the obvious gain

[jim_mich]

arthur, There are no flaws in my calculations.

Yes, the weight travels up a frictionless, smooth curve. But to calculate such a curve requires one to break it down into small integrated steps and then average the steps together.

The fall is most definitely 0.15705 meters. This is not false. Most everyone here on the forum and in the scientific world knows that the only way for gravity to do work such as extend a spring is for a weight to fall. You think that because the weight follows a constant upward path that it is not also falling. Maybe we should use different words? The fact is that gravity pushes the weight downward as the arm is lifted upward. The weight falls relative to the end of the arm that is rising.

This might seem backwards because gravity it pushing down on the weight as the arm at the same time is raising the weight. The net result is the weight rises and the spring gets stretched. Both happen simultaneously. The arm tilts upward as gravity pushes the weight downward (falls) and the path is a constant rise, until about 70º.

If gravity did not push the weight down (fall) against the force of the spring then the torque applied to the arm would lift the weight higher than it does. Thus the question here is not so much semantics of words, but rather it is whether the force to lift the weight is greater when the weight follows the outer path that causes energy to be added to the the spring as it is extended. Clearly the force is greater and the spring does not get stretched without adding more energy than is needed to just lift the weight.

I don't see how I can make it any clearer.

[arthur]

Yes, the weight travels up a frictionless, smooth curve.

I am glad you can at least acknowledge this.

You think that because the weight follows a constant upward path that it is not also falling.

I think this speaks for itself.

-----------

Jim let me ask you this.

If the arm fell from 70 degrees (when the weight is at 1 meter)

would the spring add force to the falling arm?
would the weight land with any more force than simply dropping from 1 meter?

If the spring does not affect the force of the (arm/weight) falling from 1 meter
then it does not affect the force of the (arm/weight) lifting to 1 meter.

[Oystein]

Hello, take a look at this attachement as it describes what you are asking.

So the question we are left with from reading your posts is :
What (who or how) got the spring extended ? This input is where "your" energy "gain" is coming from...

Best
Oystein

[AB Hammer]

Oystein

Good diagram of what I also believed that arthur has been saying. But you will still loose what is gained on the descending side to the ascending side, which will have the opposite reactions in a wheel device. There is also no OU due to the spring is simply being stretched and compressed in and out The best I figure is that what arthur has would be best designed into a triggering device.

[Quartz]

Hi Oystein

There is no gain, just an exchange of energy between the falling weight and the spring, as the weight drops torque increase spring potential decreases. See how nicely conserved it is.

Ken

[jim_mich]

Jim let me ask you this.

If the arm fell from 70 degrees (when the weight is at 1 meter)

would the spring add force to the falling arm?
would the weight land with any more force than simply dropping from 1 meter?

If the spring does not affect the force of the (arm/weight) falling from 1 meter
then it does not affect the force of the (arm/weight) lifting to 1 meter.

Yes, the spring would add force to the falling arm. Oystein shows it well in his diagram and formula.

When the arm is tilted up the weight gets lifted higher against gravity and then the weight drops stretching the spring.

The reverse happens when the arm is tilted down the spring lifts the weight higher and then gravity drops the weight returning more force than if the weight was dropped from the same height without the spring.

You can argue that the weight doesn't actually move to an upward position but that it only because the spring is lifting the weight at the same time that the weight is falling, much like what happens when it is being raised, only in reverse.

[arthur]

sorry I have been away from the computer for a little while.

ok.

you people are confused.
it is sad to see everyone ignore and deny this simple logic.

your diagram shows the arm falling from the vertical position.
my question was regarding the arm falling from the 70 degree postion, where the weight starts at 1 meter, and (according to jim's picture) follows a continuous path downwards.

this is so simple there is no need for formulas or computer programs.
use your brain if you have one left.

If a weight falls down a frictionless, curved path 1 vertical meter,

It will land with the same force as simple dropping straight down from 1 vertical meter.

(the force of the spring does not accelerate the fall, it is used to keep the weight on it's curved path)

go ahead and keep denying this.

[jim_mich]

It will land with more force when you have energy stored in a spring. The spring adds lift to the weight as it falls thus causing the weight to move out farther on the arm and to take longer to fall as the weight gains more speed and hits the bottom with more force.

If it is so simple then build two arms, one with a spring and one without, and have them compete with each other as to force required to lift and force produced when falling.

Arguing will get you nowhere. You need proof to back up what you state. So far you have not shown any proof.

[arthur]

to me this is common sense. it is like arguing 2+2.

the spring moves the weight out further, yes,

but it does NOT lift the weight.

[Quartz]

2+2 = conserved

this is a simple exchange of energy, no net gain no net loss except for friction, There is no obvious gain.

Conserved

Ken

[arthur]

lets go back to the original concept.
the lift.

the question here is not so much semantics of words, but rather it is whether the force to lift the weight is greater when the weight follows the outer path

http://www.besslerwheel.com/forum/download.php?id=5660

lifting 1 kilo -- 1 vertical meter up the black path requires the same force needed
to lift 1 kilo -- 1 vertical meter up the blue path.

(1 kilo x 1 vertical meter) = (1 kilo x 1 vertical meter)

the path does not matter.

you should know this jim

[jim_mich]

the path does not matter.

The path does not matter only if a spring is not being stretched. Fix the weight to the arm and the path will end much higher. If you then release the weight it will fall and stretch the spring. The path that the weight follows as it is lifted up and stretches the spring does not matter. Art, you should know that!

But saying that an arm with a spring should behave the same as an an arm without a spring is incorrect.

[AB Hammer]

@ arthur

The drop the swing are all subordinate to the lift and reset. You show nothing of any reset and in this game, without reset you have nothing. Show us how it will work with the reset and then you might see what the rest of us see.

[arthur]

Fix the weight to the arm and the path will end much higher. If you then release the weight it will fall and stretch the spring.

this method of stretching the spring is costly.

I have clearly shown a much less costly way to stretch the spring.
believe it or not it is free.

I have been repeating myself for a while now so
if noone gets this,
then I dont know what else to say.

[arthur]

The drop the swing are all subordinate to the lift and reset. You show nothing of any reset and in this game, without reset you have nothing. Show us how it will work with the reset and then you might see what the rest of us see.

AB

again, take a look at "arthurs design" to see the full circle.
the motion is there.

I just dont feel like explaining it.
it is a very simple design and should not need an explanation.