I missed the point? look again at the second video.
I was just trying to explain to you the conversion of energy in your or any rotating system is reduced if you don't let it fall as far.
The horizontal position of a mass is not the be all and end all. In fact it means nothing. IMO, until you realise this, you realise nothing about gravity wheels.
Basic Pm Idea
Moderator: scott
re: Basic Pm Idea
Tarsier,
Do you mind if I point out what your missing ?
>> I was just trying to explain to you the conversion of energy in your or any rotating system is reduced if you don't let it fall as far. <<
This is wrong. As the over balance passes past the 45 degree from the level of the axle to bottom center, it will start creating drag. And to make matters worse, it would need to be lifted robbing the system of the energy it developed. Why all previous attempts have failed except for Bessler's.
And in the thread "Conservation of Angular Momentum", it seems that they are confusing the work needed to accelerate a mass with the work a mass can do at a constant velocity.
From Wikipedia;
The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,
F s = 1/2 mv^2
Then there is
The kinetic energy of an object is related to its momentum by the equation:
Ek = p^2 / 2m
Myself, I am working with the first expression. When gravity accelerates a mass according to the acceleration of gravity a, then it is vt. and with a cross bar design or wheel, gravity can accelerate a mass to about .65 m/s in about .2s. This means that from a stand still to the velocity of .65 m/s means that it would drop about 32.5 cm's.
Also, with the attached video, you will see that when a mass of 32 oz.s is accelerated by 4 oz.s, that something odd seems to happen.
That with 4 being 1/9th of the total mass, then acceleration could be expected to be at about 11% the acceleration of gravity. But it seems to accelerate at twice that rate.
I assume this is something that you are aware of. And as you already know is that since only 1/2 of the weight being accelerated is moving in opposition to gravity, then it is this opposition which changes the rate of acceleration.
As such, it does seem that the Conservation of Momentum works in the inverse when converting gravity into mechanical energy in a wheel type concept.
This would be because the weights accelerate at a faster rate being further away from center. and this allows more energy from gravity to be converted into kinetic energy and/or momentum.
So why don't you explain to me where I am wrong instead of saying I am wrong ?
BTW, in the video, there is a reference marker using a tape measure so that a drop of about 30 cm's is observable and the time that it takes for a weight to rotate past this point allows it's velocity to be calculated.
By the way, increasing the amount of over balance does not increase the velocity at which a gravity wheel would work. Something people have mistakenly thought. In what the video shows, I did move the weights in 1/2 of the distance to center and barely got 90 degrees of rotation.
Simply put, more over balance is not necessarily better.
http://youtu.be/EXwowrLieyI
@All, some calculations which show that kinetic energy from a body or mass moving at a constant velocity is 1/2 of that of an equal mass accelerating to the same velocity.
The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,
Kinetic Energy (acceleration)
Ek = 1/2mv^2
1/2 2kg 3m/s^2 = 9
The kinetic energy of an object is related to its momentum by the equation:
Kinetic Energy (momentum)
p^2/2m
Mo^2/2m
Mo = 18/2*2 = 4.5
Long story short, it takes twice as much energy to accelerate something as it does to keep it moving. I guess this means that 1/4mv^2 might be the right expression for a moving body. It's just restating p^2/2m.
http://en.wikipedia.org/wiki/Kinetic_energy
edited for content
Do you mind if I point out what your missing ?
>> I was just trying to explain to you the conversion of energy in your or any rotating system is reduced if you don't let it fall as far. <<
This is wrong. As the over balance passes past the 45 degree from the level of the axle to bottom center, it will start creating drag. And to make matters worse, it would need to be lifted robbing the system of the energy it developed. Why all previous attempts have failed except for Bessler's.
And in the thread "Conservation of Angular Momentum", it seems that they are confusing the work needed to accelerate a mass with the work a mass can do at a constant velocity.
From Wikipedia;
The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,
F s = 1/2 mv^2
Then there is
The kinetic energy of an object is related to its momentum by the equation:
Ek = p^2 / 2m
Myself, I am working with the first expression. When gravity accelerates a mass according to the acceleration of gravity a, then it is vt. and with a cross bar design or wheel, gravity can accelerate a mass to about .65 m/s in about .2s. This means that from a stand still to the velocity of .65 m/s means that it would drop about 32.5 cm's.
Also, with the attached video, you will see that when a mass of 32 oz.s is accelerated by 4 oz.s, that something odd seems to happen.
That with 4 being 1/9th of the total mass, then acceleration could be expected to be at about 11% the acceleration of gravity. But it seems to accelerate at twice that rate.
I assume this is something that you are aware of. And as you already know is that since only 1/2 of the weight being accelerated is moving in opposition to gravity, then it is this opposition which changes the rate of acceleration.
As such, it does seem that the Conservation of Momentum works in the inverse when converting gravity into mechanical energy in a wheel type concept.
This would be because the weights accelerate at a faster rate being further away from center. and this allows more energy from gravity to be converted into kinetic energy and/or momentum.
So why don't you explain to me where I am wrong instead of saying I am wrong ?
BTW, in the video, there is a reference marker using a tape measure so that a drop of about 30 cm's is observable and the time that it takes for a weight to rotate past this point allows it's velocity to be calculated.
By the way, increasing the amount of over balance does not increase the velocity at which a gravity wheel would work. Something people have mistakenly thought. In what the video shows, I did move the weights in 1/2 of the distance to center and barely got 90 degrees of rotation.
Simply put, more over balance is not necessarily better.
http://youtu.be/EXwowrLieyI
@All, some calculations which show that kinetic energy from a body or mass moving at a constant velocity is 1/2 of that of an equal mass accelerating to the same velocity.
The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,
Kinetic Energy (acceleration)
Ek = 1/2mv^2
1/2 2kg 3m/s^2 = 9
The kinetic energy of an object is related to its momentum by the equation:
Kinetic Energy (momentum)
p^2/2m
Mo^2/2m
Mo = 18/2*2 = 4.5
Long story short, it takes twice as much energy to accelerate something as it does to keep it moving. I guess this means that 1/4mv^2 might be the right expression for a moving body. It's just restating p^2/2m.
http://en.wikipedia.org/wiki/Kinetic_energy
edited for content
re: Basic Pm Idea
Something doesn't seems right here ?Smith66 wrote:
@All, some calculations which show that kinetic energy from a body or mass moving at a constant velocity is 1/2 of that of an equal mass accelerating to the same velocity.
The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,
Kinetic Energy (acceleration)
Ek = 1/2mv^2
1/2 2kg 3m/s^2 = 9
The kinetic energy of an object is related to its momentum by the equation:
Kinetic Energy (momentum)
p^2/2m
Mo^2/2m
Mo = 18/2*2 = 4.5
Long story short, it takes twice as much energy to accelerate something as it does to keep it moving. I guess this means that 1/4mv^2 might be the right expression for a moving body. It's just restating p^2/2m.
re: Basic Pm Idea
Fletcher,
What would be confusing you is one aspect of converting gravity into mechanical energy.
The example I gave is the video I posted. The cross bar accelerates to about .65 m/s moving/dropping about 32.5 cm's. The next 65 cm's would be covered in one second as the initial 32.5 cm's were.
This is an example of why acceleration requires more energy.
With the >> Mo = 18/2*2 = 4.5 <<, this is based on a 2 kg. weight moving at 3 m/s.
If mv^2 = momentum, then 2*3^2 = 2*9 = 18 units of force.
With kinetic acceleration, I disagree with p^2/2m. If it were so, then 18^2/2*2kg = 324/4 = 81 units of force. That is 18 times the energy of a mass with a constant velocity.
With the numbers I show, it's twice the energy and with the example of gravity's acceleration, a mass may be able to accelerate to velocity in 1/2 the distance.
This latter part is because of gravity's constant rate of acceleration which makes me believe that accelerating a mass to velocity in 1/2 the distance requires twice the energy, not the 18 times that the current formula for kinetic acceleration calculates.
And this would seem to fit w = m*d while considering t (time).
There is one reason why p^2 might be considered and that is if a moving body has the same rate of spin as it does linear velocity. One of the differences between physics and mechanical engineering.
What would be confusing you is one aspect of converting gravity into mechanical energy.
The example I gave is the video I posted. The cross bar accelerates to about .65 m/s moving/dropping about 32.5 cm's. The next 65 cm's would be covered in one second as the initial 32.5 cm's were.
This is an example of why acceleration requires more energy.
With the >> Mo = 18/2*2 = 4.5 <<, this is based on a 2 kg. weight moving at 3 m/s.
If mv^2 = momentum, then 2*3^2 = 2*9 = 18 units of force.
With kinetic acceleration, I disagree with p^2/2m. If it were so, then 18^2/2*2kg = 324/4 = 81 units of force. That is 18 times the energy of a mass with a constant velocity.
With the numbers I show, it's twice the energy and with the example of gravity's acceleration, a mass may be able to accelerate to velocity in 1/2 the distance.
This latter part is because of gravity's constant rate of acceleration which makes me believe that accelerating a mass to velocity in 1/2 the distance requires twice the energy, not the 18 times that the current formula for kinetic acceleration calculates.
And this would seem to fit w = m*d while considering t (time).
There is one reason why p^2 might be considered and that is if a moving body has the same rate of spin as it does linear velocity. One of the differences between physics and mechanical engineering.
re: Basic Pm Idea
p = mv , at least in my world.
re: Basic Pm Idea
Fletcher,
As to your >> p = mv , at least in my world. << , changing the subject from kinetic energy to momentum ? How positively clever of you.
Why not explain how everyone including yourself were using the WRONG calculations for kinetic energy ?
Still, for what I am working on, it's not p =mv but p = p1 + p2 = m1v1 + m2v2. This is because I am accelerating 2 weights simultaneously. :-)
And with gravity, it does seem to have a nice 2 to 1 ratio of velocity/distance in respect to time. Could hardly say that an accelerating mass has the same momentum as one moving at a constant speed. And as such, kinetic energy is relative to the momentum of a body in motion.
It's just that with things like nucleons, photons and electrons, they all seem to have linear and angular momentum to consider. But again, with what I am working on, I am only concerned with linear momentum because my weights will not have any spin. It doesn't matter if they are moving around an axle.
@All, I have been working on the next progression of my build. As I've mentioned before, it is a step by step process which means it will take time.
One thing that does look promising is that the efficiency in which acceleration is transferred agrees pretty much with the math. this means little energy/momentum is lost due to friction, resistance, etc.
Besides, if it works, it will be the FIRST openly demonstrated working perpetual device. Something to be said for that I think.
edited to add; Fletcher, according to scientists, leibniz is the father of modern physics. Any way, something for you to think about.
It is not entirely clear why Leibniz should have chosen mv2 as this quantity rather than Descartes' mv,
http://scienceworld.wolfram.com/biography/Leibniz.html
What you might be missing is that it may take 4 times the enrgy to accelerate a given mass in 1/2 the distance. If so, it's momentum would never be calculated p =mv while it is accelerating.
This could be the fundamental difference that Leibniz might have been aware of.
As to your >> p = mv , at least in my world. << , changing the subject from kinetic energy to momentum ? How positively clever of you.
Why not explain how everyone including yourself were using the WRONG calculations for kinetic energy ?
Still, for what I am working on, it's not p =mv but p = p1 + p2 = m1v1 + m2v2. This is because I am accelerating 2 weights simultaneously. :-)
And with gravity, it does seem to have a nice 2 to 1 ratio of velocity/distance in respect to time. Could hardly say that an accelerating mass has the same momentum as one moving at a constant speed. And as such, kinetic energy is relative to the momentum of a body in motion.
It's just that with things like nucleons, photons and electrons, they all seem to have linear and angular momentum to consider. But again, with what I am working on, I am only concerned with linear momentum because my weights will not have any spin. It doesn't matter if they are moving around an axle.
@All, I have been working on the next progression of my build. As I've mentioned before, it is a step by step process which means it will take time.
One thing that does look promising is that the efficiency in which acceleration is transferred agrees pretty much with the math. this means little energy/momentum is lost due to friction, resistance, etc.
Besides, if it works, it will be the FIRST openly demonstrated working perpetual device. Something to be said for that I think.
edited to add; Fletcher, according to scientists, leibniz is the father of modern physics. Any way, something for you to think about.
It is not entirely clear why Leibniz should have chosen mv2 as this quantity rather than Descartes' mv,
http://scienceworld.wolfram.com/biography/Leibniz.html
What you might be missing is that it may take 4 times the enrgy to accelerate a given mass in 1/2 the distance. If so, it's momentum would never be calculated p =mv while it is accelerating.
This could be the fundamental difference that Leibniz might have been aware of.
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justsomeone
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re: Basic Pm Idea
Be careful Smith66 while entering a battle of wits with Fletcher, you're unarmed!
. I can assure the reader that there is something special behind the stork's bills.
re: Basic Pm Idea
Tarsier,
When Leibniz came up with mv^2, he dropped a weight onto clay and measured the depth of the impact.
To his credit, he co-invented Integral Calculus along with Newton. p = mv is another way of stating Newton's f = m* (v*t) = ma.
When people consider mv2/r (inertia), this is based on momentum being mv^2. This is one of the reasons why I mentioned that some things in physics might not translate that well into mechanical engineering.
With the acceleration tests that I've done, it does seem to verify that a mass(es) can accelerate and then maintain velocity. During acceleration, a weight will accelerate (via gravity) to a velocity twice that of it's drop. This would most likely work for the first 4.9 meters. after that, acceleration would be 9.8 m/s/s.
With what I'm working on, I've found that adding all the weights and then calculating the over balance as a percentage of net force will give a percentage of gravity times 2 seems close.
And if what I'm doing works, then the same math could be used to understand how much over balance Bessler's wheel would have needed.
The link is to a video showing where I'm at build wise. It may take another week or 2 as these last details need to be right. Fortunately it seems everything is going pretty well :-)
@justsomeone, thanks for the heads up !
If you look at the other videos, you can see the changes I've made.
http://youtu.be/cmH9VWRaUC8
When Leibniz came up with mv^2, he dropped a weight onto clay and measured the depth of the impact.
To his credit, he co-invented Integral Calculus along with Newton. p = mv is another way of stating Newton's f = m* (v*t) = ma.
When people consider mv2/r (inertia), this is based on momentum being mv^2. This is one of the reasons why I mentioned that some things in physics might not translate that well into mechanical engineering.
With the acceleration tests that I've done, it does seem to verify that a mass(es) can accelerate and then maintain velocity. During acceleration, a weight will accelerate (via gravity) to a velocity twice that of it's drop. This would most likely work for the first 4.9 meters. after that, acceleration would be 9.8 m/s/s.
With what I'm working on, I've found that adding all the weights and then calculating the over balance as a percentage of net force will give a percentage of gravity times 2 seems close.
And if what I'm doing works, then the same math could be used to understand how much over balance Bessler's wheel would have needed.
The link is to a video showing where I'm at build wise. It may take another week or 2 as these last details need to be right. Fortunately it seems everything is going pretty well :-)
@justsomeone, thanks for the heads up !
If you look at the other videos, you can see the changes I've made.
http://youtu.be/cmH9VWRaUC8
I just thought of something that might be kind of funny.Tarsier79 wrote:Jim
KE = mv^2/2
P= mv
I believe P should = mv^2, but then you would have to alter the maths in the conversion from P to KE. Then the figures would hold true ( at least in some cases).
In Wikipedia, they say p^2/2m. or as you put it, mv^2/2m.
The m's should cancel out leaving v^2/m to simplify it.
Math can be fun to mess around with.
edited to add; if things work out, then Bessler's wheel might be able to show empirical evidence of the proper calculation for inertia. This would be because if a Bessler type wheel rotates at 20 or 30 rpm or more, then with 8 rotating weights, it'd give math junkies something to think about.
Re: re: Basic Pm Idea
Chris,triplock wrote:Hello there,
Right, lets talk turkey.
How is the 'pendulum string' lengthened again once it has wrapped around the semi-circle to retract the weighted bob. ?
Chris
I thought since this is what I'm working on now that I'd let you know what I'm going to try.
The red area beneath the half circle would be a catch or know in the line. If you take a close look at the video, I am using a 1/4 round that has a 1/8th inch gap for the line to fit in or follow. When the cross bar rotates so one weight is near bottom center, a fixed piece of wood would move the knot past the end of the 1/4 round so it could be pulled around again.
The line pulling up the weight will also use a knot so it can be held by a tab so the weight will stay in place until it rotates past top center.
@Tarsier,
The black line that goes from 3 o'clock to bottom center would be the approximate path of the weight. Since it would fall no lower than the opposing weight is lifted above the axle (important detail, prevents using energy from the cross bar), they would be in balance.
Until then, there should be some over balance generated. I don't think it will be much but I guess every little bit will help.
I hope this helps you guys to understand it better. Just wanted to get some of the build done so it will be easier to see what I'm trying.
Smith66
edited to change pics :-)
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re: Basic Pm Idea
@All,
I thought I'd post a couple of pictures so it'll be easier to see some of the details.
With the line that holds the weight, there is a way it can be made slack before it is loosed from the 1/4 round. this is something I'll explain later.
The first pic is a side view of the 1/4 round. The second pic is it's front view. I made it from gluing together 3 pieces of thin 1/8th inch plywood. this allows for a nice gap to help keep the line in place. I'll probably add a couple of fish eyes later.
And the 3rd pic is the rails for the weights to roll on. As you can see, there are 2 pieces with a round cut out for the weight to nest in. It is between these 2 pieces that I am going to place another piece that will act as a guide. What the guide will do is catch the line when the weight is retracted.
I think you'll like some of the solutions I have figured out to work around things that might rob energy from the system. Just because gravity is free and unlimited doesn't mean it can be wasted in a design like this. :-)
I thought I'd post a couple of pictures so it'll be easier to see some of the details.
With the line that holds the weight, there is a way it can be made slack before it is loosed from the 1/4 round. this is something I'll explain later.
The first pic is a side view of the 1/4 round. The second pic is it's front view. I made it from gluing together 3 pieces of thin 1/8th inch plywood. this allows for a nice gap to help keep the line in place. I'll probably add a couple of fish eyes later.
And the 3rd pic is the rails for the weights to roll on. As you can see, there are 2 pieces with a round cut out for the weight to nest in. It is between these 2 pieces that I am going to place another piece that will act as a guide. What the guide will do is catch the line when the weight is retracted.
I think you'll like some of the solutions I have figured out to work around things that might rob energy from the system. Just because gravity is free and unlimited doesn't mean it can be wasted in a design like this. :-)
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triplock
re: Basic Pm Idea
Smith
Right I'll put this to you....
Setting aside all clever designs, mechanisms and a bucket load of wishful thinking, what you are trying to do is let a weight drop x-distance and then for that weight to return back up to the exact starting point.
In simple terms that must be the case.
Image a rigidly strung pendulum. You invert it to 12 and release. Will it, even totally unhindered, swing down, around and back up again to it's start point ? Again, NO.
Anything added, or put in this falling pendulum's way, ANYTHING, will only make the situation worse. Your design, although interesting, will not work ever. Sorry :(
Chris
Right I'll put this to you....
Setting aside all clever designs, mechanisms and a bucket load of wishful thinking, what you are trying to do is let a weight drop x-distance and then for that weight to return back up to the exact starting point.
In simple terms that must be the case.
Image a rigidly strung pendulum. You invert it to 12 and release. Will it, even totally unhindered, swing down, around and back up again to it's start point ? Again, NO.
Anything added, or put in this falling pendulum's way, ANYTHING, will only make the situation worse. Your design, although interesting, will not work ever. Sorry :(
Chris
re: Basic Pm Idea
Chris,
>> Anything added, or put in this falling pendulum's way, ANYTHING, will only make the situation worse. Your design, although interesting, will not work ever. Sorry :( <<
This is why I am building it, to find out otherwise ;-)
edited to add; since this is the first complete prototype, I am only going to have one weight retract.
>> Anything added, or put in this falling pendulum's way, ANYTHING, will only make the situation worse. Your design, although interesting, will not work ever. Sorry :( <<
This is why I am building it, to find out otherwise ;-)
edited to add; since this is the first complete prototype, I am only going to have one weight retract.
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triplock
re: Basic Pm Idea
Smith
Ok mate , fair enough.
Ultimately though, no matter what the method of energy extraction or manipulation of a falling weight, whether that be through ropes, pulleys, levers, cams, actuator assist, inertia, CF, CP, Scissor Jacks, pantograph, springs, tracks, bellows etc, you cannot overcome that fundamental hurdle, which is to get a pendulum to reset to its starting point.
Once that is understood , or more importantly accepted, then , indeed, roses do grow in the garden .
Regards
Chris
Ok mate , fair enough.
Ultimately though, no matter what the method of energy extraction or manipulation of a falling weight, whether that be through ropes, pulleys, levers, cams, actuator assist, inertia, CF, CP, Scissor Jacks, pantograph, springs, tracks, bellows etc, you cannot overcome that fundamental hurdle, which is to get a pendulum to reset to its starting point.
Once that is understood , or more importantly accepted, then , indeed, roses do grow in the garden .
Regards
Chris