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Champion Wheels Vibration In Rain Because Of Groove On The Inside Lip

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Old Aug 18, 2005 | 08:05 AM
  #31  
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Hello,
'
I have the 19's and I don't notice a significant handling difference.
For example, when I put 17" borbet rims on my bmw e30 m3, the difference vs stock was very noticable. The new wheels were heavier, the handling and nimble steering disappeared etc.

But I didn't notice this change when I went to the champion 19's. I did place both my stock turbo style wheels and tire combo on a scale and for each corner, the replacement champion + tire equaled the factory rim and tire exactly

Even the wide rears. Now whether or not more weight is located further from the center I don't know, but I don't notice it,
thanks
Andrew
 
Old Aug 18, 2005 | 08:10 AM
  #32  
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thanks for all the info....Ted
 
Old Aug 18, 2005 | 11:09 AM
  #33  
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Originally posted by damon@tirerack
Water is fluid, it will distribute itself evenly out over the surface of a wheel indention, not collect in one spot. How can water collect in one spot while rotating? It can't. This is basic fluid dynamics.

We've played with this theory with several of our wheels here while spinning on a balancer and can't get them to go out of balance while pouring water into lip channels. It simply can't happen.
Basic fluid dynamics? This would be true if wheels were perfectly round- which they are not. Most three-piece wheels use spun inner rims, and you'll find they are not perfectly round in this region of the rim.
Sometimes, you can even feel this with your bare hand if you spin the wheel on a balancer (without a tire) or if you use a dial indicator to check for radial runout. If the inner rims were PERFECTLY round, the water would effectively have no place to go, but...

Cetrifugal force generated at speed will cause liquid to migrate to the low spot (high spot as measured from OUTSIDE the barrel) in the channel between the drop-center and rim flange area causing an imbalance. This is true of all modular wheels that use this type of drop-center design- the American stuff, the Japanese stuff- any time there is a channel like this between the center portion of the rim and the drop center.

Typically, the water will not be significantly "thrown out" while driving, it must evaporate or be dried off. OZ has done a good job with their modular stuff, with a patented rim design that uses blind assembly screws and does not expose this channel area.

The channel is necessary in other modular wheels so the wheels can be assembled. There's no way around it.

I keep an Absorber in the trunk...
 
Old Aug 18, 2005 | 12:03 PM
  #34  
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WASNT SOMEONE SAYING TEH SAME THING ABOUT HRE'S? A WHILE BACK?
 
Old Aug 18, 2005 | 12:19 PM
  #35  
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wow....thanks for all the info....
 
Old Aug 18, 2005 | 02:57 PM
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When this issue came up here last (it rears its ugly head once in awhile), we got some help from Jerry Hinnefeld, chair of Physics at Indiana University South Bend, and Steven Shore, a professor of astrophysics and specializes in fluid dynamics at the University of Pisa. This is their collective answer condensed:


In one sense you MAY be right. If there is any irregularity in the wheel, it will trap the fluid. But let's look at the basic problem first. Consider a layer of fluid that has some internal friction (viscosity). When you spin the wheel up, the fluid in contact with the inner surface is fiorced to co-rotate while the rest of the fluid spins up on a timescale that depends on the depth of the layer (actually as the square-root of the depth, but that's not as important as the statement that the thicker the layer, the more rotations are required to bring it into rigid co-rotation).

There's also an instability that results from the shear, waves are generated at the interface in this initial step that make the layer turbulent. But, the layer should spread uniformly. It's not precisely a centrifugal effect, it's that the fluid must be dragged by the moving surface. There are also secondary circulations set up, but these should be on very small scale. Now imagine the fluid is being jostled by bouncing. Then there are modes, which aren't symmetric, that move like waves within the layer. These don't last for too long, a number of rotations each, but every time the fluid is disturbed it'll generate these waves. Since water's actually pretty heavy, if the wheels are otherwise perfectly balanced these waves will be causing a sloshing internally within the rotating tire and MIGHT be felt.

But I doubt it.

In fact, the key here may be the assumption of perfection, both of the alignment and of the symmetry. My bet is that the wheel is slightly off alignment and this is causing a sort of rocking motion of the fluid that's addeed to the mass of the wheel with a larger moment of inertia. This is where the centrifugal forces come into play. When you're rotating, there's an acceleration of the material that otherwise seems to be at rest. So in this sense, you interpret this as a force. For the tire problem, the fluid would keep moving linearly if there weren't a wall in the way but it piles up against the wall because it's being accelerated relative to the stationary "outside world". This also means it is at the largest possible distance from the rotation axis, and therefore has the largest moment of inertia (in other words, like a lever, it has the largest torque ar). So as a perturbation, a disurbance on the otherwise uniform balanced system, it may have the graetest effect

Look at the following chart


25" OD
velocity tire omega alpha
mph [ft/s] [rps] [rad/s] [m/s^2] G's
10 14.67 2.24 14.08 44.31 4.52
20 29.33 4.48 28.16 177.23 18.07
30 44.00 6.72 42.24 398.77 40.65
40 58.67 8.96 56.32 708.93 72.27
50 73.33 11.20 70.40 1107.70 112.92
60 88.00 13.45 84.48 1595.09 162.60
70 102.67 15.69 98.56 2171.10 221.31
80 117.33 17.93 112.64 2835.72 289.06
90 132.00 20.17 126.72 3588.95 365.85
100 146.67 22.41 140.80 4430.81 451.66


18x8.5 SSR GT3 radius in groove = 223.5mm = 0.2235m

At 80 mph, the "centrifugal force" is 289 G's. At this speed, the wheel is rotating close to 18 revolutions per second, or 1,080 rpm.

The water would become RIGID and co-rotate with the wheel, as in there is no relative motion between the fluid and the wheel. At 289 G's, any "sloshing" or rippling would quickly dissipate.

Despite whatever suspension geometry the wheel might have, large negative camber or zero negative camber, the water will sit "level" to the trough and parallel to the axis of rotation. So literally, the axis of rotation could be vertical, and the water would
sit in the trough with the surface vertical to the ground (ok, gravity will change the level slightly but that would be the co-tanget of 1/289 = 0.19825 degress from level)"

There is no way this perception is anything more then urban legend.
 
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Old Aug 18, 2005 | 03:06 PM
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