Mechanical Engineering Question

Young's Modulus tells you have much a piece of metal will bend under strain; it's a measure of stiffness. This website (which sounds like they know what they are doing) claims that titanium and aluminum are about 5% stiffer than steel for the same weight. Steel is much stiffer for the same thickness than aluminum, but steel is about 7.8 g/cc, while aluminum is about 2.7 g/cc. For the same stiffness as steel, you use a much thicker piece of aluminum--but you can still end up ahead on weight--and weight matters for this application.

I'm looking to rebuild Big Bertha in a form that is substantially lighter. Right now, it is unnecessarily heavy--perhaps 250 pounds, which is absurd, considering that the optics, eyepiece focuser, and mounting hardware for the optics weigh perhaps 45 pounds. If I just replaced the current wooden superstructure with Sonotube, I can get the telescope down to 110 pounds--perhaps less. But an equatorial mount that would handle it would cost $6000--and even then, it would be a bit heavy for that mount.

There's a saying in backpacking that every pound you take off your feet is equivalent to five to six pounds out of your backpack. The same is true with telescopes: for every dollar you spend taking weight off the telescope, you can save many dollars on the mount itself. As an example, the Losmandy GM-8 mount that I have is nominally capable of carrying a 30 pound instrument, and costs about $1500. The next step up is the G-11, nominal capacity 60 pounds, and costs about $2200. The next step from there is the HGM Titan, nominal capacity 100 pounds, and costs about $6000. I don't even look at the step up from that; it makes my wallet scream in agony to even think about it (and even the HGM Titan causes me to cringe at the price).

I don't know that I can get Big Bertha light enough to fit on a G11, but if I can, it is worth spending a bit of money lightening it up enough to do so. If it is just too heavy for a G-11, then it will be at least light enough to cause no strain for the HGM Titan.

So, one strategy is to build an octagonal skeleton tube, using eight aluminum tubes 81" long, cross braced with perhaps 1/4" thick aluminum flats, all bolted together with stainless steel bolts. How thin can the 81" long tubes be, and still provide sufficient stiffness once cross braced? I need it to be stiff enough that even with roughly 38 pounds at one end (where the mirror is), and about 7 pounds at the other end (diagonal mirror, its holder, and the eyepiece focuser), the deformation of those tubes by gravity will be barely measurable, or not measurable at all. (The telescope will be mounted at approximately the center of gravity, with a plate that will bolt to two of the long tubes.)

There must be a way to calculate this--I know that mechanical engineers don't do everything by experiment.

The telescope pictured here might be another strategy. It uses a single, extremely stiff member to hold the two ends in the correct positions. Perhaps this is another approach, since this very stiff member could also be the dovetail plate that slides into the telescope mount saddle. One other aspect of this design that is attractive--it is possible to disassemble it quickly into two "rings": the top one carries the eyepiece focuser, diagonal, and finder, and the bottom ring carries the primary mirror. These can attach to the base with several bolts, making it quickly come apart into two relatively light and compact parts, and one very long rail.

UPDATE: One of my readers responded with:

There's a measure of flexural rigidity for a beam which you can use to compare some alternatives (it doesn't account for twisting though). It's simply the elastic modulus of the material multiplied by the cross-sectional moment of inertia (E*I).

The moment of inertia for a solid circular rod is I = pi * (d^4) / 64. For a tube, you just subtract the inner moment from the outer one.

Circular Tube: I = pi * (d1^4-d2^4) / 64 or I = pi * (d^4-(d-t)^4) / 64
Solid Rectangle: I = b*h^3 / 12
Rectangular Tube: I = (b1*h1^3 - b2*h2^3) / 12
Square Tube: I = (w^4 - (w-t)^4)/12

For a point load (W) at the end of a weightless beam supported firmly at the other end, the vertical deflection at the loaded end is:

dy = W*L^3 / (3*E*I)

The angle of rotation at the end of the beam is:

theta = W*L^2 / (2*E*I)

You can consider the weight of the beam a distributed load (q, in lbs/in for example), so without a point load you have:

dy = q*L^4 / (8*E*I)
theta = q*L^3 / (6*E*I)

It's probably accurate enough (for small deflections anyway) to just add the results of both sets of calculations together for the net effect.

Where the ^ indicates exponentiation.