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Cody Clark

Iron Cross Training

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Razz

Yeah lol I'm using almost the whole stack and I'm a small skinny kid :lol: oh and it's with perfectly straight arms of course :D

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jpshepard
I can do cable crossovers for 5-10reps with like 50kg per hand and I'm almost, if not, able to hold a cross. at 73kg bw.

For the calculations I'm aware that the muscular force will be muuuch greater than bodyweight. In fact in a simplified physics report I did we calculated the combined adductor force of each arm to be 20x bw. Using some representative tendon insertions.. However the actual upward force and downward forces should be equal and the dumbbell held should be somewhat close to ½ bw in each arm, anything else seems like miscalculations to me. When holding a dumbbell it's also at the same leverage as your body, if you had to hold 1.5x bw in each arm then you're implying that the leverage is much greater for the upside down dumbbell exercise which I don't see is true.

You are actually right, I am talking about torque force and you are talking about weight. so half body weight in each hand should equal the same torque force at that distance from the fulcrum. I stand corrected... I still have yet to find a conversion formula for torque to linear force, when I do I can calculate what is actually happening during an iron cross. When i return to class next term I will be sure to ask my professors about this and other gymnastics skills.

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Marlon

*edit* i see that a lot of the discrepancy here was dealt with between razz and jpshepard already, making alot of what i just wrote very wordy and redundant :oops:

jpshepard, I respectfully disagree with you.

I believe you have a misunderstanding about the way the body generates force during straight arm work. You are claiming that in order to hold an iron cross a person need to generate more downward force through their hands than is equivilant to the force of gravity acting upon their body as a result of the poor leverage involved in such straight arm work (or any work in general really).

If this is true, then your example of loosening a nut with a wrench would most certainly not be accurate. If what you are saying is correct, then using a wrench / lever of any length to unscrew a nut somehow increases the total amount of torque that must be exerted on the nut itself to loosen it. This simply isn't true, the amount of torque required to loosen a nut is constant, the only thing that changes when you use a long lever to unscrew a nut is the amount of force you must apply at the end of the lever.

If we look at a human being doing an iron cross as a machine, then the part of the machine where the poor leverage involved in holding an iron cross becomes apparent is in the shoulder joint, not at the ring itself. The amount of downward force it takes to suspend a particular amount of weight from a pair of rings is constant, the force that needs to be generated at the fulcrum of the "machine" is, however, drastically more than the required downward forces on the rings.

Beyond that, I'm not trying to say that the equation we would need to use to figure this out would be simple in any way. In fact, I would venture to say that whatever equation(s) you used to calculate the force required to do an iron cross is far far too simple, and nothing close to an accurate representation of how the body generates force on rings. That isn't simply because you are missing certain variables like the inward pressure of the rings, and tendon insertion points in the arms of the person used as an example. It's because the shoulder joint just plain isn't analogeous to the example of a tight nut being loosened by a wrench or a simple lever. We aren't just dealing with roational force. There IS purely downward pressing involved. You merely need to shrug your shoulders to understand that they can produce purely downward force even when you arms are held out at your sides. You are dealing with some combination of squeeing inward, pushing directly down, and generating rotational force through you shoulders while doing an iron cross.

If you don't believe that, just look at other straight arm strength positions held on rings. A maltese for example, is by definition an iron cross with the body held horizontal to the ground. your body generates force through a lever of the same length, with the "fulcrum" being in the same place (at the shoulder), but no one is going to argue that when you do a maltese you push down. Sure there is rotational force involved (although I sure couldn't tell you what axis or axes it is being exerted accross) but the majority of whats going on is direct downward pressure, generated at the shoulders through the hands.

/end rant

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Joshua Naterman

If you take into account the slight outward angle of the rings and the slight inward pressure this creates we may end up needing very slightly LESS than half bodyweight in each hand to hold a cross. Probably not a very significant difference though...

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jpshepard
If you take into account the slight outward angle of the rings and the slight inward pressure this creates we may end up needing very slightly LESS than half bodyweight in each hand to hold a cross. Probably not a very significant difference though...

Slizz,

Ah, I think we are getting some great conversation about bio-mechanics and physics here. So I agree the ft/lbs of torque may translate to about 50% bodyweight. I have yet to find a rotational force to weight formula/calculation. I think the fact that the body shape is pushing out the rings and thus must pull them back in with torque, it will require more not less force. Because if you are parallel with the floor and you don't produce enough force the rings wont keep you up, they will continue the path of the arms and you will fall downward. so I think it is slightly more rather than less. if the rings had more weight to them i would see how the angle of them could give help to the individual but because of their little weight i don't see it as an aid. I would sure love to see the notes of the guy who invented them. Any Physics buffs or PTs on the forum?

JP

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Blairbob

Simply contact Roger Harrell from Drills&Skills about the calculations.

As well, false grip in crosses probably changes the numbers as well.

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Newguy

Hey, as a side note, do you know if Drills&Skills is ever updated anymore???

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Blairbob

Not really a priority but it stays as an archive as people still like to read it.

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Newguy

Yeah, I love the site, Just its never updated anymore

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tyciol

If someone has access to an adjustible cable pulley, wouldn't it be simpler to just do straight-arm pulldowns (but with hands to sides instead of front) rather than going to all that complication of hanging upside down just to use freeweights?

I mean yeah, since the pulley origins are probably going to be wider out and not directly above the hands, there may be some outward traction on the shoulder and people may need to actively adduct their scapulae a bit while doing this, but it doesn't seem like a huge factor.

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Joshua Naterman

It is totally different, a regular cable crossover machine won't work. You would want to use a freemotion-type machine that allows you to use a more realistic angle.

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