Calculate Braking Torque for Winch Emergency Stop

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The discussion focuses on calculating the braking torque required for a winch emergency stop using the formula T = (W*K^2*N)/308t, where W is the weight of the rotating member, K is the radius of gyration, N is the speed of the rotating shaft, and t is the stopping time. Participants emphasize the importance of considering both the rotating mass of the drum and the linear kinetic energy of the load and wire being unspooled. Additionally, they highlight the need for a factor of safety and the implications of over-speed conditions on winch operation.

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sd1606
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Hoping someone on here could help; I'm looking for a point in the right direction for a winch braking query. I have a mass moving horizontally, unspooling a free-wheeling winch as it moves. I'm looking to select a suitable pneumatic caliper disc brake to act as an emergency stop for this winch.

From a supplier's website, I've found a fairly general calculation, which appears to allow me to calculate the braking torque required for the rotating mass:

T = (W*K^2*N)/308t

Where W = weight of the rotating member (lbs)
K = Radius of gyration of rotating member (ft)
N = Speed of rotating shaft (rpm)
t = Stopping time required (seconds).
308 - constant, unsure where this has came from?

Now my understanding from this would be that I can take the weight of the rotating member to be the gross weight of the drum, unspooled, which will give me a value for the braking torque required to stop this member at a particular time.

Is this a suitable method of identifying braking torque required? If not, does anyone know of any more suitable literature/calcs available I can have a look through?

Greatly appreciate any help, many thanks.
 
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I would think you also need the rotating mass of the max weight of the spool of wire on the drum as that has to stop as well. And I would add the energy in the cable. If you just stop the drum, the force that is unspooling the cable and its cable spring tension will override the brake, or is that considered to be light as the object is horizontal? Like a person pulling out a free spooling truck winch?

you have an interesting equation from your supplier, the 308 is a sum of constants, not sure which. Basically it is about torque
Net torque = I * α, then the brake force is the torque / radius.
I is moment of inertia for your drum and spool of wire. a dimensional constant, likely 1/2 m r^2
α - is angular acceleration = ((rpm / 60 s) - end rpm) * 2π / time, your end rpm is 0, time is your desired stopping time.

I would then add the cable tension force and apply a factor of safety.
 
sd1606 said:
I have a mass moving horizontally, unspooling a free-wheeling winch as it moves.
By “free-wheeling” do you mean with the gear train disengaged or with the clutch released. It is important that you do not allow the winch to unspool with gear train engaged at a significantly greater RPM than it would normally pull the wire in. Over-speed can disintegrate the clutch or the motor.

sd1606 said:
Now my understanding from this would be that I can take the weight of the rotating member to be the gross weight of the drum, unspooled, which will give me a value for the braking torque required to stop this member at a particular time.
Not quite. The equation you have is for a rotating mass. But you also have the load mass that is pulling out the wire, plus the mass of wire that has been pulled out. Those non-rotating components have a linear kinetic energy that must be braked. It comes down to management of that kinetic energy.

I am ignoring the rotational energy of the drum and remaining wire because we do not know radius or moment of inertia.
power = torque * angular velocity.
power = energy / time.
( torque * angular velocity ) = ( energy / time )

The required braking torque is = energy / ( time * angular velocity )
So you will need to know; the total kinetic energy of the system, (load, wire and drum), the maximum time allowed to stop, and the angular velocity of the drum at the start of braking.
 

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