Measuring Large Diameter Parts

I'd look into extruded aluminum. You can get just about any shape you want, and some are extremely rigid, and still lightweight enough to handled easily.

I have to plug Rexroth.

But many others are out there which are similar _{(if not identical for less ).}

This is going to be a good thread. At the lengths you are measuring, thermal expansion is going to be a factor to the measuring device construction. The heat from your hands can make a difference. Check the expansion of the aluminum. How about glass tube? Same question, though. Fiberglass or carbon fiber tube may be contenders. I remember my windsurfer mast being stiff, and it was very light.

Why not use a laser measuring device?

What I was suggesting was the use of a laser measuring device to do the actual measurement rather than for setting the target for the tool. A direct laser measurement would seem much more accurate than the micrometer set up.

/tb/

We have a laser tracker but the method we are using (setting large pin micrometers) is preferred over using lasers, it is far faster, cheaper and in my opinion...more trustworthy. We account for thermal expansion. We cannot use galvanized, zinc, lead, aluminum or other soft metals.

A thin wall 1 3/4 inch aluminum pipe would sag (under its own weight) less than 1/2 in 200." A Carbon fiber or fiberglass tube would work fine in this respect also. In each case, the larger the diameter and thinner the walls, the better.

I'd be somewhat concerned with humidity affecting length with the fiberglass, less concerned with the same effect in carbon fiber, but concerned about temperature with the aluminum. I would think it would be very difficult to get consistent, repeatable measurements to within .001"

If you go to "Engineer's Edge" on the web, they have a number of quick, free calculators that can help you identify your requirements. For instance, with the dimensions you gave (5/8" ID x 1-3/4" OD, Round Tube), the moment of inertia (I) is 0.45957 in¨4, and the cross-sectional area is 2.40529 in¨2. Maximum deflection at the center of a simply-supported beam with uniform loading is given by y=%/384*(W*L¨3/EI), where W is the area times the length times the density of the material. We have reduced the problem to two unknowns- you want a material that will give you a ratio of density to Young's Modulus of 4.59e-9 (assuming we are working in inches, Young's Modulus in psi and density in pounds per inches¨3). If you can not find a material that meets this requirement, you can play around with moment of inertia for various cross-sectional areas (several options are provided on the Engineer's Edge) that could perhaps give you a ratio that would be doable with available materials. Or you could find an old copy of Roarke's Stress and Strain and look up the formulas from the myriad collection offered in this fine volume.

I agree with post #1, I use AL extrusion or channel designed as edge trim or supports for large glass panels.

You can use an aircraft grade aluminum extrusion if the temperature is stable and you can be slightly off in your readings (a generous tolerance) otherwise you'll need to use carbon fiber but that will be expensive.

We've had a discussion about this topic in the past here. The comments in that post may be useful.

Why not use diameter tapes and forget about the unimportant details?

They're measuring inside, not easy with a tape.

Why not use a pi tape which measures the O.D. and subtract 2 metal thicknesses. Pi tapes can be made for any diameter.

Diameters may have irregular surfaces...

I'd measure it vertically if possible .Straight up and down. But that's because I'm a straight up kind of guy.