Mechanics of an Iron Cross

[freaky trickster]

Howdy,

I am a recreational gymnast, and I was trying to discuss the mechanics of an iron cross with one of my training partners. He proclaimed that a taller person should have a harder time doing an iron cross, and he even set a limit and stated that it was virtually impossible for anyone over 5'8" to do an iron cross. He tried to tell me that the force experienced by the performer increases with the square of his wingspan.
First of all, I know that torque is directly proportional to distance from the center of rotation, not proportional to the square of that distance. I tried to disprove his theory by explaining to him the mechanics of a fulcrum.

Later on, I reconsidered my position, and realized this: While it is true that torque increases linearly with respect to distance from a fulcrum, that is only part of the problem. A taller person will not only have a center of mass that is farther from the rings, he will also be heavier. And it seems to me that as a person gets taller, assuming he maintains the same proportions, his mass will increase proportionally to the cube of his height. Hence... the torque that an athlete experiences while performing an iron cross is proportional to the 4th power of his height.

Am I right about this? Would a 6 foot man really need to be twice as strong as a 5 foot man to be able to hold an iron cross?

Here is a pic to help you visualize the problem / (to show off) :) (http://www36.websamba.com/jonathanbarlow/trickri_old/pic_gym4.jpg)

Thanks for the help!

-Jonathan

[ZapperZ]

Personally, I LOVE questions such as these. :)

The problem here is that there can be way too many variables when one tries to compare a 5 ft person with a 6 ft person, such as difference in weight (mass), different in muscle mass, etc. So let's try to simplify it by doing just a simple hypothetical comparison. We assume that both of them weigh the same. The only difference being that the 6 ft person has a longer wingspan than a 5 ft person.

If you do this, then it is clearly obvious that there is more torque (or moment) required by the 6ft person to hold himself up because T = r x F = rF, where r is the length of his arm, and F is his weight. So if you go by just this, then we already can see that the 6ft person requires a lot more force to maintain himself up.

If you remove other assumptions one at a time, then it gets a bit more complicated. The 6ft person will tend to be heavier, so that compounds the problem and the weight. But he could have larger muscle mass, so he could still maintain the position.

Now, the square of the length thing is a but puzzling, unless your friend is thinking of the moment of inertia, which does depends on the square of the "radius". However, this would only come into play if you are attempting a spinning move, and thus, need to generate an angular momentum. I don't think this is what you're doing in maintaining the iron cross.

Zz.

[freaky trickster]

Thanks for the response Z!

You confirmed most of my thoughts. However, let us assume that the athletes are of the exact same shape. I think that the mass would increase with the cube of the height, since height would increase, as well as the radial thickness of the limbs, etc. So, with:

T = r * F

r increases, and F increases with the cube of r, so T increases with the 4th power of r, or:

T2/T1 = (r2/r1)^4

If we plug in the numbers 6 feet and 5 feet we get:

(6/5)^4 ~= 2

This means that the 6 foot man experiences 2 times the torque of the 5 foot man.

This seems unreasonable to me, yet my reason has led me to this conclusion.

[ZapperZ]

Thanks for the response Z!
You confirmed most of my thoughts. However, let us assume that the athletes are of the exact same shape. I think that the mass would increase with the cube of the height, since height would increase, as well as the radial thickness of the limbs, etc. So, with:
T = r * F
r increases, and F increases with the cube of r, so T increases with the 4th power of r, or:
T2/T1 = (r2/r1)^4
If we plug in the numbers 6 feet and 5 feet we get:
(6/5)^4 ~= 2
This means that the 6 foot man experiences 2 times the torque of the 5 foot man.
This seems unreasonable to me, yet my reason has led me to this conclusion.

I think you did what a number of physicists sometime do - assume that a cow is a sphere.

:)

There's nothing to prevent you from doing that, of course, but wouldn't estimating the volume of a person to be a cylinder to be more accurate? If that is the case, then the height and the lateral size of the person are decoupled, and the volume depends on the square of the lateral radius.

However, note that this is not the radius that I used earlier, which is the wing span. This is the radius of the cross-section of the cylinder. You also used the same "radius" of the volume of the person to be equal to the wingspan. Is this realistic? A person THAT fat would never be able to suspend himself on the rings, much less doing an iron cross!!

:)

Zz.

[pervect]

The "Body Mass Index" formula assumes that a person's weight should be proportional to the square of their height. I couldn't find out much about the justification or origin of this formula, but it does seem to support Zapperz's "cylinder" perspective.

see for instance
http://info.cancerresearchuk.org/healthyliving/bodyweight/bodymassindex/

If we assume that height is proportional to arm length (reasonable), then the moment (strength required to do an iron cross) should go up with arm-length^3.

[freaky trickster]

Ha! Yes, you are correct. I was unclear before. I was assuming that both people were of the same proportions. We could assume the people to be cylinders, so that the ratio of the height of the cylinder to the radius of the cylinder is always the same. If this were the case, then an increase in the height would also increase the radius, and therefore the volume would increase with the cube of the height.
In other words, I'm assuming that the 2 people look exactly alike

[freaky trickster]

Oh, weight is proportional to square of height? That is interesting. Wouldn't that make taller people look skinnier?

[pervect]

If we compare someone who is 4 ft high with someone who is 6 ft high, the BMI formula says that their waist measurement should grow by sqrt(1.5) = 1.22 times larger. This seems closer to being right to me than the linear formula which says that their waist would be 1.5x larger. I haven't been able to find any theoretical justification of the BMI formula, but it does seem to be widely used.

[freaky trickster]

Thanks Pervect.

Then, according to the BMI, taller people are required to be skinnier for optimal health. This is probably due to the fact that bones, muscles, and connective tissue have a maximum amount of weight that they can support. Also, the heart and other organs probably have a limit on the amount of matter that they can support.

Interesting stuff.