[ATM] Re: Second Thoughts on Mirror Support

[mikell at optonline.net (Michael Lindner)]

On Saturday 07 February 2004 01:58 pm, Steve Houlihan wrote:
> This is hard to figure without a picture.  I guess you are looking at the
> actual force vector through the pad and I was only looking at the force
> perpendicular to the back of the mirror.  Is that right?

Not at all, and I did make a mistake in the math. Look at it this way. You 
computed a torque of 7 inch-pounds on each bearing, which is 1.75" behind the 
mirror, with pads 1" above and below the bearing.

That 7 inch-pounds is spread over the 2 pads, so each one sees 3.5 inch 
pounds. However, the pads are 2 inches from the bearing ( the bearing, pad, 
and mirror back form a right triangle with legs 1" and 1.75", the hypotenuse 
is about 2"), so the force on the pad is 3.5 inch pounds / 2 inches = 1.75 
pounds.

Furthermore, since this is torque (rotation), the force is *not* applied 
perpendicular to the mirror, but applied perpendicular to a line form the 
bearing to the pad, at an angle to the mirror. That force can be broken down 
into the sum of 2 vector forces ( one perpendicular to and one parallel to 
the mirror). The force perpendicular to the mirror is the force time the sine 
of the angle the imaginary line from bearing to pads makes with the normal to 
the mirror,, which is 1/2 the force on the pad, so the actual force against 
the back of the mirror is about 0.825 pounds, or about 14 ounces.

Now Mel says you can detect that much force on the back, but I'm not sure what 
the magnitude of the deformation would be.

-- 
Michael Lindner
http://www.starastronomy.org *** http://home.att.net/~mikel
http://www.atmsite.org *** http://www.atmlist.net