Use a radioactive eyepiece!

"Quadibloc" wrote in message
...

On Jan 30, 8:46 pm, Quadibloc wrote:
On Jan 30, 5:22 pm, RichA wrote:

The Kodak Ektamate and Ektar lenses all use thorium glass. Pretty
harmless just sitting around but I wouldn't want to press my eye to
one for any length of time.

http://www.surplusshed.com/pages/item/l3759.html

There once were some World War II lenses that used radioactive glass,
but that is long gone. Any surplus from the days of digital Group III
fax machines would not use lenses made from that kind of glass -
because the glass isn't made any more.

I see that some radioactive glass was used even in the 1960s, so I'm
mistaken...

http://www.bnphoto.org/bnphoto/LostS...adioactive.htm

John Savard
=================================================
Good that you can admit it, Savard. It's really quite painless, isn't it?
Nobody is going to beat you to death over it now. You are mistaken
about relativity, too, but that's because you are hopeless at algebra.
Your pal Bill Owen hasn't come to your rescue, either. He's gone away
to think about it. He's only been gone three weeks. Perhaps he's gone
away to forget about it.


"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
===============================================
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

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