interesting physics article on iron cross

'>http://www.geocities.com/bright_effect/ironcross.html

btw newtons = (kg * m)/s^2

gravity = 9.8m/s^2

so 655 newtons = 66.9 kg

67kg = 147 lbs

so these numbers were calculated for a 150 lbs gymnast

That article got it all wrong!

Reasoning that way you couldn't even hold a pound in your outstretched arm, because the tension would still approach infinity....

The mistake is applying the reacting forces an armlength away (literaly!) without considering the torque.

Actually when the arms are horizontal, the "tension" is ZERO and the torque on the armpit/shoulder reaches its maximum.

Here are the formulas:

P = weight of the gymnast

L = armlength

a = angle between arms and horizontal line

T = tension in the arm (it's actually a compressing force as long as the gymnast is above the iron cross position)

M = torque on the shoulder

T = P/2 * sin a

M = P/2 * L * cos a

Assuming the 67kg gymnast has an armlegth of 60cm

a(deg)---T(N)-----M (Nm)

45--------232------139

30--------164------170

15---------85------190

0-----------0-------197

The perfect iron cross would place a 197Nm torque on each of the gymnast's shoulders, which is not surprisingly exactly the same as holding a weight half his bodyweight in each hand with arms stretched out to the sides. (33.5kg * 9.81m/s2 * 0.6m = 197 Nm)

Yes, I think Regaman is correct on this one, I'm not to much of a physics savvy person, but I do remember my brother who does know his stuff saying something along the lines of what Regaman has said.

One thing I do not completely agree on is

e perfect iron cross would place a 197Nm torque on each of the gymnast's shoulders, which is not surprisingly exactly the same as holding a weight half his bodyweight in each hand with arms stretched out to the sides.

heh, i thought this was off, because people pull butterflies from hang

im taking AP Physics right now

the misconception this article makes is that the tension T generated by the arm is directly out to the side, or only in the x direction or is like the classic tightrope problem

in this problem you are to solve for the tension in a tightrope some length, to support a ten pound weight, so that there is no bend in the rope

when you solve this your answer as h (angle) approches 0 the tension approaches infinity

however, gymnasts are able to generate a downwards force, or a force in the y direction that undoes this

so basically the misunderstanding is that the gymnast will only be generating force in 1 direction outwards, when in reality he generates in two, which is why that article is so incorrect

the article is correct in the fact that it is impossible to generate a force so great in the x direction that it compensates for the y