I was a believer in Pequades momentum gain theory and had the idea which I have attached below. I actually got a fair way through construction then decided that i no longer think you can gain momentum with a heavy balanced wheel and a small overbalance weight.
What do you think? Should i complete the construction or am i wasting my time?
The big light blue wheel is a heavy balanced flywheel, say 20kg. The swing arm assembly shares the same axle as the flywheel but turns independently and contains the small overbalancing weight (the black circle say 1kg).
Sorry the attachment sequence should be read bottom to top 1-4
Idea based on Pequades momentum theory
Moderator: scott
Idea based on Pequades momentum theory
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Clay - you say you "no longer think you can gain momentum with a heavy balanced wheel and a small overbalance weight."
I don't understand that statement. As long as the small weight is connected to the flywheel as it falls, it's weight (the force of gravity) will act on the flywheel to accelerate it. You can see this with any wheel if you attach a small overbalance weight - the wheel will slowly accelerate. As long as you have mass and velocity, you have momentum. So there is no way you could possibly NOT gain momentum. Maybe what you doubt is whether you gain enough momentum? If that is your question - then I would say two things:
1 - how much momentum do you need? and
2 - do you realise that the amount of momentum you can generate from a fixed fall of a fixed mass with this type of setup is a variable? By varying the ratio of the flywheel mass to the driver mass, you can take as long as you need to create as much momentum as you need. (On paper anyway - the practical realities of friction etc make it impractical to push it too far - but the point is, if you calculate how much momentum you would need to swing the same sized pendulum back up to TDC, you could design your system to create 2 or 3 times that amount of momentum.
The big catch is that the Momentum thus created needs to be transformed from heavy/slow momentum to light/fast. If you overlook this transformation, you can't achieve anything more than an ordinary pendulum (which is the simplest example of an overbalanced flywheel - a one-spoke flywheel).
The comforting fact is that Momentum is considered to be a conserved quantity. So in theory, the necessary transformation is possible.
The resident smart alecs in this forum will attempt to tell you that Momentum cannot return a pendulum - only energy can return a pendulum. That's just being stupid. If you think about a pendlum falling from TDC to BDC, by the time it hits BDC it has reach a certain velocity. At that point, it is a Mass with Velocity. Therefore you can calculate both Momentum and Energy, which are merely formulas playing with the Mass and Velocity numbers. It is just semantics as to whether you say it is the Momentum of the pendulum that causes it to rise back up (nearly) to TDC, or whether it is the Energy that causes it to rise.
So from simple textbook physics, you can calculate how much momentum is required to swing a pendulum back up again. I recommend you do this for yourself, because it's more satisfying seeing the numbers for yourself.
Then calculate what ratio of mass you need to create sufficient Momentum. This is important. For example, there are plenty of cowboys who can make a flywheel experiment that "proves" this idea doesn't work. But because the amount of Momentum generated is a variable, you have do ensure that you have designed a system that makes more than what you need, otherwise you have proved nothing. If you don't understand the maths, you can't expect to make this work.
Skeptics also like to point out that there is a difference between linear momentum and angular momentum. This is why Pequaide likes to refer to an Atwoods system, which keeps everything as linear momentum. This is to satisfy the academics who like to confuse everything. But there is no reason why a flywheel can't work, or why we can't approximate the effect using linear momentum maths. If a flywheel disintegrates, the pieces fly off on a linear tangent: the point being that the distinction between linear and angular momentum is not as black & white as some would suggest. For practical purposes, if you have a moving mass, it's got some punch behind it - whether you constrain it to a circular path or not.
I can't quite follow your sequence. I don't see how the driver mass can ride the rim of the flywheel as it falls down? And once the driver mass has reached the bottom, and the flywheel is slowly turning as fast as it is able to, how do you intend to accelerate the driver mass back up again? The idea of using a tether to jerk it back up again is a popular one - and very probably what Bessler used. But how to best do this, i'm not sure. I would be happy just to see a one-shot action that demonstrated a fast rise that exceeds the fall height. I'm thinking a leveraged percussive hit might be required - similar to Bessler's toypage pictures.
Just my one-eyed opinion of course.
I don't understand that statement. As long as the small weight is connected to the flywheel as it falls, it's weight (the force of gravity) will act on the flywheel to accelerate it. You can see this with any wheel if you attach a small overbalance weight - the wheel will slowly accelerate. As long as you have mass and velocity, you have momentum. So there is no way you could possibly NOT gain momentum. Maybe what you doubt is whether you gain enough momentum? If that is your question - then I would say two things:
1 - how much momentum do you need? and
2 - do you realise that the amount of momentum you can generate from a fixed fall of a fixed mass with this type of setup is a variable? By varying the ratio of the flywheel mass to the driver mass, you can take as long as you need to create as much momentum as you need. (On paper anyway - the practical realities of friction etc make it impractical to push it too far - but the point is, if you calculate how much momentum you would need to swing the same sized pendulum back up to TDC, you could design your system to create 2 or 3 times that amount of momentum.
The big catch is that the Momentum thus created needs to be transformed from heavy/slow momentum to light/fast. If you overlook this transformation, you can't achieve anything more than an ordinary pendulum (which is the simplest example of an overbalanced flywheel - a one-spoke flywheel).
The comforting fact is that Momentum is considered to be a conserved quantity. So in theory, the necessary transformation is possible.
The resident smart alecs in this forum will attempt to tell you that Momentum cannot return a pendulum - only energy can return a pendulum. That's just being stupid. If you think about a pendlum falling from TDC to BDC, by the time it hits BDC it has reach a certain velocity. At that point, it is a Mass with Velocity. Therefore you can calculate both Momentum and Energy, which are merely formulas playing with the Mass and Velocity numbers. It is just semantics as to whether you say it is the Momentum of the pendulum that causes it to rise back up (nearly) to TDC, or whether it is the Energy that causes it to rise.
So from simple textbook physics, you can calculate how much momentum is required to swing a pendulum back up again. I recommend you do this for yourself, because it's more satisfying seeing the numbers for yourself.
Then calculate what ratio of mass you need to create sufficient Momentum. This is important. For example, there are plenty of cowboys who can make a flywheel experiment that "proves" this idea doesn't work. But because the amount of Momentum generated is a variable, you have do ensure that you have designed a system that makes more than what you need, otherwise you have proved nothing. If you don't understand the maths, you can't expect to make this work.
Skeptics also like to point out that there is a difference between linear momentum and angular momentum. This is why Pequaide likes to refer to an Atwoods system, which keeps everything as linear momentum. This is to satisfy the academics who like to confuse everything. But there is no reason why a flywheel can't work, or why we can't approximate the effect using linear momentum maths. If a flywheel disintegrates, the pieces fly off on a linear tangent: the point being that the distinction between linear and angular momentum is not as black & white as some would suggest. For practical purposes, if you have a moving mass, it's got some punch behind it - whether you constrain it to a circular path or not.
I can't quite follow your sequence. I don't see how the driver mass can ride the rim of the flywheel as it falls down? And once the driver mass has reached the bottom, and the flywheel is slowly turning as fast as it is able to, how do you intend to accelerate the driver mass back up again? The idea of using a tether to jerk it back up again is a popular one - and very probably what Bessler used. But how to best do this, i'm not sure. I would be happy just to see a one-shot action that demonstrated a fast rise that exceeds the fall height. I'm thinking a leveraged percussive hit might be required - similar to Bessler's toypage pictures.
Just my one-eyed opinion of course.
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nicbordeaux
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re: Idea based on Pequades momentum theory
LOL Greendoor, you calling me a cowboy :=)there are plenty of cowboys who can make a flywheel experiment that "proves" this idea doesn't work
If you think you have an overunity device, think again, there is no such thing. You might just possibly have an unexpectedly efficient device. In which case you will be abducted by MIB and threatened by aliens.
re: Idea based on Pequades momentum theory
The swing arm/small weight assembly is latched to the flywheel from the tdc to bdc. When the arms swing out they take the momentum of the flywheel and stop it like the cylinder and spheres.
The swing/small weight assembly unlatches from the flywheel and rotates to the top again.
It's similar to Broli's design here http://www.youtube.com/watch?v=lnnM0rCN31E[/url]
The swing/small weight assembly unlatches from the flywheel and rotates to the top again.
It's similar to Broli's design here http://www.youtube.com/watch?v=lnnM0rCN31E[/url]
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nicbordeaux
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re: Idea based on Pequades momentum theory
Clay, nice design, followed it 4 thru 1, seems good. except when i read
" The swing arm assembly shares the same axle as the flywheel but turns independently and contains the small overbalancing weight (the black circle say 1kg). " Not sure I get that, if the big OB weight and the tethers are not interacting, how does it work ?
Anyway, if that is a sim, before building (because it will do just what you say, eg arms deploy and stop big wheel at certain mass ratios), how about we say it's rotating CW, and that the deployment arms are attached to big wheel with a tension spring. Action asks for counteraction, but by using a spring tether, the brutal deployment will overcome spring easy. If the springs are attached to wheel(disk) say 10 cm (random figure) behind deployed arm, the flung mass should drag on the wheel in the CW direction ? Then as all this disturbance subsides, the spring tension is enough to reset your arms.
I definitely wouldn't throw any money at it yet, but if it is a sim, I'd be curious to see what this spring tether (actually, length of string + spring seems best scenario).
Just some thoughts...
EDIT: didn't see your post Clay which explains the axle stuff. Get it now. Still of the same opinion.
" The swing arm assembly shares the same axle as the flywheel but turns independently and contains the small overbalancing weight (the black circle say 1kg). " Not sure I get that, if the big OB weight and the tethers are not interacting, how does it work ?
Anyway, if that is a sim, before building (because it will do just what you say, eg arms deploy and stop big wheel at certain mass ratios), how about we say it's rotating CW, and that the deployment arms are attached to big wheel with a tension spring. Action asks for counteraction, but by using a spring tether, the brutal deployment will overcome spring easy. If the springs are attached to wheel(disk) say 10 cm (random figure) behind deployed arm, the flung mass should drag on the wheel in the CW direction ? Then as all this disturbance subsides, the spring tension is enough to reset your arms.
I definitely wouldn't throw any money at it yet, but if it is a sim, I'd be curious to see what this spring tether (actually, length of string + spring seems best scenario).
Just some thoughts...
EDIT: didn't see your post Clay which explains the axle stuff. Get it now. Still of the same opinion.
If you think you have an overunity device, think again, there is no such thing. You might just possibly have an unexpectedly efficient device. In which case you will be abducted by MIB and threatened by aliens.
re: Idea based on Pequades momentum theory
Yeh sorry, the attachments need to be read bottom to top. I had considered the use of a small spring more for resetting the swing arms than anything else.
For giggles, heres my half made model without the weights attached and without the swing arms.
For giggles, heres my half made model without the weights attached and without the swing arms.
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re: Idea based on Pequades momentum theory
Hi clay .. I'd follow it thru & try & complete the build, for a couple of reasons.
1. you already started & it doesn't look to expensive or time consuming.
2. as greendoor has pointed out there are some most fundamental physics principles at stake here that proved one way or another would deflate a lot of people & give hope to others, worthy of experimentation I'd say if you believe in the possibility of momentum as capacity to do work rather than energy calculations - just one experiment could prove that & shatter the current paradigm.
A word of caution - as I see it the greatest potential show stopper is the transformation process i.e. taking that huge momentum [no argument that it has large momentum] & harnessing it to do real work - this involves a physical interaction between mechanical parts or as greendoor says large slow to transform into light fast - experiments currently show that something is lost [won't say energy] in that interaction process so that you never get out more than you put in - that's why energy is the currency for work done rather than momentum math - just one successful experiment that returns greater potential than started with would prove the case conclusively.
I applaud you for getting your hands dirty & trying to validate certain claims & theories with a complete experiment - I predict that harnessing that momentum to create greater potential will be a harder nut to crack than most imagine because of a misunderstanding of what momentum is & its usefulness - good luck !
P.S. momentum is a vector quantity - rotating objects have angular momentum i.e. inertia from mass in motion which requires a force to stop it - that same rotating object also has translational & rotational kinetic energy [Ke is scalar] requiring a force to stop it [the same force x time].
1. you already started & it doesn't look to expensive or time consuming.
2. as greendoor has pointed out there are some most fundamental physics principles at stake here that proved one way or another would deflate a lot of people & give hope to others, worthy of experimentation I'd say if you believe in the possibility of momentum as capacity to do work rather than energy calculations - just one experiment could prove that & shatter the current paradigm.
A word of caution - as I see it the greatest potential show stopper is the transformation process i.e. taking that huge momentum [no argument that it has large momentum] & harnessing it to do real work - this involves a physical interaction between mechanical parts or as greendoor says large slow to transform into light fast - experiments currently show that something is lost [won't say energy] in that interaction process so that you never get out more than you put in - that's why energy is the currency for work done rather than momentum math - just one successful experiment that returns greater potential than started with would prove the case conclusively.
I applaud you for getting your hands dirty & trying to validate certain claims & theories with a complete experiment - I predict that harnessing that momentum to create greater potential will be a harder nut to crack than most imagine because of a misunderstanding of what momentum is & its usefulness - good luck !
P.S. momentum is a vector quantity - rotating objects have angular momentum i.e. inertia from mass in motion which requires a force to stop it - that same rotating object also has translational & rotational kinetic energy [Ke is scalar] requiring a force to stop it [the same force x time].
re: Idea based on Pequades momentum theory
Fletcher, having thought about it a bit more I think I will go ahead and complete it. As you say, theres not that much more to be done on it and at least I wont be left wondering.
As I posted in another topic my belief in Pequades "Principle of excess momentum" :) was shattered when i read from this website http://www.lightandmatter.com/html_book ... /ch04.html that momentum .. "cancels with momentum in the opposite direction." and that "A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy."
As I posted in another topic my belief in Pequades "Principle of excess momentum" :) was shattered when i read from this website http://www.lightandmatter.com/html_book ... /ch04.html that momentum .. "cancels with momentum in the opposite direction." and that "A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy."
re: Idea based on Pequades momentum theory
Fletcher: energy is force times distance, not force times time.
Clay: good work. You are in close proximity to the truth; but the swinging arms don’t need to reset, they need to ‘release’ the small weights that have stopped the motion.
Let’s say that the rotational inertia of your plywood wheel is 1,440 grams, and all the motion is given to the two small weights on the end of your arms that have a total mass of 72 grams. If the 1,440 grams were first moving 1 m/sec then the 72 grams would then be moving 20 m/sec (according to the Law of Conservation of Momentum). If you directed the motion of the end masses ‘up’ they would rise 20.38 meters. After it has risen place; the 72 grams on the end of a string that is wrapped around a 1,440 gram rim mass wheel. The 72 gram and the rim of the wheel will accelerate at .4671 m/sec/sec (1/21 * 9.81). After the 72 grams has dropped the 20.38 m the rim of the wheel and the 72 grams will be moving 4.36 m/sec. This is greater than the starting velocity of 1 m/sec. And it is greater momentum 4.36m/sec * 1512 g to 1440 g at 1 m/sec
The 72 grams in free fall for a distance of 20.38 meters would give you 14.39 joules of energy (72g/1000g * 9.81 newtons/kg * 20.38 m). Joules is newton meters
The 72 grams in reverse free fall (rise) for a distance of 20.38 meters would require 14.39 joules of energy (72g/1000g * 9.81 newtons * 20.38 m).
72 grams wrapped around a 1440 gram rim and dropped 20.38 meters would also give you (1/2 * 1.512 kg * 4.363 m/sec * 4.363 m/sec) 14.39 joules.
The energy produced is the same but the momentum produced is not, for the time over which the same force acts is much greater. 9.34 sec / 2.038 sec is equal to 4.363 m/sec * 1512 g / 1 m/sec * 1440 g
So who is correct: Newton or Leibniz?
I suggest you make a very shallow cut in the center of your plywood wheel along the circumference. Place a pin sticking out from the circumference of the wheel. Place the looped end of a string over the pin. Wrap the string completely around the circumference, in the grove you made in the plywood wheel, and on the other end of the string connect a 36 gram mass. This would allow the mass to almost touch the same pin but from the opposite side. Spin the wheel and release the mass so that the 36 g mass unwraps the string from the circumference of the wheel. Oh; move the mirror, the released mass will come off the pin and destroy it, or put holes in the drywall, etcetera. Maybe you should move the machine outside. You will need a diameter times pi distance to the ground.
Don’t quit you are to close.
A mass in linear motion can enter a circle and then leave the circular motion without losing or gaining linear motion. A thousand objects in linear motion can enter and be caught in the same circle; they can then be released from that same circle and continue traveling in their original direction. They will not have lost any linear Newtonian momentum. What in the world makes you think that momentum cancels in a circle? Don’t follow; think.
Clay: good work. You are in close proximity to the truth; but the swinging arms don’t need to reset, they need to ‘release’ the small weights that have stopped the motion.
Let’s say that the rotational inertia of your plywood wheel is 1,440 grams, and all the motion is given to the two small weights on the end of your arms that have a total mass of 72 grams. If the 1,440 grams were first moving 1 m/sec then the 72 grams would then be moving 20 m/sec (according to the Law of Conservation of Momentum). If you directed the motion of the end masses ‘up’ they would rise 20.38 meters. After it has risen place; the 72 grams on the end of a string that is wrapped around a 1,440 gram rim mass wheel. The 72 gram and the rim of the wheel will accelerate at .4671 m/sec/sec (1/21 * 9.81). After the 72 grams has dropped the 20.38 m the rim of the wheel and the 72 grams will be moving 4.36 m/sec. This is greater than the starting velocity of 1 m/sec. And it is greater momentum 4.36m/sec * 1512 g to 1440 g at 1 m/sec
The 72 grams in free fall for a distance of 20.38 meters would give you 14.39 joules of energy (72g/1000g * 9.81 newtons/kg * 20.38 m). Joules is newton meters
The 72 grams in reverse free fall (rise) for a distance of 20.38 meters would require 14.39 joules of energy (72g/1000g * 9.81 newtons * 20.38 m).
72 grams wrapped around a 1440 gram rim and dropped 20.38 meters would also give you (1/2 * 1.512 kg * 4.363 m/sec * 4.363 m/sec) 14.39 joules.
The energy produced is the same but the momentum produced is not, for the time over which the same force acts is much greater. 9.34 sec / 2.038 sec is equal to 4.363 m/sec * 1512 g / 1 m/sec * 1440 g
So who is correct: Newton or Leibniz?
I suggest you make a very shallow cut in the center of your plywood wheel along the circumference. Place a pin sticking out from the circumference of the wheel. Place the looped end of a string over the pin. Wrap the string completely around the circumference, in the grove you made in the plywood wheel, and on the other end of the string connect a 36 gram mass. This would allow the mass to almost touch the same pin but from the opposite side. Spin the wheel and release the mass so that the 36 g mass unwraps the string from the circumference of the wheel. Oh; move the mirror, the released mass will come off the pin and destroy it, or put holes in the drywall, etcetera. Maybe you should move the machine outside. You will need a diameter times pi distance to the ground.
Don’t quit you are to close.
A mass in linear motion can enter a circle and then leave the circular motion without losing or gaining linear motion. A thousand objects in linear motion can enter and be caught in the same circle; they can then be released from that same circle and continue traveling in their original direction. They will not have lost any linear Newtonian momentum. What in the world makes you think that momentum cancels in a circle? Don’t follow; think.
re: Idea based on Pequades momentum theory
Pequade, in my mind this is just a scaled up vertical version of the cylinder & spheres with rigid swing arms instead of string attaching the spheres. Man I wish I knew how to create GIF animations.
When i say the swing arms need to be reset they need to be reset back to the "inwards" position at the end of the whole sequence so the sequence can start again.
Incidentally, I work on it outside, it was inside for the photo cause I dont want moisture getting into the wood. Thankfully it's easy to relocate.
When i say the swing arms need to be reset they need to be reset back to the "inwards" position at the end of the whole sequence so the sequence can start again.
Incidentally, I work on it outside, it was inside for the photo cause I dont want moisture getting into the wood. Thankfully it's easy to relocate.
Re: re: Idea based on Pequades momentum theory
First time I ever heard of you. You seemed to be a mystery on this discussion form. If you want to make your idea work and have strong beliefs, then go out and gather investors to make your dream come true! Otherwise, continue talking. You are the so called person to discover this idea? Congratulations. I am just flared up with people disrespecting Nick whom found the same discovery after you.
pequaide wrote:Fletcher: energy is force times distance, not force times time.
Clay: good work. You are in close proximity to the truth; but the swinging arms don’t need to reset, they need to ‘release’ the small weights that have stopped the motion.
Let’s say that the rotational inertia of your plywood wheel is 1,440 grams, and all the motion is given to the two small weights on the end of your arms that have a total mass of 72 grams. If the 1,440 grams were first moving 1 m/sec then the 72 grams would then be moving 20 m/sec (according to the Law of Conservation of Momentum). If you directed the motion of the end masses ‘up’ they would rise 20.38 meters. After it has risen place; the 72 grams on the end of a string that is wrapped around a 1,440 gram rim mass wheel. The 72 gram and the rim of the wheel will accelerate at .4671 m/sec/sec (1/21 * 9.81). After the 72 grams has dropped the 20.38 m the rim of the wheel and the 72 grams will be moving 4.36 m/sec. This is greater than the starting velocity of 1 m/sec. And it is greater momentum 4.36m/sec * 1512 g to 1440 g at 1 m/sec
The 72 grams in free fall for a distance of 20.38 meters would give you 14.39 joules of energy (72g/1000g * 9.81 newtons/kg * 20.38 m). Joules is newton meters
The 72 grams in reverse free fall (rise) for a distance of 20.38 meters would require 14.39 joules of energy (72g/1000g * 9.81 newtons * 20.38 m).
72 grams wrapped around a 1440 gram rim and dropped 20.38 meters would also give you (1/2 * 1.512 kg * 4.363 m/sec * 4.363 m/sec) 14.39 joules.
The energy produced is the same but the momentum produced is not, for the time over which the same force acts is much greater. 9.34 sec / 2.038 sec is equal to 4.363 m/sec * 1512 g / 1 m/sec * 1440 g
So who is correct: Newton or Leibniz?
I suggest you make a very shallow cut in the center of your plywood wheel along the circumference. Place a pin sticking out from the circumference of the wheel. Place the looped end of a string over the pin. Wrap the string completely around the circumference, in the grove you made in the plywood wheel, and on the other end of the string connect a 36 gram mass. This would allow the mass to almost touch the same pin but from the opposite side. Spin the wheel and release the mass so that the 36 g mass unwraps the string from the circumference of the wheel. Oh; move the mirror, the released mass will come off the pin and destroy it, or put holes in the drywall, etcetera. Maybe you should move the machine outside. You will need a diameter times pi distance to the ground.
Don’t quit you are to close.
A mass in linear motion can enter a circle and then leave the circular motion without losing or gaining linear motion. A thousand objects in linear motion can enter and be caught in the same circle; they can then be released from that same circle and continue traveling in their original direction. They will not have lost any linear Newtonian momentum. What in the world makes you think that momentum cancels in a circle? Don’t follow; think.
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justsomeone
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re: Idea based on Pequades momentum theory
276 Posts is not much of a mystery. Search Pequaide's posts for some interesting reading. His posts as well as the replies.
re: Idea based on Pequades momentum theory
There you go clay - pequiade has stepped into the discussion to help with your build - now you have the option of attempting to finish your original design or building another to pequaides specifications - the problems will still be the same but at least you can work together towards solutions & eventual conclusions.
IIRC, pequiade was building his own vertically arranged flywheel & bola experiment ? - incorporate a solid deployment arm arrangement & you should be able to toss a mass vertically - if, as pequaide suspects, there is excess energy then the mass should achieve higher Pe & be able to roll back to rejoin the mech at the start conditions.
N.B. don't worry about the arms not coming back to a closer radius at this early stage - the mathematicians should be able to work out the new Pe of the mass at its highest position & work out an approximation of the energy required to reset the arms in close etc, to see if indeed there is a surplus as the theory suggests.
It would of course help immensely to have a picture to build to but I guess you know that ?!
IIRC, pequiade was building his own vertically arranged flywheel & bola experiment ? - incorporate a solid deployment arm arrangement & you should be able to toss a mass vertically - if, as pequaide suspects, there is excess energy then the mass should achieve higher Pe & be able to roll back to rejoin the mech at the start conditions.
N.B. don't worry about the arms not coming back to a closer radius at this early stage - the mathematicians should be able to work out the new Pe of the mass at its highest position & work out an approximation of the energy required to reset the arms in close etc, to see if indeed there is a surplus as the theory suggests.
It would of course help immensely to have a picture to build to but I guess you know that ?!
re: Idea based on Pequades momentum theory
Yes, it should be easy enough to route a slot around the perimeter of the flywheel then run a string around the circumference as pequade suggested. I can load up the wheel with 20kg weights so it is heavy balanced, then compare the "fling" results with those without the weight attached. ie heavy balanced v's light balanced.
I predict the result will be the same, but i suppose we will see.
I predict the result will be the same, but i suppose we will see.
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nicbordeaux
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Better hurry guys, I will soon move my lazy ass and demonstrate that wheel with boucy ball gizmo flinging a tethered weight 2+ x. Bike wheel driven by mass identical to flung. Or if heavier driver, distance flung directly proportionate. Still not sure what it has to do with peq/greendoor, but it's a cinch, just a question of spooling/directing/release.