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Fitness Theory and Practice. CrossFit's rationale & foundations. Who is fit? What is fitness?

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Old 02-03-2007, 09:00 AM   #1
John Schneider
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So, ever since the CF seminar I attended, I've been thinking about workouts in terms of work load and increasing my work capacity. I've recently started a new training log with columns for the workout and the score, but I've also included work FxD and work capacity (FxD)/t.

This was looking good until the 2FEB WOD. I'm deadlifting 225# 45 times which moves the bar maybe 2ft? So that comes to 20,250ft-lbs. And I run a mile and a half, which means I'm moving my 195# overweight butt 7920ft. Does that really mean running 1.5miles=1,544,400ft-lbs? And I finished in 16:30 so that would come to 94,827ft-lbs. That's jsut rediculous and I have to be missing something. Otherwise, wouldn't that make running the ultimate exercise for increasing work capacity?

I know I'm obviously missing something here, so please be tolerant of my mental short comings. I'm surely not the sharpest tool in the shed. I'm just trying to get the concept correctly.
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Old 02-03-2007, 09:10 AM   #2
Craig Van De Walker
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No I don't know exactly how to calculate running work but ft/lbs is how much vertical distance the weight is lifted.

Your calculation would only be correct if you climbed a 1.5 mile high ladder. I only know this because I tried to figure out work done by running and gave up.

You might figure out the work of rowing the same distance on a C-2 and claim it as an approximation.

My tool sharpness is also not up to this task.
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Old 02-03-2007, 11:47 AM   #3
Steven Low
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Lets think about it logically. When you run, you are pushing with full extension through your legs to elevate you maybe a few inches off the ground in your next stride.

Therefore, you need to determine 3 things to find your answer (with calculations to find it, of course):
1. the length of the stride
2. how much your center of gravity is moving up in the stride
3. how many strides in the given distance

With 1 & 2 we can calculate the force vector per stride given that we have the (1) vertical and (2) horizontal components of the force. Then use the number of strides to calculate the total force used and then divide by time for your answer.

(Message edited by braindx on February 03, 2007)
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Old 02-04-2007, 01:23 AM   #4
Andres Diaz
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> Does that really mean running 1.5miles=1,544,400ft-lbs?
Of course not.

Steven, i don't think your model suits this physics scenario, the center of gravity fluctuation would be minimal.
Most part of the energy would be wasted on friction between the leg and the floor, one composite of which is proportional to the weight of the moving object and the roughness of these two surfaces.
Anyways, calculating this way you would get Joules (or it's non-metric work equivalent).

That was a pretty formalized short-time scenario.
Of course, it would work only when you are walking (in contact with the surface all the time), otherways friction energy got to be substracted.

I think what you really need is a reference:

My lifting workout makes me do ~30000 Joules. And according to this site, you would make ~1000 Joules in a mile run. So i think it's rather accurate. They also take into account the running time, cuz it's a very different story when you are floating on air half of the time.

(Message edited by epsa on February 04, 2007)
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Old 02-04-2007, 12:02 PM   #5
Steven Low
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Hmm, I thought the model would fit because the center of gravity fluction isn't minimal... it's "average" is basically each strides combination vector of the movement of the horizontal and vertical change in your center of mass.

Although now that you found/brought up friction, your model is probably right. I am going to try to calculate something based on my scenario though cause I still think it would be somewhat close, lol.
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