Re: [ATM] telescope vibrations

["Mel Bartels" <mbartels@bbastrodesigns.com>]

> Mel, I would have thought that the reduction would have been 1/4 of the
> force.  Where am I wrong on this?  The tube, being 1/2 the length, would
> take 1/2 the wind load and that load would be working over 1/2 the 
> distance.
> Then again, that would be the simplistic view and it may be that you have 
> to
> figure the average loading which gets reduced as you go closer to the
> mounting point.


Well I was thinking that doubling the length adds a cross section at 3x the 
moment arm (moment arm of original length=1/2 that length, and moment arm of 
new length=3/4 original length) as a quick BOE calc.  I think if one 
integrates force over area, that the ratio comes out something like 1.4 : 
.35 or more like 4x difference.  Oh I see, that's what you're suggesting. 
Ok, I agree.  I think that the center of gravity can be ignored here.

Mel Bartels

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