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#1
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Re: Chiaphua Gear Box
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Now, this is not 100% true, actually only about 75% true. Your typical spur gear has a 14.5º pressure angle. This means that the force is not actually transmitted tangent to the shaft but 14.5º off from the tangent direction. Thus, whatever force is being transferred (motor torque ÷ gear's pitch radius), you need to multiply by sin(14.5º), which yields the "radial force" (force perpendicular to the circumference of the shaft). The radial force on the gear must be supported by the shaft's bearings. Since the motor already has two bearings in it, it is designed to support a radial force at some distance. You can check the motor specs to see if your radial force at your given distance is within the motor's specifications. The other issue is deflection. If your shaft is too thin and the force is too far away from the motor, you could have significant deflection. If it deflects too far, the gears will slip. If it deflects just a little, the gears will temporarily become misaligned, and the gears will wear down quicker. If it deflects insignificantly, then Bob's your uncle (there will be no problems). Having worked with these motors (and seeing that the output shaft is not that long), my guess is that the deflection will not be significant at all. But if you really are concerned about deflection, go ahead and put the third bearing on. But it MUST be perfectly aligned, otherwise the third bearing is deflecting the shaft, and applying a constant torque. Even if it is just a couple thousands of an inch off center, that may be enough to bust one of the bearings in the motor. My recommendation is to not use the third bearing. But you can do the calculations and find out for sure. Quote:
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- Patrick |
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"The Chiaphua does not need a 3rd bearing" -- Dr. Joe
At 2 N-m stall torque and min. a radius of 4mm (a bit less if you carve the a gear into the 8mm diameter shaft like the ones that they put in the kit the first year the Chiaphua motors were in the kit), the tooth load would be 500N.
If we double this number to account for dynamic effects and the force from the pressure angle of the gear, we get a max load on the shaft of 1000N. If we assume that the load is cantilevered 15mm from the bearing and we get a bending moment on the 8mm shaft of just under 15N-m. Now, if we get out our old Beer & Johnston textbook, we see the following formulas (I am assuming a cantelivered beam, which is not strictly true, but when you arbitrarily put in factors of 2 for this or that reason, it is pretty hard to get picky at this point in the calculations ;-) Max Deflection: -M L^2 / (3EI) Max Stress: Mc/I M = 15N-m (from above) L = 15mm (from above) E = 30,000 psi = 200,000 MPa (property of steel -- look it up in a ref. book) I = (1/4)pi r^4 (also from Beer & Johnson -- I LOVE that book) I = 2E-10 m c = r = 4mm Max Deflection: .03mm Max Stress: 300 MPa The deflection is practically non-existant. The stress is getting high for crummy steels, but that shaft should have a yeild stress of something like 400 MPa or so I am not terribly worried especially since this is about twice higher that we expect under stall conditions of the motor. Fatigue should not be a problem since the number of cycles at that relatively high stress should be pretty low (note that if we are running at 4000 RPM or so then the shaft must not be loaded very much at all -- it is hard to rack up tens of thousands of cycles at over the stall condition on a FIRST robot). Bottom line: I do not think supporting the shaft is required. Joe J. P.S. Beer & Johnston is perhaps the most valuable textbook I have from my college days... ...actually, I foolishly sold it back to my college bookstore while I was an undergrad. Only later did I realize my error and buy another one. Last edited by Joe Johnson : 06-01-2004 at 10:31. |
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