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Acceleration load
- Thread starter ag89
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Try this (working in ft, second units only) -
The final velocity is 450/60 = 7.5 ft/sec.
That gives a = 7.5/1 = 7.5 ft/sec^2, which is approximately 23% of gravity. (7.5 ft/sec^2 = 27000 ft/min^2 - the time in trainguy's post should have been 1/60 in minutes).
I don't know the total mass of your mover (including its 3000 lb load), so I will assume that it weighs 4000 lb. in all. Then the mass = 4000/g, where g is approx 32.2 ft/sec^2.
Thus the average force over the 1 second duration = 4000/32.2*7.5 = 932 pounds force.
Were this my project, I would add a healthy 'ignorance allowance' so that the mechanism had sufficient capacity for the peak acceleration.
The final velocity is 450/60 = 7.5 ft/sec.
That gives a = 7.5/1 = 7.5 ft/sec^2, which is approximately 23% of gravity. (7.5 ft/sec^2 = 27000 ft/min^2 - the time in trainguy's post should have been 1/60 in minutes).
I don't know the total mass of your mover (including its 3000 lb load), so I will assume that it weighs 4000 lb. in all. Then the mass = 4000/g, where g is approx 32.2 ft/sec^2.
Thus the average force over the 1 second duration = 4000/32.2*7.5 = 932 pounds force.
Were this my project, I would add a healthy 'ignorance allowance' so that the mechanism had sufficient capacity for the peak acceleration.
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