There are two ways to drive the hips up: one where you try to keep the knee position stationary, touching the TUBOW the whole time; the other where you allow the knees to move away from the TUBOW during the hip drive.
Are you supposed to try to minimize the change in the angle of the tibia? Or does it matter, and if so why?
Most people's knees go back a little as they initiate the squat. You don't want too much knee extension without some concomitant hip extension, however. As long as the person isn't losing their balance or doing a good morning, knees going back at the start is not a problem. You also don't want them to rebend or go forward in the middle of the ascent, either.
You mean "as they initiate the hip drive," right?
So ideally the knees are perfectly still and the only angle you are changing at the start of the hip is at the knee vertex, and not (also) at the hips. I remember Rip saying it is normal for the back to become more horizontal as you start the hip drive. So I'm guessing that the ONLY thing that is supposed to change at the start of the hip drive is the angle at the knee vertex. Or is the hip angle supposed to increase as well at the start of the hip drive to keep your back angle relative to the ground fixed at 45 degrees?
"Initiate the ascent" would have been clearer yet. Hip drive is a component of the ascent on the squat.
If the knees stayed in place with respect to the toes, but extended and the hips did not also extend, the bar would move forward of the midfoot. Also, given that almost every human initiates the squat by pushing the knees and hips back slightly, I don't know that trying to do something else is advantageous.
As mentioned above, I don't think keeping the knees perfectly in place on the beginning of the ascent is necessary. I am not sure if it is even possible. The hip should also extend enough to keep you in balance, even if the torso becomes slightly more horizontal initially.
Related question, if I may.
I've always wondered why the model recommends establishing the knee position 1/3 to 1/2 of the way down the ascent. Surely it's possible to instead gradually bring the knee forward continuously so that it ends up in the correct position by the end of the descent, no?
I'm not saying the model is wrong, in fact I think it's a good idea, but I've always wondered exactly why. [ed note: Reasons are discussed in Starting Strength]
My guess at an answer is this:
Are there any other reasons?
You are correct. You can bring the knees forward continuously during the descent and many people do. It often results in non-ideal positioning at the bottom when that happens, however. Your second enumerated reason is largely why. You might also get a stronger stretch reflex out of the bottom when the knees are set early, too.
I'm in an introductory to mechanics course (calc based) and my class is in a chapter pertaining to work.
Now, W=F*(Delta)r*cos(theta) so work needs not only a force but a displacement.
Both the book and my professor go on to give examples like this:
"Say you picked up a chair with one arm and held it in the air, with your arm completely straight and perpendicular to you body, for some time x; you are performing no work because while you are you applying force for some time x on the chair, the displacement of the chair in some time x is zero, so the work done is zero."
I understand the definition and the explanations due to the formula and I know minds much smarter than mine came up with this shit but I just have a hard time believing that when I see weightlifters hold 200kg+ bars over their heads for a few seconds and wait for the signal to drop the bar, that the lifters are doing no work.
Maybe I am missing something, maybe I'm not thinking hard enough on this...I don't know. I just don't like it.
I think you don't like the word "work" as applied to an activity where ATP is being expended even as no motion is occurring. Work is indeed motion-based, and just because something is hard doesn't mean it's "work." The work the lifter did against gravity was the vertical displacement of the bar x the weight of the bar ("weight" being the term I'm using precisely). Holding it there for the down signal is "hard" but not "work."
Tell me about a Farmer's Walk.
If you're carrying something at a constant velocity, all the displacement is entirely perpendicular to the force(Farmer's Walk), our angle is 90, and cos(90) returns zero as well.
So just remember, the next time you do heavy ass farmer walks, you're doing zero work.
Yeah, that's hard for lots of people to accept. But it also is the basis for our vertical bar path rule, which is very hard to refute. Because of the math.
How to Power Snatch –Josh Wells
Fear in Barbell Training (and How to Overcome it) –Steve Ross
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