268 
MINUTES OP PROCEEDINGS OF 
each other at the point 0. From 0 draw 01 parallel to GC. Then the 
z 1110=\= Z DCK= z NCM. Join CB, B being the point of contact of 
the inside limber wheel with the friction plate on trail, and draw BA 
perpendicular to GC; BA represents the half-width of the trail = 5. 
Then, in right-angled triangle BCD, in which 
BD = a — a, CD — w\ put Z BCD = <£, 
tan <p - 
BD _ s — a 
CD w 
In right-angled triangle ABC, in which 
AB = b and Z ACB = 7 —(</> + X), 
4 
BC= . 
sin ACB cos (<£ + X) 
.(i). 
Again, in right-angled triangle BCD , in which we have the values of 
DC and z BCD as before. 
^ ^ 
sin CBD cos cp . ^ ' 
Equating these two (I. and II.) values of BC, 
b w 
cos (X + </>) cos cp 3 
hence w cos (X + <£) = b cos </>, 
and cos (X -f <p) = — cos <£; 
w 
from whence having calculated cp, we can deduce X. 
On consideration of the figure the least width in which the carriage can 
turn in, or B, is the sum of the straight lines LH and 1IF, the latter being 
the least radius of circle in which the carriage can make a complete circuit, 
and the former the distance between the centre H and the point L , the 
outside edge of the wheel of the gun or wagon. 
In right-angled triangle 1011 , in which we have 
10 = and Z 1110 = X 3 
IH = d tan Z IOH = d cot X, 
and BO = -A- . 
sin X 
