Dudley
New member
OK, I will post one more time. I sent the information to a friend of mine to calculate the physics of this mod. He spent the last 20 years teaching physics. This is what he wrote back:
The answer can only be approximated because there is no coefficient of friction or air resistance and we don't know the rate of negative acceleration. Assuming it is instantaneous:
70 mph *5280/3600 = 102.67 ft/sec If the keg weighs 40 lbs then the total instantaneous force would be 102.67 ft/sec * 40 lbs = 4106.8 ftLbs/sec
Divide that by the number of straps, each strap would have to hold 4106.8 / 2 = 2053.4 lbs
Do you still want this on your back? That's up to you. Looks like I was about 3950 footpounds off.
The answer can only be approximated because there is no coefficient of friction or air resistance and we don't know the rate of negative acceleration. Assuming it is instantaneous:
70 mph *5280/3600 = 102.67 ft/sec If the keg weighs 40 lbs then the total instantaneous force would be 102.67 ft/sec * 40 lbs = 4106.8 ftLbs/sec
Divide that by the number of straps, each strap would have to hold 4106.8 / 2 = 2053.4 lbs
Do you still want this on your back? That's up to you. Looks like I was about 3950 footpounds off.
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