341 
1911-12.] The Railway Transition Curve. 
than the third as correction terms, and remembering that the space rate of 
change of curvature might have been taken as a quadratic function of 
s or x, a suitable form of equation for the transition curve is seen to be 
y = lx 3 + mx* + nx 5 , where l, m, and n are all unknown. To determine these 
values there are the three equations : — 
lx 3 + mx^ + nx 5 - y = 0 , 
3 lx 2 + 4 mx z + 5 n# 4 - y' = 0, 
6 lx + 1 2 mx 2 + 20 nx 3 - y" = 0 , 
where x, y, y\ and y" now denote known quantities at the point of com- 
pounding. 
Solving this linear system, we have 
l_y" 4 y' 10 y 
2x x 2 x s 
y" 7 y 15?/ 
„=i!_ 3 V + 6 2'. 
Ozy»3 /y*4 <v5 
Rj tAj JU 
These values of l, m, n require to be calculated and substituted in the 
equation y = lx 3 + mas 4 + xix 5 , and a transition curve giving exact compound- 
ing at specified points is obtained. 
As the calculations are more complicated than in the case of the cubical 
parabola they may be simplified by assuming £= a, and expressing x, y,' y", 
in terms of R and \fr, though it is probably best to express merely y' and y" 
in terms of R and \fr. This gives for the values of l , m, and n : — 
-A 
10 y 
m= w 
| '25 sec 3 ifr cosec i/a j> — 2 cosec 2 i/a 
| - -25 sec 3 i/a cosec 2 i/a j + ^jseci/A 
fH \ T~c se ° 3 cosec 3 i/a l - j A sec i/a cosec 3 i/a 
R 4 L (16 j (16 
cosec" 
Example. — 3° curve, angle between the tangents 129° 8'. 
Then 
£=a=140', y = 6-85'. 
R = ^- = 1909*86'. 
6 
sin i/a = 4 /. i/a = 4° 12' 14" 
R 
77 = 5-13847 
$ = y-v)= 1-71153 
SB = (R + 1 3 ) cot o) = 909-044'. 
