1919 .] 
Jenkinson.—Balancing of Locomotives. 
21 
Proceeding to deal with the ordinary two-cylinder locomotive, we will 
assume that the right-hand crank leads and that the balance-weight is cast 
in the wheel so that in the left-hand wheel it lies at an angle 180° + 8 ahead 
of the crank, and in the right-hand wheel at an angle 180° — 8 ahead of 
the crank (see fig. 1). 
The following notation is used (see fig. 1) :—* 
M = sum of the revolving masses (at crank-radius) on one side, in 
pounds; 
Mu = sum of the moments of these revolving masses on one side 
about the longitudinal central vertical plane, in inch- 
pounds ; 
R = sum of the reciprocating masses on one side, in pounds ; 
2c = transverse distance between cylinder centre-lines, in inches ; 
B = equivalent mass at crank-radius of the balance-weight, in 
pounds; 
2b — transverse distance between centres of gravity of balance- 
weights in opposite wheels, in inches ; 
2 g = transverse distance between planes of opposite wheel-contact 
with rail, in inches. 
Resolving horizontally for any instantaneous crank-angle (0) of the 
left-hand crank, we have— 
Surging-force in pounds (forward) 
V 2 r 
= 3"207 ££ [(M + R) {cos 0 + cos (90° + 0)} -f B cos (180° -f- 8 -}- 0) 
+ B cos (270° - 8 + 6) + mR {cos 2 <9 + cos 2 (90° + 0)} ] 
V 2 r 
= 3*207 -^ 2 (cos 0 — sin Q) {M + R — B (cos 8 + sin 8)} 
Y 2 r 
= 4*535 -pr cos (0 + 45°) {M + R — B (cos 8 + sin 8)} .(1). 
Taking moments of the horizontal components about the longitudinal 
central vertical plane, we have— 
Nosing-couple (clockwise) in inch-pounds 
Y 2 r 
= 3’207 [(Mu -j- Rc) {cos 6 — cos (90° -j- 0)} + B6 cos (180° -f- 8 + 6) 
— B6 cos (270° — 8 -f- 6) -J- mRc {cos 2 0 — cos 2 (90° -j- 0)} ] 
Y 2 r Y 2 r 
= 4-535 -jy sin (0 -f- 45°) {Mu + Rc — B5 (cos 8 — sin 8)} -f- 6 - 414 -^r cos 2 0 
(mRc).. (2). 
Taking moments of the vertical components about the plane of the 
right-hand wheel contact— 
Hammerblow downwards in pounds 
Y 2 r 
= 3-207 
2 gd 2 
[(Mu -f M</) sin 0 — (Mu — M#) cos 0 -j- B (b — g) cos (0 — 8) 
— B (b + g) sin ((9 + 8)] 
= 1*6035 
Y 2 r 
d 2 
. (Mu — B b (cos 8 — sin 8) ) 
sin 0 j-+ M — B (cos 8 + sm 8) [ 
9 
i Mu — B6 (cos 8 — in 8) _ . ^ . .. ) 
— co 0 . ——-t—— r . -.. . . M + B (cos 8 + sm 8) 
9 J 
(3). 
Considering equation (1) for the surging-force, it will be seen that the 
term in cos 2 9 disappears, this being the algebraical expression of the 
fact that the secondary forces balance each other for a two-cylinder engine 
