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Hi,
thanks for the replies.
the stepper motor i'm using is able to rotate the load with a total reduction ratio of around 670.
But I'm changing the gear train and am using a geared stepper with a reduction of 125:1, and the output torque rating of the gearbox is 100Ncm or 1000mNm.
This will further be reduced by a ratio of 4:1 with timing pulleys at the output stage.
so I just wanted to get a rough idea of what the torque required would be for my load or how to calculate it.
just wanted to understand the concept if we ignore friction at the bearings, mass of the gear, etc. just the basic calculation first and then consider friction and other factors.
or, since moment of inertia is dependant on mass and distance of the mass from the axis of rotation, how do you calculate force required to overcome the moment of inertia.
in practical terms, if the gearbox output torque rating as above at 100Ncm, how does the subsequent reduction ratio alter the final output torque. eg. if the reduction stage after the gearbox is 4:1, what would the final output torque rating be after reduction.
just wanted to get my head around the math involved or parameters to be considered.
another reason is that if i replace the stepper that came with the gearbox with one that has a higher holding torque, how do I calculate if the gearbox can handle it.
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