MontanaMan Posted January 22, 2020 Share Posted January 22, 2020 Hello: I'm working on a difficult problem. Imagine a wheel, dangling from an axle, allowed to spin freely. If you tape a weight to the rim of the wheel, then that section of the rim will rotate to the bottom, right? Now ... Imagine the axle is not perfectly in the middle of the wheel. That "Offset" (or runout) will cause the wheel to rotate, even if the wheel is perfectly balanced around it's true center point. My question is this: How much offset (runout) will be equal to adding 3 grams of weight to the wheel, 0.216 meters from the exact center of the wheel. Weight of wheel: 19000g (19 kg). Diameter of wheel: 0.725 meters Assumptions: No friction from axle. Wheel is uniformly dense across it's structure (the wheel is uniformly constructed with wheel weight evenly distributed). Calculating the rotational force of the added weight is very simple: 3g x 0.216m = 0.648 meter-grams of torsion. Calculating the rotational force caused by the axle offset is much more difficult, since it's a round wheel, not a simple horizontal bar. I'm sure there will be integrals involved in this, but I just can't seem to set up the problem for it. Any help you could provide would be much appreciated. Link to comment Share on other sites More sharing options...
studiot Posted January 22, 2020 Share Posted January 22, 2020 17 minutes ago, MontanaMan said: Hello: I'm working on a difficult problem. Imagine a wheel, dangling from an axle, allowed to spin freely. If you tape a weight to the rim of the wheel, then that section of the rim will rotate to the bottom, right? Now ... Imagine the axle is not perfectly in the middle of the wheel. That "Offset" (or runout) will cause the wheel to rotate, even if the wheel is perfectly balanced around it's true center point. My question is this: How much offset (runout) will be equal to adding 3 grams of weight to the wheel, 0.216 meters from the exact center of the wheel. Weight of wheel: 19000g (19 kg). Diameter of wheel: 0.725 meters Assumptions: No friction from axle. Wheel is uniformly dense across it's structure (the wheel is uniformly constructed with wheel weight evenly distributed). Calculating the rotational force of the added weight is very simple: 3g x 0.216m = 0.648 meter-grams of torsion. Calculating the rotational force caused by the axle offset is much more difficult, since it's a round wheel, not a simple horizontal bar. I'm sure there will be integrals involved in this, but I just can't seem to set up the problem for it. Any help you could provide would be much appreciated. I'm not totally clear on what you are trying to do but it sounds similar to the analysis of stability in ships. https://en.wikipedia.org/wiki/Metacentric_height Link to comment Share on other sites More sharing options...
MontanaMan Posted January 22, 2020 Author Share Posted January 22, 2020 (edited) I'm designing a wheel balancer which will mount the wheel on a horizontal axle, using cones to center the wheel hub on the horizontal shaft. Any heavy spot on the wheel/tire combo will rotate to the bottom, which will reveal the location of the imbalance. It's called a static balancer. Due to manufacturing realities, some runout of the axle and cones is unavoidable. The runout will cause the wheel to rotate around a point that is not the exact center of the wheel. Make sense? So ... I need to calculate how much runout is equal to 3 grams of weight applied to a 17" rim (that would put the weight 0.216 meters from the center of the wheel). The shipping link you provided is interesting, but it doesn't apply to my situation, since there is no buoyancy in my setup. Any help provided with the wheel problem would be much appreciated. Thanks!! Edited January 22, 2020 by MontanaMan Link to comment Share on other sites More sharing options...
dimreepr Posted January 22, 2020 Share Posted January 22, 2020 you could buy one. 1 Link to comment Share on other sites More sharing options...
MontanaMan Posted January 22, 2020 Author Share Posted January 22, 2020 (edited) 21 minutes ago, dimreepr said: you could buy one. I'm designing one to manufacture for resale. So I need to calculate the runout tolerance. Also ... the kind that I am designing does not exist for car wheels ... only for motorcycle wheels, which have a different hub design, requiring a different system for mounting to the rotating shaft. Does anyone know how to set up and solve the physics problem as I've described it? If anything about my question is unclear, please let me know and I will try to clarify. Thanks!! Edited January 22, 2020 by MontanaMan Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted January 22, 2020 Share Posted January 22, 2020 2 hours ago, MontanaMan said: Hello: I'm working on a difficult problem. Imagine a wheel, dangling from an axle, allowed to spin freely. If you tape a weight to the rim of the wheel, then that section of the rim will rotate to the bottom, right? Now ... Imagine the axle is not perfectly in the middle of the wheel. That "Offset" (or runout) will cause the wheel to rotate, even if the wheel is perfectly balanced around it's true center point. My question is this: How much offset (runout) will be equal to adding 3 grams of weight to the wheel, 0.216 meters from the exact center of the wheel. Weight of wheel: 19000g (19 kg). Diameter of wheel: 0.725 meters Assumptions: No friction from axle. Wheel is uniformly dense across it's structure (the wheel is uniformly constructed with wheel weight evenly distributed). Calculating the rotational force of the added weight is very simple: 3g x 0.216m = 0.648 meter-grams of torsion. Calculating the rotational force caused by the axle offset is much more difficult, since it's a round wheel, not a simple horizontal bar. I'm sure there will be integrals involved in this, but I just can't seem to set up the problem for it. Any help you could provide would be much appreciated. It's actually equally simple...19 kg X the offset will equal 0.648 meter grams as well The shape doesn't matter as it is the centre of gravity that counts Link to comment Share on other sites More sharing options...
MontanaMan Posted January 22, 2020 Author Share Posted January 22, 2020 8 hours ago, J.C.MacSwell said: It's actually equally simple...19 kg X the offset will equal 0.648 meter grams as well The shape doesn't matter as it is the centre of gravity that counts Yes!! Of course. It's been a long time since my physics classes. Thanks for reminding me. Link to comment Share on other sites More sharing options...
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