Re: [ATM] Re: ATM Digest, Vol 17, Issue 10

["David Harbour" <scarab2@cox.net>]

This sounds reasonable to me- it is in the right "order of magnitude"- Arjan
must have just made a mistake somewhere with some term. Anyone can do that
(I am so stupid with math that no one would want to trust me with
calculations for a trip to the moon- we'd smash on the rocks).-

Davey


----- Original Message ----- 
From: "Koehler, Steve" <Steve_Koehler@securecomputing.com>
To: "ATM List" <atm@atmlist.net>
Sent: Wednesday, May 11, 2005 9:13 AM
Subject: RE: [ATM] Re: ATM Digest, Vol 17, Issue 10


Arjan wrote:
> > Equivalent formalae (equal paraxial ROC = R) would be:
> > sphere:     x^2 + (y-R)^2 = R^2, or y = R - sqrt(R^2 - x^2)
> > parabola:  y = (x^2)/2R
> >
> > so for x = radius = 5.5m and R = 33, the sag would be
> > sphere:    929.8 mm
> > parabola: 458.3 mm
> >
> > This deduction may, of course, be wrong.

David Harbour responded:
> I have confidence in you as a mathemetician, but the value seems to
> awfully
> high, to me.

Here's what my R tools come up with.  First, here is my sagitta function,
from
"Telescope Optics".  Y is radius on the mirror.  r is ROC.  sc is the
Schwartzchild constant (-1 for paraboloid, 0 for sphere).

> .sagitta
function (Y, r, sc) Y^2/(r + sqrt(r^2-(sc+1)*Y^2))

Let's define a Y and r for this problem:

> Y <- 11/2 * 1000
> r <- 11 * 1.5 * 2 * 1000
> Y
[1] 5500
> r
[1] 33000

Here's the sagitta for the paraboloid:

> .sagitta (Y, r, -1)
[1] 458.3333

... and the sphere:

> .sagitta (Y, r, 0)
[1] 461.5612

So, the difference is 3.228 mm

-- Steve Koehler
   steve_koehler@securecomputing.com


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