Bar Whip on Heavy Sets of Squats Bar Whip on Heavy Sets of Squats - Page 2

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Thread: Bar Whip on Heavy Sets of Squats

  1. #11
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    Quote Originally Posted by Kyle Stevens View Post
    This lifter is not using ALL Bumper plates to get 505 on the bar, as I don't think that is possible. I had to double check after I posted this as I realized that it's not possible. He has 2 red 50lb bumpers and sandwiches iron plates in between. At his level of experience under the bar I would imagine he wouldn't use the bumpers if he didn't have to. He is planning on starting a home gym this summer when he moves out of his apartment, that in my opinion is the best route.
    Andrew your post made my brain hurt, but you summed it up nicely at the end. If you're lifting that much you need cast iron or calibrated plates.
    50# bumper plate doesn't sound right. You mean 55lb/25kg?
    One bumper plate and 4 iron plates isn't going to shift the CG outwards THAT terribly much.
    If you mean 2 plates per side, with irons inbetween those. I'm not sure that would do it (that bad) either.
    Sure, bar whip is a thing. Plenty of people have squatted 500-600 pound without calibrated plates, and with bumpers.

    Even if he HAS to put those 2 reds on there, seems to me to use up the 3 iron plates first closest to the collars (inside).
    Then put the reds outside.

    He more than likely has a small technique problem in the bottom.

    But yeah, in his position to going to a public gym, buying a ~$300 bar is the best solution rather than buy a bunch of (thinner) iron plates for the gym.
    When he starts his home gym, he'll already have the/a bar.

  2. #12
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    Quote Originally Posted by TommyGun View Post
    Andrew, we have been waiting for your article about this topic. Let’s get after it.
    It's been in the back of my mind for a while. I'm doing testing at the moment, so it's going to take a little more time.

  3. #13
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    Quote Originally Posted by AndrewLewis View Post
    Deflection of a bar is generally governed by the following equation:

    Deflection = (FL^3)/("Some Constant"*E*I) where I is the second area moment of inertia. In the case of a rod, I = Pi*(r^4)/4, so we can see that the equation becomes

    Deflection = (FL^3)/("Some Constant"*E*r^4).
    E (elastic modulus or Young's modulus) is essentially the same for all steels.

    Therefore, the only factors you can controls are:
    1) F, Force - which we are not going to modify for the sake of reduced deflection
    2) L, Length - which is the distance from the center of mass of the plates (sort of) to the center of the bar.
    3) r, Bar radius


    Let's examine the Ohio Power Bar.
    You can buy the Ohio Power Bar at 29 mm bar instead of a 28.5mm bar which is a 1.8% difference. A 7ft bar with collar to collar length of about 54 inches will have a length from collar-to-center of 26". If you can go from 5 rubber plates to 5 cast iron plates, you reduce the distance between the plate center of mass and center of bar from about 34.1in to 29.3 (a 4.8in difference or 14% change).

    When we account for the power factors, it is shown that increasing the radius will decrease the deflection by 6.7% and going from rubber plates to cast iron plates will decrease the defection by 57.6%.


    The conclusion being that the biggest realistic factor in affecting the whip of the bar is going to be the distance from center of bar to center of mass of plates.
    An exacerbating factor is the weight of the collar, which has to be increased* on the smaller diameter bar in order to maintain the 45 lbs/20kg nominal weight. Thus, the smaller diameter bar gets a second penalty in terms of deflection.

  4. #14
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    Quote Originally Posted by Satch12879 View Post
    An exacerbating factor is the weight of the collar, which has to be increased* on the smaller diameter bar in order to maintain the 45 lbs/20kg nominal weight. Thus, the smaller diameter bar gets a second penalty in terms of deflection.
    I don't think that would matter much. The added weight to each side at the collar would be a little under half a pound per side if my math is right. The added distance would only be 1.2mm per side. This is all assuming you're going from a 29mm bar to a 28.5mm bar.

    And I didn't even account for the fact that the bar itself bar is a distributed load, not a series of point loads.

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