SPEED OF GEARING.

The ratio of the numbers of teeth in a pair of wheels, must be the same as that of their diameters.

To find the speed of the driving wheel:-Multiply the number of teeth in the driven wheel, by the number of revolutions it makes per minute, and divide the product by the number of teeth in the driving wheel.

To find the speed of the driven wheel.-Multiply the number of teeth in the driving wheel, by the number of revolutions it makes per minute, and divide the product by the number of teeth in the driven wheel.

To find the final speed of a train of wheels.-Multiply the number of revolutions per minute of the first driving wheel, by the product of the number of teeth in the driving wheels, and divide the result by the product of the number of teeth in the driven wheels.

To find the number of teeth in the driving wheel.-Multiply the number of teeth in the driven wheel, by the number of revolutions it makes per minute, and divide the product by the number of revolutious of the driving wheel.

To find the number of teeth in the driven wheel.-Multiply the number of teeth in the driving wheel, by the number of revolutions it makes per minute, and divide the product by the number of revolutions of the driven wheel.

To find the relative numbers of teeth in a pair of wheels, when the speeds of the driving and driven shafts are given. Divide the speed of the driven shaft, by the speed of the driving shafts; the quotient is the ratio of their speeds; and the numbers of teeth in the wheels must be in the same ratio.

To find the diameters of a pair of wheels, the distance between the centres, and also the speed of each shaft being given. Multiply the speed of one shaft by the distance between the centres in inches, and divide the product by the sum of the speeds of the two shafts, the result will give the radius of one wheel, which doubled, will give its pitch diameter in inches. The radius of this wheel subtracted from the distance between the centres, will give the radius of the other wheel.

To find the pitch of a wheel.-Divide the diameter of the wheel at the pitch circle, by the number of teeth, and multiply the quotient by 3'1416. To find the number of teeth in a wheel.-Divide 3'1416 by the pitch, and multiply the quotient, by the diameter of the pitch circle in

inches.

To find the diameter of a wheel at the pitch circle.-Divide the pitch by 31416, and multiply the quotient by the number of teeth.

Wheels and Pinions.-A wheel should not have more teeth than for 1 of its pinion. Large pinions are desirable, because when a large

wheel drives a small pinion rapidly, the teeth of the pinion moving in a small circle, abruptly meet the teeth of the wheel, and cause an uneven jolting motion. When wheels drive pinions, no pinion should have less than 20 teeth, and in millwork not less than from 35 to 45 teeth, to enable them to work properly, and have a sufficient number of teeth in gear at the same time. When pinions drive wheels no pinion should have a less number than 13 teeth; rather 16 or 18. When quick speed is required instead of using a large wheel and very small pinion, it is better to get up the speed by using an intermediate shaft with wheel and pinion, and the friction will not be materially increased thereby.

POWER OF WHEEL GEARING.

Power of Wheel Gearing.—The pressure on the teeth of wheels varies inversely as the number of revolutions and directly as the power transmitted. Thus, if the same power be transmitted by two wheels at different velocities, say one at 30 and the other at 120 revolutions, the strain on the former will be four times that of the latter; or if one wheel transmits 10 horse-power and another 20 horse-power at the same velocity, the strain on the latter will be double that of the former. Again, the power transmitted by a wheel depends upon the number of teeth in gear at one time and also upon its velocity.

Power of Spar Wheels.-The horse-power of the ordinary spur wheels used in machinery and millwork is given in table 19, which has been deduced from cases in practice. In cases where wheels are subject to unusually great strains they are made of other materials than cast iron. Good Malleable Cast-Iron Wheels have double the strength of castiron wheels.

Good tough Gun-metal Wheels, have double the strength of cast-iron wheels.

Wrought-Iron Wheels, are three times as strong as cast-iron wheels, when made of best iron, with the grain of the iron in the direction of the circumference of the wheel.

Good Cast-Steel Wheels, are four times as strong as cast-iron wheels. Shrouded Wheels, or wheels with two flanges, are from one-third to one-half stronger, according to the form of tooth, than plain wheels.

The Power of Bevel and Mitre Wheels may be taken from table 19, but instead of the maximum, the mean diameter and pitch must be taken; for instance, a bevel wheel 36 inches maximum diameter, with 6 inches face, has a minimum diameter of 30 inches, the mean diameter is therefore 33 inches, the pitch is 3 inches, but the rainimum pitch is in pro

portion to the diameter; thus 3 x 30

2

36

= 2·5 minimum pitch, and the mean

pitch will therefore be 3+25= 2.75 mean pitch, and in looking for the horse-power in the table, it must be called 33 inches diameter × 2 inches pitch.

Power of Mortice-Wheels.-When running at a good speed, morticewheels are quite as strong as iron toothed wheels, but at a low speed they are weaker than iron wheels.

Power of Crane Gearing.—When wheels work at very low velocities lifting heavy weights, as in cranes, the safe working load should not exceed of the breaking weight, and the strength of the teeth should be calculated accordingly. A bar of good cast-iron 1 inch long, and 1 inch square, loaded at the end, will break with 6000 lbs., and the tooth of a wheel is similar to a beam loaded at one end and fixed at the other, hence the following rule :

To find the Breaking Strain of each Tooth in a Wheel.-Multiply the square of the thickness of one tooth by its width, then by 6000, and divide the result by the length of tooth, the product will be the breaking weight in lbs. of each tooth.

Example: A crane to lift 4 tons, has a wheel 4 feet diameter, with a barrel 12 inches diameter, measuring to the centre of the chain. The pressure at the pitch-line of the wheel will be the weight to be lifted in lbs., multiplied by the diameter of the barrel in feet, and divided by the diameter of the wheel in feet: then 8960 × 1 = 2240 lbs. actual strain, and suppose

the teeth to be thick × 1 long × 4 inches wide, then 75a × 4 × 6000

1'125

= 11,786 lbs. the breaking weight of one tooth, and if two teeth are in gear at the same time, the breaking strain of two teeth will be 11,786 × 2

23,572 lbs., the ratio of which to the actual strain is 23572 = 10 to 1,

2240

which is ample for safety. Machinery subject to shocks from sudden. change of speed and irregular strains, must have an excess of power in the gearing to provide against accidents. This rule for obtaining the actual strength of teeth applies to wheels working slowly and lifting heavy weights -the following rule is used for ordinary gearing.

Horse-Power of Gearing.-To find the horse-power of ordinary irontoothed spur wheels, used in machinery and millwork. Rule: Multiply the square of the pitch of the teeth in inches, by the width of face of the teeth in inches, multiply the product by the diameter of the wheel in feet at the pitch circle, and multiply that product by the number of revolutions per minute, and divide the result by the constant number 240, the result will be the actual or indicated horse-power which that wheel will properly transmit.

CRANE GEARING.

To find the strains at the pitch-lines of a train of wheels, such as the

then

Fig. 100.-Crane-Gearing.

Diameter of the barrel at the

centre of the chain=20 inches.

Radius of the handles=16 inches,

60 lbs. strain at the handles x 16 radius of the handle 3 inches radius of first pinion A strain at the pitch lines of wheels A and B. 320 lbs. x 18 inches radius of first wheel B 5 inches radius of second pinion C

pitch lines of wheels C and D.

=320 lbs.

= 1152 lbs. strain at the

1152 lbs. × 30 inches radius of second wheel D 10 inches radius of third pinion E

pitch lines of wheels E and F.

3456 lbs. 40 radius of third wheel F

10 inches radius of barrel at the centre of the chain' the chain,

ог

13824 2240

=

3456 lbs. strain at the

=13824 lbs. strain on

=6126 tons, and as the snatch block, or running pulley, will double the power, about 12 tons would be lifted by the 4 men.

Double Helical Toothed-Wheels, shown in the engraving on the next page, possess a strong and durable form of tooth; they work smoothly and almost noiselessly, without vibration, and the teeth always keep in the right plane of revolution. As angular teeth of this form approach to, and recede from each other more gradually than ordinary straight teeth, a more perfect rolling motion is obtained. A good angle for the teeth is 30°, from the straight line or 60° from the side of the wheel, but the angle may be varied.

The Strength of Double Helical Toothed-Wheels with teeth at an angle of 30°, is 20 per cent. greater than the strength of ordinary toothedwheels of the same pitch and width.

The Horse-power of Double Helical Toothed-Wheels, having teeth at the above angle, may be found by this Rule. Multiply together the square of the pitch in inches, the width of face in inches, the diameter of the wheel at the pitch circle in feet, the number of revolutions per minute, and divide the product by the constant number 200, the quotient will be the actual or indicated horse-power which that wheel will properly transmit. Frictional Gearing. The pitch of frictional gearing varies from inch to 1 inch; the driving power is one sixth of the interpressure between the wheels. Fig. 111 is a full size section of teeth 1 inch pitch-which is the pitch generally used for hoists-and represents the exact form of tooth found to answer best in practice for this purpose. Thickness of tooth =

T

Fig. 111.-Frictional Gearing.

ths the pitch: width of space between the teeth = 3ths the pitch: depth

of tooth ths the pitch: angle of point of tooth = 70°.

=

4 inches circumference for small powers, and 5 and 6 inches circum