The iron cross is one of the most difficult skills on the rings. To practice, we sometimes use a pulley strap to help support. Or, one could use the strap that hold the rings and pull it out so that it would support part of our arm while attempting the iron cross. When one sees gymnasts in the Olympic do the Iron Cross, do they really hold the iron cross with a ninety degree angle formed with the arms and his torso? A physics investigation of the iron cross would explain.

Here is an article from Boston University:

Although many people do not realize it, physics is involved in every aspect of life. Gymnastics is one of many sports that is performed by combining the machinery of the human body and basic physics. One particular example is the ring exercise. The body of a gymnast performing the ring exercise is subject to extreme stress, coped with by an integration of physics and biology. Although many elements of physics are incorporated into performing this procedure, I will primarily focus on one in particular- tension. By applying Newton's second law under equilibrium conditions, I will examine the question of whether or not the perfect "iron cross" maneuver on the ring exercise is physically attainable. Do you think a gymnast can support his weight on the rings having his arms form a perfect ninety degree angle with his torso?

The basic formulas I will use to answer the aforementioned question are as follows:

The net force in the x direction = 0

The net force in the y direction = 0

By combining these simple equations with basic trigonometry, I will solve the problem.

The conditions for the problem I will solve are as follows: a 655 Newton gymnast is suspended in air by holding on to the rings. However, the rings are not the focus of the problem, because the object upon or with which the gymnast supports himself is not under examination. (In fact, one could substitute two solid surfaces or two bars for the rings.) If the gymnast were to hold on to the rings so as to create 60 degree angles between his armpits and a plane just below his armpits which is parallel to the ground, what would the tension have to be in his arms to support his weight? Assuming equilibrium in the y direction:

Tsin 60 + Tsin 60 -655= 0--->2Tsin 60 =655-->T=378.2 N

What if the gymnast were to create 35 degree
angles between his armpits and the imaginary plane, what
then would be the necessary tension in each arm?

Tsin 35 + Tsin 35 - 655=0----> T= 570.98 N

Now
pose the same question with 15 degree angles.

Tsin 15 + Tsin 15 - 655=0----> T= 1265.4 N

As you can see, the closer the gymnast gets to forming angles of zero degrees between his armpits and the imaginary plane, the stress (in the form of tension) applied to his arms increases significantly. This is simply because as the angle decreases, so does the value of the sine function of the angle. In order to compensate for the decreased sine value, the tension force is increased. Eventually, the tension required in each arm will become impossible to sustain. I will solve for the tensions required for two more angles to demonstrate this point. If the angles were 2 degrees:

Tsin 2 + Tsin 2 -655=0----> T= 9384.1 N

If the angles were .001 degrees:

Tsin .001 + Tsin .001 - 655=0----> T=18764367.8 N

In conclusion, it is obvious that this enormous tension is unsustainable by the human arm. This fact, however, does not diminish the fact that gymnast's do possess superior strength. Although it is physically impossible to perform the perfect "iron cross," the human eye cannot easily perceive this inability. For when one watches a world-class gymnast perform this maneuver, it is often far from obvious that his arms are not exactly perpendicular to his torso.