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THE PARALLEL MOTION.

ce.

n 1

The Beam being given, the Length of Stroke and Link,

to find the Length of Radius Rod. Fig. 1. A B - cd = BC; then, as A Bicf:: BC:dk, cf

dk
=ch; and
2

2 Assume the right-angled triangle c d e, then sc d2 ce2 = de. Now dc- de = the versed

dn2 tnr2 sine nr; then,

= the diameter of the circle described by the radius rod d x. Thus, let the beam equal 12 feet, link 2 feet 9 inches, and stroke 4 feet; required the length of radius rod d x.

A B will = 6 feet, or 72 inches ; B C = 39 inches, and cf= 48 inches; then, as 72 : 48 :: 39 : 26 inches, or d k, and 42 = 24, also a = 13, that is, 24 =ch and 13 = dn. And 24 13 = 11 or c

e,

then dc= 33 inches, and 332 - 112 = 31.11, and 33

132 + 1.892 91.3 31.11 = 1.89 or n r, then

= 45.65 1.89

2 B C2 inches nearly.—Or,

CA

= d x nearly. And mul. tiply C A by the decimal .55, the product will be the length of front and back straps for that motion.

Fig. 2. In this motion G f may either be the whole or the half length of beam, as may be required. But, in this instance, let b a = the length, bd = any length at pleasure, and A a = the length of stroke;

A a

Dd then, as b a : Aa:: bd: d D.

= a f and 2

2 d e; then a f de = ca, whence arises the rightangled-triangle a cd; then, ✓d a2 - ca2 = c d.

de? +Be Nowda - cd=Be, the versed sine; then the diameter of the circle described by the radius rod

Be

d F. Or, let b a = 67] inches, b d, 324. and A a = 36, to find d F. As 67.5 : 36 :: 32.25 : 17.2 inches, or d D, 36

17.2 then = 18 or a f, and = 8.6 or d e, and 2

2 18 8.6 = 9.4 or c a, then da = 354 inches, and ✓35.252 9.42 33.97; then, 35.25

33.97 = 8.62 + 1.282 59.06 1.28 or B e, then

= 29.53 inches 1.28

2 nearly, length of radius rod, or d F.

Fig. 3--As the calculation of this motion is rather tedious, on account of the different angles formed by the side rods, drawing is preferable.

And with the radius G a, equal to half the length of the beam, describe the arc a A e, equal to the length of stroke; draw the line G a, G A, and Ge; also, from G as a centre, with about i of G a, describe the arc 6 B. Erect a perpendicular, from half the versed sine described by the end of the beam, which will be the centre of the cylinder ; then from a A and e as centres, with the length of the side rods, cut the perpendicular line at the top, middle, and bottom stroke; draw lines to each, and from the end of the cross-head, or top of the side rod, set off the distance for the pin, or the end of the link, as e C, &c. and from c C as centres, with the distance a b, describe arcs at d D; also, from 6 B, with the distance A C, cut the former arcs in D d, draw the lines D C, dc, &c. then will d f be the semichord, and D f the versed sine of the circle described by the pin of the crank d k.-Then the length of the crank may be found either by the sixth problem in Geometry, or the eighth problem in Mensuration.

THE CONNECTING ROD.

The proportionate length of connecting rod is three times the length of stroke; which determines the perpendicular distance between the centre of the beam and centre of fly-wheel shaft. Or, if the engine is erected, the length of connecting rod is the perpendicular distance between the centre of the flywheel shaft, and centre of the beam.

THE FLY WHEEL.

To find the Weight of the Rim or Ring of a Fly

wheel proper for a Steam-engine. RULE.-Multiply the constant number, 1368, by the number of horses' power that the engine is equal to; divide the product by the diameter of the wheel, in feet multiplied by the number of revolutions per minute; and the quotient is the weight of the ring in cwts. nearly.

EXAMPLE.-Required the weight of the rim of a fly-wheel, proper for an engine of 20-horse power, the wheel to be 16 feet diameter, and make 21 revolutions per minute ? 1368 x 20

= 81.4 cwts. nearly. 16 x 21

Note.—The fly-wheel of an engine for a corn or flour mill ought to be of such a diameter that the velocity of the periphry of the wheel may exceed the velocity of the periphry of the stones, to prevent, as much as possible, any tendency to back lash, as it is termed.

The necessary weight and diameter of the wheel being found, suppose a breadth of rim, and the thickness to make the weight in cast iron will be found by the following

RULE. —Divide the required weight in lbs. by the area of the ring in inches multiplied by .263, and the quotient is the thickness of the ring in inches ?

ExamplE.-What thickness must a ring be to equal 81.4 cwts. when the outer diameter is 16 feet, and inner diameter 14 feet 8 inches ?

81.4 cwts. = 9116.8 lbs. And, by problem 12, in Mensuration, the area of

the ring = 4624.43 inches. Then, 9116.8

= 7.496 inches nearly. 4624.43 x.263

And if the ring is to be of a cylindrical form, find the diameter of a circle, (by problem 9 in Mensuration,) having the same area as the cross-section of the ring found.

Thus, suppose the ring in the last example, be required to be cylindrical.-Required its cross-sectional diameter to equal 81.4 cwts., the diameter of the wheel being 16 feet.

7.496 x 8= 59.968 inches, cross-sectional area of the ring found. And 59.968 X 452

= 8.73 inches diameter nearly. 355 Or, as an approximate, multiply the required weight, in lbs., by 1.62; divide the product by the diameter of the wheel in inches, and the square root of the quotient will be the diameter of the crosssection of the ring, in inches, nearly. Thus, 9116.8 x 1.62

= 8.77 inches. 16 x 12

Sometimes (for various reasons) it is necessary to have the fly-wheel upon a second mover; for instance, there is a 6-horse engine making 50 revolutions per minute, having a fly-wheel of 7 feet diameter, and 9 cwt., but, by the rule, it ought to be 23.46 cwt. Now, a larger wheel cannot be got in, but the same may be put upon a second motion, -required the

cnt.

cwt.

velocity that will increase its momentum equal to 23.46 cwt. on the first motion,

7 feet diameter = 21.9912 feet circumference, and 21.9912 x 50 revolutions = 1099.56 feet velocity. velocity.

velocity. Then, as 9 : 1099.56 :: 23.46 : 2866.1864 = 21.9912

= 130 revolutions, per minute, nearly.

To find the Centrifugal Force of a Fly-wheel. Rule.-Multiply the decimal, .6136, by the diameter of the wheel, in feet, and divide the product by the square of the time of one revolution; the quotient is the centrifugal force when the weight of the body is 1.

EXAMPLE.-Required the centrifugal force of a fly-wheel, 15 feet diameter, and making 40 revolutions per minute, the weight of the ring being 3 tons.

60 = 40 = 1.5 time of one révolution. And, .6136 x 15

= 4.09 x 3 = 12.27 tons, the cen1.52 trifugal force. The centre of percussion, in a fly-wheel, or wheels in general, is ths distant from the centre of suspension nearly.

Note.--The centrifugal force is that power, or tendency, which all revolving bodies have to burst, or fly asunder in a direct line.

And the centre of percussion, in a revolving body, is that point where the whole force, or motion, is collected, or, that point which would strike any obstacle with the greatest effect.

THE GOVERNOR OR REGULATOR. The Length of Pendulums given, to find the Number of

Revolutions

per

Minute. RULE.—Divide 375 by the square root of the pendulum's length, and half the quotient will be the velocity required.

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