Difficulty vs Angle of Arms in Iron Cross

[hooligan8200]

TL;DR My model of an iron cross is D = sin(Θ)[1-cos(Θ)]. See graph below. Post more accurate model.

This was being discussed in another topic, but I didn't want to derail the thread with my rambling. In said thread, people were discussing the difficulty of an Iron Cross versus the Angle at which your arms are at. The general consensus is that at angle 0, arms next to the body in a support position, the difficulty is 0% and at 90 degrees, arms perfectly straight out in a perfect iron cross position, the difficulty is 100%. Difficulty just refers to a persons relative difficulty of performing the iron cross, since variables such as weight, arm length, etc will all change from person to person.

What I didn't agree with in said thread was that at 45 degrees, the difficulty is at 70.71...%. If anybody has tried comparing a 45 degree cross to a 90 degree cross, they can feel this is not true. When I try to compare the two, it almost feels like the 90 degree hold is more than twice as hard as the 45 degree hold. So I'm just going to try and throw out a slightly more accurate model. Not going to be the most accurate thing in the world, also probably has some things wrong with it.

D = 100*sin(Θ)*[1-cos(Θ)]

lolwut where did I pull that from?

The 100 is just for converting from .707 to 70.7%, no big deal. Just makes things easier to read when it's graphed.

The sin(Θ) is what everybody has been using originally. It refers to the strength required to hold yourself up in an iron cross. The further your arms go out, the longer the lever arm is, the more strength is required. Again, no big deal.

But then there's the [1-cos(Θ)] part. That refers to the effective weight you are holding up. As your arms go further out, the weight used in calculation effectively increases. At 0 degrees, you essentially weigh nothing. At 90 degrees, you are 100% of of your actual weight.

Weight changing... dafuq?

Well, not really. Personal trainers, anatomy and physiology majors are all going to need to bear with me. I am not in a field even remotely related to health and fitness so you guys will all need to contain your rage at my explanation. Basically, there's 2 "connections" (in my little world) to the rings. Not 2 hands, but rather 2 types of connections. "Boney" and "Muscley" connections.

A "Boney" connection basically means that the skeletal structure of our body is holding us up. Eg. in a support hold, the force is driven straight through the skeletal structure of our arm, and the rest of our body sort of hangs off of our shoulders. This means that NO strength is required in a support hold. It's easier to look at us standing up. While standing, the force to keep us up is driven directly through the bone structure in our legs, and to the rest of our body. It basically requires no strength to keep us standing up.

Now it should be fairly obvious that you need SOME strength to maintain stability and keep correct posture and form in either a support hold or while we are standing, but bear with me. I'm simplifying. We are only looking at the strength required of an iron cross, or even inverted cross as this model should work for that as well.

A "Muscley" connection basically means that the muscular structure of our body is holding something up. Imagine you're holding a weight in your hand. You are standing and your arm is just sagging next to your body. Now imagine you are holding the weight directly out to the side, like an inverted cross. The skeletal structure of your body is doing nothing to hold this weight up. Rather, it is the muscular structure of your body holding it up completely. This would be considered a muscley connection. It is 100% pure strength and you are receiving no assistance from the skeletal structure of your body. In between, at 45 degrees, it is equal parts bone and muscle doing the work. Since we're assuming your skeletal structure can handle this weight, we factor it into the equation by subtracting out the force it provides. This results in you holding a weight that, well... weighs less. You can represent this by the [1-cos(Θ)]. As you increase the angle from your body, the effective work by your skeletal system lessens, and the weight "increases".

Now keep in mind that we are talking about difficulty. Not total weight or torque or whatever you want to label it as. We are talking about difficulty since from person to person, it will vary. Maximum difficulty just means that it is the most difficult a person can make it on their muscular system.

If you still are struggling to understand, imagine you are holding yourself up between two walls using only your arms. The walls have an infinite friction, so you won't slip, and the walls also don't hurt your wrists due to the angle your holding yourself up at. The thing is, you can't push downwards like an iron cross because you can't produce a torque about your shoulders. So you are forced to push straight into the wall. When they are close together, it should feel extremely easy to hold yourself up. Now they begin to slide apart. As they get further and further apart, you need to start driving your hands into the wall to hold yourself up. Once they are far enough apart such that your arms are at a 90 degree angle, you can't push into them anymore, thus resulting in you falling. This is the "boney" connection part. Now imagine that same scenario except there are handles that you can hold on to and your shoulders are working properly. This allows you to push downwards like an iron cross. This adds the muscley component, and gives you the ability to apply torque to keep yourself up at 90 degrees. It's sort of like a slow transition from bone to muscle as you go from 0 degrees to 90 degrees.

And here's a graph of the equation. D = 100sin(Θ)[1-cos(Θ)]

Obviously not a perfect model as rings add a whole new level of instability. My model sort of assumes that the rings are replaced with 2 walls with handles on them. Hopefully it's a step in the right direction from the old model of D = sin(Θ). This model helps explain things such as "Why is a perfect cross MUCH harder than a 45 degree one?" and "Why are the last 30 degrees the most difficult?". Feel free to offer improvements and point out mistakes, because they're probably in there. Let's hammer out an accurate Iron Cross model that can be used from person to person. Someone more intelligent than I am could probably find a way to sneak arm length, weight, and all that other good stuff in there. Offer your own models and reasoning behind them, aaaaaand... FIGHT!

[Ian Myers]

I'm a sucker for making models of stuff like this, but Wouldn't it be easier to just train by putting the rings on your forearms and over time moving the rings closer to the hands? This way you could simply measure difficulty as d=a/r where r is the distance from the ring to the attachment of the lat to the upper arm and a is the distance from the palm to the attachment of the upper arm.

For angle measurement, I'm tempted to try my hand at an equation :wink:

[hooligan8200]

I'm a sucker for making models of stuff like this, but Wouldn't it be easier to just train by putting the rings on your forearms and over time moving the rings closer to the hands? This way you could simply measure difficulty as d=a/r where r is the distance from the ring to the attachment of the lat to the upper arm and a is the distance from the palm to the attachment of the upper arm.

Noted, and updated equation. My equation only assumed you were holding the rings on your hands. Your equation assumes the person is in a perfect cross, but slowly moves the rings out. I combined them and got this. While it WOULD be easier to move the rings out as you progress on the cross, finding this stuff out for the angles is still fun. Also, since I don't use any cross trainers, finding this stuff out for angles helps me a bit. I'm currently at like 10 degrees high on an inverted cross for a 2 second hold.

D = [a/r]*sin(Θ)*[1-cos(Θ)]

r = 100. Using percent instead of actual lengths. Same difference.

For angle measurement, I'm tempted to try my hand at an equation :wink:

Do it! Mine right now shows that at 45 degrees, it's like 25% difficulty. My guess from experimentation is that it's closer to 35-40%. I couldn't come up with anything better so I just settled for what I had.

[Joshua Slocum]

'Exerted force' might be a better label than 'difficulty.' In order to calculate difficulty you'd need to have some sense of how the mechanical advantage varies over angle, due to different shoulder positions and different muscles being engaged. Unfortunately my knowledge of body mechanics is woefully inadequate, so I can't make further predictions of how this would affect the chart. If you had a counterweight system, you could calculate the force an athlete needs to exert based on their bodyweight, and have them try to hold a cross at different angles with different weights to get a sense for how the arms' maximum exerted force varies with shoulder angle.