torque calculations
Submitted: Thursday, Feb 24, 2011 at 15:31
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Member - bungarra (WA)
Trying to calculate the torque requirements for a worm drive / dc motor gearbox calculation if I build this boat loader
Google tells me to multiply distance by the weight to shift to get pounds/foot then convert to Nm (which is what most of the specs. seem to use on the net loking for the dc motor
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using this diagram i get 3085 ponds/foot. Google conversion gives me 4,192 Nm
am I doing something wrong?..........where do I source a suitable worm drive reduction gearbox with a 12V dc motor suitable?
Thanks .
.hopefully someone can point me in the right rirection
Reply By: briancc - Thursday, Feb 24, 2011 at 16:21
Thursday, Feb 24, 2011 at 16:21
Not that easy unfortunately. Line pull is what you need to start. If you were to lift the pointy end vertically, you would be looking at about half the 95kg. Worst case, call it 95 which is about 950N. Now because the lift isn't vertical, you need the component in the direction of the rope. That angle with the vertical is around 57 degrees, with cosine (trig) of around 0.54. This will make the line pull 950/.54 = ~1760N. Now the torque required is going to depend on the drum diameter / radius of the winch. This is where the T=FxD formula kicks in. Say a 50mm radius (wild guess) 1760x0.050 = 88Nm. I think. Needs a
check. Be aware of the different radii that the winch can work at. Also the number may change as the boat goes onto the roof, although I don't think it would get larger.
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