Here is a nice over head shot of the spheres at highest energy. Their velocity is consistent with momentum conservation. The spheres are at the point of 450% of the original energy. The cylinder and spheres is floating or falling if you choose. The rotational motion is independent of the fall.
Soon the spheres will share their motion with the cylinder and will return the entire system to it original rate of rotation. I still find it fascinating that the spheres will not hit the side of the cylinder as they return to a very near position by its side. The spheres just float near the side of the cylinder as they both rotate.
energy producing experiments
Moderator: scott
re: energy producing experiments
From what I can gather about your set-up,it seems as though there is an energy gain due to the gradual shortening of the tethers.The other day at work I was processing some 8 foot long steel beams and as I swung them around,they slid through my gloves and gradually lengthened the amount of beam below my hand.When this lengthening occurred,the beam moved almost effortlessly through it's arc with no additional input from me.I wonder if this reaction is similar to what you are working on? Thanks so much for taking the time to share your ideas.
Trying to turn the spinning in my brain into something useful before moving on to the next life.
As far as I understand that's Centrifugal. (reminds me I still wanted to do some Dawn Yo-Yo sim)
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
It would be nice to have another opinion ME.
re: energy producing experiments
A sim would certainly be an opinion; but experiments are facts. It is not my opinion that momentum is conserved it is an experimental fact.
I already determined (an opinion) that momentum remains constant, and energy is basically some unknown variable which will be crossed out anyway - but could be calculated when one really wants.
A sim would be just out of curiosity what's happening in a yo-yo-despin situation.
Pequaide, could you post a diagram or a video of your set-up?
A sim would be just out of curiosity what's happening in a yo-yo-despin situation.
Pequaide, could you post a diagram or a video of your set-up?
re: energy producing experiments
Me; The half circumference tether can stop the cylinder from rotating in about 13/240th of a second.
This is the same cylinder with a tether length of (½ pi d) or pi r. The actual physical length is 6.28 inches. With this half circumference tether the same spheres can stop the cylinder twice and restart it twice before it reaches the ground.
The cylinder has a mass of about 468 g and the two spheres have a mass of about 130 g for a total mass of 598 g.
The 1.2 r tether length can stop the cylinder from rotating in 22/240th of a second. Its physical length is 2.4 inches.
This is to be expected because the longer tether length maintains a tangent with the cylinder until the cylinder is stopped. The short tether spends some time around 90° to tangent. At 90° it applies zero rotational force upon the cylinder.
The shorter tether was chosen so that it would only stop the cylinder once and restart it once instead of twice. Everything other that tether length remained the same. I adjusted the length to a perfect stop so I could compare the two sides (from the perfect center) of the shared momentum. The momentum is transferred from the shared motion of the cylinder and spheres to the spheres; until the spheres have all the motion. Then the spheres transfer most of the motion back to the cylinder until the system again has its original velocity. The two side are mirrors of each other which means the momentum changes are identical.
The spheres can only give their momentum to the cylinder because ballistic pendulum experiments prove that. Therefore the system can only give their momentum to the spheres because the overall motion of the system is conserved.
I don't think this site does videos.
This is the same cylinder with a tether length of (½ pi d) or pi r. The actual physical length is 6.28 inches. With this half circumference tether the same spheres can stop the cylinder twice and restart it twice before it reaches the ground.
The cylinder has a mass of about 468 g and the two spheres have a mass of about 130 g for a total mass of 598 g.
The 1.2 r tether length can stop the cylinder from rotating in 22/240th of a second. Its physical length is 2.4 inches.
This is to be expected because the longer tether length maintains a tangent with the cylinder until the cylinder is stopped. The short tether spends some time around 90° to tangent. At 90° it applies zero rotational force upon the cylinder.
The shorter tether was chosen so that it would only stop the cylinder once and restart it once instead of twice. Everything other that tether length remained the same. I adjusted the length to a perfect stop so I could compare the two sides (from the perfect center) of the shared momentum. The momentum is transferred from the shared motion of the cylinder and spheres to the spheres; until the spheres have all the motion. Then the spheres transfer most of the motion back to the cylinder until the system again has its original velocity. The two side are mirrors of each other which means the momentum changes are identical.
The spheres can only give their momentum to the cylinder because ballistic pendulum experiments prove that. Therefore the system can only give their momentum to the spheres because the overall motion of the system is conserved.
I don't think this site does videos.
Ah, now I get it (I think): It's like the yo-yo-despin, but without the release (almost a rattle drum)?
My current (terribly slow-)running simulation of such is as follows: a 1200 kg disc rotating 360°/s (60 RPM) with a diameter of 1m is attempted to be stopped by two half-wound tethers (3 gr each) on each end a mass of 3 kg, with a result of (...still running...)
Anyhow, the ideal situation for a full restart would be when the cylinder's velocity equals zero when completely unwound (90° to tangent, and symmetrical in action), if they weren't loosing there axial relation (and can't complete their half cycle) due to their falling by gravity.
Perhaps it's an idea to compensate gravity by having those sphere's hung by some very long thread connected to the ceiling?
(Next time -when this is finished- I'll try to enter you numbers, to see what sim-world has to say)
My current (terribly slow-)running simulation of such is as follows: a 1200 kg disc rotating 360°/s (60 RPM) with a diameter of 1m is attempted to be stopped by two half-wound tethers (3 gr each) on each end a mass of 3 kg, with a result of (...still running...)
Anyhow, the ideal situation for a full restart would be when the cylinder's velocity equals zero when completely unwound (90° to tangent, and symmetrical in action), if they weren't loosing there axial relation (and can't complete their half cycle) due to their falling by gravity.
Perhaps it's an idea to compensate gravity by having those sphere's hung by some very long thread connected to the ceiling?
(Next time -when this is finished- I'll try to enter you numbers, to see what sim-world has to say)
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
Well I get 1256.6 m/sec for momentum conservation and 62.8 m/sec for energy conservation; for the 3 kilograms. But ask your sim to explain how (62.8 m/sec * 3 kg) 188.4 units of momentum, contained by the 3 kilograms, can give the 1200 kg 3769.9 units of momentum when the motion is returned to the craft.
Can 95% of the momentum just get invented? And experiments prove that the motion is returned.
Can 95% of the momentum just get invented? And experiments prove that the motion is returned.
re: energy producing experiments
My sim is in some sort of protest-mode.
I've read in some paper the weight of the tether itself was important for calculating the deceleration. I wanted to skip that part, and make it as light-weight as possible: probably wrong. I suspect the weight of the tether is important to keep it steady. Too light and it breaks, a little heavier and it whips the mass, so I make the tether about on tenth of the attached weight.
So that sim is taking its time. (I'll try not to mess your topic up too much with this stuff)
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From the point of symmetry it should return all the momentum.
The system start with some speed, yo-yo releases, for symmetry-sake the cylinder is stopped when fully extended (90°), then it goes back in symmetry.
From a centrifugal perspective the yo-yo is actually a pendulum swinging from the ceiling(the rim) to the ground(fully extended). As a pendulum it should just swing back "up" to the rim. And the tether jerks the cylinder around in the same symmetry.
This all in a perfect situation, and yours is 95% perfect - close enough.
So I simply suspect the used formulas are wrong; especially because it all (except 5% losses) pans out eventually.
I tried some formula but it seems a bit tricky and needs some more attention than currently available, I'll get back to that.
Marchello E.
I've read in some paper the weight of the tether itself was important for calculating the deceleration. I wanted to skip that part, and make it as light-weight as possible: probably wrong. I suspect the weight of the tether is important to keep it steady. Too light and it breaks, a little heavier and it whips the mass, so I make the tether about on tenth of the attached weight.
So that sim is taking its time. (I'll try not to mess your topic up too much with this stuff)
---
From the point of symmetry it should return all the momentum.
The system start with some speed, yo-yo releases, for symmetry-sake the cylinder is stopped when fully extended (90°), then it goes back in symmetry.
From a centrifugal perspective the yo-yo is actually a pendulum swinging from the ceiling(the rim) to the ground(fully extended). As a pendulum it should just swing back "up" to the rim. And the tether jerks the cylinder around in the same symmetry.
This all in a perfect situation, and yours is 95% perfect - close enough.
So I simply suspect the used formulas are wrong; especially because it all (except 5% losses) pans out eventually.
I tried some formula but it seems a bit tricky and needs some more attention than currently available, I'll get back to that.
Marchello E.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
I get the idea you're are comparing Linear Momentum of a single object to the Angular Momentum of the whole system.
That satellite/cylinder is not moving through space, but just spinning in place.
The Linear momentum of the whole system equals 0 as both weights move in opposite directions.
The Angular momentum of those extended weights (still rotating) plus the residual Angular Momentum of the cylinder/satellite should equal the Angular Momentum when first spun-up.
Two 3 kg weights on a half-circle tether cannot stop the 1200 kg satellite from halting its rotation.
According to my sim (which eventually needed a 3 kg tether) it takes about two 17 kg weights.
According to my calculation (and a half circumference weightless tether of 1.57 m) it actually needs two weights of 18.5726 kg.
When this tether is made twice as long (full circumference: 2*pi*r=3.14 m) then it needs two times a 5.7643 kg to stop a 1200 kg satellite at 90° tangent.
The linear momentum of a weight starts to have meaning when a weight smashes into something, the line simply breaks or gets detached.
How fast do you make your 4-inch diameter cylinder spin-up?
Marchello E.
That satellite/cylinder is not moving through space, but just spinning in place.
The Linear momentum of the whole system equals 0 as both weights move in opposite directions.
The Angular momentum of those extended weights (still rotating) plus the residual Angular Momentum of the cylinder/satellite should equal the Angular Momentum when first spun-up.
Two 3 kg weights on a half-circle tether cannot stop the 1200 kg satellite from halting its rotation.
According to my sim (which eventually needed a 3 kg tether) it takes about two 17 kg weights.
According to my calculation (and a half circumference weightless tether of 1.57 m) it actually needs two weights of 18.5726 kg.
When this tether is made twice as long (full circumference: 2*pi*r=3.14 m) then it needs two times a 5.7643 kg to stop a 1200 kg satellite at 90° tangent.
The linear momentum of a weight starts to have meaning when a weight smashes into something, the line simply breaks or gets detached.
How fast do you make your 4-inch diameter cylinder spin-up?
Marchello E.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
They are actually going in the same direction because the masses never work against each other.
The linear momentum is equal to arc momentum; a clockwise motion can be transferred to a 100% counterclockwise rotation, no loss. If the experiment (the real world) does not care about direction then do not place artificial barriers in the way of your thinking. You will only confuse yourself; but nature is not confused.
Galileo's pendulum proves that angular momentum conservation and the concept (the concept in the lab) is false. An infinite number of tether lengths gives you an infinite number of angular momentums, for the same moving object. It is a false concept 'drop it'.
Know what you know; and we know how ballistic pendulums work.
We know that the 3 kilogram masses will restart the 1200 kilograms, and 188.5 units of momentum can not give you 3770. So you must have 3770 units of momentum in the 3 kg, either it is a head on collision or a re-wrap, you still need 3770 units.
In fact NASA admits to a partial restart that is greater that 188.5 units of momentum. Their backward motion is 8% of the original momentum and they say you can only have 5%.
I spin the 4 inch cylinder to cross a 20 mm distance in between 3/240th sec and 4/240th sec, that would be 1.6 m/sec to 1.2 m/sec.
From my experiments; a half circumference tether is more in the range of a 1 to 18 mass ratio. That would be in the 65 kilogram range for 1200 kg. A full circumference tether is in the range of 1 to 40, if I recall correctly, and that would be in the 30 kilogram range. Note that the 3 kilograms required several wraps.
Your calculation are fairly close though; what formula did you use? I don't include the mass of the tether.
The linear momentum is equal to arc momentum; a clockwise motion can be transferred to a 100% counterclockwise rotation, no loss. If the experiment (the real world) does not care about direction then do not place artificial barriers in the way of your thinking. You will only confuse yourself; but nature is not confused.
Galileo's pendulum proves that angular momentum conservation and the concept (the concept in the lab) is false. An infinite number of tether lengths gives you an infinite number of angular momentums, for the same moving object. It is a false concept 'drop it'.
Know what you know; and we know how ballistic pendulums work.
We know that the 3 kilogram masses will restart the 1200 kilograms, and 188.5 units of momentum can not give you 3770. So you must have 3770 units of momentum in the 3 kg, either it is a head on collision or a re-wrap, you still need 3770 units.
In fact NASA admits to a partial restart that is greater that 188.5 units of momentum. Their backward motion is 8% of the original momentum and they say you can only have 5%.
I spin the 4 inch cylinder to cross a 20 mm distance in between 3/240th sec and 4/240th sec, that would be 1.6 m/sec to 1.2 m/sec.
From my experiments; a half circumference tether is more in the range of a 1 to 18 mass ratio. That would be in the 65 kilogram range for 1200 kg. A full circumference tether is in the range of 1 to 40, if I recall correctly, and that would be in the 30 kilogram range. Note that the 3 kilograms required several wraps.
Your calculation are fairly close though; what formula did you use? I don't include the mass of the tether.
re: energy producing experiments
When I assume the main object will be stationary when the weights are fully extended then we can use the conservation of angular momentum.
Actually I have a whole page of formula's (making sense or not), but it's basically L = I·w, and the initial angular velocity(w) equals the final angular velocity of the masses.
I calculate a single yo-mass (for this situation) as:
m = I / (2·s²)
where 's' is the deployed distance (radius+tetherlength), and 'I' the moment of inertia in the initial situation.
We could also rewrite this to get the tether-length for a given mass.
Marchello E.
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add:
hmm, I see I oversimplified things: because the inertia 'I' includes 'm', still solvable though...
Enjoy: m=M·r²/(4*s·(1+2*r²)), for a disc - replace that 4 by a 2 when it's a cylinder or ring.
Well, at least we can calculate the tether length for:
M=1200 kg (disc), m=1.5 kg, r=0.5 m--> Length= 6.59 m (about 2 times the circumference)
or, when there are two times 3 kg which I assume was not the case:
M=1200 kg (disc), m=3 kg, r=0.5 m--> Length= 4.525 m (about 1.44 times the circumference)
Actually I have a whole page of formula's (making sense or not), but it's basically L = I·w, and the initial angular velocity(w) equals the final angular velocity of the masses.
I calculate a single yo-mass (for this situation) as:
m = I / (2·s²)
where 's' is the deployed distance (radius+tetherlength), and 'I' the moment of inertia in the initial situation.
We could also rewrite this to get the tether-length for a given mass.
Marchello E.
-----
add:
hmm, I see I oversimplified things: because the inertia 'I' includes 'm', still solvable though...
Enjoy: m=M·r²/(4*s·(1+2*r²)), for a disc - replace that 4 by a 2 when it's a cylinder or ring.
Well, at least we can calculate the tether length for:
M=1200 kg (disc), m=1.5 kg, r=0.5 m--> Length= 6.59 m (about 2 times the circumference)
or, when there are two times 3 kg which I assume was not the case:
M=1200 kg (disc), m=3 kg, r=0.5 m--> Length= 4.525 m (about 1.44 times the circumference)
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
Angular momentum is equal to the 'moment of inertia' times 'angular velocity'. The moment of inertia for a mass at a single radius is mrr. Angular velocity is arc velocity divided by r. L = mrr * arc v/r so the one r in the denominator and one r in the numerator drop out for L = mr * arc v. Arc velocity is equal to linear tangent velocity; thus, mrv.
At 60 rpm for a .5m r circle and a 1200 kg cylindrical mass; the initial angular momentum is, 1200 kg * .5 m * 3.14159 m/sec = 1885. The final angular momentum is 3 kg * 12 m (r) * arc velocity. So for angular momentum conservation the final arc velocity is; 1885 / (3 kg * 12 m) = 52.4 m/sec. That is for 3 kg so the momentum is 52.4 m/sec * 3 kg for 157 units of momentum.
We know from experiment that the spheres can return all the motion to the cylinder; which is in this case 3770 units of arc momentum. So where can angular momentum conservation find the other (3770 - 157 =) 3613 units of momentum?
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Angular momentum conservation in the lab is a false concept. And energy is not a conserved quantity.
Of the formulas on your pages of formulas; it is F = ma that actually works; and the other two (KE, L) don’t.
It took 3770 newton seconds to make the 1200 kg cylinder spin at 3.14159 m/sec; and it will take 3770 newton seconds to make the mass stop. Even if the 3 kilograms has all the motion it will still take 3770 newton seconds to make the mass stop; not 188 or 157 newton seconds. Sixteen newtons applied for 10 second is insufficient to start the spin and it would be insufficient to stop it.
I use 12 meters (paragraph 2) for the Dawn Mission tether; but it was actually too long. NASA got a 3 RPM backward spin (8%); down from 36 RPM forward. The backward spin can only occur if the rotation stops before 90° and the tether then pulls the cylinder backwards. This means a shorter tether could have been used for the 90° stop.
The linear equivalent to arc velocity is present whether the tether is attached or unattached. Getting in the way of an attached tethered mass is no less dangerous than getting smacked by a loose one.
At 60 rpm for a .5m r circle and a 1200 kg cylindrical mass; the initial angular momentum is, 1200 kg * .5 m * 3.14159 m/sec = 1885. The final angular momentum is 3 kg * 12 m (r) * arc velocity. So for angular momentum conservation the final arc velocity is; 1885 / (3 kg * 12 m) = 52.4 m/sec. That is for 3 kg so the momentum is 52.4 m/sec * 3 kg for 157 units of momentum.
We know from experiment that the spheres can return all the motion to the cylinder; which is in this case 3770 units of arc momentum. So where can angular momentum conservation find the other (3770 - 157 =) 3613 units of momentum?
Â
Angular momentum conservation in the lab is a false concept. And energy is not a conserved quantity.
Of the formulas on your pages of formulas; it is F = ma that actually works; and the other two (KE, L) don’t.
It took 3770 newton seconds to make the 1200 kg cylinder spin at 3.14159 m/sec; and it will take 3770 newton seconds to make the mass stop. Even if the 3 kilograms has all the motion it will still take 3770 newton seconds to make the mass stop; not 188 or 157 newton seconds. Sixteen newtons applied for 10 second is insufficient to start the spin and it would be insufficient to stop it.
I use 12 meters (paragraph 2) for the Dawn Mission tether; but it was actually too long. NASA got a 3 RPM backward spin (8%); down from 36 RPM forward. The backward spin can only occur if the rotation stops before 90° and the tether then pulls the cylinder backwards. This means a shorter tether could have been used for the 90° stop.
The linear equivalent to arc velocity is present whether the tether is attached or unattached. Getting in the way of an attached tethered mass is no less dangerous than getting smacked by a loose one.
re: energy producing experiments
It can't.At 60 rpm for a .5m r circle and a 1200 kg cylindrical mass; the initial angular momentum is, 1200 kg * .5 m * 3.14159 m/sec = 1885 N·m·s. The final angular momentum is 3 kg * 12 m (r) * arc velocity. So for angular momentum conservation the final arc velocity is; 1885 N·m·s/ (3 kg * 12 m) = 52.4 m/sec. That is for 3 kg so the momentum is 52.4 m/sec * 3 kg for 157 N·s.
So where can angular momentum conservation find the other (3770 unit?- 157 N·s=) 3613 units? of momentum?
Where did the 3770 come from?
I guess you used: m*r*w (?) = 1200 kg * 0.5 m * 6.283 rad/s [N·s]
Sorry, but you can't use it that way.
The term "r*w" is meant to be the velocity 'v' at radius 'r' at angular velocity 'w'.
But this 'v' is the absolute value of the vector on the rim, where the direction is tangent (or perpendicular, I guess that's the same) to the axis of rotation.
When you 'sum' all those vectors with all their vector direction of all the points on the rim, then that results in 0.
The linear velocity of the whole system is 0 m/s.
It rotates on a fixed point in space and not translates or moves through space.
Therefore p=m*v=(1200+2*1.5) * 0 = 0 N·s
There's no-way to get a factor of Linear Momentum
- - -
Perhaps interesting when early Linear Momentum does happen:
When one weight gets loose and disconnects at its point on the rim (leaving the tether) then things start to move and momentum arises:
The weight at the rim has a locally linear velocity of 3.14 m/s
Its momentum will be p=m*v = 1.5 * 3.14 = 4.71 N·s when it breaks free.
The rest (the other weight + cylinder) gets a kick-back because of conservation of Linear Momentum:
So that's M*V =-4.71 = 1201.5 * V --> V= - 0.0039 m/s (and wobbles around a new CoM)
And both the free-weight and cylinder+weight don't experience any change in angular velocity [rad/s].
Angular momentum and Linear momentum are just two separate things.
-Sorry if this spoiled the fun, or a case of misunderstanding-
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---