Wayne jeez look at my reply in the thread on bb.com at the top of page 26. That all but debunks your whole idea.
Not the reply 'ewe' were expecting I'll bet...
Eccentric to concentric force.
I was trying to work out the force needed to lift a 100kg weight, however the lifter first is lowering this weight at .5 of a second, for 1m, and at the last instant he has to accelerate this weight up 1m at .5 of a second.
Someone worked out for me the lifting and lowering of a 100kg for 1m each way at .5/.5 and 2/4, but from zero movement at the start of the lift. However, as the force needed is going to be higher in the transition from the eccentric to the concentric, as the weight is being lowered at .5 of a second down 1m, so the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load, as if something is falling slightly faster would not it appear heaver to the muscles, and in each repetition of the lifting and lowering as in weightlifting.
Say we lowered the 100kg at .5 of a second down 1m, and at the last 10 or 20% we had to use as much force the lift the weight up again 1m in .5 of a second, what force would this take. Thx in advance for your time if anyone is able to work it out, here are the calculations a friend worked out, hope they are right not sure.
Let's assume we're comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets.
Known: Mass(m) =100kg (220lbs) Acceleration(a)=?? Distance(d)=1m
First let's solve for the acceleration required to move the weight 1m in the time frames of the sets.
Calculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):
d=1/2at^2
1m=1/2*a*(0.5s)^2
a=2(1m)/(.25s^2)
a=8 m/s^2
Calculate a, to travel 1m in 2s (Raising set 2):
d=1/2at^2
1m=1/2*a*(2s)^2
a=2(1m)/(4s^2)
a=0.5 m/s^2
Calculate a, to travel 1m in 4s (Lowering set 2):
d=1/2at^2
1m=1/2*a*(4s)^2
a=2(1m)/(16s^2)
a=.125 m/s^2
Now lets solve for the forces required to accelerate the weight
Calculate the force required to raise 100kg, 1m, in 0.5s:
Sum of forces=ma
F1-mg=ma
F1-(100kg)(9.81m/s^2)=(100kg)(8m/s^2)
F1=(981+800) kg m/s^2
F1=1781 kg m/s^2 required to raise the weight 1m in 0.5s
Now compare it to the force required to raise 100kg, 1m, in 2s
Sum of forces=ma
F2-mg=ma
F2-(100kg)(9.81m/s^2)=(100kg)(0.5m/s^2)
F2=(981+50) kg m/s^2
F2=1031 kg m/s^2 required to raise the weight 1m in 2s
***As predicted, it takes more force to raise the weight 1m in 0.5s than it does to raise it in 2s. It takes F1/F2=1.73 times as much force to do so. You're right about this but nobody is disagreeing with you here
However, now let's look at what happens on the way down.
Calculate the force required to lower 100kg, 1m, in 0.5s:
Sum of forces=ma
mg-F3=ma
(100kg)(9.81m/s^2)-F3=(100kg)(8m/s^2)
F3=(981-800) kg m/s^2
F3=181 kg m/s^2 is required to lower the weight 1m in 0.5s
Now compare it to the force required to lower 100kg, 1m, in 4s
Sum of forces=ma
mg-F4=ma
(100kg)(9.81m/s^2)-F4=(100kg)(.125m/s^2)
F4=(981-12.5) kg m/s^2
F3=968.5 kg m/s^2 is required to lower the weight 1m in 4s
***Contrary to your belief, it takes MORE force to lower the weight 1m in 4s than it does to lower it in 0.5s. In fact, it take A LOT more. It takes F5/F3=5.35 times as much force to do so! That's 5.35 times more force to go slow than fast!!!![/u][/B]
Here is a video of me doing both lifts.
YouTube - waynerock999's Channel
Wayne