The size of the rope wheels can now be fixed at 32 inches
diameter. Suppose the cabin to weigh 14 cwt., and to be
supported by four §-inch ropes. The first step in the calcu-
lation is to determine the efficiency of the ropes working
over five 32-inch wheels upon 2|-inch pins. Consulting
Table VIII., we find the efficiency of a f-inch rope on a



174 HYDRAULIC POWER ENGINEERINCJ.
30-inch wheel .99, which is also the efficiency of four ropes
on the wheel. The efficiency of the sheave on the pin = .96.
Therefore efficiency of the sheaves, rope, and pins of the
hoist only = (.96 x .99)^ = .95^ = .76.
The next step is to ascertain how much of the cage weight
must be left unbalanced to enable it to overcome the friction
to descend empty. The overhead pulleys may be 32 inches
diameter, and as the ropes only lap one quarter of the
circumference on each, the efficiency of two pulleys is equal
to the efficiency of one = .99, and the efficiency of sheaves
on pins .96.
Therefore, the weight to overcome this =
{i-(.99 X .96)} X weight of cage = .os X 14 X 112 = 78 lbs.
Again, the cage has to overcome the friction
of the pulley sheaves of the hoist, having an
efficiency of .77. Then the weight to overcome
this is (i - .77) X weight of cage = .23 X 14 X 112 =360 „

Finally, the cage has to pull the ram into the

cylinder against the friction of its stuffing box,

and, from Table IV., this 24 inches diameter ram

. , 90 100
requires 90 lbs. to move it, so that -, x = 20 „



Total 458 lbs.
