74 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[MARon, 



RAILWAY CURVES. 



Ox reconsidering (his snlijocf, we tliink, as our correspondent R. W. 

 T. SMSfiTests, t,li;i( tlie cngini'er wlio has (o set out the line of a railway 

 upon the gronnd sliouM, in general, confine liinisell' (o the curves, and 

 in every respect to (he line laid down upon the plan, in whicli case no 

 such f|nestion as that proposed by " An Assistant Engineer," conlil 

 occur. But, since a devia(ion from the plan may in sonio instames 

 bo allowed, as "An Assistant Engineer's" question ])roves, we shall 

 endeavour to solve R. \V. T.'s dillicnlty; and (or (his we must first 

 consider what may have been the cause of (he failure. Now there 

 are two cases: cil/ier the curve has been commenced at a wrong point 

 of the tangent, or (lie operadon of setting it out has been iiraccurately 

 ))erlormed. In tlic first case (he error can be rectitied by referring to 

 the plan and ascertaining the right point uf contact, and then setting 

 out (he curve afresh. This method would, no doubt, be exceedingly 

 troublesome, and it appears to be (he object of "An Assistant Engi- 

 neer" to dispense with tlie lalio\n' attending such a proceeding; the 

 quesdon tlien is, wliat is tlie best method of getting over the difficulty 

 without returning to the plan laid down, when a deviation from the 

 laKer is allowable. Now when two curves wore intended to meet and 

 form an .S, and the engineer employed to set tliem out has not suc- 

 ceeded in efi'ecting their pmction, there are two cases: ei/kr the two 

 curves intersect each other, or they do not. In tlie first of these cases, 

 it is (rue, (he two curves may be joined by a third, tangent to the two 

 loriner, and of less radius than (lie one which it touches on the con- 

 cave side, though, in onr opinion, it would be preferable to correct 

 the curve in accordance with (he plan. Now there are an infinity of 

 circular arcs which will satisfy the condition of being tangent to (he 

 two given curves, so that another cfmdition must be imposed before 

 the connecting curve can lie determined ; it may therefore be required, 

 either (hat (his curve shall touch the concave or the convex curve at a 

 given i)oint, or tliat its radius should be of a given length, which la(ter 

 is the O(pnili(ion ;issuined by " An AssistantEngineer." We shouhl 

 recommenil solving the problem on the plan, and not on the ground, 

 believing (he former mode (o be much more facile (han the latter ; 

 we shall (herefore adapt onr solution to (hat inetitod. 



Fig. 1. 



Let A n and C D, (fig. 1) be the two given curves fsay of 130 chains 

 radius), an<l let it be reipiired to unite them by a third curve of less 

 radius, tangent to A H on its concave, and to C D on its convex side. 



1st case. Tlie required curve is to pass through the point G of the 

 curve A Li. 



Erinn I', the centre of A B, and (hrough G, (he required point of 

 contact, draw the straight line F (t H, equal to the sum of the radii of 

 (he two given curves, or twice the railius I' G (both curves being sup- 

 ]iosed to have the same radius); from H draw the straight line H <) 

 lo the ceuire of (he curve C D ; and from K, the miildle point of H O, 

 draw the perpenflicular K (j, intersecting the straight line 1' H at (he 

 point Q. (j will \h: the centre of the required curve, and its radius 

 \vill be equal to Q G. .Toin tj (), and the jHiint E, where Q O inter- 

 sects the curve C 1), will be its poiiil of con(act wi(li the required curve. 



For, the right angled triangles H K Cj, tj K () being eepial, tj H = 

 Q i) ; and, if from these c-cpials we t;dve the eipials (j H and E O, the 

 remainders (,) G, tj E will also be equal; and, since they are situated 

 on normals to the given curves, the circular arc G E will be tangent to 

 both these curves. 



2nd case. The required curve must pass through the point E in 

 the curve C D. 



From the cenfre O, and (hrough (he given point of contact E, draw 

 the straight line () K (j ; and froiu (lie cende F draw P L parallel to 



U (^ and equal to the smn of the radii of the given curves i from P as 



a centre, and with a radius equal to P L, describe ;ui arc of u circle in 

 the direction in which the connecting curve G E is expected to meet 

 (he given curve A B, and from L draw the straight line L () H, inter- 

 seciing (hat arc at (he point II; join HP. The point Q, where 

 11 P intersects () Q, will be the centre of (he reqinnul curve, and the 

 )ioint G, where it intersects the curve A D, will lie (he point of con- 

 (ac( of (he required curve with A B. 



For, since O Q is parallel to L P, the triangles O H (J, L H P are 

 similar; and consequently, L P being equal to P H, <) Q = (^ H; and, 

 if from these equals we take (he equals E O, (i H, the remainders 

 (j G, Q K w ill also be e(|ual. And an arc of a circle passing through 

 (he p()in(s G and E, and having Q for its centre, will be tangent (o the 

 two given curves, as we ]u-oved for the first case. Or after having 

 drawn O Q an<l L P, construct the isosceles triangle O L ]\I, of which 

 the side O M=M L; then from Pas a centre, and wi(h a radius =M L, 

 describe an arc of a circle, in(ersecting O Q at the point (j, which will 

 be the centre of the required curve, as before. 



For, if (hrough (he point <.j wo. draw (he straight line P H ecpial to 

 P L, we shall ha\e, by reason of the similar triangles H Q O, H P L, 

 (^ H=Q O ; and, taking away the equals G H, E O, we have Q G = 

 Q E, as before. 



3rd case. The required curve is to have a given radius (say 100 chains). 



From (he point O as a centre, and wi(li a railius eipial to the sum of 

 (he radii of (he given curve C t) and of the required connecting curve, 

 describe an arc of a circle in the direction in which the centre of tlie 

 latter is expected to be found, and from the centre P, with a radius 

 etpial to the difl'erence between the radius of the curve A B and that 

 of the required curve, describe anotlicr arc, intersecting the former 

 Q. Q will be the centre of the required curve. 



For, drawing the radius P G through the point Q, the part Q G is 

 equal to the radius of the connecting curve, since P t^ is the difference 

 between (hat radius and P G ; also Ij E is equal to the radius of the 

 required curve, because O Q is equal (o tha( radius, plus the radius 

 of the curve C D, which is equal to the part O E, (herefore the re- 

 maining part Q E is equal to the radius of the required curve. And 

 it mav bo proved, as in the former cases, that the arc G E, described 

 with that radius and with the centre Q, will be tangent to both tlje 

 given curves. 



In the case when tlie two curves intended to liax e met do not inter- 

 sect each other, we should certainly recommend connecting them by a 

 tangent, if it should not be required to make thera meet, as in the plan. 



Fipr. 2. 



Let A B, C D (fig. 2) be the given curves, tlie former being tangent 

 to tlie straight Hue 1 A at the point A. To draw' a common tangent to 

 the two given curves. 



Join (on the plan) their centres () and P by a straiglit line, and on 

 O P as ii diameter, descrHie th(( circumference O H P K; than from 

 t) and P as centres, and with radii I'qual to (he sum of (he radii of (he 

 (wo given curves, describe two arcs uf circles intersecting the curcnm- 

 ference O H P K in the poiuls M and K respeclively ; draw the radii 

 t) H and P K, and the ])oiuts L and M, where they intersect the given 

 curves, will be their points of contact with the tangent, that is (o say, 

 a s(raightliue L M, drawn through these points, will be tangent to both 

 the given curves. 



F(U' O H and P K are parallel, and L H=P M ; therefore, joining 

 P H, L M :uid P H are equal and parallel ; but P H is perpendicular to 

 O H, therefore L M is perpendicular to the radius O L, and coiise- 



