I have been looking at getting a rollbar for my '72 K5. I have looked at different places. Most offer the standard hoop 4 pt design. I had some questions about tubing size and thickness and wondered what is the best to go with. Just a little background here, I am an electrical engineer and asked a couple of the mechanical engineers that are buddies/coworkers of mine about this from a technical standpoint. I will list below my question and the response. It's pretty theoretical, but I found it interesting nonetheless and thought I would pass it along. I hope that someone is interested in this babble. ---------- My note the to the mechies -------- Question: If I were to determine which of two tubes was a better support structure, how would I go about it? I'll just cut to the chase. I'm looking at roll bars. Bar #1 is 2.00" OD with a wall thickness of 0.120". Bar #2 is 1.75" OD with a wall thickness of 0.134". Now Bar #1 has a cross-sectional area of 0.7087 in^2, while Bar #2 has a cross sectional area of 0.6803 in^2. Now lets assume that the material is identical. Is it better to have a thicker tube (albeit with less material overall due to the smaller cross-sectional area) or the bigger diameter? Or, is there no real way to determine this empirically? Thanks, Jeff. ---------- The Response ----------- Tubes supporting a structure: Failure mode: column buckling. Euler Column Buckling equation: Pcr = (pi)^2*E*I ---------- (Le)^2 where E = Modulus of Elasticity (fixed by material) I = moment of inertia of the cross section wrt the buckling-bending axis Le = equivalent length depending on end conditions (sliding, hinged, fixed, etc.) Pcr = critical force to failure pi = delicious I is the important [censored], for you, young EE. More I means it can handle more force (note, [censored]'s different if you want to design for stress, but I don't think you do). ~*~*~*~ I = A(rho)^2 where A = cross-sectional area rho = radius of gyration rho(slender rod) = sqrt((Dod^2+Did^2)/16) A1 (OD=2", t=0.12") = 0.7087 in^2 rho1 = sqrt((2^2+1.76^2)/16) = 0.666 in I1 = 0.314 in^4 A2 = 0.6802 in^2 rho2 = 0.573 in I2 = 0.224 in^4 If my math's correct (and it rarely is), then go big and go home. ~N (I had to pull out Juvinall. I'm not enough of a nerd to remember the one day of Euler Buckling off the top of my head.) ------------- The weanie ---------- I didn't realize that thicker wall tubing is not necessarily better. OD counts for quite a bit in structural rigidity. Time to go have beer before my brain explodes. Jeff.