Re: ATM mirror grinding mechanics & marshmello test

[Cary Chleborad <bennu@ns.net>]

Scott Rychnovsky wrote:
> 
> >Scott Rychnovsky wrote:
> >
> >> You have ignored the overhanging mirror.  That big lump of glass doesn't
> >> just float out there in space, it puts disproportionate pressure on the
> >> EDGE of the tool and the CENTER of the mirror.  This increase in pressure
> >> increases the rate grinding on those parts of the surface.
> >>
> >> --
> >> Scott Rychnovsky
> >> srychnov@uci.edu
> >
> >
> >Well, I really hate to keep giving all of these ddeas the buzzer
> >but.....
> >
> > The glass is essentially a rigid body, any force that is applied to the
> >center of the mirror is equally distributed through the area of cantact
> >between the mirror and tool assuming they are of equal diameter.
> >
> >Scenario #1
> >
> >An 8" (324.2 cm^2 in area) mirror is sitting on top of an 8" tool,
> >center over center.
> >We observe that the force, due to gravity, of the mirror on the tool is
> >1kg.
> >Since the glass is a rigid body, the force is equally and evenly
> >distributed upon the area of contact with the tool.
> >The force per unit area is 1kg/324.3 cm^2
> >
> >Scenario #2
> >
> >Same mirror and tool.
> >We push the mirror's center to the tool's edge - exactly
> >The force due to gravity is still the same.
> >Once again, since the glass is rigid the force is equally and evenly
> >distributed upon the area of contact with the tool.
> >The force per unit area (pressure) is now 1kg/126.7cm^2
> >
> >We observe that the pressure has gone up 2.5 times but, it is equally
> >distributed throughout the area of contact.  Given this, the grinding
> >action is the same throughout the area of contact.  This still leaves
> >the question of why the top piece goes concave.
> 
> Cary,
> 
> I am just repeating the analysis you can find in amateur telescope making
> volume 1 section A.1.
> 
> Why do you think the force will be evenly distributed?  Lets take example
> 3.  A sloppy amateur (like me!)  uses too long a stroke and pushes the
> center of the mirror all the way to the edge of the tool.  Now suddenly the
> mirror teaters or the edge of the tool and then falls to the ground where
> by some miricle it does not shatter into a thousands pieces but bounces to
> safety.
> 
> So what?  The point is that when the mirror center reaches the edge of the
> tool ALL of the pressure is focused on a very small area of contact right
> at the center of the mirror and the edge of the tool.  Although a much
> larger area overlaps,  the force is clearly distributed very unequally.
> Try it if you don't believe me  (just kidding).  This building up of force
> at the edge is a gradual thing that depends on how much overlap ther is
> between the two discs.  With the 1/3 center-over-center stroke you do get
> increased pressure towards the end of the stroke on the mirror center and
> tool edge.
> 
> I have too admit I don't have an instinctive feel for this behavoir, but
> the basic explanation sounds sensible.
> 
> Scott Rychnovsky


Alright, I've finally resolved this... I guess I need to work on my
understanding og Physics a bit.

There IS a differential pressure from the center of the mirror to it's
edge (located over the center of the tool).  I was discussing this
problem with my dad, who is a rocket engineer at Aerojet, and he devised
a simple test to measure the distribution of force on the area of
mirror/tool contact.

I placed minerature marshmellows on the surface of the tool and placed
the mirror on top of them.  The mirror overhang was close to 1/2
diameter.  The marshmellows at the edge of the tool compressed
considerably more than the ones at the center of the tool.  When viewed
from the edge you could see the mirror leaning away from the center of
the tool.  This indicates that there is more pressure at the edge of the
tool than the center.

Question answered!  Now the next question - Why is the pressure higher
at the edge of the tool? - back to the physics books I guess.

Thanks for all your input.

-Cary Chleborad