324 
MINUTES OF PROCEEDINGS OF 
Prom a draw ad parallel to Ik, meeting kc produced in d . Then, 
he = hd — cd 
ca . sin cad 
= ab 
sin adc 
n= ir—R . sm ~ ^ 
COS 0 
= TP'S 1 ^ » sin (9 (f) ") 
\ a cos (f> j 
Also from the triangle adc, 
since bh = ad, 
ac . sin acd 
and ad = 
we have T — 
sin adc 5 
R . cos 6 Wl cos 6 
COS 0 
a cos 0 
(3) 
(4) 
The above may also be found by taking moments round B (Pig. 14). and 
resolving horizontally and vertically. We have here neglected the weight of 
the spar forming the derrick, but it can be taken in account by supposing 
half its weight to act at either end, as in the case of the sheers. 
When there is not time, nor sufficient timber at hand to erect a proper 
derrick, a single spar may be used. (See Pig. 16). If we draw AK vertical, 
the sides of the triangle ABK will evidently be proportional to the three 
forces keeping the point B at rest. There must be several side guys in this 
case, but by resolving all the tensions parallel to the plane of the paper, and 
then along the line BH, we can reduce the forces to three. 
January, 1873. 
