Circumference of large groove 11"·84, of small
groove 10"·36; velocity ratio 16; mechanical
efficiency 6·07; useful effect 38 per cent.;
formula _P_ = 3·87 + 0·1508 _R_.
+-----------+---------+----------+----------+---------------+
| | | | P. | Difference |
| Number of | R. | Observed |Calculated|of the observed|
|Experiment.| Load | power | power | and calculated|
| | in lbs. | in lbs. | in lbs. | values. |
+-----------+---------+----------+----------+---------------+
| 1 | 56 | 10 | 12·3 | + 2·3 |
| 2 | 112 | 20 | 20·8 | + 0·8 |
| 3 | 168 | 31 | 29·2 | - 1·8 |
| 4 | 224 | 38 | 37·7 | - 0·3 |
| 5 | 280 | 48 | 46·1 | - 1·9 |
| 6 | 336 | 54 | 54·6 | + 0·6 |
| 7 | 392 | 64 | 63·1 | - 0·9 |
| 8 | 448 | 72 | 71·5 | - 0·5 |
| 9 | 504 | 80 | 80·0 | 0·0 |
| 10 | 560 | 86 | 88·4 | + 2·4 |
+-----------+---------+----------+----------+---------------+
The first column contains the numbers of the experiments, the second
the weights raised, the third the observed values of the corresponding
powers. From these the following rule for finding the power has been
obtained:—
216. To find the power, multiply the load by 0·1508, and add 3·87
lbs. to the product; this rule may be expressed by the formula
_P_ = 3·87 + 0·1508 _R_. (See Appendix.)
217. The calculated values of the powers are given in the fourth
column, and the differences between the observed and calculated values
in the last column. The differences do not in any case amount to 2·5
lbs., and considering that the loads raised are up to a quarter of a
ton, the formula represents the experiments with satisfactory precision.
218. Suppose for example 280 lbs. is to be raised; the product of 280
and 0·1508 is 42·22, to which, when 3·87 is added, we find 46·09 to be
the requisite power. The mechanical efficiency found by dividing 46·09
into 280 is 6·07.
219. To raise 280 lbs. one foot 280 foot-pounds of energy would be
necessary, but in the differential pulley-block 46·09 lbs. must
be exerted for a distance of 16 feet in order to accomplish this
object. The product of 46·09 and 16 is 737·4. Hence the differential
pulley-block requires 737·4 foot-pounds of energy to be applied in
order to yield 280 useful foot-pounds; but 280 is only 38 per cent. of
737·4, and therefore with a load of 280 lbs. only 38 per cent. of the
energy applied to a differential pulley-block is utilized. In general,
we may state that not more than about 40 per cent. is profitably used,
and that the remainder is expended in overcoming friction.
220. It is a remarkable and useful property of the differential
pulley, that a weight which has been hoisted will remain suspended
when the hand is removed, even though the chain be not secured in
any manner. The pulleys we have previously considered do not possess
this convenient property. The weight raised by the three-sheave
pulley-block, for example, will run down unless the free end of the
rope be properly secured. The difference in this respect between these
two mechanical powers is not a consequence of any special mechanism;
it is simply caused by the excessive friction in the differential
pulley-block.
221. The reason why the load does not run down in the differential
pulley may be thus explained. Let us suppose that a weight of 400 lbs.
is to be raised one foot by the differential pulley-block; 400 units of
work are necessary, and therefore 1,000 units of work must be applied
to the power chain to produce the 400 units (since only 40 per cent. is
utilized). The friction will thus have consumed 600 units of work when
the load has been raised one foot. If the power-weight be removed, the
pressure supported by the upper pulley-block is diminished. In fact,
since the power-weight is about ¹/₆th of the load, the pressure on
the axle when the power-weight has been removed is only ⁶/₇ths of its
previous value. The friction is nearly proportional to that pressure:
hence when the power has been removed the friction on the upper axle is
⁶/₇ths of its previous value, while the friction on the lower pulley
remains unaltered.
We may therefore assume that the total friction is at least ⁶/₇ths of
what it was before the power-weight was removed. Will friction allow
the load to descend? 600 foot-pounds of work were required to overcome
the friction in the ascent: at least ⁶/₇ × 600 = 514 foot-pounds would
be necessary to overcome friction in the descent. But where is this
energy to come from? The load in its descent could only yield 400
units, and thus descent by the mere weight of the load is impossible.
To enable the load to descend we have actually to aid the movement by
pulling the chain D (Figs. 36 and 37), which proceeds from the small
groove in the upper pulley.
222. The principle which we have here established extends to other
mechanical powers, and may be stated generally. _Whenever more than
half the applied energy is consumed by friction, the load will remain
without running down when the machine is left free._