RE: ATM effective balancing weight for a dob

["Arjan te Marvelde (ELN)" <Arjan.te.Marvelde@eln.ericsson.se>]

OK, forget it. Of course it IS equivalent, just did the trig myself.
/AtM

> -----Original Message-----
> From: Arjan te Marvelde (ELN) 
> Sent: Tuesday, January 22, 2002 1:52 PM
> To: 'atm@shore.net'
> Subject: ATM effective balancing weight for a dob
> 
> 
> What is the most effective place on the bottom of the mirror 
> box to put the balancing weights?
> If you consider a mirrorbox with weights in the two bottom 
> corners on both sides of the alt axis, what would then be the 
> equivalent weight if only put in the center of the box?
>  
> My gut feeling is that it is just the sum of these two 
> original weights, but i don't trust my guts lately (and i'm 
> too lazy to calculate it myself ;). 
> Possibly i overlook the fact that the center of the bottom is 
> closer to the axis than the corners, otoh this effect cancels 
> out between corners.
> 
> Thanks,
>  ... Arjan te Marvelde 
>