Topic: Ever heard of dynabeads?

[mlinder]

He's not a troll. I'm convinced he actually believes what he's saying, and that his ideas and explanations are not only comprehensible, but valid!
Logic can never find it's way through walls of insanity or stupidity without fundamental changes made by the person who erected  said wall.
There's no point arguing. Trolls do this kind of thing to cause unrest and strife, as they feed off these kinds of interactions.
Crazy people, however, are just crazy.

[dilbone]

The beads will tend to make the tire mass distribution even around the circumfrence.  Once they are in place and moving at high speed the friction force holding them where they are will be pretty large.  Motion of the tire other than rotational motion will have to be pretty abrupt to cause a force large enough to dislodge the beads from where they were.  I wouldn't be worried much about small bumps and things of that nature, the larger the bump the more trouble it will cause.

[mlinder]

The beads will tend to make the tire mass distribution even around the circumfrence.  Once they are in place and moving at high speed the friction force holding them where they are will be pretty large.  Motion of the tire other than rotational motion will have to be pretty abrupt to cause a force large enough to dislodge the beads from where they were.  I wouldn't be worried much about small bumps and things of that nature, the larger the bump the more trouble it will cause.

I also said this... however, no one has been able to validly explain the force that causes the beads to travel to, and remain, to the spot of the tire that is lightest.

[bistromath]

Turbogrimace asked me in a PM to do the math for him, and I had some spare time, so I did. I hate being troll fodder, but I like math more, so here you go. Dilbone, please feel free to comment on my math. Even if I got some magnitudes wrong, the signs are correct as far as I can tell, and point to the forces moving beads in the direction of balance.

So consider a wheel rim rotating with a compound motion, with a linear period equal to its rotation rate: that is, for every time it moves up and down along the shocks, it is rotating once around the axle. This is the case for an unbalanced wheel, since the weighted side will "throw" the wheel up and down a bit.

For fun, let's first calculate the centrifugal acceleration of the beads: this is the "false force" imposed on the beads by their rotation with the rim. It's easy to do: v-squared-over-r. If you want to use real units, let's say we're running a 15 inch wheel (we'll simplify here and say the wheel is the same ID and OD), and we're moving at 50MPH. The acceleration toward the rim is then given by (50MPH^2)/(15/2 inches), and is 2622m/s^2. Dang, that's 267g of acceleration. Not kidding, are we? That's a lot of force! (edit: fixed embarrassing magnitude mistake)

Incidentally, that's the reason I think SiliconDoc's belief the beads will fly about inside the tube when the contact patch shifts them up is crap. Wait! Actually, this is pretty easy! Since the radius of curvature of the tire as it transitions to the contact patch will always be less than the radius of curvature of the wheel, and since there's a constant tangent rate (road speed), the acceleration due to contact patch will always be less than the acceleration due to wheel rotation. Therefore, the beads will never lift off of the surface of the inner tube due to the contact patch.

Having the centrifugal force on the beads, we then note the eccentricity of the wheel's rotation along the shocks. This has a few effects. The one we're concerned with is the change in acceleration experienced over a single rotation of the wheel. The trick to depicting elliptical motion is that R, the radius, is constantly changing. Let's say the shocks are moving a quarter inch in each direction over the course of a rotation. This seems like a pretty serious imbalance to me. So when the heavy side of the wheel is up, the shock is most compressed. When the light side is up, the shock is most extended.

Note that "R" now depends on where the bead is located in relation to the light or heavy side of the wheel. When the bead is located at the heavy point, the change in R over the course of a rotation produces a variation in force that is completely radial: that is, there is no tangential component of force on the bead. When the bead is 90 degrees offset to the heavy point, the force changes both in radial and tangential components over the course of a rotation. It's those tangential forces that do the balancing work of the DynaBeads. I don't care much about the variation in radial force for the purposes of this analysis, both because they are insignificant compared to the static radial force of 134g, and because they don't work to change the position of the beads, which is what I'm interested in. Therefore, we focus on the tangential forces for now.

Let's make this as easy as possible, and use the rotating reference frame of the wheel as our inertial reference frame. The heavy side of the wheel will be at 0 degrees, for reference. The light side is then at 180 degrees. A good mental image right about now is to picture the rotation of the wheel as it moves up and down along the shock axis. (I'm not worrying about fork angle for this discussion, since it's the travel itself that matters, not the direction.)

The tangential force is then equal to the sine of the radial angle between the heavy point and the bead in question, times the acceleration produced by the linear shock travel. For the case of a bead located exactly at the heavy point, given that the sine of zero is zero, there is no tangential force, and the bead will not move along the rim. The same applies to a bead located exactly at the lightest point. For the case of a bead located halfway between the heavy and light points, the sine of 90 is 1, and maximal tangential force results. Let's find that force.

In the "90 degree" case, the acceleration of the shock is always tangent to the rim, producing a tangential force equal to the shock acceleration. If the shock is moving sinusoidally at 18.7Hz (15 inch rim moving at 50MPH) with a displacement of 1/2 inch (as postulated above), we can say its displacement from zero becomes D = 0.25 inch x cos(18.7*2*pi*t)

The velocity of the shock, being the derivative w.r.t. time of the displacement, becomes V = 18.7 x 0.25 inch x sin(18.7*2*pi*t).

The acceleration, which is what we're interested in, becomes the derivative w.r.t. time of the velocity, or A = 18.7 x 18.7 x 0.25 inch x -sin(18.7*2*pi*t). Of particular note is the sign of the acceleration, which is opposite our reference point (the heavy point), producing a force that accelerates beads away from the heavy point. This is true for the tangent force all the way around the wheel: at every point that is not exactly at the heavy point or exactly at the light point, a tangent force is produced that accelerates the beads toward the light point.

The change in tangent acceleration is synchronized (and in phase) with the rotation of the wheel itself, and because of this, both these accelerations can be combined into a static tangent acceleration that does not vary over the rotation of the wheel. This acceleration is A = shock displacement x (radial rate ^ 2) x -sin(angle between heavy point and point in question). Multiply this acceleration by the mass of a DynaBead to find the tangent force acting on the bead in the direction of the light side.

Of course, as the beads shift toward the light point, the shock displacement decreases due to the lessening imbalance. This decrease in travel causes a corresponding decrease in tangential force, until the system comes into balance. Then, with zero shock displacement, the beads no longer move.

It's a cute problem! I had a lot of fun figuring out the math. I'm sure SiliconDoc will have something pithy to say, but this is for Turbogrimace, not him. And in any case, I've got him ignored.  

[Inigo Montoya]

Maybe after he picks up the pieces of his head after trying to absorb all that!

[mlinder]

Turbogrimace asked me in a PM to do the math for him, and I had some spare time, so I did. I hate being troll fodder, but I like math more, so here you go. Dilbone, please feel free to comment on my math. Even if I got some magnitudes wrong, the signs are correct as far as I can tell, and point to the forces moving beads in the direction of balance.

So consider a wheel rim rotating with a compound motion, with a linear period equal to its rotation rate: that is, for every time it moves up and down along the shocks, it is rotating once around the axle. This is the case for an unbalanced wheel, since the weighted side will "throw" the wheel up and down a bit.

But this is not correct. It is not one movement up and down per rotation of the wheel, except, possibly, at one very specific speed. Herein lies the rub. At most speeds, there will be multiple up and down motions inbetween the extremes of compression and extension.

For fun, let's first calculate the centrifugal acceleration of the beads: this is the "false force" imposed on the beads by their rotation with the rim. It's easy to do: v-squared-over-r. If you want to use real units, let's say we're running a 15 inch wheel (we'll simplify here and say the wheel is the same ID and OD), and we're moving at 50MPH. The acceleration toward the rim is then given by (50MPH^2)/(15 inches), and is 1311m/s^2. Dang, that's 134g of acceleration. Not kidding, are we? That's a lot of force!

Incidentally, that's the reason I think SiliconDoc's belief the beads will fly about inside the tube when the contact patch shifts them up is crap. Wait! Actually, this is pretty easy! Since the radius of curvature of the tire as it transitions to the contact patch will always be less than the radius of curvature of the wheel, and since there's a constant tangent rate (road speed), the acceleration due to contact patch will always be less than the acceleration due to wheel rotation. Therefore, the beads will never lift off of the surface of the inner tube due to the contact patch.

Having the centrifugal force on the beads, we then note the eccentricity of the wheel's rotation along the shocks. This has a few effects. The one we're concerned with is the change in acceleration experienced over a single rotation of the wheel. The trick to depicting elliptical motion is that R, the radius, is constantly changing. Let's say the shocks are moving a quarter inch in each direction over the course of a rotation. This seems like a pretty serious imbalance to me. So when the heavy side of the wheel is up, the shock is most compressed. When the light side is up, the shock is most extended.

Again, here's the problem. If the tire itself is the cause of the imbalance, the heaviest side will extend further from the axis. With 134 g's of acceleration, centrifugal force will cause the beads to move to that point, adding to the imbalance.

Note that "R" now depends on where the bead is located in relation to the light or heavy side of the wheel. When the bead is located at the heavy point, the change in R over the course of a rotation produces a variation in force that is completely radial: that is, there is no tangential component of force on the bead. When the bead is 90 degrees offset to the heavy point, the force changes both in radial and tangential components over the course of a rotation. It's those tangential forces that do the balancing work of the DynaBeads. I don't care much about the variation in radial force for the purposes of this analysis, both because they are insignificant compared to the static radial force of 134g, and because they don't work to change the position of the beads, which is what I'm interested in. Therefore, we focus on the tangential forces for now.

Let's make this as easy as possible, and use the rotating reference frame of the wheel as our inertial reference frame. The heavy side of the wheel will be at 0 degrees, for reference. The light side is then at 180 degrees. A good mental image right about now is to picture the rotation of the wheel as it moves up and down along the shock axis. (I'm not worrying about fork angle for this discussion, since it's the travel itself that matters, not the direction.)

The tangential force is then equal to the sine of the radial angle between the heavy point and the bead in question, times the acceleration produced by the linear shock travel. For the case of a bead located exactly at the heavy point, given that the sine of zero is zero, there is no tangential force, and the bead will not move along the rim. The same applies to a bead located exactly at the lightest point. For the case of a bead located halfway between the heavy and light points, the sine of 90 is 1, and maximal tangential force results. Let's find that force.

In the "90 degree" case, the acceleration of the shock is always tangent to the rim, producing a tangential force equal to the shock acceleration. If the shock is moving sinusoidally at 18.7Hz (15 inch rim moving at 50MPH) with a displacement of 1/2 inch (as postulated above), we can say its displacement from zero becomes D = 0.25 inch x cos(18.7*2*pi*t)

The velocity of the shock, being the derivative w.r.t. time of the displacement, becomes V = 18.7 x 0.25 inch x sin(18.7*2*pi*t).

The acceleration, which is what we're interested in, becomes the derivative w.r.t. time of the velocity, or A = 18.7 x 18.7 x 0.25 inch x -sin(18.7*2*pi*t). Of particular note is the sign of the acceleration, which is opposite our reference point (the heavy point), producing a force that accelerates beads away from the heavy point. This is true for the tangent force all the way around the wheel: at every point that is not exactly at the heavy point or exactly at the light point, a tangent force is produced that accelerates the beads toward the light point.

The change in tangent acceleration is synchronized (and in phase) with the rotation of the wheel itself, and because of this, both these accelerations can be combined into a static tangent acceleration that does not vary over the rotation of the wheel. This acceleration is A = shock displacement x (radial rate ^ 2) x -sin(angle between heavy point and point in question). Multiply this acceleration by the mass of a DynaBead to find the tangent force acting on the bead in the direction of the light side.

Of course, as the beads shift toward the light point, the shock displacement decreases due to the lessening imbalance. This decrease in travel causes a corresponding decrease in tangential force, until the system comes into balance. Then, with zero shock displacement, the beads no longer move.

It's a cute problem! I had a lot of fun figuring out the math. I'm sure SiliconDoc will have something pithy to say, but this is for Turbogrimace, not him. And in any case, I've got him ignored.  

I fear that disagreeing with you will only add more fuel to silicons fire, though, so I apologise in advance.

[bistromath]

But this is not correct. It is not one movement up and down per rotation of the wheel, except, possibly, at one very specific speed. Herein lies the rub. At most speeds, there will be multiple up and down motions inbetween the extremes of compression and extension.

it is, dude. the wheel imbalance is what causes the movement in the first place.

[mlinder]

But this is not correct. It is not one movement up and down per rotation of the wheel, except, possibly, at one very specific speed. Herein lies the rub. At most speeds, there will be multiple up and down motions inbetween the extremes of compression and extension.

it is, dude. the wheel imbalance is what causes the movement in the first place.

That's true, but you are missing some very important stuff here.

I promise you that an out of balance wheel will cause fork action that looks something like this, dependent on speed and degree of imbalance.

[bistromath]

Care to tell me the important stuff I'm missing? Resonance with the fork springs and sprung mass, which is what I think you're getting at, will cause a change in total displacement, but never frequency. The frequency will remain the same, at 1 displacement cycle per rotation. If the frequency could change, then you go dig up Joseph Fourier and tell HIM that, 'cause I don't want to be the one to have to tell him!

Put another way, if you take multiple cycles of rotation and average them, the "noise" in your image will wash out. Put a third way, the total energy of the noise will average to zero over time, and the displacement component due to the wheel mass will dominate.

[mlinder]

I don't know the math for it, but I've seen it, and understand it.

[bistromath]

No, I think I know what phenomenon you're talking about, fork spring resonance. It's a valid concern, too. But as far as balancing wheel motion, the resonance noise, if not equal to the rotational rate, will average out over time with respect to the larger displacement caused by the wheel imbalance. In fact, it could be that that resonance could set a lower bound to how accurate the balancing will get. I'll leave that one to others....

But even if the resonance noise is high, there is still a component of displacement at the wheel rotation frequency, produced by the imbalance, and the beads' aggregate motion over time will move to counteract that.

[mlinder]

No, I think I know what phenomenon you're talking about, fork spring resonance. It's a valid concern, too. But as far as balancing wheel motion, the resonance noise, if not equal to the rotational rate, will average out over time with respect to the larger displacement caused by the wheel imbalance. In fact, it could be that that resonance could set a lower bound to how accurate the balancing will get. I'll leave that one to others....

But even if the resonance noise is high, there is still a component of displacement at the wheel rotation frequency, produced by the imbalance, and the beads' aggregate motion over time will move to counteract that.

OK, fair enough. I wasn't sure it was an problem with the issue at hand, but please read my other concern that I posted.

Again, here's the problem. If the tire itself is the cause of the imbalance, the heaviest side will extend further from the axis. With 134 g's of acceleration, centrifugal force will cause the beads to move to that point, adding to the imbalance.

[dilbone]

I didn't go through all of the post but there is at least one problem...

Your value of g's should be half of your value unless we're talking about a 30inch diameter wheel.  You used 15 inches as a radius and not diameter which would make it 67g's.

I'm not convinced that making up numbers will get us what we want it to get us.
I would like to see some physical testing of this to see what would happen in the many places our wheels would find themselves...I'm not going to pretend to know exactly what would happen in any of these situations.  Many times we think we've covered all of the bases but leaving out one small detail wrecks the whole thing...

[bistromath]

Your value of g's should be half of your value unless we're talking about a 30inch diameter wheel.  You used 15 inches as a diameter and not radius which would make it 67g's.

Dang! Knew I messed that up somewhere. Still a hell of a lot of force. And the beads still won't lift at the contact patch. 

I'd like to see serious testing done, too. But despite that, I'm very convinced that this is the mechanism by which they work. Seeing widespread testimonials and the youtube video validating the scheme, combined with the theory, gives me a high degree of confidence that this isn't just snake oil. Their use at the limits, though, I can't testify to, and that's where all the hard-to-model bits crop up. I'm just saying they work in an ideal world, not guaranteeing they'll work in every real world. 

[bistromath]

Again, here's the problem. If the tire itself is the cause of the imbalance, the heaviest side will extend further from the axis. With 134 g's of acceleration, centrifugal force will cause the beads to move to that point, adding to the imbalance.

You're confusing axes again. The axis of rotation is still exactly about the hub. It's just that the hub moves. Since the moving hub is the axis of rotation, the distance from the hub to the axis will always remain the same.

[dilbone]

oops...I was wrong...

It should be double what you have...

(22.35m/s ^2)  /  (.19m) = 268g's

The 15inch is kinda small though for the diameter the bead will spanning...

There is some validity to it I think, but I wouldn't try it on my bike.  Not without some testing.

[mlinder]

Again, here's the problem. If the tire itself is the cause of the imbalance, the heaviest side will extend further from the axis. With 134 g's of acceleration, centrifugal force will cause the beads to move to that point, adding to the imbalance.

You're confusing axes again. The axis of rotation is still exactly about the hub. It's just that the axis moves. Since the hub moves with the axis of rotation, the distance from the hub to the axis will always remain the same.

I'm not confusing anything.
If the tire extends at the heaviest point away from any axis, it is furthest from the point of rotation the majority of time. If 67g's is enough to keep the beads from moving in relation to the tire if the tire is in balance, it's enough to force the beads to the point furthest from the axis, therefore sending more weight to what is already the heaviest point of the tire.

[bistromath]

If the tire extends at the heaviest point away from any axis, it is furthest from the point of rotation the majority of time. If 67g's is enough to keep the beads from moving in relation to the tire if the tire is in balance, it's enough to force the beads to the point furthest from the axis, therefore sending more weight to what is already the heaviest point of the tire.

If you mean the eccentricity due to shock travel, that's incorrect, since the hub stays in the same place relative to the axis of rotation, which is after all about the hub. If you mean the tire actually stretches a bit at the heavy point, you're right, but I doubt it is substantial enough to cause an issue. I believe this theory was part of SiliconDoc's rant.

[bistromath]

You know what, mlinder, you're right. The heavy side WILL have a larger radial acceleration. Sorry, I got confused there for a minute. The extra radius is equal to the shock displacement distance at the heavy point, and varies with the cosine of the angle between the heavy point and the point in question. I'm guessing it's not enough to offset the tangential force.

[mlinder]

Actually, it's the opposite of silicons rant...

Are you implying that the heavy side of the out of balance tire is not further away from the center? Because if that was the case, balance wouldn't matter, as centrifugal force apparently isn't working in your model. If the extra weight did not cause the heaviest part of the wheel to try to get away from the center with more force than the rest of the wheel, we wouldn't notice the imbalance in the first place.

[mlinder]

You know what, mlinder, you're right. The heavy side WILL have a larger radial acceleration. Sorry, I got confused there for a minute. The extra radius is equal to the shock displacement distance at the heavy point, and varies with the cosine of the angle between the heavy point and the point in question. I'm guessing it's not enough to offset the tangential force.

Oh, last post was before you replied. Sorry.
The math should be easy for you. Again, what g's, and difference between g's at the lesser radius and supposed friction would be required to keep them from traveling to the most extreme radius of our wheel?

[bistromath]

Oh! This isn't too bad, after all. Since the radial component of force is greatest at the heavy point, and the radial component is always directly toward the outside of the inner tube, there's no tangential force to cause migration of the beads. Just a change in their apparent weight within the rotational inertial reference frame. The change is 1 part in 30 for a 15-inch wheel. Not much.

In other words, the wheel is not elliptical, just the axis. So while the radial force changes over the ellipse, the radial force is always pointing straight out of the wheel, so the beads won't move.

[mlinder]

Oh! This isn't too bad, after all. Since the radial component of force is greatest at the heavy point, and the radial component is always directly toward the outside of the inner tube, there's no tangential force to cause migration of the beads. Just a change in their apparent weight within the rotational inertial reference frame. The change is 1 part in 30 for a 15-inch wheel. Not much.

In other words, the wheel is not elliptical, just the axis. So while the radial force changes over the ellipse, the radial force is always pointing straight out of the wheel, so the beads won't move.

I uh... I don't think so, Bistro. If that were the case, nothing would motivate the beads to the outside of the wheel in the first place....

/edit: eh, I see what you are saying. But this implies that the beads were stuck where they are before the wheel starts spinning. The centrifugal force will overpower the frictional force as you accelerate to cruising speed, causing the beads to seek the point furthest away from the rotational center, then the same force will keep them there, as well as your now supposed frictional force that would cause them to not move at said speed.

[dilbone]

yes, friction is required for the beads to get to where we want them to be...but then it is friction that holds them there once they get there.

Friction always opposes motion...but motion isn't possible without it...it is the catch 22 of the universe...

The friction force is directly proportional to the square of the velocity of the wheel.  At slower speeds the friction force isn't equal to or greater than the weight of the beads, but at higher speeds it's more than enough to keep the beads in place.  Remember the beads want to travel in a straight line tangent to the wheel, but the wheel won't let them.  That force times the coefficient of friction is the actual friction force holding the beads in place.  Once that force is larger than the weight of the beads then they "shouldn't" be moving around.  Just like being in a rollercoaster in a loop.  As long as the centripetal force on you is at least the size of your weight you won't fall out of the car(with or without shoulder harnesses).

[bistromath]

Naw, I don't care about friction for this discussion. Besides, the beads roll instead of sliding, so there's very little friction in there. I'm just saying that the radial force will cause them to move outward to the wall of the inner tube, and the tangential force will cause them to migrate to balance. The only reason the beads don't move after balance is there is no tangential force to cause them to move, since there's no shock displacement after balancing. That's all. The radial force will keep the beads stuck to the side of the tube, but does NOT cause them to try to migrate to the heavy side of the tire. There is no force in the system causing movement toward the heavy side. Changes in radial force due to elliptical motion do NOT cause bead migration, they just cause a change in radial force of the bead pressing on the inner tube. Remember, radial force is acting straight INTO the tube, pressing the bead into the tube surface. A ball doesn't move if you push straight down on it. I've decoupled the motion of the shocks into radial (in-phase) and tangential (out-of-phase) components, and the tangential component is the one doing the bead-moving.