Stretch

This was mentioned once in this (Front Loaded Squats) thread:

If I am 6'6" tall and can squat 225 for reps, and my friend is 5"8" and can squat 365 for reps, how much more work is he doing than I am? I've heard comments like "Yeah, but look how far he has to move that weight", and not necessarily only during squats. Is there any truth in this?

P-funk

you would have to measure your torso, more importantly the placement of the bar on your back (resistance arm) to your lower back (fulcrum) and then the length of your hamstring tendons and where they attaches at the tibia and fibula (force arm).

then you do the math with the weight you are lifting (resistance) where "f" is the force being applied (what you are measuring), ra= resistance arm, r= resistance (the weight lifted) and fa= force arm.

f x fa=r x ra

I think that is right. It is tough because somewhere in there you would have to account for the quad muscles extending the knee so they would also be in there as force arms.

http://pwtraining.blogspot.com/.....come and see what is on my mind!

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Stretch

So, because the distance from my knees to my hips is longer than most, I would have a longer force arm, right? So with a longer force arm, less force is need to overcome a given resistance (weight)?

P-funk

It is more then that. it is basically a lever system. You have a force arm and a resistance arm both are seperated by a fulcrum, in this case the lower back which serves as the axis of rotation. It is a first class lever (like a see-saw). If you are to much in one direction (ie, longer toros or short muscle attachment) you don't have optimal force aplication.

Just plug numbers in....You are 6'6". So say you have a torso length of 30in. (bar to lower back) and a hamstring tendon attachment length of 5in. While your buddy at 5'8" has a torso length of 24in. (bar to lower back) and a hamstring attachment length of 5in also since we are talking about the tendon length from its start to its attachemnt point. You both go to lift 220lbs.

so for you
f= what we are trying to find out
RA= 30in
R= 220lbs
FA= 5in

f x 5 = 220 x 30
f x 5= 6600
f= 6600/5
f= 1320lbs force inches


your friend
f x 5= 220 x 24
f x 5= 5280
f= 5280/5
f= 1056lbs force inches.

You lifted 264 more lbs in force incehs then he did.



*disclaimer* I think I did this correctly but I am not 100% sure. I was not that great in physics. I know the equation is right but I am not 100% sure that I set up the proper numbers. At any rate you can see where I am going with it and someone that has studied physics more than I will be able to come up with a similiar answer. I will come back to this thread in a few years (after I work on my masters in kineseology...lol).

http://pwtraining.blogspot.com/.....come and see what is on my mind!

Ivonne's Blog on Health and Wellness!

Optimum Sports Performance

"In the beginners mind there are many possibilities, in the experts there are few."
-Buddha's Little Instruction Book