and OB and O'C' are the equal radii of the serpentine curve B G C, the common normal of the portions B G, G C of the curve, being OG O′ = 2B0 = 2 C O'.*

2. To find the common radius of the two portions of the Serpentine Curve by calculation, the same things being given as in the preceding case, and the angles TB C, T'C B.

Put BC= 8, B0 = C 0′ = r, ▲ TBC = a, and ▲ T'CB=a';

* Demonstration.

Draw o' b parallel to O B; then by similar triangles,

Co: o'o' = 2 Co: C O': 2 C O'

=

0 0',

BOC O'. Q. E, D.

then _r=sin. a + sin. a' + 2 sin. of arc to cos. (cos. a+ cos. a ́')'

Ex. Let 200 chains, a = 27°, a'= 50° : then from a table of nat. sines sin a = .45399, its cos. = .89101; sin. a'=.76604, its cos. = .64279: whence (cos. a + cos. a′) = 1⁄2 (.89101 + .64279) = .76690 =cos. and 2 sin. of arc to cos. .76690 is 1.28334; therefore 200 200 =79.89 chains, or nearly .45399.76604 + 1.28334 2.50337

r=

80 chains, the radius required.

=

NOTE. The method of forming the serpentine curve with a common radius is much to be preferred to any other, when the nature of the ground will admit of its being done; and more especially so, when the data, as in the preceding example, will only just give a common radius of 80 chains, whereas, if the radius of one of the portions of the curve had been taken greater than 80 chains, the other radius would have necessarily been less than 80 chains.

PROBLEM IX.

To make a given deviation HQ from a straight portion of a Line of Railway, A H D, by means of three Curves, BG, G Q Gʻ, G' C, having their radii, OB, O' Q, O' C, all equal; in order that the lateral Works of the Line may avoid the Building, or other Obstruction, b, which is close to the centre of the straight portion of the Line.

=

=

* Investigation. Draw PO, P' O'L to BC; through O draw O Q parallel to B C, meeting O' P' prolonged in Q: then O O' = 2 r, B P r sin. a, O P = P' Q = r cos. α, C P = = r sin. a', O' P = r cos. a', O' Q : O' P' + P' Q = r (cos. a + cos. a'): whence, from the right-angled triangle O QO', sin. ≤ O' O Q (cos. a + cos. a′), which is therefore given, whence the comp. of ≤ O' O Q 40 O'Q is known, and may be thus expressed, sin. ≤ OO' Q = sin. of arc to cos. (cos. a+ cos. a). Whence OQ PP' 00' x sin. 40 O'Q : = 2 r x sin. of BCBP + PP' + P' C = r sin, a + r sin. cos. a′), from which

=

are to cos. (cos. a + cos. a′), d
a' +2r sin. of arc to cos. (cos. a +

r =

=

δ

=

sin. a + sin. a' + 2 sin. of arc to cos. (cos. a + cos. a')

Q. E. I.

=

When the radii are unequal, and one of them, as r = O' C is given, and

=

=

=

R+r, BP

=

=

the other R = BO is required; then O' O R sin. a, PO: PQ: R cos. α, 0' Q = R COS. arcos. a', and o Q = P P' = √0'02 O' Q2 √(R+r)2 - (R cos. a +r cos. a′), d R sin, a +r cos. a' + √ (R+r)2 − (R cos. membering that sin.2 +

=

a + r cos. a')2. By transposing and squaring, and re1, &c. there results 8-28 (R sin. a + r sin. a')=

- COS. α

This formula is used for finding

the value of B O', Note 2., Prob. VII., where the symbols are defined. Q. E. I.