Chapter XXXIX.—The Mechanical Powers. Levers, Pulleys, Gear Wheels, Etc

Power is distinguishable from force or pressure in that the term power means force or pressure in motion, and since this motion cannot occur without the expenditure of the force or pressure, power may, with propriety, be termed the expenditure of force or pressure.

If we suppose a piston to stand in a vertical cylinder sustaining a weight upon its surface and compressing the air within the cylinder, so long as there is no motion no work is done, as the term “work” is understood in a mechanical sense, and the weight merely produces a pressure. If, however, the weight be removed, the compressed air will force the piston upward, performing a certain quantity of work which may best be measured by the amount of power exercised or expended.

The mechanical value of a given amount of power cannot be either increased, diminished or destroyed by means of any mechanical device or appliance whatsoever through which it may be transmitted.

It may be concentrated, as it were, by decreasing the amount of its motion. It may be distended, as it were, by increasing the distance through which it moves, or it may be expended in giving or producing motion, but in either case the amount of duty or work done is the exact equivalent of the amount of power applied.

A gain or increase in speed is not, therefore, a loss of power, but merely a variation in the mode of using or utilizing such power.

For instance, 1 lb. moving through a distance of 12 inches in a given time represents an amount of power which may be employed either as 1 lb. moving a foot, 2 lbs. moving six inches, or ⁄2 lb. moving through 24 inches, in the same space of time, the amount of the power or duty remaining the same in each case, the method of utilization merely having differed.

It is an inexorable law of nature that power is concentrated in proportion as the amount of its motion is diminished, or distended in precise proportion as such motion is increased.

Suppose, for example, that in Fig. 3351 l is a lever having its fulcrum at f, which is 4 inches from end a, and 8 inches from end b, and (leaving the weight of the lever out of the question) if we place an 8 lb. weight on a it will just balance 4 lbs. at b.

If the lever is moved, the amount of motion will be twice as much at end b as it is at end a.

If we apply the power at a, the lever has become a means of converting 8 lbs. moving a certain distance into 4 lbs. moving twice that distance, and nothing has been either gained or lost.

If we apply the power at b, the lever has merely been used as a means of converting 4 lbs. moving a certain distance into 8 lbs. moving one half that distance, and nothing has been gained or lost.

Suppose that end a was moved an inch, and the power at that end will be 8 inch pounds or 8 lbs. moving an inch, whereas at the end b the power is 4 lbs. moving 2 inches; we have, therefore, reduced the weight in the same proportion that we have increased the distance moved through.

Suppose now that the lever is moved to the position denoted by the dotted line m m, and the leverages will be altered; that at end a becoming that denoted by the distance from f to the vertical c, and that for end b being denoted by the distance from f to the vertical d.

This occurs because we are dealing with gravity, which always acts in a vertical line.

A crow bar is an excellent example of the application of the lever. In Fig. 3352, for example, we have a 1 lb. weight on the long end of the lever, and as we are dealing with a weight, the effective length of the long end of the lever is from the fulcrum f to w, which is divided into 10 equal divisions. The short end of the lever is from f to p, which is equal to one division, hence the 1 lb. is balanced by the 10 lbs.

A simple method of distending power is by means of pulleys or gear wheels. Suppose, for example, that in Fig. 3353, we have a weight of 12 lbs. suspended from a shaft or drum, whose radius a is 10 inches, and that on the same shaft there is a pulley, whose radius b is 20 inches, and the two weights will balance each other.

In this case the falling of either weight would not effect the leverage, because the distance of both weights would remain the same from the centre of the shaft. The leverage of the 12 lbs. is denoted by the line a, and that of the 6 lbs. by b.

So far as the transmission of power is concerned, therefore, pulleys are in effect revolving levers, which may be employed to concentrate or to distend power, but do not vary its amount.

Suppose we have two shafts, on the first of which are two pulleys, b and c, Fig. 3354, while upon the second there are two pulleys d and e. A belt h, connecting c to d. Let the pulleys have the following dimensions:

If we take the first pair of wheels b and c, we have that the velocity will vary in the same ratio or degree as their diameters vary, notwithstanding that their revolutions are equal.

Radius. Diameter. Circumference.

B = 5⁄8 inches. 10⁄4 inches. 32.2 inches.

C = 10⁄4 „ 20⁄2 „ 64.4 „

D = 7⁄8 „ 15⁄4 „ 47.9 „

E = 15⁄4 „ 30⁄2 „ 95.8 „

The velocity is the space moved through in a unit of time, and as it is the circumference of the pulley that is considered, the velocity of the circumference is that taken; thus, if we make a mark on the circumferences of the two pulleys, b and c, Fig. 3354, the velocity of that on c will be twice that upon b, or in the same proportion as the diameters.

Let there be suspended from the circumference of b 10 lbs. weight, and let us see the degree to which this power will be distended by this arrangement of pulleys, supposing the weight to rotate b, and making no allowance for the friction of the shaft.

Suppose the weight to have fallen 32.2 inches, and we have 10 lbs. moving through 32.2 inches, this power it will have transmitted to pulley b.

To find what this becomes at the perimeter of c, we must reduce the number of lbs. in the same proportion that the perimeter of c moves faster than does that of b; hence we divide the circumference of one into the other, and with the sum so obtained divide the amount of the weight; thus, 64.4 (circumference of c) ÷ 32.2 (circumference of b), = 2; and 10 lbs. ÷ 2 = 5 lbs., which, as the circumference of c is twice that of b, will move twice as fast as the 10 lbs. at b, hence for c we have 5 lbs. moving through 64.4 inches.

Now c communicates this to d by means of the belt h, hence we have at d the same 5 lbs. moving through 64.4 inches.

Now e moves twice as fast as d, because its circumference is twice as great, and both are fast upon the same shaft, hence the 5 lbs. at d becomes 2⁄2 lbs. at e, but moves through a distance equal to twice 64.4, which is 128.8 inches. To recapitulate, then, we have as follows:

The weight gives 10 lbs. moving through 32.2 inches.

Pulley b „ 10 „ „ „ 32.2 „

„ c „ 5 „ „ „ 64.4 „

„ d „ 5 „ „ „ 64.4 „

„ e „ 2 ⁄2 „ „ „ 128.8 „

That the amount of power is equal in each case, may be shown as follows:

For c, 5 lbs. moving through 64.4 inches is an equal amount of power to 10 lbs. moving through 32.2 inches, because if we suppose the first pair of pulleys to be revolving levers, whose fulcrum is the centre of the shaft, it will be plain that one end of the lever being twice as long as the other, its motion will be twice as great, and the 5 at 10⁄4 inches just balances 10, at 5⁄8 inches from the fulcrum, as in the common lever.

In the case of d we have the same figures both for weight and motion as we have at c, because d simply receives the weight or force and the motion of c. In the case of e, we have the motion of the weight multiplied four times; for the distance e moves is 128.8 inches, which, divided by 4, gives 32.2 inches, which is the amount of motion of the weight, hence the 10 lbs. of the weight is decreased four times, thus 10 lbs. ÷ 4 = 2⁄2 lbs., hence the 2⁄2 lbs. moving through 128.8 inches is the same amount of power as 10 lbs. moving 32.2 inches, and we may concentrate or convert the one into the other, by dividing 128.8 by 4, and multiplying the 2⁄2 lbs. by 4, giving 10 lbs. moving 32.2 inches.

If, therefore, we make no allowance for friction, nothing has been lost and nothing gained.

Thus far, we have taken no account of the time in which the work was done, more than as one wheel is caused to move by the other, and all of them by the motion of the weight, they must all have begun and also have to move at the same time. Suppose, then, that the time occupied by the weight in falling the 32.2 inches was one minute, and the amount of power obtained may be found by multiplying the lbs. of the weight by the distance it moved through in the minute, thus 10 lbs. moving 32.2 inches in a minute gives 32.2 inch lbs. per minute, being the amount of power developed by the 10 lb. weight in falling the 32.2 inches.

We may now convert the power at each pulley perimeter or circumference into inch pounds by multiplying the respective lbs. by the distance moved through in inches, as per the following table:

Distance moved.

Lbs. Inches. Inch lbs. of power.

Weight at b 10 × 32.2 = 322

„ „ c 5 × 64.4 = 322

„ „ d 5 × 64.4 = 322

„ „ e 2 ⁄2 × 128.8 = 322

If we require to find the power in foot lbs. per minute, we divide by 12 (because there are 12 inches in a foot), thus 322 inch lbs. ÷ 12 = 26.83 foot lbs. per minute.

Now suppose that b was moved by a belt, with a pull of 10 lbs. at its perimeter, and made 100 revolutions in a minute instead of one, then the pull at the perimeters of c, d, and e would remain the same, but the motion would be 100 times as great, and the work done would therefore be increased one hundred fold. It will be apparent, then, that the time is as important an element as the weight.

The velocity and power of gear wheels are calculated at the pitch circle.

Now suppose the gear a in Fig. 3355 has 30, gear b 60, gear c 10 and gear e 80 teeth, and that 5 lbs. be applied at the pitch circle of a; to find what this 5 lbs. would become at the pitch circle of e, we multiply it by the number of teeth in b and divide it by the number of teeth in c, thus:

Lbs.

At pitch of circle a 5

Number of teeth in b 60

Number of teeth in c 10 ) 300

30

Answer, 30 lbs. at the pitch circle of e.

Now suppose that on the shaft of a there is a pulley 20 inches in diameter, and that on this pulley there is a belt exerting a pull of 5 lbs., while on the shaft of e there is a pulley 16 inches in diameter, and to find how much this latter pulley would pull its belt, we proceed as follows:

2 ) 20 = Diameter of pulley on a.

10 = Radius of pulley on a.

5 = Pull on pulley a.

Number of teeth in a = 30 ) 50 = Pull at centre of shaft of a.

1.666

60 = Number of teeth on b.

Number of teeth on c = 10 ) 99.960 = Pull at axis of shaft of b.

9.996

80 = Number of teeth on e.

Radius of pulley on shaft of e 8 ) 799.680 Pull at axis of shaft e.

99.96 Pull at perimeter of last pulley.

We have in this case treated each pulley as a lever whose length equalled the radius of the pulley, while in the case of the wheel we have multiplied by the number of teeth when the power was transmitted from the circumference to the shaft, and divided by the number of teeth (the number of teeth representing the circumference) when the power was transmitted from the shaft to the teeth.

We thus find that power is composed of three things, first, the amount of impelling force; second, the distance that force moves through; and third, the time it takes to move that distance.

If we take a number of pulleys, say four, and arrange them one after another so that they drive by the friction of their circumferences, then the amount of power transmitted by each will be equal and the velocities will be equal, whereas, if we arrange them as in Fig. 3354, the power will be equal for each, but the velocities or space moved through in a given time will vary.

What is known as the unit of power is the foot lb., being the amount of power exerted in raising or lifting one lb. one foot, and from what has already been said, it will be perceived that this is the same amount of power as 12 lbs. moving a distance of one inch.

Watt determined that the power of a horse was equal to that necessary to raise 33,000 lbs. one foot high in a minute, and this is accepted, in English speaking countries, as being a horse power.

An engine or machine has as much horse power as it has capacity to lift 33,000 lbs. a foot high in a minute.