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Aerospace Engineering
Help Calculating Forces on engaged sprocket teeth
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[QUOTE="Baluncore, post: 5906055, member: 447632"] Consider the chain without tension fits neatly in the roots of the sprocket. With tension in the chain, the chain pitch increases, so the chain rollers are progressively laid higher up the tooth, where the increased radius matches the stretched chain pitch. The rollers therefore start higher on the tooth and progressively roll down to the root as more chain is fed onto the sprocket, with tension being progressively reduced until the chain is lifted out of the root and so departs contact with the sprocket. Since there will be a change in chain direction at every roller, there will be a radial force and a circumferential force applied by each roller against the contact angle of the sprocket tooth. The tension in the chain will equal the sum of all the circumferential forces on the sprocket teeth. If the chain transfers that tension over say 12 teeth, then you have a series of terms that reduce the chain tension to zero over those 12 teeth. I believe chain stretch will be significantly greater than sprocket deformation, so I would initially ignore sprocket deformation. You can tell how high the tense chain will ride at the first contact because you know the stretch in chain pitch and so the radius of the circle required. You can apply that rationale to each link of the chain as tension is progressively reduced as it crosses each of the 12 teeth. [/QUOTE]
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Help Calculating Forces on engaged sprocket teeth
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