Torque calculation

[pequaide]

If you are referring to the 3.8 kilograms on the circumference 'as in an Atwood's'; then yes they still have leverage advantage. Just because they are balanced does not take their mass or leverage away. They have a leverage advantage of 2 over the drive mass position. And the drive mass is less than 150 grams.

[eccentrically1]

Pequaide,
Trevor is talking about clay's setup, not yours. The two 1 KG weights are both 1 meter from the pivot, so neither one is leveraging the rod. Maybe it would help if you read Jim's answer.

[bobriddle]

If we have a rod with an equal weight at both ends and a pivot point in the middle (see attached pic), how do you calculate the amount of weight required to spin it to say 10RPM, and how long it will take to reach the 10RPM from standstill (I believe its called breakaway torque)?

clay,
One thing I kind of came up with is a little different.
I had built a 4 weighted wheel that I used for studying acceleration and movements of rotating and swinging weights, including pendulums.
One thing I noticed was that if 2 weights are at 90 degrees to each other, then if it rotates clock wise, the weight at 3 o'clock comes pretty close to stopping at 6 o'clock. Pretty much an even exchange of enrgy with a little loss due to friction.
When I tried using 4 weights and had a percentage of over balance, things seemed to change quite a bit.
One thing I have considered is along the lines of f = ma. And with this, the weight doing the work needs to have sufficient over balance to account for the other weights.
Going with a 4 weighted design, if the potential of the over balanced weight is increased by 100% (like having a second weight there), then acceleration or velocity would be 25% of gravity. Simply put, the extra force is divided equally among all weights. It's a basic rule of thumb I use when working on something.
And most mechanical designs have trouble achieving any signifigant amount of over balance. Hope this helps.

Jim

[pequaide]

Same answer. Of course they are leveraging the rod. The rod is a lever.

If the drive force is at .1 meters and the driven mass is at 1 meter then the leverage is 10.

You need to consider how close the calculations are to reality. The leverage method of calculation is nearly exactly mimicking experimental reality. And the other method would never let you do such an experiment because you would need a tower 120 feet high. I have done dozens of these type of experiments in an 8 foot room.

[bobriddle]

Clay,
The 4 weighted wheel I mentioned is something that might work perpetually.
This is because it's source of over balance is known as well as how that momentum would be conserved.
With something like the idea I am mentioning, what people would over look is that after the wheel rotates 90 degrees, it would need to be stopped and held in place. This is because it's balance would be in opposition to the over balance that caused the wheel to rotate.
What would happen when the wheel is stopped is the weights that are perpendicular to gravity could shift restoring over balance in the same direction the wheel is designed to rotate.
There is a simple device that would be used with the wheel that does work without requiring an expenditure of energy. This is where "free energy" would come into play.
While this type of wheel is a basic design, it is something that would help people to understand how torque, over balance, acceleration and also a pause of movement all can be used to generate potential energy in a gravity type wheel.

edited to run spell check

[cloud camper]

Bob - I believe this concept is the real deal!

We need to analyze this concept closely. It is basically the same idea I am working with although I believe I have found a mechanism that accomplishes what you're proposing "on the fly".

You are demonstrating the same necessity for commutation and latching that I have arrived at.

What you are actually doing with this concept is essentially "splitting the system" so that we can thereby maintain separate rotational PE values for the descent and lift sides of the system throughout the 360 degree cycle.

In your concept, I think you are contemplating a four cycle system which would mean stopping the wheel every 90 degrees to horizontally shift the weights. This is fine.

So every 90 degrees we stop the wheel, horizontally reposition the necessary weights to maintain the overbalance and restart. This process essentially "reassigns" the rotational PE value back to what we started with. This occurs without losing any overall vertical PE.

Every 90 degrees we terminate the old closed system as the rotational PE has diminished, reassign the rotational PE back to the start position and then initiate a new closed system.

If we speed this process up so everything happens "on the fly", we see we are getting an abrupt instantaneous increase in rotational PE that cannot occur in a continuous smooth curve closed system.

Thus we bypass Emmy's law since we are making the changes at the boundaries of separate closed systems.

With only a single inertial system that is typically tried, we can only engineer a single rotational PE value that can be assigned to the system at any point in the cycle.

This is because no matter how the weights flop around, the overall system has a vertical CoM that is continually descending. This is the only way the wheel can turn with a single closed inertial system.

But when we split or bifurcate the system, we can then maintain separate separate rotational PE values between lift and descent sides.

We can then maintain a higher overall rotational PE value to the descent side than on the lift side.

This is what we are doing each time we stop the wheel in your concept and reconfigure the rotational PE value to our advantage. We are creating a functional binary system in which neither system knows what the other is doing and thus cannot "average" the rotational PE value.

If our system is allowed to "average" the rotational PE values then our system can only keel to the lowest vertical PE position and stop.

We need a continual "split" in the rotational PE values to rotate the wheel. In this way we are forcing the system to operate strictly on this differential in rotational PE and disallowing the system to use up vertical PE.

Nice work Bob! I think you've nailed it!

[bobriddle]

Thanks cloudcamper.
One of the reasons why it needs to be stopped every 90 degrees as you rightly understood is to give the weights time to shift. And if the horizontal plane is off by a few degrees, then gravity could cause the weights to roll on their own.
This would mean that as one weight rolls away from center (creating the over balance), another weight rolls towards center, all the way to the hub.

[Furcurequs]

Thanks for the detailed reply Jim. Lets assume for theory that there is no friction, the rod, rope and pulley are weightless and the following dimensions:
Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

Hello again,

Now that my headache has eased up, I'm a bit more clear headed. I would probably approach the problem like this:

We can calculate the speed of each of the masses when the rotation rate has reached 50 RPM and from that determine the kinetic energy of each.

Then, we could simply divide the total kinetic energy of all the masses by the weight of the 500 gram mass to get the vertical distance through which it has to have moved.

I would then work backwards to find the acceleration rate and time.

So, at 50 RPM the 1 kg masses would each have a speed of 2 * pi * 1 meter * 50 rotations / minute * 1 minute / 60 seconds or about 5.23 meters / second.

The 500 gram mass would have a speed of 2 * pi * 0.1 meter * 50 rotations /minute * 1 minute / 60 seconds or about 0.523 meters /second. ...which, of course, is just one tenth of the speed of the outer masses.

The total kinetic energy is then 1 / 2 * 1 kg * (5.2 meters / second)^2 * 2 + 1 / 2 * 0.5 kg * (0.523 meters /second)^2 or about 27.5 Joules.

(1 Joule = kg * meter^2 / second^2 = 1 Newton * meter) ...btw

The distance the 500 gram mass needs to have moved, then, is 27.5 Newton * meter / (.5 kg * 9.8 Newton / 1 kg) or about 5.6 meters.

...and since we should have had a constant acceleration...

From the equation v=at (velocity equals acceleration times time) and the equation s=1/2at^2 (distance equals one half acceleration times time squared) we can get the equation t=2s/v (time equals two times distance divided by velocity).

So, plugging in the distance and velocity of our 500 gram mass we can determine how long it took to come up to speed...

t = 2 * 5.6 meters / 0.523 or about 21.4 seconds

The acceleration rate of the 500 gram mass, then, can be found using either of the above two equations to be about 0.0245 meters / second^2

It appears that about 99.6% of the energy has gone into the spinning masses.

Anyway, these numbers don't seem to agree with those before. So, please feel free to correct me if I've made any errors.

Thanks.

Dwayne

[cloud camper]

Thanks cloudcamper.
One of the reasons why it needs to be stopped every 90 degrees as you rightly understood is to give the weights time to shift. And if the horizontal plane is off by a few degrees, then gravity could cause the weights to roll on their own.
This would mean that as one weight rolls away from center (creating the over balance), another weight rolls towards center, all the way to the hub.

OK Bob - time to start building!

The only real difference between your concept and mine is that in your concept the wheel must stop.

There are mechanisms that would only cause the weights to stop while the wheel keeps turning. This would then be consistent with MT138! The additional advantage secured here is that all CF is eliminated at the point the weights change direction.

[clay973]

Ok, thanks jim_mich, pequade and Furcurequs. I'll have to give the 3 different answers some thought.

[Furcurequs]

Hey Clay973,

I, of course, assumed conservation of energy when doing the calculations for this problem. So, the kinetic energy gain comes from the conversion of the gravitational potential energy of the drive mass during its "fall" - as per a conventional scientific understanding.

I've since checked my math by plugging the numbers into Wubbly's formula found here:

http://www.besslerwheel.com/forum/download.php?id=11164

...and I get the same rates of acceleration using both methods.

Also, if you were to investigate, you might find that pequaide is known for using unconventional math formulas and making perhaps unsubstantiated claims. ...though to be fair, he does seem to have his share of followers/believers.

I'm not sure why the discrepancy with jim_mich's calculations. I've not looked at them closely. If I've ever done similar calculations using radius of gyration, it's been a very long time ago, too, I guess.

Maybe others can chime in and point out any errors.

Thanks.

Dwayne

[bobriddle]

OK Bob - time to start building!

The only real difference between your concept and mine is that in your concept the wheel must stop.

There are mechanisms that would only cause the weights to stop while the wheel keeps turning. This would then be consistent with MT138! The additional advantage secured here is that all CF is eliminated at the point the weights change direction.

It'll probably be end of the month. Might do some work on the mechanics between now and then. You know, figure out how to go
about making it.

I think one thing everyone is missing on torque and rpm's is that gravity accelerates at a max of 9.8 m/s and if the percentage of over balance is
a percentage of acceleration, then dividing the circumfrence of a wheel by m/s could give a rough idea on rpm.
A basic example is with a net force of 10% of total mass, then acceleration may be .98 m/s (allowing for f = ma). And if 2R*Pi = is something like
2 meters * 3.142, then 6.28 divided by .98 equals a little over 6 seconds per single rotation or a little less than 10 rpm.
But without a workable design, it's just math.
And math can pretty much predict how much potential a design might have.

@cloudcamper, the fishing might work, but would take understanding the math before building. kind of goes to Mt 26. But there is something missing in that drawing.

edited to add; have started building. with wood, every time I glue something together, it takes 2 to 3 hours of drying time.
Will be using 1 lb. weights with about a 10 inch radius (swing) with a conservation movement (retraction of the weight on it's up swing) of about
2 1/4 inches.
Might take a week or 2 to get'er done.

[pequaide]

I myself would not call F = ma unconventional math. And my experiments are pictured and can be easily repeated.

[bobriddle]

pequaide,
what is unconvential is the application. A falling weight is f = ma. Within tne concept of 4 weights, 3 of the weights are calculated as drag.
By considering their potential as torque, the further their CoG is from center, the more energy tbey will consume.
This is one reason why weights not performing work be as close to center as possible.
With 2 weights next to the hub, then what is left is one weight performing work and one weight being acted upon.

Bob

[cloud camper]

It'll probably be end of the month. Might do some work on the mechanics between now and then. You know, figure out how to go
about making it.

edited to add; have started building. with wood, every time I glue something together, it takes 2 to 3 hours of drying time.
Will be using 1 lb. weights with about a 10 inch radius (swing) with a conservation movement (retraction of the weight on it's up swing) of about
2 1/4 inches.

Might take a week or 2 to get'er done.

I can't wait Bob!

You have shown you understand that a gain in energy cannot be obtained once a closed system has initiated. At this point, the system can only smoothly convert rotational PE into KE or the reverse. This keeps Emmy happy.

But by halting the system every 90 degrees, we can then convert the available vertical PE back to maximum rotational PE by commutating the weights horizontally back out to the rim. This amounts to an instantaneous and abrupt increase in rotational PE four times per wheel rev.

I believe this is exactly what JB did except he found a mechanism that only stopped the weights while the wheel kept turning, trapping and preserving inertia.

We are effectively creating here four separate closed systems that are each sequentially switched closed and then open four times per wheel rev, sixteen times in total, with the horizontal commutation cycles occurring at the boundaries.

Of course the kicker is that we must not lose any vertical PE in the process. I believe this is where the "pairs of pairs" concept must be integrated. We want to force the system to operate strictly off the differential in rotational PE and prevent the system from using any vertical PE.

Your project really deserves it's own thread with it's own catchy name, maybe something like the Discontinuous Rotational PE Gyrator!

Well, that's a start!