Monday, December 17, 2007

Aluminum Tube Vendor Needed

The people that I was planning to use are concerned that they won't be able to weld an aluminum tube that is perfectly round--as much as 1/4" out of round. Since they can only make tubes in 1/4" increments, this means that I have to go to a 20.75" outside diameter tube to get what I want. Maybe the problem is that they don't have a mandrel the right diameter to wrap the aluminum around when they weld it.

Any suggestions? The people that did the plates for the mirror cell are high on my list, since they did a good job and fast. Any other suggestions?

UPDATE: More details, in case you run a welding shop! 20.40" ID +- .005", .090" wall, 8.00 +- .05" length. It can be roll formed. Cheap is important; an extrusion is unnecessary expense.
Grind marks on the weld are just fine; I will polish and paint it myself. I could consider somewhat thicker tubing, but even going to .125" wall increases the weight from 4.52 to 6.28 pounds. I'm reluctant to go thinner out of concern about deformation. One of the tubes is going to carry 30 pounds, supported by three screws, although that weight is such a close fit that it will strengthen the tube (at least in that section); the other tube will carry less than ten pounds.

UPDATE 2: At least one vendor says that .090" wall aluminum tubing won't be stiff enough for my purposes. And yes, he is out of .090" anyway at the moment; wouldn't I like it made of .125" instead? There's obviously a formula for computing how much a tube deforms under load, depending on thickness and Young's modulus. My guess is that the formula is pretty much the same as a solid rectangle. Right or wrong?

I have an intuitive sense that because the top half of a tube is essentially an arch, which handles load far better than a flat surface (hence, the importance of the arch to creating Roman architecture) that if you use the deformation formula for a plate supported on four sides with a point load that you won't be too far off the same results for a point load at the center top of a tube, since much of the point load would be distributed down through the "arch" into the rest of the tube. If so, a .090" wall tube of the dimensions specified above with a ten pound load (one third of the mirror cell plus mirror weight) would produce a deformation of about .3 mm at the surface. This is acceptable.

A .125" wall would knock that deformation down to .1 mm. It's tempting to go for the extra thickness, even though it increases the weight by a bit more than 1 1/2 pounds per tube.

Part of what makes the equation so interesting is that most of the weight of the telescope is in the lower part of the tube, where the mirror and mirror cell are located. The upper assembly weighs almost nothing. When computing the deformation of the square tubes on which both assemblies are located, the weight of the lower assembly takes precedence; it is far heavier, and induces far more deformation.

I have been working on the assumption that from the balance point of the telescope to the lower end would be 23 inches, but the closer that I get the balance point to the mirror cell assembly, the shorter the distance over which the mirror cell deformation takes place. Reducing the weight in the upper assembly moves the balance point closer to the mirror cell, reducing the deformation in the square tubes, and allowing me to go to smaller, lighter tubes. For example, if I moved the balance point from 23 inches from the mirror end to 15 inches, it reduces the worst deformation from .014" to .0037". At that point, I could consider going to a thinner set of square tubes, reducing the weight a bit more.

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