I am trying to wrap my head around the unbalanced wheel. I made a very fast sketch in paint to illustrate it in the extreme, where the one weight follows the circumference of the wheel as it descends, and follows the vertical axis on its way up.
Questions:
1. Does this path create the maximum off balance (highest power output)? If not what is the theoretical Ideal path for the weight to follow.
2.is it still in balance? The weight on the way down is has less downward force because it is following an incline except for a brief moment when it reaches the height of the axial while the full downward force is in effect on the upward travelling weight.
If you have more questions please add them and hopefully some of you guru's who have thought this through will be able to shed some light on this.
Thanks in advance everyone for your contributions :)
Reserved for future use (will consolidate information from discussion here)
The amount of power delivered to the rotational shaft center depends on how much weight and how far from center it is. The more the value, the more the output.
thaelin
In that case...
Imagine a lifting tower as sketched, and a very slightly sloping plain right to it.
Once way out, the weight transfers onto a lever coming from the base of the tower.
The tower could be central and the plain circular, directing each next weight to another lever, organised like a sunflower around it.
If indeed the long lever makes a difference, there should be sufficient energy stored, in a wind-up kind of engine (toy car engine for proto) to bring the weights up.
A very slow sytem, but it could have a high through-put.
I am sceptical though about this. A longer lever can deliver more focused pressure I would say, not do more work for a given weight and height overcome. This has been tried so often, there have to be better ways.
Thanks guys for the replies.
Lets take the example picture (...E1) it consists of a wheel with teeth (blue dots) for a chain (green) to run on (think bicycle chain). The two black circles are sprockets which at the top help lift the chain over the teeth and allow the chain to seat back on the wheel, while the bottom sprocket pops the chain off the wheel.
(For the sake of simplicity lets assume a frictionless system) Now as I analyze the system I find that the gravity pulling the chain down is balanced. Even though more links of chain are on the downward travelling side, the force is diluted by the 'slope' as the chain follows the circumference of the wheel. Thus we can expect the chain to effortlessly spin as the force needed to raise the chain is exactly equal to the force created by gravity pulling down on the chain (no surprises here).
The magic comes into play when we look at the balance of the wheel and see that it has more weight on the the right than the left. This should cause the wheel to twist clockwise (see ...E2).
The problem is I am fairly confident that this has been tried many times as it would be such a simple build... I can't begin to understand what makes it not work (or does it????). Can someone please explain why this has not worked for others.
I've also started to believe that IF there's a way to get a weight system to run, it's because how acceleration works. Or rather, decelleration. (distance travelled of the weight per video frame). A chain doesn't allow speed variances through it.
A weight rolling up a ramp, covers the first half of its potential height (from horizontal speed at 6:00) in a fraction of the time it would take the second (top) half, and faster than when following the wheel, and especially faster than a ramp shaped as the wheel.
If coming up to the vertical chain would allow the weight to carry it's momentum as om a ramp, gradually losing speed, it might work. You just need it to get more height than the downward weight loses height, so the latter has enough potential energy to spare to lift the former up and over the top.
What if the chain were near weightless in comparison, and running at the rim speed, but slowing down as the climbing weight does, with a minimum (a spring system) could take up that difference) before the speed would be catching up with the rim again.
The upwards weight just needs to get some height down to chase the downward one.
What helps: the downward one loses little height at the start of the downstroke, due to a cirdle starting out flat.
So the up weight gains height while the down weight has little to help the up weight with. In the second half of the half-cycle, this relationship flips. The down weight has been able to get some extra enertia into the wheels during the time the up weight was on it's own. Seems totally plausible to me. Just needs a simple set up. Abeling/Dusty have one, so we need another.
Hi Cloxxky !
I agree with the trajectory.
For me,the (semi) circular fall is a long time storage of a gravity fall into a rotational inertia(an "other kind " to balance energy).
The linear up trajectory,it's a simple jump,due to a short discharge of (stored) rotational energy.
I wish you success!
All the Bests! / Alex
Quote from: Flyboy on May 06, 2009, 02:26:30 AM
I am trying to wrap my head around the unbalanced wheel. I made a very fast sketch in paint to illustrate it in the extreme, where the one weight follows the circumference of the wheel as it descends, and follows the vertical axis on its way up.
Questions:
1. Does this path create the maximum off balance (highest power output)? If not what is the theoretical Ideal path for the weight to follow.
2.is it still in balance? The weight on the way down is has less downward force because it is following an incline except for a brief moment when it reaches the height of the axial while the full downward force is in effect on the upward travelling weight.
If you have more questions please add them and hopefully some of you guru's who have thought this through will be able to shed some light on this.
Thanks in advance everyone for your contributions :)
If the difference inn hight between the top via the half circle, and the top stright down to the bottom is greater (Well, which it isn't) you will gain energy. But in this case there is no difference in hight from the ground and to the top no matter what path the weight is following. When you start and finish at the same hight, there cant be any difference in energy. Therfor you cannot get energy out of a overbalanced wheel.
Sorry.
Vidar
This kind of idea has been around for a long time. It does not work.
Hans von Lieven
Quote from: Low-Q on May 11, 2009, 05:19:32 PM
If the difference inn hight between the top via the half circle, and the top stright down to the bottom is greater (Well, which it isn't) you will gain energy. But in this case there is no difference in hight from the ground and to the top no matter what path the weight is following. When you start and finish at the same hight, there cant be any difference in energy. Therfor you cannot get energy out of a overbalanced wheel.
Sorry.
Vidar
I must agree after thinking about it a bit I now can see intuitively that as I originally drew (or hansvonlieven's illustration above) is in balance as the system is broken down into two seperate systems, one being the wheel itself which is balanced, and two, the chain which is balanced as well... the fact that it is half on a wheel is just an illusion.
Now lets have a more in depth look into the classic unbalanced wheel, where we force the wheel to be unbalanced via differing torque. (see illustration attached). To clarify, each spoke has a weight on it, on the way down is at its maximum radius (max leverage) and on the way up the radius shortens (less leverage). Theoretically the balls take the same amount of energy to go up as it does to go down, so in our frictionless thought experiment the device will spin effortlessly. However the 'MAGIC' comes into play in this model when we compare the distance from the axis for each matched pair of weights... thus a spinning motion is created on the wheel.
So my questions now are:
1. Have I accurately summed up the essence of the unbalanced wheel theory?
2. Since I'm sure this has been tried many times... what is the flaw in the theory?
3. If it works what is the optimal theoretical path for the weights to take giving the max power?
Quote from: Flyboy on May 11, 2009, 11:41:12 PM
2. Since I'm sure this has been tried many times... what is the flaw in the theory?
Here is the simplest way I can put it. You are just adjusting degrees of leverage. Do you think there is any energy to be gained via a lever? Clearly there is not. While you can lift one ball using another of the same weight, by extending one ball farther from the fulcrum, the price you pay for this lift is that the ball being lifted does not rise as far as the other ball falls.
So that is all your wheel is, a series of levers. At each step (each lever), there is no advantage, no overunity. So you cannot add them up into one device and hope to achieve an energy gain.
Quote from: Flyboy on May 11, 2009, 11:41:12 PM
I must agree after thinking about it a bit I now can see intuitively that as I originally drew (or hansvonlieven's illustration above) is in balance as the system is broken down into two seperate systems, one being the wheel itself which is balanced, and two, the chain which is balanced as well... the fact that it is half on a wheel is just an illusion.
Now lets have a more in depth look into the classic unbalanced wheel, where we force the wheel to be unbalanced via differing torque. (see illustration attached). To clarify, each spoke has a weight on it, on the way down is at its maximum radius (max leverage) and on the way up the radius shortens (less leverage). Theoretically the balls take the same amount of energy to go up as it does to go down, so in our frictionless thought experiment the device will spin effortlessly. However the 'MAGIC' comes into play in this model when we compare the distance from the axis for each matched pair of weights... thus a spinning motion is created on the wheel.
So my questions now are:
1. Have I accurately summed up the essence of the unbalanced wheel theory?
2. Since I'm sure this has been tried many times... what is the flaw in the theory?
3. If it works what is the optimal theoretical path for the weights to take giving the max power?
That theory is a classical mistake. In your 2. question: The theory isn't complete. There isn't more torque in one side of a classical overbalanced wheel. It can't because all weights is allways starting and ending at the same hight. The difference is the space between the weights, the distance from center, and TIME it takes to lift it up. Using these three factors in the theory, the sum will allways be 0. TIME is the most forgotten factor in such wheels.
Quote from: Low-Q on May 12, 2009, 01:10:38 AM
That theory is a classical mistake. In your 2. question: The theory isn't complete. There isn't more torque in one side of a classical overbalanced wheel. It can't because all weights is allways starting and ending at the same hight. The difference is the space between the weights, the distance from center, and TIME it takes to lift it up. Using these three factors in the theory, the sum will allways be 0. TIME is the most forgotten factor in such wheels.
Thank you both for the replies, but while I'm sure the reason this wheel does not work is clear to you, your explanation has left me no closer to understanding the 'flaw' as I was unable to follow your logic (sorry if I'm not as sharp as most :P).
if we simplify it more by breaking the system down further to just two spokes with one weight each directly opposite each other (forming a diagonal line that passes through the axis)
1. if you stop the wheel at any point and do a vector analysis of the forces you quickly see that torque on the wheel exists. Intuitively we can see the wheel will want to spin.
2. Vertical speed is varying constantly on both size in opposite quantities... as on increases, the other slows down and visa versa. yet the sum total of all accelerations (positive and negative) equal the other side resulting in weight reach top at same time the other weight reaches the bottom. We can safely assume that the wheel is capable of storing any momentary energy gains and losses in the form of kinetic energy (momentum) thus at the end of each half revelation (180°) we net zero gain or loss of energy.
3. As the weight begins its upward climb the radius is shortened causing the wheel's spin to increase. This is due to the weight wanting to maintain it's velocity, but due to shorter radius it must slow down. Similarly as it passes half way up, the radius increases causing wheel to slow, due to the weight needing to speed up its velocity as the radius increases. These two effects should exactly balance each other.
4. Energy required to shorten radius of weight should equal the energy released as the radius lengthens. these should also equal.
5. ...I must have missed something because point 1 is out of balance causing wheel to spin, while points 2-4 are all balanced leaving point 1 unchecked... As I mentioned earlier, we know this system does not work because others have tried it over and over... So what is countering the unbalanced force in point 1?
The classic theory is that the EXACT same amount of energy gained from the descending weight is used to bring the weight back up... they claim the "MAGIC" in in creating usable torque in the wheel due to the cleaver setup causing the wheel to naturally spin.
as a static set up your wheel will rotate 90 degrees clockwise.
The "magic" that's missing is the bit that makes the weights move.
The weights in the right hand side are static, so there is no change in energy there.
The weights in the bottom left quarter of the wheel are moving toward the hub.
The weights in the top left quarter of the wheel are moving away from the hub
To move a weight in the bottom left of the wheel towards the hub, you need to move it right, and upwards - moving it right is fine, but upwards takes energy.
To omve a weight in the top left quarter away from the hub, you need to move it left, and upwards. moving left is fine, but upwards takes energy.
Quote from: stgpcm on May 13, 2009, 06:49:46 AM
as a static set up your wheel will rotate 90 degrees clockwise.
...
The weights in the bottom left quarter of the wheel are moving toward the hub.
The weights in the top left quarter of the wheel are moving away from the hub
To move a weight in the bottom left of the wheel towards the hub, you need to move it right, and upwards - moving it right is fine, but upwards takes energy.
To omve a weight in the top left quarter away from the hub, you need to move it left, and upwards. moving left is fine, but upwards takes energy.
I think I covered this before... the distance the weight drops is equal to the distance the weight rises... Same energy difference on both sides... so I agree energy is needed to raise the weight but should equal the energy gained from the dropping of the weight... this leaves point one still out of balance, with the wheel constantly wanting to rotate 90° clockwise.
Its frustrating that everyone else can see why this model does not work... but I'm still seeing viable model.
OK, the system as it stands has a potential energy of one quarter turn clockwise. This potential energy is created when you first build it. if you let the whole setup turn it will do so, but so what - you're now at E2.jpg
what your diagram doesn't show is the linkage that provides the upward force to the weights on the left - it's the energy you need to power that linkage that EXACTLY BALANCES the wheel.
you've agreed that all of the energy from the energy from the balls dropping on the right is needed to raisie the balls rising on the right. Once they are doing that, they are no longer applying any force to the wheel itself, so the wheel no longer wants to turn
Quote from: stgpcm on May 14, 2009, 03:18:55 AM
OK, the system as it stands has a potential energy of one quarter turn clockwise. This potential energy is created when you first build it. if you let the whole setup turn it will do so, but so what - you're now at E2.jpg
what your diagram doesn't show is the linkage that provides the upward force to the weights on the left - it's the energy you need to power that linkage that EXACTLY BALANCES the wheel.
you've agreed that all of the energy from the energy from the balls dropping on the right is needed to raisie the balls rising on the right. Once they are doing that, they are no longer applying any force to the wheel itself, so the wheel no longer wants to turn
Thanks, I think I now understand what it is your saying. My next step is to now try and show what you are explaining using vector diagrams as proof... If anyone knows any work already done that would help a lot, preferably with formula to show work :)
The reason for this is I had a thought some time ago that is similar to this... as of yet I can't think of any reason my idea wont work so now I guess I should bite the buliet and figure out the math... :-\ Who knows... it might be the holy grail or it might be one more 'balanced' wheel ;D
yes, the weights on the right are on longer arms than the weights on the left, so there will be more torque - lets say there are 10 weights of 1 Kg and the wheel is of radius 1 metre, and by a miracle of engineering the weights on the left manage to make the upward journey 1 centimeter from the hub.
at the horizontal position, Rr is the resultant weight of the right weight, and Rl is the resultant weight of the left weight the clockwise torque is:
((Rr x 1kg) x g x 1m) - ((Rl x 1Kg) x g x 0.01m)
or
(Rr - 0.01Rl) Nm
Nice torque!
now, what are the resultant weights of the left and right sides.... as discussed all of the weights on the right are matched by all of the weigts on the left, so the resultant weight is 0, so our torque is
(0-(0.0x0)) Nm
or
0 Nm
Quote from: stgpcm on May 15, 2009, 04:17:20 AM
... as discussed all of the weights on the right are matched by all of the weigts on the left, so the resultant weight is 0, so our torque is
(0-(0.0x0)) Nm
or
0 Nm
Thanks for trying to help but your math is a little incomplete as you made an assumption and then assigned values based on the assumption. I was hoping to get some schooling in how to calculate all the forces in a simple system like this one that would then give a mathematical result showing balance or out of balance... this way I can then apply it to other systems.
Flyboy
So you want to figure out the power? Simply take the advantage of overbalance and do the math. For instance if a wheel has a 100lb advantage of falling weight (this is over the top for balances) and you want to figure out the distance/leverage from the axle and then do your math at, weight times distance to the axle and you will have your basic torque figured out, and go from there.
for the right hand path, it's quite easy to calculate the torque at any particular point.
let f be the downward force of the weight due to gravity,
let r be the radius of the wheel
let a be the angle from TDC
the distance (d) of the weight from the axle will be
r*sin(a)
so the torque will be
f*r*sin(a)
Which would vary between 0 and f*r, the average being around 0.6366*f*r, which is actually
f*r*(pi/2)
for your 16 weight diagram, there would be 8 weights on the right hand side, 22.5 degrees apart.
The torque on the system due to the 8 weights would be
f*r*(sin(a)+sin(a+22.5)+sin(a+45)+sin(a+67.5)+sin(a+90)+sin(a+112.5)+sin(a+135)+sin(a+157.5))
which would vary between 5.0273*f*r and 5.126*f*r (approx), the average torque would be 5.093*f*r approx, or
(8*f)*r*(2/pi)
now, the energy is torque x angle (in radians), so a half turn (which is PI radians) of the 8 weight system gives us
(8*f)*r*(2/pi) * pi
or 16*f*r energy. (if r is in meters and f in newtons, that gives you joules)
The mathematically optimal route of the left hand weights to take is to move to the hub at the bottom (a distance r), and stay there while their spoke rotates to the top, then slide up the spoke to the rim. in order to do that, we will have to push them each time. To move a weights of mass f/g through the distance r against gravity, to get it to the hub, we need to do f/g*r*g work, and to get it to the rim, we also need to do f/g*r*g work. In half a turn, we need to do do this for 8 weights, so we need
8 * ((f/g*r*g) + (f/g*r*g))
or 16*f*r energy. The question is, where do we get it from?
Quote from: stgpcm on May 16, 2009, 06:16:34 PM
for the right hand path, it's quite easy to calculate the torque at any particular point...
Thanks will have a play with these numbers and see how I go.
Here are the forces in the vector diagrams that were missing to explain why it doeas not work.
I hope this helps Flyboy.
The enrgetic explanation is very logical and easy. The vector explanation is some more complex.