SimonsWheel

[S1m0n]

Hi guys. I've been messing round with a PMM concept and have been advised to post here.
If you fancy taking a look and passing on your thoughts they would be appreciated.
www.SimonsWheel.co.uk

[Tarsier79]

Welcome to the forum.

re the video on your website :

1. The more distance, "maximizing gain" the more energy it takes to move the weights into position.

2. The more weights you add, the more you have to lift. It takes the same energy to lift a weight as you gain from that weight from rotation of the wheel. If your wheel was going to work with 128 weights, it would work with 2.

3. Reduce friction. That is a given.

Also, shifting weights at the periphery of the wheel gives no more torque than shifting them the same distance at the center of the wheel.

For a similar attempt, google "Sjack Abeling". Preston Stroud from this forum did a superb build and video on his design.

[S1m0n]

Hey thanks for looking and posting a reply.

With due respect though I feel I may not have explained myself sufficiently either on the site or in video.

Preston Strouds build and video, although fantastic would never work as the torque on the RHS is balanced by the weight on the LHS.

He has 16 weights, but it takes the distance of 4 weights or 1/4 of the wheels diameter for a weight to be extended and the same for a weight to be retracted.

I feel this is one of the key design failures in all PMM attempts and falls outside the design parameters that I've set.

[AB Hammer]

Greetings S1m0n

Welcome to the forum.

Most of us have done or explored similar approaches to what you have and found problems to overcome. The near exact one was done by an X member to this forum by the name of Darrel. Here is his wheel.

[Silvertiger]

Hey thanks for looking and posting a reply.

With due respect though I feel I may not have explained myself sufficiently either on the site or in video.

Preston Strouds build and video, although fantastic would never work as the torque on the RHS is balanced by the weight on the LHS.

He has 16 weights, but it takes the distance of 4 weights or 1/4 of the wheels diameter for a weight to be extended and the same for a weight to be retracted.

I feel this is one of the key design failures in all PMM attempts and falls outside the design parameters that I've set.

Simon, if you haven't already, look up Aldo Costa's wheel. :) You both have pretty much the same thing.

[S1m0n]

Ooh ABH, what a valiant attempt Darrel has made.

The weights are raised at o and 180 degrees, so produce work close to their entire cycle, but with only a few weights to create torque it would never work.

Aldo on the other hand (who I'd never heard of before today) has hit the nail on the head.

Brilliant.

I'll link his Youtube vid on my site.

[AB Hammer]

The problem with Aldo is it is the size of a large fairest wheel and can not be proved for the wind effects the wheel. In the days before I was born (reference my Father who as a Navy medic who worked with the first astronauts and NASA. Then altitude chambers for the US Air force later in his career.) NASA stated that the wheel would have to be the size of a fairest wheel to light a light bulb, and unpractical. Aldo has also proven this by trying this method. There are much better ways. ;-)

P.S. I was named after Alan Shepard for Dad liked the name. He also worked with ejection seats and high altitude suits as well as the space suits the MK2 MK3 and MK4


Alan

edited to add: I looked at your sight and see one my old wheels at the top. Good looking sight.

[primemignonite]

S1m0n,

The more traditional greeting as usually done by the Old Great Ones (they are currently preoccupied with
another issue having to do with me little self elsewhere) is as follows:

Welcome to the Nut House.

or a variation I once saw

Welcome to Bedlam

and yet another classic

More FRESH MEAT for the grinder!

I'm sure you get the general idea with those few, no?

As your career here progresses, you will learn the actual, secret rules and, hopefully, accede to them for
if not . . . . . . . . . . . . . . . <<< Here fill-in-the-blank as your more perfervid imagination may direct.

CHEERS!

James

[S1m0n]

If anyone’s more familiar with Costas wheel than me and spots an error in my interpretation, please get in touch.

His giant wheel was 17068mm in diameter
Weighed over 9000killo
He was lifting 2.4killo weights
Height of weights raised was 34mm

34mm is 0.2% of the length of the diameter
2.4killo is 0.25% of the weight of the wheel

My forth rule states.
’Consider what your net gains are trying to achieve'

Maybe it's all about the net gain....

If there were two 2.4kg weights on a seesaw both 8500mm from the fulcrum.
One weight was moved 34mm (Ooooh, and the seesaw weighed sooo much)
What would happen?

Could the height that the weights be easily increased using ramps? and would it make a difference?
Is a 2.4kg weight enough? After all it's only a 1/4 of 1 percent of the overall weight.

Mmmmm. Off the walk the dog.

[Silvertiger]

No ramps. Break up the wheel into two halves. One side has a set of weights leveraging more torque than weights on the other side. I'll use your number: 64 sets. So lets say you have 31 sets extended on one side and 31 retracted on the other. The remaining 2 sets are being compresses and released at the same time; one set at the nadir, and the other at the zenith. Each set requires a lever to retract it at the nadir and lock it in place, compressed against a spring, and once again at the zenith to unlock the mechanism so the spring can lift the weight again. The question is, what is the minimum number of extended sets required to lift one weight against a compression spring, plus the resistance of the lifting lever against a stationary roller, plus the torque of the retracted sets on the opposing side? If the numbers are not equal, then it will not work.

[AB Hammer]

To add to Silvertiger's

When you have your ramps starting at 6:00 and 12:00 you create a lot of resistance to the actions for the reaction needed. If this would work? Most of us here would have had the wheel for a long time. We have become experts on what doesn't work and always looking for new untried methods and designs.

Alan

[Silvertiger]

https://youtu.be/MHYODn98yPI

A visual representation for Simon: roughly 55 to 60 Lbs of a linear force vector are required to lift this 10 Lb weight roughly 6 inches and compress the spring. I simulated the contact between the lifting lever and a stationary roller. (In the sim, it's actually the roller that's moving, but there is virtually no difference, so I changed the POV to the roller.)

Addendum: Since the mass of the apparatus as a whole, when arrayed in several radially symmetrical sets about a fixed axis, balances out on both sides of the wheel, that mass can be thrown out of your calculations, with the exception of the weight to be lifted. Only use the entire mass when calculating the flywheel storage potential. That being said, I'll try some quick math here. Let's say that the radius from the axis of rotation to one apparatus is 6 ft. The rough torque required at the contact between the nadir-positioned static roller and the lifting lever is 180 ft-Lbs. The torque of one weight on the descending side, extended 6 inches (yielding a moment arm of 6.5 ft), at it's outermost position along the X-axis, is 65 ft-Lbs. On the opposing side, the same weight at its retracted position at 6 ft is 60 Ft-Lbs, thus yielding a net bias of 5 ft-Lbs. But you don't really know how many weights you need on the descending side yet, on account of the diminishment of the length of the apparent moment arm relative to any of the weights' positions on the wheel. You would need to draw out a polar array of how many sets you think you should start with; then draw out the free-body diagrams, specifically your torque triangles, to calculate each diminishment of the moment arm length. Then add all torques together. Wash, rinse and repeat until you have enough torque to lift 10 Lbs against the additional load of a compression spring. The one I used I set to 24 inches (relaxed) with a K value of 10 Lbs/in. Have fun.

[Silvertiger]

Update here: https://youtu.be/LIfKcl0X5p4

The Nadir Lever does all the heavy lifting, and it only needs to lift one at a time. The Zenith Lever requires far less work, since it is merely releasing the compression spring to reset the weight. The red blocks are 10 Lb weights, The red ball is a stationary roller. The blue rectangles are the levers, and the green mechanisms are the locking clamp calipers.