## A Treatise on Mills: In Four Parts. Part First, On Circular Motion. Part Second, On the Maximum of Moving Bodies, Machines, Engines, &c. Part Third, On the Velocity of Effluent Water. Part Fourth, Experiments on Circular Motion, Waterwheels, &c |

### Common terms and phrases

16 feet 20 feet accelerating force angular velocity aperture axle beam buckets central force centre of gravity centre of gyration centre of oscillation centre of percussion centrifugal force circumference cog-wheel compared with gravity completing the square computed cubic feet cylinder depth descend distance divided Dq² effect produced EXAMPLE feet diameter feet per second floats fluxion following experiments given power given space given stream gives hence hole impulse locity maximum Michelotti minute mn² mn³ momentum moves twice moving power multiplied by 16 number of revolutions number of turns orifice piston pr² Prob PROBLEM quantity of water radius ratio shorter end space passed square root suppose surface suspended Table tance Theo Theorem tiplied trundle tube velo velocity per second water is applied water wheel weight raised wheel moves whole resistance whole weight

### Popular passages

Page 152 - The effect produced by a given stream in falling through a given space, if compared with a weight, will be directly as that space ; but if we measure it by the velocity communicated to the wheel, it will be as the square root of the space descended through, agreeably to the laws of falling bodies. "Experiment 1. A given stream is applied to a wheel at the centre ; the revolutions per minute are 38.5 " Ex. 2. The same stream applied at the top, turns the same wheel 57 times in a minute. " If in the...

Page 49 - The centres of percussion and of oscillation are generally treated separately, but the two centres are in the same point, and therefore their properties are the same. As in bodies at rest the whole weight may be considered as collected in the centre of gravity, so in bodies in motion, the whole force may be considered as concentrated in the centre of percussion ; therefore, the weight of the rod multiplied by the distance of the centre of gravity from the point of suspension, will be equal to the...

Page 96 - ... is equal to that which a heavy body would acquire in falling through a space equal to the depth of the opening below the surface of the fluid, and is expressed as follows: v—i/lgh.

Page 108 - Measure the depth (of the vessel, &c.) in feet, extract the square root of that depth, and multiply it by 5.4, which gives the velocity in feet per second ; this multiplied by the area of the orifice in feet, gives the number of cubic feet which flows out in one second.

Page 52 - Multiply the distance of the centre of oscillation, from the centre or point of suspension, by the distance of the centre of gravity from the same point, and the square root of the product will be the aaswer.

Page 152 - ... as that space ; but if we measure it by the velocity communicated to the wheel, it will be as the square root of the space descended through, agreeably to the laws of falling bodies. . . , Experiment 1. A given stream is applied to a wheel at the centre ;. the revolutions per minute are 38.5. Ex. 2. The same stream applied at the top, turns the same wheel 57 times in a minute. If in the first experiment the fall is called 1, in the second it will...

Page 11 - If a fly 2 tons weight and 16 feet diameter, is sufficient to regulate an engine when it revolves in 4 seconds ; what must be the weight of another fly of 12 feet diameter revolving in 2 seconds, so that it may have the same power upon the engine ? Here, by art 8, Central Forces, we must have — — =— — ; . vf D t* 40cwtXl6X4 160 , , therefore w = — ; - = -- = - = 135 cwL, the d T" 12X16 12 weight of the smaller fly.* Note.

Page 105 - ... wave is great in comparison with the depth of the canal, and when the maximum height of the wave is small in comparison with the same quantity, was given long ago by Lagrange, and is now well known. A wave of any form, subject to the above conditions, is propagated unchanged, and with the velocity which would be acquired by a heavy body in falling through half the depth of the canal. The velocity of propagation here referred to is of course relative to the undisturbed water. If we attribute to...

Page 110 - ... side of a vessel or dam, and open at the top, will be found by multiplying the velocity at the bottom by the depth, and taking f of the product for the area ; which again multiplied by the breadth of the slit or notch, gives the quantity of cubic feet discharged in a given time. EXAMPLE. Let the depth be 5 inches, and the breadth 6 inches ; required the quantity run out in 46 seconds ? The depth is .4166 of a foot. The breadth is .5 of a foot.