the obvious gain

[bluesgtr44]

Here's two snaps of the start and finish...as the beam is lifted, the weight starts to drop and the red part is what shows the difference in potential gained from stretching the spring. So, I see where you are at this point....now, how can we apply this in a closed loop system. These are interdependant on one another....all connected together and the removal of any one will void this supposed gain? Basically, how can you release this potential and then, reconnect and start it over again?


Steve

[arthur]

look at "arthurs design" and you will see the closed loop system.

there is a fixed point 45 degrees from the axle that suspends the moving weight.

It should be easy to understand.

[arthur]

Jim,
by the way,

your graph does NOT show the weight dropping 0.157080 meters!!!

the weight does not drop until about 80 degrees,
and the drop in the blue line from 80-90 degrees is much less than 0.157 meters!!!!

the actual weight drop is closer to 0.04 meters.

As you can see from the graph you must lift the weight higher so that it can fall and stretch the spring.

again this is not true.
from 0-80 degrees the weight stretches the spring, yet it does NOT fall.

[bluesgtr44]

look at "arthurs design" and you will see the closed loop system.

there is a fixed point 45 degrees from the axle that suspends the moving weight.

It should be easy to understand.

I see a pendulum....the fixed point holds the rope that is connected to the weight and the weight is attached to a beam and it can slide up and down. There is a spring connected from a point on the beam and then to the weight......no fixed point suspends the weight.....the beam and the spring suspend the weight. The rope connected to the fixed point will actuate the weight in and out depending on the rotational position of the beam, this will extend and retract the spring....is this the whole design (system)? or is this to be a cross beam and the pendulum is just a simple example?

I tried this the way you have shown and at the 10:30 position it can go basically either way, so I tilted a bit more up and then down to activate it in one or the other direction....it takes pretty good because I have the spring loaded to a favorable starting position....it will not, however maintain a bit. So I made the cross beam and it actually launched a bit when started in a loaded position....it will not, however overcome the amount of energy required to reload the extension of the spring....not matter how I move the weight up or down and manipulate the length of the rope.....I got about 3/4 of a revolution out of it and that's it! So, what the heck am I missing, Art?


Steve

[jim_mich]

arthur, please take another look at the graph! Do you see the saw toothed steps? Each tooth represents the weight falling. The weight must be rotated upward before it can fall and stretch the spring. The blue line that looks like a simple curve is actually 90 individual steps. The only way that the spring can be stretched by gravity is for it to be lifted upward first. There is no force to stretch the spring before the weight is lifted. Only when you lifting the weight by rotating the arm does the angle change allowing the vectored weight force to increase so that the weight can then slide down the tube and stretch the spring. This happens in little tiny increments all along the curved path. My computer program allows me to adjust the incremental steps. I set the steps to 10 degrees for the black line and I set it to 1 degree for the blue line.

The weight is being lifted during all of the 90 degrees of rotation. You stated that this lifting required the same input energy (1 kilo x 1 meter) as would be required without the spring. Your statement was not true. You continue to make false statements. Please go back to class and learn how things actually work before trying to tell me that I am wrong. I don't take kindly to being told I'm wrong. Yes, I do make mistakes at time; we all do. But I seldom make mistakes.

You cannot just talk about lifting from 70º to 90º as that is not how it happens.

If you give me specs on spring strength then I can give you actual real numbers. For instance the graph that I posted uses a spring that starts at zero tension and increases linearly to 1 kilo at a stretched length of 0.2 meters. I prefer to work in inches and pounds but I have no problem converting between different systems of units.

When you add the rope as in "arthur’s design" then things get more complicated. Why don't you learn to use WM2D? It should be able to handle your design. Of course it's my belief that WM2D will never show a working wheel even if you gave it one. That is because it assumes that input must match output. If by some design or accident you were to give it a working wheel, it would force the output into equaling the input.

But for simple designs it works great as a learning tool. If you don't believe WM2D then you need to build a real live working prototype or you need to learn how to calculate the forces involved.

I've wasted too much time debating your idea, so this will probably be my last posting on this thread. The rain has stopped and the sun is now shining and I have yard and garden work to do.

[arthur]

Jim I'm sorry but

given that there is negligent friction for the weight to slide along the beam,

there are no sawtooth steps

the weight moves frictionlessly along the the curved blue line.

from 0-80 degrees the weight stretches the spring,
but NEVER falls.

the energy stored in the spring is free energy.

you fail to acknowledge this.
whatever.

[arthur]

steve,

to start the 'wheel'
edit: (I'm talking about "arthurs design")

lift the arm to the 12:00 position
and give it a good push counter clockwise.

maybe you can't do that with your program.

[bluesgtr44]

Apply it....I'm not saying you haven't shown where a movement can store potential energy....but, you are just taking a "snapshot" is what I see. IMHO, you are lacking the vision to see this thing through the whole cycle and then some. Now, if you have a complete mapped out plan as far as how to apply this...I'm willing to listen...


Steve

[ovyyus]

so, I really could care less if people believe it or not.

Quite right, belief has nothing to do with how things actually work. The "free" energy stored in the spring was put there by the effort of your initial lift. Humility often precedes an easy lesson.

[arthur]

I'm not saying you haven't shown where a movement can store potential energy

I can pick a rock up off the ground and store potential energy.
big whoop

this arm movement can store more potential energy than the cost of the original movement. that's over unity.

in "arthurs design",

the 'free' spring energy is used to propel the arm from 10:30 to 4:30.
with less cost on the other side- (4:30 to 10:30)

take a closer look.

[AB Hammer]

@arthur

Springs have unique reactions in wheels. That doesn't make them over unity. They are simply energy in and energy out. I myself have designed a spring wheel. Ralph can confirm this, for as my mentor he has a copy of the design, but he will not show it, but he can confirm that he has a copy of it. I will build my spring wheel someday but it is down on the list for now. It was one of my earlier designs.

[bluesgtr44]

this arm movement can store more potential energy than the cost of the original movement. that's over unity.

Prove it! You look at one side.....show what happens when it's time to pay back the piper! You show a storage of potential is all......and then when it comes around....you stop at 4:30. Why? Show us the rest........

the 'free' spring energy is used to propel the arm from 10:30 to 4:30.
with less cost on the other side- (4:30 to 10:30)

If what you have has any merit, there is going to be some tight tolerances going on.....where are they? If that "free" energy you talk about is going to be able to apply itself.....where? when? How is that fixed point going to be able to adapt to an increase in acceleration? And just how fast do you think this could go?

You say to take a closer look....man, you don't have anything....no specifics whatsoever.....none....the length of that rope would be critical, would it not? The position of the weight on the beam in contrast to this would have to be pretty on the money, right? So, show us something.....I think you lose this "free" energy to the piper.....

Look, I'm not all that and a bag of chips.....but, I think I am at east a bag of chips....nothin' fancy....little salt, that's all....Show me, Art....just show me the full revolution with a little bit of acceleration....


Steve

[jim_mich]

arthur,

The basis of your idea is that the energy in the spring gets put into the spring freely without any extra effort. This just ain't so! Attached is a screen shot of the previous program upgraded showing calculation of average torque required to lift the weight 90 degrees while stretching the spring (shown as black) and without stretching the spring (shown as blue). It shows 0.637 Kilo/Meters average torque to lift the 1 kilo weight 90º to a height and radius of 1 meter. My program is calculating this correctly because I know that the average torque needed to lift 1 unit of weight 90 degrees 1 unit radius high should always equal 2 / Pi, which it is.

The average torque needed to lift the same weight to a same height while stretching the spring is a little higher at 0.735 Kilo/Meter.

Someday you will look back on this event and realize how much you still needed to learn about gravity wheels.

Instead of telling us to take a closer look I suggest that you take a closer look.

[arthur]

Jim
there are clearly some flaws in your calculations.

first of all I will remind you that the weight travels up a frictionless, smooth curve.

there are no sawtooth steps.

you have somehow calculated that the weight falls 0.15708 meters.
this is clearly false.
your picture shows the path of the weight dropping from 80-90 degrees,
but this drop is obviously much less than your "calculated drop" of 0.157 meters.

sure, the weight travels along the arm 0.2 meters,
the spring extends 0.2 meters,
but the drop according to your picture is less than 0.05 meters.

as to your latest torque calculation--

I will repeat what I said earlier and calculate the force required to lift the weight to 70 degrees, (1 meter) for simplicity.

+++++

the force needed to lift the arm (the torque requirement)
is the same as
the force of the arm falling.

if the arm fell down from it's 70 degree position,
the spring would not add force to the fall...
.. and the arm would land with the force (1 kilo x 1 meter)

.....the same force needed to lift the arm to 70 degrees.

+++++

the spring energy is not even part of this calculation,
because it is free !!

[AB Hammer]

What I think arthur is trying to say that he gets gain from the fall and the CF making the spring compress which becomes his energy gain. But I don't see where this will help him in a wheel due to it has to extend that energy back, not unlike the super balls that hit the ground and bounce. But in a wheel the compression of the spring will not start to release until it would be back above the 8:30 mark in a clockwise spin. This will kill any advantage of the springs energy from the compression. Thus the energy gained is lost and no over unity.