208 
MINUTES OF PROCEEDINGS OF 
Draw SV 3 ca and CA perpendicular and cB parallel to MN } let us take 
the co-ordinates of the centre of gravity, 
aV—cF=x’ i VS =?/, 
the radius of the front wheels ca — r , that of the hind wheels CA = R 
and the distance between the axles cC — a, then is F$ — y — r, and in the 
right-angled triangle cBC } putting the angle BcC — a • 
R — r 
Through B and F draw the horizontal straight lines JDE and GH and 
through c the vertical straight line cH, then the angle RIFC = FSG = <^and 
HcB =90° — <p + a 3 and we have in the right-angled triangle SGF 
FG = FS sin </> = (y — r) sin 
and in the right-angled triangle cFIF 
FH — cF cos <p = x cos <jf>; 
therefore 
FG + FH = GH = DE — x cos <f> + (y — r) sin 
Again, in the right-angled triangle cFD, 
cD = 
DE _ x cos </> + (y — r) sin <£ 
COS (</> — a) cos (<jf> — a) 
consequently, 
CD = „-cD = a-^l±±^zlU^i. 
cos (</> — a) 
If this expression be substituted in the above determined equation of 
moments, we have 
f x cos <£ + (y — r) sin 6 ~) TT , 
ap = < a -— -j-j- - - 
k cos (<£ — a) J 
or x cos <j) + (y — /•) sin </> = a ^1 - cos (<£ — a). • 
Tor any other angle of inclination of the platform with which the 
pressure on the front axle is y, we have similarly, 
x cos cf)' + (y — r) sin — a cos (<£' — a). 
By the combination of these last two equations we finally find as values of 
the co-ordinates of the centre of gravity of the carriage less the wheels 
