Wubbly quote: Once I have that cocked spring, I can take it and point it in any direction I want and release it and I just broke the conservation of momentum, so the spring was not storing momentum.
A spring loaded gun will deliver the same amount of momentum no matter what direction you release it in. North, west, south or east; it doesn’t matter. If it gives you 5 m/sec north it will give you 5 m/sec south, direction canceling momentum is another idea in the category of silliness.
Vectors are for objects that are interacting; the vector concept is not for individual objects. A spinning wheel has plenty of momentum; it is equal to the same mass moving the same speed in a straight line. You can transfer all the momentum of a flywheel to an object moving in a straight line. In fact the cylinder and spheres concept can be done with objects moving in a straight line.
Springs don’t store momentum because they store in a force times distance relationship. Momentum is force times time.
Wubbly quote: Momentum
mass x velocity = force x time
If you look at an Atwoods, the time variable increases (it takes longer for the mass to fall).
You can also see that the velocity variable gets smaller (the mass falls slower than freefall).
So how can we balance the equation? Obviously you have to decrease the force. This is exactly what happens.
Sure, gravity did not decrease, but the tension in the string creates an almost equal and opposite force in the other direction, so the net force on the mass did decrease.
If you find a way to stretch out time, you are not accumulating momentum, you are decreasing the force. End quote.
Atwood’s machines are proof for F = ma. Of course F remains constant. And the mv is a product of force times time, and the momentum created can be huge. Wubbly makes all these statements without even knowing how an Atwood’s works. Atwood’s do a near perfect F = ma; that would not happen if F changed.
energy producing experiments
Moderator: scott
pequaide can you give your opinion on the following.
What if you have a certain setup where the heavy mass wheel is not allowed to stop. Meaning even if the small masses swing out and cause a counter torque you kill this force by forcefully keeping the wheel in motion. Would it be correct to assume that the small weights will end up with even more motion than when they caused the wheel to stop? Surely then you'ld still end up with an energy gain? If so this is handy to know since it opens some doors to new setups and designs.
What if you have a certain setup where the heavy mass wheel is not allowed to stop. Meaning even if the small masses swing out and cause a counter torque you kill this force by forcefully keeping the wheel in motion. Would it be correct to assume that the small weights will end up with even more motion than when they caused the wheel to stop? Surely then you'ld still end up with an energy gain? If so this is handy to know since it opens some doors to new setups and designs.
re: energy producing experiments
I enjoyed stopping the cylinder just as the tether string entered a slit in the cylinder. While the string is in the slit the cylinder remains stationary. If the cylinder was too heavy then the cylinder would still be moving forward while the string was in the slit. Or if the cylinder was too light then the cylinder will have been forced backwards before the string enters the slit and the cylinder will be moving backward while the string is in the slit.
If the cylinder has been forced backwards the spheres have lost momentum and energy from their maximum quantity. If the cylinder is still moving forward then the spheres have not gained all the momentum of the cylinder. Have the spheres gained enough to increase the energy of the system? Probably.
Other thoughts: The F in F = ma in an Atwood’s is not a variable. Once you choose the quantity of over balanced mass then F remains constant. Atwood’s are deadly accurate in proving F = ma. I have three Atwood’s and I time them with photo gates. They make huge amounts of momentum. All the force of the overbalanced mass of an Atwood’s is used to accelerate all the mass of the Atwood’s.
The F in an Atwood’s is the force exerted by the over balanced mass, if you are using one kilogram of over balance then the force is 9.81 N, if a string holds the over balance then the force in the string will be 9.81 newtons when the Atwood’s is held at rest.
The m of F = ma in an Atwood’s is the total mass; we will assume that all the mass is moving at about the same speed as the over balanced mass. Let us say we build a balanced Atwood’s with 9 kilograms of balanced mass. This gives us an m of ten kilograms.
Once we have the Atwood’s built; acceleration (a) is the quantity that is to be determined by the other two factors. In this Atwood’s (a) would be .981 m/sec/sec. The force in the string suspending the over balanced mass drops (to 8.829 N) when motion begins because a portion of the force (.981 N) is being used to accelerate the overbalanced mass. But all of the force is being engaged upon all of the mass, we must remember that the overbalanced mass is a portion of all the mass.
If the overbalanced mass (1 kg) drops one meter the velocity of all the system is 1.4007 m/sec. This is 14.007 units of momentum. A 1 kilogram mass moving 14.007 m/sec also has 14.007 units of momentum and it will rise 10 meters.
If the cylinder has been forced backwards the spheres have lost momentum and energy from their maximum quantity. If the cylinder is still moving forward then the spheres have not gained all the momentum of the cylinder. Have the spheres gained enough to increase the energy of the system? Probably.
Other thoughts: The F in F = ma in an Atwood’s is not a variable. Once you choose the quantity of over balanced mass then F remains constant. Atwood’s are deadly accurate in proving F = ma. I have three Atwood’s and I time them with photo gates. They make huge amounts of momentum. All the force of the overbalanced mass of an Atwood’s is used to accelerate all the mass of the Atwood’s.
The F in an Atwood’s is the force exerted by the over balanced mass, if you are using one kilogram of over balance then the force is 9.81 N, if a string holds the over balance then the force in the string will be 9.81 newtons when the Atwood’s is held at rest.
The m of F = ma in an Atwood’s is the total mass; we will assume that all the mass is moving at about the same speed as the over balanced mass. Let us say we build a balanced Atwood’s with 9 kilograms of balanced mass. This gives us an m of ten kilograms.
Once we have the Atwood’s built; acceleration (a) is the quantity that is to be determined by the other two factors. In this Atwood’s (a) would be .981 m/sec/sec. The force in the string suspending the over balanced mass drops (to 8.829 N) when motion begins because a portion of the force (.981 N) is being used to accelerate the overbalanced mass. But all of the force is being engaged upon all of the mass, we must remember that the overbalanced mass is a portion of all the mass.
If the overbalanced mass (1 kg) drops one meter the velocity of all the system is 1.4007 m/sec. This is 14.007 units of momentum. A 1 kilogram mass moving 14.007 m/sec also has 14.007 units of momentum and it will rise 10 meters.
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re: energy producing experiments
Wow: now that is an Atwood’s picture. The bag is a bag of BB’s, it is better to thump the equipment (or the experimenter) with a bag of BB’s than with a steel sphere. This Atwood’s is a wheel, it does an almost perfect F = ma. You have to calculate the mass distribution at the different radii, but that is doable.
Re: re: energy producing experiments
Why are you mentioning the weird "drop" in force. Your final numbers don't take them in account. It's best to no spread confusion with meaningless things. Something like:pequaide wrote: Once we have the Atwood’s built; acceleration (a) is the quantity that is to be determined by the other two factors. In this Atwood’s (a) would be .981 m/sec/sec. The force in the string suspending the over balanced mass drops (to 8.829 N) when motion begins because a portion of the force (.981 N) is being used to accelerate the overbalanced mass. But all of the force is being engaged upon all of the mass, we must remember that the overbalanced mass is a portion of all the mass.
If the overbalanced mass (1 kg) drops one meter the velocity of all the system is 1.4007 m/sec. This is 14.007 units of momentum. A 1 kilogram mass moving 14.007 m/sec also has 14.007 units of momentum and it will rise 10 meters.
1kg mass pulls atwood down with gravity force:
- a=F/m=9.81N/10kg=0.981 m/s/s
d=1m
since v=sqrt(2*d*a) then v=1.4 m/s
momentum=m*v=10kg*1.4m/s=14 units
If all is transfered to 1kg then energy is increased by factor 10
But peq I thought we were done with these math riddles a few pages back. I think you should concentrate on documenting your personal experimentations properly rather than repeating the basic math over and over.
Re: re: energy producing experiments
pequaide wrote:A 1 kilogram mass moving 14.007 m/sec also has 14.007 units of momentum and it will rise 10 meters.
broli wrote:1kg mass pulls atwood down with gravity force:
a=F/m=9.81N/10kg=0.981 m/s/s
d=1m
since v=sqrt(2*d*a) then v=1.4 m/s
momentum=m*v=10kg*1.4m/s=14 units
If all is transfered to 1kg then energy is increased by factor 10
It will rise 10 meters, verses, if all [momentum] is transferred to 1 kg then energy is increased by factor 10.
How you do that is what this discussion is about isn't it & what Neptune is trying to engineer on another thread ?
Re: re: energy producing experiments
I don't think that's the discussion. If you want to shoot the small mass in the air you just release it from the tether onto a ramp when the big wheel has stopped like peq said. You can do this just like a trebuchet. But then what? It would only prove to be more energetic to John Doe. It's not a compact unit that can be powering a house or even that of an effective method to prove overunity.Fletcher wrote:
How you do that is what this discussion is about isn't it & what Neptune is trying to engineer on another thread ?
re: energy producing experiments
You are right Broli. Keep it simple. I will keep saying that to myself.
The weird reference to a drop in force came from another thread; I was trying to correct faulty thinking.
I don’t know how many times I need to say things over and over. Some are catching on but then you have others that say that they don’t know how to transfer all the momentum to a smaller mass. To quote Fletcher: “if all [momentum] is transferred to 1 kg then energy is increased by factor 10. How you do that is what this discussion is about isn't it & what Neptune is trying to engineer on another thread ?� There are pictures galore of machines that work and they are trying to slam seesaws together.
Neptune; The second seesaw in your double seesaw experiment does not secure all the momentum of the first seesaw and it adds a sizable amount of mass (at rest) to the motion. Plus; how much motion is lost when the two seesaws clunk together. If you were interested in the truth you would do a proper experiment, as you have been shown how many times.
Okay we will go over it again. On page 29 on the “energy producing experiments� thread are ten frames of a video of a working machine. You don’t need to slam seesaws together we already have a working device that transfers all the motion of a large mass to a much smaller object. The large mass is the cylinder and the small mass is the spheres.
Obviously a cylinder in freefall does not lend itself to a repeating cyclic function, but a horizontal wheel attached to a vertically mounted wheel does. I have set these two wheels up and they function just like the cylinder and spheres. These two wheels are pictured on page 3 and 4 of the same “energy producing experiments� thread.
Sorry about the last pictures being huge; but they still show real experiments. Doing experiments gives you a clear picture of the concept and experimenting gives you ideas on how to improve it.
The weird reference to a drop in force came from another thread; I was trying to correct faulty thinking.
I don’t know how many times I need to say things over and over. Some are catching on but then you have others that say that they don’t know how to transfer all the momentum to a smaller mass. To quote Fletcher: “if all [momentum] is transferred to 1 kg then energy is increased by factor 10. How you do that is what this discussion is about isn't it & what Neptune is trying to engineer on another thread ?� There are pictures galore of machines that work and they are trying to slam seesaws together.
Neptune; The second seesaw in your double seesaw experiment does not secure all the momentum of the first seesaw and it adds a sizable amount of mass (at rest) to the motion. Plus; how much motion is lost when the two seesaws clunk together. If you were interested in the truth you would do a proper experiment, as you have been shown how many times.
Okay we will go over it again. On page 29 on the “energy producing experiments� thread are ten frames of a video of a working machine. You don’t need to slam seesaws together we already have a working device that transfers all the motion of a large mass to a much smaller object. The large mass is the cylinder and the small mass is the spheres.
Obviously a cylinder in freefall does not lend itself to a repeating cyclic function, but a horizontal wheel attached to a vertically mounted wheel does. I have set these two wheels up and they function just like the cylinder and spheres. These two wheels are pictured on page 3 and 4 of the same “energy producing experiments� thread.
Sorry about the last pictures being huge; but they still show real experiments. Doing experiments gives you a clear picture of the concept and experimenting gives you ideas on how to improve it.
re: energy producing experiments
Our posts overlapped; so I want to make something clear. The “you are correct� refers to keeping the math simple.
I think the raising masses concept is small enough to power a house. And I think velocity (1/2mv²) is absolutely enough to prove overunity.
I think the raising masses concept is small enough to power a house. And I think velocity (1/2mv²) is absolutely enough to prove overunity.
re: energy producing experiments
That last picture is the underneath side of the horizontal wheel, showing the seat for the puck mass.
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Your vertical flywheel setup is quite clever. I've realized that if you want to increase the COP it's best to keep the mass of the radius as small as possible. But doing this on a standard setup is not easy, for instance having 50kg mass on a radius of 0.1m is near impossible unless plutonium is used.
But your vertical flywheel setup solves this problem. You can have an as big as you want flywheel hooked to a small radius through a pulley system. This will act as if a lot of mass is concentrated on a small radius, and thus will require a relative small tether length to bring it to a stop. For instance if the flywheel pulley was hooked at 2cm from your small white plastic setup. You could have a 25kg flywheel which could be stopped by two 0.25 kg masses at final radius of 1m giving an energy increase by a factor of 50 or 5000%.
But your vertical flywheel setup solves this problem. You can have an as big as you want flywheel hooked to a small radius through a pulley system. This will act as if a lot of mass is concentrated on a small radius, and thus will require a relative small tether length to bring it to a stop. For instance if the flywheel pulley was hooked at 2cm from your small white plastic setup. You could have a 25kg flywheel which could be stopped by two 0.25 kg masses at final radius of 1m giving an energy increase by a factor of 50 or 5000%.
re: energy producing experiments
Thank you: and the vertical wheel can be used as an Atwood's machine.