THE liOYAL ARTILLERY INSTITUTION. 
317 
is applied; Q is a screw which, when the weight is to be lowered, opens a 
passage for the water to return into the upper cistern. 
inches. 
Total length of handle, AC . 24 
Effective length of u ... 22 
Distance, BC, of fulcrum from pivot of plunger . 1 
Diameter of plunger, BE .. Of 
n piston, 1IK . 2£ 
Then, we have by leverage, 
Z = Jl. 
W 22 ; 
by hydrostatic pressure, 
9 
7r — 
P _ area of plunger _ 64 
W area of piston ^25 
16 
Therefore, for total advantage gained, 
P _ 9 
W 2200* 
9 
Too* 
Hence, by a power exerted on the handle of a little over 90 lbs. we can 
obtain a lifting force of 10 tons—which is the greatest weight this machine 
is constructed to raise. 
We have neglected the friction of the water-collars, which, however, is 
quite trifling compared with the loss by friction in the screw-jack; the latter 
will not raise nearly so great a weight, although it appears the more powerful 
instrument when friction is neglected. 
Triangle-Gyn. 
Let AD, BD, CD (Fig. 6), represent the three spars or “legs” of a 
triangle-gyn, bolted together at D, and so placed on the ground that their 
feet form an equilateral triangle, ABC. 
Prob. —To find the horizontal force required to keep the foot of each spar 
from sliding , the normal reaction, and the whole pressure on the ground. 
It is evident, from the symmetry of their positions, that the conditions of 
equilibrium of each of the three legs of the gyn are the same; let us, there¬ 
fore, consider the case of the spar AD. 
Let l — length of each leg. 
a = distance between the feet, 
z a = angle of inclination of each leg to the vertical, 
W — weight of gun and tackle, 
N = normal reaction of the ground at each foot, 
F = horizontal force (however applied) keeping each foot in position. 
B = resultant, or whole pressure on the ground. 
Bisect BC in the point F, join AB, and draw AZ vertical; then the spar 
AD lies wholly in the vertical plane EAZ. 
