Even more simplified experiment to show OU using artificial gravity

[telecom]

Hi Lumen,
I presume the blue disk is rotating ACW?

Sorry,
I ment CCW.

[lumen]

Hi Lumen,
I presume the blue disk is rotating ACW?

It could rotate either direction but it makes sense to rotate this setup ACW because of where the weights are.

I have done some simulation testing and this is at least a very interesting configuration in that the torque on the output is based solely on the weights, without them there is no output torque.

I plan to build a small test setup with some small DC motors to test the idea.

[telecom]

This concept of yours looks very similar to UE.
http://www.universalengines.com.au/how-ue-works
Please keep us posted on the progress.

[lumen]

I agree it is almost the same, but his is overly complicated in that he is trying to time out the imbalance.
(actually after reading some of his work, I believe it's the same principal)

The same thing can be done even easier than I have shown using just four sprockets and some chain on the wheel. (+ the weights)

I have been an engineer for over 30 years and though I am sure this can't work, I don't see why!
The torque on the output is totally decoupled from the drive. If this was not so than a similar setup could be use to build a gravity wheel.
It's like the weights can be continuously falling as the wheel turns, driving the output, and the faster it turns is like stronger gravity.
If anyone knows why this does not work, then speak up so I don't need to build it!

[nybtorque]

IINM .. the 0.2 joules of energy invested to maintain equilibrium of forces is the amount of energy inputted to overcome windage [drag forces] of the bottle opener & rotating wire ?!

N.B. for a constant velocity [rpm] if we know the drag force [proportional to the square of the speed] & the circumference of the circle at radius 20cm then f x d = joules of energy expended - since the drag force is a function of air mass [e.g. Force in Newtons = Cd.1/2(air density).velocity^2.frontal area] then according to Newton's Laws we must input muscle energy equivalent to the energy lost to air drag.

The 0.2 J of kinetic energy E=( m * w^2 * r^2 ) / 2, where w is angular velocity (rad/s),  i.e.   (0.05 * 0.2^2 * 14.1^2 ) / 2 =0.2 J. I assume no friction from windage.

Force [N's] = mass x acceleration

Centrifugal force = Centripetal force = m.v^2/r

Gravity force = Weight force = m.g where 'g' = 9.81 m/s^2

1. A small increase in muscle input energy will break the equilibrium of forces & increase rpm & velocity of the bottle opener [bo] - this will also increase air drag proportionately since it is a factor of air velocity squared.

2. the Cf's acting on the bo will increase by the square of the velocity so that the Cf is greater than the weight force of the suspended mass.

3. since Cf's [mv^2/r] > mg the suspended mass will rise upwards gaining Pe [mgh in joules].

4. the radius of rotation will increase to greater than 20 cm.

5. the Cf will reduce because of a greater radius.

Questions:

a. does the Potential Energy [Pe] gained by the suspended mass ever exceed the excess muscle energy input required to maintain the new & higher rpm ?

b. is there ever a situation using a mechanical assembly of any sort where Pe gained is greater than the input energy ?

It is easy to calculate an example to answer the questions. If we start with my example of equilibrium, where 0.2 J of kinetic energy is invested for force equilibrium. Then, lets say we invest another 0.2 J in the rotation using finger and arm muscles, total kinetic energy is now 0.4 J. What will happen?


As Fletcher stated RPM will increase, the radius of the rotating BO will increase and the GP will move upwards in search for a new force equilibrium. This will be found sooner or later because energy is proportional to radius squared and force to radius (non squared).


Now we have two unkowns, w (new rpm) and r (new radius), but we have two equations as well: 


(1)The new kinetic energy of our system: E(new) = w(new)^2 * r(new)^2 * m(bo) / 2 = 0.4 J    and 


(2)The new force equilibrium: w(new)^2 * r(new) * m(bo) = m(gp) * g


=>   r(new) = 2*E(new) / ( m(gp) * g )  =   0.4 m


The adding of 0.2 J  in kinetic energy actually lifted the m(gp) 0.2 m (since the radius increased from 0.2m->0.4m). How much potential energy is that?


Well, by lifting the garlic press 0.2 m we get 0.2kg*g*0.2m = 0.4 J...




The result is by investing another 0.2 J of kinetic energy in the rotation we get TWICE back in gained potential energy. This is done by using the "artificial" gravity caused by rotation (centrifugal/centripetal force). 


I believe pumping water the same way is even more interesting since we do not have to deal with increased radius and rpm...