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134. And, generally, the forces are as the distances or radii of the circles directly, and the squares of the times inversely.

135. The squares of the times are as the distances directly and the forces inversely.

136. Therefore if the forces are inversely as the square of the distances, the squares of the times are as the cubes of the distances.

137. The centrifugal force of a body revolving in a circle is equal to the square of the velocity divided by the radius.

Let F and f represent the forces, T and t the times, and D and d the distances; then,

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Suppose the diameter of a grindstone to be 44 inches, and its weight half a ton; suppose also that it makes 326 revolutions in a minute; what will be the centrifugal force, or its tendency to burst?

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The radius of the circle of gyration is 2, or the

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2

=r √2 22 √2 31.11 inches
d being the radius of gyration,

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2 d (3·1416) 2 2.59 × (3-1416)9

161 to

=

weight of the stone; hence,

47.18 =23.59 tons, the centrifugal force.

2

Example 2.

. If a fly-wheel, 10 feet diameter and 2 tons weight, perform a revolution in 6 seconds, and suppose another of the same weight revolves in 4 seconds, what must be the dia

meter of this last, so that their centrifugal forcés may be equal?

D d

By Art. 137, F:f; but, in this case, the

T2

centrifugal forces are equal, that is F = f; hence,

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If a fly-wheel, 10 feet diameter, revolve in 6 seconds, and another of the same diameter in 4 seconds, what proportion do the weights bear to each other, when their central forces are equal?

The weights are as the square of the times; that is, w: W:: 16 : 36 :: 1 : 21.

Example 4.

If a fly, 2 tons weight and 16 feet in diameter, is sufficient to regulate an engine when it revolves in 4 seconds, what must be the weight of another fly, of 12 feet diame→ ter, revolving in 2 seconds, so that it may have the same power upon the engine?

For each fly to have the same power on the engine, their centrifugal forces must be equal.

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138. Let A P and AQ represent the arms of the governor, P and Q the balls; and let ∞ = B P, the variable horizontal distance of each ball from that shaft, a = the

corresponding altitude, t the time of one revolution of the shaft, and p =3·1416. (See Fig. 1 to Steam Engine.)

Then the velocity of each ball = 2; and since the

t

centrifugal force is equal to the square of the velocity di

2

vided by the radius, we have (2 p *)*

t

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The balls are acted on by two forces, the centrifugal force and gravity; the former acting in a direction parallel to the horizon, and the latter in a direction perpendicular to it. The resultant of these two forces is evidently always in the direction of the arms A Por A Q. It follows, therefore, that the ratio of the gravity to the centrifugal force is as A B to B P.

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Hence the periodic time varies as the square root of the altitude of the conical pendulum, let the radius of the base be what it may.

When A B = B P, the centrifugal force of each ball is equal to its weight.

PART II.

ON THE STEAM ENGINE.

THE Steam Engine is, beyond all doubt, one of the no-
blest inventions which any age or country can boast of;
and many disputes have arisen concerning its origin.
Some have considered the famous Denys Pepin as its in-
ventor; others ascribe it to the Marquis of Worcester.
But be this as it may, we are quite certain that it is by
the
great skill, powerful genius, and unremitting assiduity
of the celebrated Watt, that it has attained its present
state of usefulness and perfection.

FUEL, &c.

Mr. Watt found that, with the most judiciously constructed furnaces, it required 8 feet of surface of the boiler to be exposed to the action of the fire and flame to boil off a cubic foot of water in an hour; and that a bushel of Newcastle coals, so applied, will boil off from 8 to 12 cubic feet; and that it requires about a cwt. of Wednesbury coals to do the same.

A bushel of coals, which is the consumption of a 10 horse engine for one hour, grinds and dresses about 10 bushels of wheat, Winchester measure.

The consumption of a bushel of good Newcastle coals (= 84 lbs.) will raise 18.34 millions pounds of water one

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foot high. Or 1.08 bushels per hour will supply an engine of 10 horse power. And, at the same rate, each horse power will require 9.07 lbs. of such coals per hour.

=

For a bushel of coals (= 84 lbs.) may be assumed to evaporate 10.08 cubic feet of water into steam; and every cubic inch of water, so evaporated, will produce a cubic foot of steam, of the same elasticity as the atmospheric air. Hence, a bushel of coals will produce (10-08 x 1728) 17418 cubic feet of steam; and if each cubic foot of steam raises one cubic foot of water 16-86 feet high, then the consumption of a bushel of coals must raise 17418 × 16.85 293500 cubic feet of water, or 18340000 lbs. of water, one foot high.*

When bodies are heated above the temperature of the surrounding media, the heat escapes from them by radiation; that is, by passing from the body in straight lines through that surrounding media.

Professor Leslie has proved, by a variety of experiments, that the heat which is propagated by radiation from different bodies, varies with the nature of their external surfaces; the quantity which flows in a given time from a body with a polished surface being much less than would flow from the same body with a rough surface. It, therefore, follows, that the external surfaces of the steam pipes of steam engines and steam cylinders should be as smooth as possible, and should be covered with any body which is a bad conductor of heat.†

Dr. Hook found that if water were put into any vessel and set on the fire, the temperature of the water would continually increase until it began to boil; after which, its temperature would increase no further, as much heat being carried off by the vapour as is received from the fire. But Papin, by using enclosed vessels, found that the tem

*Farey on the Steam Engine, page 488.

+ Experimental Enquiry into the Nature and Propagation of Heat, page 17.

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