Fly-wheel power

A fly-wheel is in effect a mechanical battery. In the early 1950’s the Swiss company Oerlikon introduced the Gyrobus simultaneously in Switzerland, Leopoldville in the Congo and Belgium. The flywheel had a diameter of 1.5 metres (5ft.approx.), weighed 1.6 metric tones and was charged to 3000 rpm. The fly-wheel was charged electrically by overhead electric booms. Once fully charged the gyrobus could travel from 6 Km – 12 Kms at speeds of 50 Kms/hr. – 60 Kms/hr. (on flat ground). Unfortunately none of the buses lasted more than a few years, they were extremely difficult to drive due to precession of the fly-wheel and so were almost impossible to steer, also they were extremely heavy about 10 tonnes. Braking was done electrically and was fed back to power the fly-wheel. You can see the details of these gyrobuses here :

The kinetic energy of such a flywheel can be calculated using the equation (at least I’ll give it a try):

E _{ k } = ( ½ * I * w)

Where I = the inertial momentum of the disc which can be calculated by I = ½ * mr2

and w = the angular velocity in radians. r = radius.

Taking the figures which have been given above the Kinetic energy of the fly-wheel on the gyrobus can be worked out :

M = 3,520 lbs.

3000 rpm = 50 rps

r = 2.46 ft.

Then I = mr^^2/2 = 3520 * 6.05 / 2 = 10,650.82

The Kinetic energy would then be : (1/2 * 10,650.82 * (50 * 6.28)^^2

= (5,325.41 * 98596 ) = 525,023,700 ft. poundal/sec = 16,406,990 ft lbs/sec

= 22,247 Kw

= 6.18 kwh.

In metric :

M= 1600

R = 1.5

*w* = 50 x 6.28 = 314 rads/s

I = mr^^2/2 = 1600 * 0.56 /2 = 448 Kg m^^2.

K.E = I * *w* = 448 * 314^^2/2 = 22,085,504 Joules = 6434 Kw

= 6 Kwh and assuming it ran for 10 minutes it would have an output of 50 hp.

So the fly-wheel in the gyrobus had an energy of 6.1 Kwh approx. which equals an output of 50 hp for 10 mins.

If you take a smaller fly-wheel :

M = 110 lbs

Rpm = 30,000 = 500 rps

R = 0.75 ft

Then I = (1/2 * 110 * 0.56) = 30.8

E _{k} = (1/2 * 30.8 * (500 * 6.28)^^2 = 151,837,840 ft. poundal/sec = 4,744,932 ft lb/sec

= 6434 Kw = . 1.78 Kwh which would give 14.38 hp over 10 minutes.

In metric:

M = 50

R = 0.23m

*w* = 3140 rad/sec

I = 50 * 0.25^^2 = 50 * 0.05 = 2.64/2 = 1.32 Kg m^^2.

K.E = (1.32 * 9859600)/2 = 6,519,660 w/sec

= 6519.7 Kw = 1.8Kwh

If you increase r to 1ft.

M= 110 lbs

*w* = 3140 rad/sec

r = 1

I = 55 lb ft sq

K.E = (55 x 9859600) /2 = 271,139,000 poundal/sec. = 8,473,093 ft lb/sec =

11,489 Kw/sec = 3.1 Kwh.

In metric :

M= 50

R = 0.3

*w* = 3140 rad/sec

I = 2.25 Kg sec sq.

K.E = (2.25 * 9859600)/2 = 11,092,050 w/sec = 3.09 Kwh. Or 24 hp over 10 minutes.

30,000 rpm is a relatively low speed for a flywheel, and precession can be overcome by the use of gimbals. The Rotary Pulse Jet would be the ideal solution to power such a system since it can be small enough to place the whole system on gimbals, with leads from the generator, going to electric motors. The Auragen electrical generator would be ideal for such an application used in multiples since it has very small physical dimensions and has already been tested extensively by the military.

[MOD: subject altered to make it a clear question. CS]