the obvious gain

[arthur]

ok brainiacs

take a look at this little picture
------------------

again the red dot is the axle,
the blue line is an extension spring,
and the round weight can move along the arm

lets say:
*the weight of the arm + spring is negligible
*the round weight is 1 kilo
*the weight travels vertical distance of 1 meter.

the energy input to move the arm as shown in the picture
is simply: weight x vertical distance (1 kilo x 1 meter)

the potential for energy output = (1 kilo x 1 meter) + force now stored in the spring

the energy output is greater
the gain is the force now stored in the spring

[AB Hammer]

@arthur

In allot of gravity wheels there are gains and the equal or more losses. This is what has to be overcome. To have more gain than loss. I am not sure of what you are trying to achieve. ???

[jim_mich]

No! The energy input is the same as the energy output. In order to stretch the spring the weight must be lifted higher and then it falls stretching the spring. There are a number of ways to visualize this in your mind. The simplest is to envision the weight not moving in the tube until the tube has been lift up vertically and then the weight falls extending the spring. Another way is to envision some force other than gravity extending the spring before the weight is lifted up. The way that will actually happen is a complex movement where the torque required to rotate the tube from horizontal to vertical varies all along the weight path. The torque is greatest at the beginning of rotation dropping to zero at the finish of rotation. When the total rotational torque averaged over rotational distance is calculated then the amount of energy needed to lift the weight while compressing the spring is (1 kilo x 1 meter) + force now stored in the spring.

[arthur]

In order to stretch the spring the weight must be lifted higher and then it falls stretching the spring.

no.

it can be set up so that the weight never falls, and is only lifted
as I have drawn.

the spring extends without the weight falling.

envision the weight not moving in the tube until the tube has been lift up vertically and then the weight falls extending the spring.

umm this does not happen in the scenario I have drawn,
moving the arm requires much less force than what you describe.

[jim_mich]

Arthur, yes your scenario does not show the weight being moved higher and then falling nor does it show the weights being moved against the spring in the tube before lifting. As I stated, these are ways to visualize (the end result) in your mind. Calculating the torque requirement degree by degree is cumbersome unless you write and use a computer program or use WM2D.

Only by measuring the torque required to lift the weight while stretching the spring can you determine the input energy that is needed. It takes more torque to rotate the tube when the weight is farther out (as it is at the beginning of the lift) than it would take to lift the weight when it is closer to the axle. This extra torque is the energy that indirectly stretches the spring.

You state that, "the spring extends without the weight falling." Before the weight can fall it must be lifted higher. Gravity only does work when a weight falls.

It takes a certain amount of energy to lift the weight up 1 meter. If you stretch a spring in the process then you will not be able to lift the weight to 1 meter unless you add the extra energy needed to stretch the spring.

What proof do you have that, "moving the arm requires much less force" than what I describe?

[arthur]

Arthur, yes your scenario does not show the weight being moved higher and then falling nor does it show the weights being moved against the spring in the tube before lifting. As I stated, these are ways to visualize (the end result) in your mind. Calculating the torque requirement degree by degree is cumbersome unless you write and use a computer program or use WM2D.

no actually I believe it is pretty simple.
(assuming the axle has negligent friction) -
the weight is moved up the frictionless path of the green arrow.

this requires the same force that would be needed
to roll the weight up a hill with the same path.

however, rolling a weight up a hill with the same path as the green arrow-
would not reward us with the bonus energy stored in the spring.

Only by measuring the torque required to lift the weight while stretching the spring can you determine the input energy that is needed. It takes more torque to rotate the tube when the weight is farther out (as it is at the beginning of the lift) than it would take to lift the weight when it is closer to the axle. This extra torque is the energy that indirectly stretches the spring.

the "torque" that you are describing is simply the force required to move the arm, which moves the weight up the path of the green arrow.
this force is separate and perpendicular from the force that stretches the spring.
so yes, it is indirect.

You state that, "the spring extends without the weight falling." Before the weight can fall it must be lifted higher. Gravity only does work when a weight falls.

once again the weight does not fall.

It takes a certain amount of energy to lift the weight up 1 meter. If you stretch a spring in the process then you will not be able to lift the weight to 1 meter unless you add the extra energy needed to stretch the spring.

the energy that stretches the spring is supplied "free" from gravity,
and does not detract from the force needed to lift the arm.

What proof do you have that, "moving the arm requires much less force" than what I describe?

you were describing lifting the weight higher than 1 meter.
in my scenario the arm lifts the weight no more than 1 meter.
lifting the weight higher than this would take more force,
this is common sense.

[arthur]

In allot of gravity wheels there are gains and the equal or more losses.

I think you mean to say in a lot of gravity wheel attempts there are energy outputs, and equal or more energy inputs. -- (when I say gain it implies the opposite)

This is what has to be overcome.

no kidding.

I am not sure of what you are trying to achieve. ???

wow.
you must be confused. I think I already spelled it out for you.

[greendoor]

This idea about compressing springs with falling weights is very common and basically flawed.

The simple non-maths thing to consider is this: usually in a gravity wheel you are using the potential energy of a mass at a hight to balance another weight that is at a lower height that needs to be raised. One kilogram, 1 meter above another kilogram has the PE to raise that 1 kg as it falls (ignoring friction). OR, using leverage - it has the ability to raise a smaller weight higher than 1 meter, or a greater weight less than 1 meter.

BUT - if you start compressing springs while you are trying to do this, you will find that you can either do one or the other. Or a little bit of one and the other. There is no free lunch.

[arthur]

Calculating the torque requirement degree by degree is cumbersome unless you write and use a computer program or use WM2D.

Jim it is actually a very simple calculation.

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

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

.....the same force needed to lift it.

[arthur]

usually in a gravity wheel you are using the potential energy of a mass at a hight to balance another weight that is at a lower height that needs to be raised. One kilogram, 1 meter above another kilogram has the PE to raise that 1 kg as it falls (ignoring friction). OR, using leverage - it has the ability to raise a smaller weight higher than 1 meter, or a greater weight less than 1 meter.

greendoor what I am talking about here is not the generic 'gravity wheel' concept of weights balancing/lifting each other.

look at "arthurs design" and you will see the concept is weights in free fall
more like a pendulum but with a spring in the radius.

please be more specific to address the flaw in my logic.

[bluesgtr44]

Hey Art, OK......the weight is in a tube with a spring connected between the weight and the top of the tube. Starting in the horizontal position, we swing the arm upwards 90 degrees to the 12 o'clock position and in doing so the spring is stretched and energy is stored.

The gain is not there....see, the spring starts stretching as soon as the arm moves upward and this will change the position of the weight within that arm and laws of leverage enter into the equation.....this is where you think you are getting a gain.....in the system....the system has to continue it's 360 degree path and when it comes around to the downside....the spring will not stretch and the position of the weight is not going to be treated nicely by the laws of leverage and it will eat up the supposed gain you got from that first quadrant of lifting the arm.

If I have missed something.....my apologies. I am assuming that this would make a complete revolution at some point. We can all pretty much see the starting torque in many ideas....it's when it has to pay back what it took to get that extra burst that we can overlook sometimes. So, if this is not how it goes completely around....I'd sure like to know what you have in mind for that.


Steve

[arthur]

the energy gain is stored in the spring.

are you denying that there is energy stored in the spring?
or do you just think it cannot be used?

the energy is there, and it can be used.

look at "arthurs design" and you will see how this spring energy can be used to add rotational force.

+++++

as I have said,

the potential for energy output of the arm in the vertical position
is (1 kilo x 1 meter) + energy stored in the spring.

agreed?
(maybe not if you keep pretending this spring energy cannot be used)

[bluesgtr44]

the energy gain is stored in the spring.

What caused it? The movement from a horizontal to vertical position, gravity pulls down on the weight and viola!....the spring stretches and thus we have stored energy as long as the weight maintains it's connection and position....am I right so far?

are you denying that there is energy stored in the spring?
or do you just think it cannot be used?

See above....there is stored energy! Now, the second question....sure it can be used...how do you plan to apply this, not use....but apply....in a situation that has the potential to produce excess energy within a closed loop system? I admit I'm kind of stuck on "what comes around, goes around..." in my full vision of this....and that might be my problem.

the energy is there, and it can be used.

Show us how to apply it. I saw "Arthurs design" and I just can't see the excess energy in this....again, what comes around goes around....

the potential for energy output of the arm in the vertical position
is (1 kilo x 1 meter) + energy stored in the spring.

agreed?
(maybe not if you keep pretending this spring energy cannot be used)

Sure it can be used, it already is being used to hold the weight up.....remove the weight and there will still be a minimal amount of stored energy just from the weight of the spring itself, would you not agree? So, how do you plan to apply it as far as "Arthurs design" goes? I wouldn't mind trying to do a sim of this if you can provide me a little more detail as to the movement and positions.....length of rope and spring and if there is a specific angle of the end point for the rope...

I guess I just don't see the whole picture the way you do....willing to give it a shot, though....


Steve

[jim_mich]

arthur,

When the tube is tilted upward 10º then the 1 kilo weight will exert 1.5% of its weight to stretch the spring. If we assume that the spring tension varies from 0 Kilogram to 1 kilogram over a distance of 0.2 meters then the 1.5% of the weight will stretch the spring 0.003 meters.

At 20º upward tilt gravity will stretch the spring 0.0120 meters
At 30º upward tilt gravity will stretch the spring 0.0268 meters
At 40º upward tilt gravity will stretch the spring 0.0468 meters
At 50º upward tilt gravity will stretch the spring 0.0714 meters
At 60º upward tilt gravity will stretch the spring 0.1000 meters
At 70º upward tilt gravity will stretch the spring 0.1316 meters
At 80º upward tilt gravity will stretch the spring 0.1652 meters
At 90º upward tilt gravity will stretch the spring the full 0.2 meters

Attached is a graph of the path of the weight in steps of 10º shown as black.
The amount of extra lift needed to stretch the spring is calculated as about 0.165 kilo/meters when using a very rough 10º step.

Also shown is a blue graph line using 1º calculation steps. It calculates out to about 0.158 kilo/meters of extra lifting.

If the steps were infinitely small then the actual extra lifting would be about 0.157080. This is equal to Pi / 4 • 0.2 kilo/meters when the weight is 1 kilo and the spring stretch is 0.2 meter.

Because the spring is not constant and has only about 1 kilo of tension when extended to 0.2 meter it will not lift the weight. But it will just hold the weight at that height until the tube starts to tilt down again.

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

If we were to use a constant tension spring, say with a tension of 0.826 Kilo then the weight will not even begin to stretch the spring until the tube has tilted 80º upward. This is because at 80º the 1 kilo weight mass will exert 0.826 kilo of force on the weight due to the tilt angle. Only then will the weight start to drop and stretch the spring.

I think you need a towel. Maybe you were born just yesterday and are still a little wet behind your ears? Some of us have been around for a very long time and we understand these PM wheel proposals very well. We know very well what will not work and we keep seeking something that might work.

I suggest that you don't take such a bully attitude. Members with high reputations usually are experts when it comes to gravity wheels.

[arthur]

ok,

good work Jim.
your graph is correct
the weight does fall slightly.

however,
as you have shown,

the weight does not fall before the arm reaches 80 degrees,
it is only lifted.

so, for simplicity,
lets just talk about the lifting from 0 - 70 degrees,
where the weight reaches 1 vertical meter.

from 0-70 degrees the weight is lifted 1 vertical meter
up the frictionless path of the blue line.

this would require the same force as rolling the weight 1 vertical meter up a hill with the same path as the blue line.

can we agree on this....??