Smith66 should be banned.

[Fletcher]

JIM_Mich, cloud camper, daan,

Jim wrote:

When the weight moves outward its velocity increases. But the rotational speed of the weight and the wheel are tied together. The rotational speed of both decrease.

Your missing one major item that I explained about a pivotal weight dropping from 12:00. The weight is hanging straight down from the pivot and actually lags behind the radiating outward moving pivot, Cf does not apply here. It is inertia attempting to keep the bob in a straight linear vertical drop.

Once the rod hits the stop then and only then does Cp come into play forcing the bob inward (which inertia is already holding inward) without Cf ever showing up until contact is made.

It is not that hard to pivot a pendulum at the rim with a rod allowing the bob to hang slightly below the axle. let it go and watch the weight fall vertically, accelerating as the pivot moves outward as it falls with the likewise accelerating wheel. Once it hits the stop, more acceleration is noted as Cp has met with no Cf to overcome. actually Cp has a "heads up" as inertia is still attempting to hold to a straight line fall.

Under these conditions, the wheel does not decelerate as the weight moves inward.

Ralph

Is this what you mean Ralph ?

[rlortie]

Fletcher,

Yes you have a very good head start!

Now let the wheel rotate until the pendulum is just about to hit the rim (which if you will note is hanging inward as it falls)... At this point attach a protruding fulcrum to the rim that the lever hits above the bob. Do not worry about the elasticity of the bob, it is just a weight.

Your bob or weight is now cantilevered. Upon contact between the rod and fulcrum which still has a tendency to hang inward,(No Cf here) you should see an increase in acceleration with the pivot point ending up in the 130-135 degree area where the machine will come to rest.

There are other modifications that can be made to augment this, bringing the pivot point travel to approximately 230 degrees of rotation. We can address those if your happy with the results of the above.

My point being here in debate is that Cf does not always play a key roll and is dependent upon orientation of applied Cp.

Ralph

[Fletcher]

I think you had better put up a quick picture of the fulcrum etc Ralph.

[rlortie]

Fletcher,

Having second thoughts after farther viewing, believe you should lengthen the rod, lowering the bob about 1/2 the bob size. You may have to play with this until finding the sweet spot. Rather than lowering the fulcrum, make it longer extending the leverage on the weight. It took me quite some time to find it the hands on way!

Ralph

[rlortie]

Fletcher,

Attempted to pull up a picture only to have windows explorer freeze up, cannot open my picture album.

Let the pendulum fall as the wheel rotates until the bob is just about to hit the rim, at this point install a stop (fulcrum) located about half-way between top of bob and axle, about 1/3 of the lever length should overhang the weight.

Hope this helps!

Ralph

[jim_mich]

Ralph, is this the general idea? Except the weight is probably way too small.



Edit, so you're the one that almost froze Scott's server a few minutes ago. :))

[rlortie]

Definitely have a windows XP explorer problem. every time I attempt to open my pictures it freezes up telling me that windows explorer must close.

I have to end task via Ctrl-Alt-Delete to get out of it.

Ralph

[rlortie]

Jim,

Yes, you have the concept right, however your weight is a wee bit small and your lever is to short, you do not have the stop or fulcrum allowing about 1/3 of the lever and weight to overhang.

In phase four of your drawing the idea is to have the pendulum fall inward cantilevering over the axle and in turn prying up on the ascending rim.

Ralph

[jim_mich]

Ralph, is this the one you're looking for:

[rlortie]

No!

That looks closer to ABHammer's work, cannot verify but it most certainly is not mine.

I will check my forum pictures here on line to see if I can come up with the right one.

Ralph

[jim_mich]

So does this look more like it?

Then the weight falls over the axle causing more rotation.

[rlortie]

Possibly a wee bit overkill on the weight, a little to long on the rod and no stop or fulcrum keeping the weight inside the rim.

I think that I should remind all those of interest, I am not attempting to show you how to build a wheel. I am only trying to prove that angular momentum inward without CF will accelerate a wheel. Fletcher has already proved that Cf does not effect the falling weight with his WM2D sim.

Ralph

[rlortie]

I have attempted to show the fulcrum using a screen shot of Fletcher's sim.

On top of my pictures freezing up, now "Norton" anti-virus program is telling me to dump my WM2D app. as it is being monitored!

I cannot post a screenshot from "Paint" with explanation and depiction of fulcrum. In fact after making it, now I cannot find it.

[jim_mich]

In fact after making it, now I cannot find it.

Open paint and then ask it to open a file. Paint should list your most recent picture files. This should tell you where the file is located. I keep a Bessler folder right on my desktop for all Bessler forum related files.

[jim_mich]

and no stop or fulcrum keeping the weight inside the rim.

There is a small stop pin in the drawing. I made it brown like the axle.

My quick analysis...

The actual location of the stop pin does not matter. It will work the same with the stop pin hitting the weight or with the stop pin hitting the rod up near the fulcrum. The important thing would be the angle of the rod when it is stopped.

The weight hanging out beyond the wheel rim makes no difference. Simply imagine the same mechanism on a larger diameter wheel.

CF does nothing while the weight is dropping from 12 o'clock until the rod hits the pin. Depending upon the wheel mass and rod angle produced by the stop pin, the weight might be swung to the left of straight down or it might be swung to the right as it hits the pin. It will be swung left if the wheel is light and fast with a heavy slow weight. It will be swung to the right if the wheel is heavy and slow with a light and fast weight, because the weight will begin moving to the right, then as the curved path of the pivot moves more downward rather than sideways, the momentum of the weight will swing the weight to the right of its pivot.

These are general observations and all depends upon the specific conditions.

Once the weight or rod is against the stop pin, then gravity and CF will hold it against the pin, depending upon rotation speed. If the wheel is given an initial push start (or is running at some presumed operating speed due to the mechanism actually working) then the CF is keep the weight outward against the pin beyond the position where the weight would normally fall inward. Thus, in this case CF would increase latency and hinder operation, which is exactly the reverse of what is wanted.

If/when the weight does start to swing inward/downward and assuming the wheel is still rotating, then momentum of the weight as it moves inward to a smaller radius will accelerate the wheel as the weight decelerates to match the slower wheel velocity at the more inward radius.

In this wheel design, CF only comes into play by preventing the weight from moving inward as soon as might be desired.

Just my observations. Maybe I don't understand something.