Physics
12th October 2007
12th October 2007
A Guide to Danger
The following still won't improve your lifting one iota.
In a Force I covered the basic force equation;
Force = Mass's Weight + Mass's Acceleration = FT = FW + FA = mg + ma = m(g + a)
I further noted that it was the muscles that generated the force FT, i.e. FT is not mysteriously generated from acceleration.
Here I will discuss the 'dangers' of fast reps.
In the following video I do 183 kg × 10 reps;
Some might take this to be 'dangerously' fast and caution me to slow the rep at the top... let's see.
Take my 3rd rep;
t = 1.200 seconds, s = 0.466 metres, g = 9.807 m.s-1
The time and distance figures are for the positive part of the rep. The camera runs at 30 fps so I was able to determine the time for the rep. From a still on the camera I was able to measure the distance in pixels that the bar moved and convert it to distance.
I can thus calculate average velocity;
v = s/t = 0.466/1.200 = 0.388 m.s-1
Lets assume that I must decelerate at the top of the move to zero velocity. Under the influence of gravity alone it would take the following distance to do this;
s = v2/2g = 0.3882/2*9.807 = 0.008 m = 0.8 cm
Yes, it only takes 0.8 cm for the weight to decelerate from 0.388 m.s-1 to 0 m.s-1!
But, you might say, the peak velocity is higher than average. Let's say peak velocity is 0.5 m.s-1 (29% higher than 0.388 m.s-1)
s = v2/2g = 0.52/2*9.807 = 0.013 m = 1.3 cm
Yes, it only takes 1.3 cm for the weight to decelerate from 0.5 m.s-1 to 0 m.s-1!!
Thus when I lockout I make no attempt to slow the weight because the 0.8 cm to 1.3 cm of distance is absorbed by bar recoil and the body's natural elasticity.
In fact if you look at powerlifter do speed reps with ~50-60% of max they make no attempt to slow at lockout.
There is another reason not to fear these 'dangerous' reps... the reversibility of Newtonian mechanics.
Take a jump squat, surely the forces on landing are the epitome of 'dangerous';
If the lifter bends the knees by the same amount on landing that he did on jumping then the forces are the same because;
a = s/t2
You can experience this quite easily. Take an empty bar and do a small jump squat. You can do a small jump squat can't you? Make sure to dip by a similar amount on landing to that of taking off. So if you dip by 0.2 metres to jump then bend by 0.2 metres on landing.
Try a higher jump, try a bigger weight.
I admit that when I first tried it I was apprehensive about 'dangerous' landing forces, but in fact (just as Newtonian mechanics shows) I found the jump squat to be very smooth and not at all 'jarring' even with large 'irresponsible' weights...
12th March 2008