Thanks for the answer and for showing the math to get there.Well I assume that the cord is wound round securely at one and and the other hangs down, supporting a pan.
Then weights are added until it stalls and the question is what weight stalls it?
So the shaft has diameter 0.25 inches = 6.35mm = .00635m
So the radius is half this or .003175m.
The cord hangs down with tension T equal to the weight in the pan.
This develops a torque of T x radius which we are told stalls at 20 kgcm = 0.2 kgm
So max T x R = 0.2
T = 0.2/0.003175 = 63 kg force, or 63 x g Newtons force = 618N.
So your motor shaft will wind up a load of just under 63kg mass.
The tangential force is the torque divided by the radius. .125"=.3175cm. 20/.3175=62.9 Kg.Suppose a motor stalls atThe 20kg.cm and its shaft is .25" in diameter. What is the maximum load that can be exerted on a cord wound directly on the shaft? (Be nice; remember I am a marketing guy.)
I converted 63 kg to 139 pounds, which tells me what I need to know, which is that a 100 pound test cord would be suitable.The only real thing is that kilograms force are a non standard unit.
In the ISO metric system the kilogram is a unit of mass and the unit of force that corresponds is the newton.
So working is kgcm is definitely likely to lead to trouble.
since you had a 1/4 inch shaft and are in America, why not keep it imperial and use poundsinches poundsfeet?
by Jake Hertz
by Aaron Carman
by Aaron Carman
by Duane Benson