Another claim to a working device...

[Grimer]

Sorry about all the confusion Mark. I was so keen to get the whole thing out into the public arena that I missed the fact that the y axis had become stretched which is rather fatal when one is talking about circular paths. :-)

[Mark]

No problem, Frank. The confusion lasted only a moment.

Which is unusual. Usually my confusions last much longer.

[Grimer]

I've prepared the above graph showing where some of the gravitational potential gets stored as strain energy when the pendulum swings from 8 to 6.

More on this topic later.

[rlortie]

Grimer,

Your above drawing brings to mind the torsion bar suspension used by the now defunct Packard automobile of the 1950's.

Your compression- tension arrows are in error, they should be at a right angle to the applied twisting force which is equal the diameter of the rod or bar.

In fact there is no tension, only compression as you twist the rod not unlike a piece of licorice twist. You arrows depict modular elasticity which would indicate a right angle load on a beam.

Ralph

[Grimer]

You obviously don't understand moment distribution, Ralph.

Mind you, my first boss, R.F.Galbraith, President of the Institution of Structural Engineers, 1953, couldn't do moment distribution either so you are in good company. He designed all his bridges with pin joints so they were structurally determinate.

The moments are both in the plane of the paper (screen).

The tensions and compressions are shown correctly.

Get a flexible ruler. Hold it at arms length with your hands at each end. Twist your left hand clockwise and your right hand anti-clockwise and you will see the ruler curve down in the middle. There is compression at the top of the ruler and tension at the bottom.

And in your piece of liquorice there is tension along the axis.

[Grimer]

I've prepared the above diagram showing where some of the gravitational potential gets stored as strain energy when the pendulum swings from 8 to 6.

More on this topic later.

The generation of a bending couple under the action of gravity and the consequent input of rotational strain energy (conservation of angular momentum energy) gives us an understanding of why the circular pendulum has a slower period than the brachistochrone pendulum.

Because some of the gravitational energy of the falling arm is being bled off into strain energy the arm is effectively experiencing a reduced downwards acceleration.

The concept of a falling weight experiencing a reduced acceleration is neatly illustrated by a pulley example.

Consider a overhead frictionless pulley with a 32 pound weight hanging down on one side and a one pound weight hanging down on the other.

The acceleration of the 32 pound weight will be reduced from the free fall acceleration of 32 ft/sec to 31 ft per second because some of the gravitational energy is bled off into raising the 1 lb weight.

This example is easy to understand because both movements are visible and involve vertical change in gravitational potential.

The bending of a pendulum shaft on the other hand is below the threshold of perception and involves transducing one form of energy (gravitational potential) into another (strain energy).

The beauty of strain energy is that its mass is insignificant and therefore one can move it around in a gravitational field without gravity being involved.

In the Proof of Principle example the strain energy was input between 8 and 6, then output between 6 and 4. The shaft absorbed gravitational energy on the downward swing and released it on the upward swing.

[daxwc]

Grimmer:

In the Proof of Principle example the strain energy was input between 8 and 6, then output between 6 and 4. The shaft absorbed gravitational energy on the downward swing and released it on the upward swing.

What difference is this from a normal spring under gravity?

[rlortie]

You obviously don't understand moment distribution, Ralph.

Mind you, my first boss, R.F.Galbraith, President of the Institution of Structural Engineers, 1953, couldn't do moment distribution either so you are in good company. He designed all his bridges with pin joints so they were structurally determinate.

Oh! As a certified structural engineer of both IBCO and UBC, I do believe I understand moment distribution. I also am familiar with the effects of pinned joints, here is some links explaining better and quicker than I can.

https://engineering.purdue.edu/~ce371/D ... xample.pdf

http://en.wikipedia.org/wiki/Moment_distribution_method

I once gave a member who did not understand moment and modular elasticity the suggestion to research simple types of roof trusses. Both ends grounded with the use of moment joints in all interior walls. Once "hurricane clips are added to the outer wall support they become determinate. All moment happens within the chord.

http://en.wikipedia.org/wiki/Truss#Truss_types

The tensions and compressions are shown correctly.

Not for a round rod with equal opposing torque applied at the ends, or for that matter one end grounded.

Get a flexible ruler. Hold it at arms length with your hands at each end. Twist your left hand clockwise and your right hand anti-clockwise and you will see the ruler curve down in the middle. There is compression at the top of the ruler and tension at the bottom.

A flexible ruler is not a rod it is an elongated beam. Twisting it gives the same results as holding it flat and bending it. And yes you have compression on the top and tension on the bottom.

Comparing a ruler to twisting a rod of symmetrical roundness is like comparing apples to rotten oranges. It does not work!

And in your piece of liquorice there is tension along the axis.

No! there is compression along the axis, it wishes to twist inward not outward.

I agree with daxwc , What you are attempting to do is no different than a spring under gravity.

Ralph

[Grimer]

I think we will have to both go our own way on this Ralph. Anyway, thanks for your comments.

[Grimer]

Grimer:

In the Proof of Principle example the strain energy was input between 8 and 6, then output between 6 and 4. The shaft absorbed gravitational energy on the downward swing and released it on the upward swing.

What difference is this from a normal spring under gravity?

For the beam to stay straight on the left hand side (8 t0 6) and freely give up it stiffness on the right hand side (6 to 4) it has to have asymmetric properties.

A reinforced concrete beam is a good example. The concrete is in compression and the steel is in tension. The concrete has negligible tensile strength so the beam is stiff the right way up when supported at both ends but falls to pieces if you turn it upside down.

In the case of a cantilever rigidly supported at the end the opposite is the case.

A normal leaf spring is not asymmetric in its properties.

Now obviously I'm not suggesting that you carry out an experiment with a reinforced concrete beam as a pendulum shaft. That was just a familiar example to illustrate the principles involved.

If you wanted to experiment one would need to make a shaft with the same asymmetric properties - and one would need the facilities of a good university engineering or physics laboratory to get a significant result.

However the trig proof is all one really needs to understand the interaction of the variables involved and steadily move on to generation of greater energies.

[rlortie]

I think we will have to both go our own way on this Ralph. Anyway, thanks for your comments.

Well that is one way of avoiding a debate! But if I am right and you are wrong don't you think readers would like to know?

As a forum member, I feel obligated to rebut what I consider wrong information. If I am right and you are wrong or vice-verse, one of us is leading members down the wrong path. Rebuttal and discussions is what this forum is for, stating "we will have to go our own way" does not clear the issue.

Even with your concrete reinforced beam you are making it clear that rods are under tension and not compression. Twisting a rod is not the same as being used in a beam formula.

Once again I rely on Euler and Bernoulli: Euler–Bernoulli beam theory
http://en.wikipedia.org/wiki/Euler%E2%8 ... eam_theory

If you wish to discuss twisting a rod then you should refer to torque values. How much torque will a specified diameter rod withstand before snapping. Torque is adding tension to a threaded rod not compression.

Ralph

[Grimer]

I don't wish to discuss twisting of rods. It's a total red herring you have dragged in when you wrote,
"Your above drawing brings to mind the torsion bar suspension used by the now defunct Packard automobile of the 1950's."

I'm not accusing you of doing it intentionally, Ralph. It's an easy mistake to make.

Twisting rods involves moments in parallel planes. The moments I am talking about are in the same plane. If you applied my moments to a rod they would bend it, not twist it.

[justsomeone]

Worthy of your own thread Frank?

[rlortie]

Definitely worthy of its own thread, this could easily be another cloud camper verses Jim-Mich scenario, if I could last so long!

I am accused of dragging in a "Red Herring" when only referring to a hard physical fact.

Now he does not want to talk about twisting rods after he started the issue.

All I am saying is that the arrows in his twisting rod depiction are aimed the wrong way! But that's OK, my point has been made.

If I shut up, maybe we can get this thread back on topic. I am still interested in learning more about the unit allegedly being built in Michigan.

Ralph

[Mark]

Speaking of the RAR build...
Did I just plain miss it before, or have they back-dated an announcement regarding the in-progress modification they've made?
It's between pictures #40 and #41.