THE ROYAL ARTILLERY INSTITUTION* 
209 
? = a f cos a _ i 3 sin <ft'. cos ((f) —a) — P' sin <j>. cos (<£' — a) 
1 TPsin (</>'—0) y 
y — r + a 
-£sin a 
+ 
P cos <£'. cos (</> — a) — jt/cos <f ). cos (</>'— • a) ' 
W sin (<£' — ’ <f>) 
= P -f* & 
P COS (f >'. cos (<£ — a) —j(/cOS (jf> . cos (<£'— a) 
fVsm(4>' — (j)) 
If the pressure P' on the front axle be determined when the platform is 
horizontal, then <j/ = 0; and hence we get from the formulae I, the simple 
formulae 
x = a cos a ^1 — , 
y — R + a - B ™ a {(/—p) cot (/). cot a —p] 5- . n * 
= P + : 
P —r 
W~ 
{{p' — p) cot . cot a — p] 
Further, if the carriage is equirotal, then R = r and a = 0, and we \ 
from Equations II. 
*= a ( l ~w)’ 
y = R + a cotcf-^y- 
III. 
2. To ascertain the centre of gravity JJ of the carriage with all four 
wheels (fig. 2), let 8 be that of the carriage and its load but without the 
LZ 
Fig. 2. 
wheels; let c and C be the centres of gravity, and ([ and Q the weights of 
the fore and hind wheels, we shall have in the plane ac of the moments 
aM{W + Q + q) = ccN. W + clA. Q, 
and for a plane Aa of the moments 
M U( W+ Q + q)=ac.q + NS.W+ AC.Q; 
