Re: Need Celestron (or Meade) 8 SCTs Data

On Feb 10, 8:57 am, "dkel...@xxxxxxxxxxx" <dkel...@xxxxxxxxxxx> wrote:

Errata. I mean as the mirrors are move farther not closer.
Farther such that the secondary is nearer to the
focus of the primary producing between 250mm-300mm
focal length. If focused at infinity, this couldn't have
happened because the focus plane would be inside
the tube near the corrector at front. But because the
focus is moved backward due closer target object,
one could then view it at the eyepiece. To make an
exact formula out of this double varible. I think one
must create a differential equation, isn't it. I tried
perturbation method to solve for the magnification
which isn't exactly 250/25 = 10X (if using 25mm
eyepiece) because the focus moving back can
increase the magnification further.

Paul

The equation is not all that difficult. If you assume the moving
primary, you know that the focal points of the secondary.
It would just be the formula:

 1/(f1+A) + 1(f2+A) = 1/f

You know f1, that is your 1000mm to the object.
f is the primaries focal length. f2 is the length of the
primaries focus to the secondaries focus.
The above equation reduces to a quadratic equation.
Dwight

Let's take the case of the 4" Rubinar with focal length
of 1000 (F/8). When target object is 1 meter away.
What would be the magnification provided 20mm
eyepiece used. If it's focused at infinity, magnification
is 1000/20=50X. Now to focus at 1.3 meter, thin
lens equation 1/f=1/f(image) + 1/f(object) would
produce f(image) = 4333mm. That is, you have
to move the focus back 4.3 meters or use an
eyepiece extension tube of 4333mm - 1000mm =
3333mm or 3.3 meter extension tube. So how
do you focus the rubinar at the object 1.3 meters
away. You increase the distance of its mirrors
such that the secondary would be closer to the
primary focal point producing not 5X but lesser
magnification like 1.5X or 2X. At 1.5X, focal
length becomes 200mm x 1.5 or 300mm. Now
at 300mm focal length, thin lens equation
would produce 428mm at 1 meter target
object, now 428-300=128mm. In the rubinar
tube where the distance is at least 200mm
(from the primary 100mm size multiply by the
primary focal ratio f/2 = 200mm), one can
now fit the 128mm extended focus from the
focal point due to target object at one meter.
Key here is that the focal point has to be put
inside the tube near to the corrector. In the
case of the SCT, one simply has to pull the
corrector out by some kind of hybrid tube
modification. Back to the rubinar example,
total magnification would be 200/20mm (eyepiece)
= 10X plus the extended focus 200/128 or 1.5626
so the 10X would become 15.626X magnification.
Well. The reason its not exact is because I don't
know the focal length of the rubinar secondary
mirror but I wanted to derive the equation using
an SCT with known secondary but with hypothetical
corrector adjustable.

Anyway.. I wonder how to plug the above in your
equation. What's the A in your equation? What's
it solving for assuming the target is 1.1 meter
away and primary focal length 200 and secondary
unknown focal length in the 4" Rubinar or target
for the 8" SCT 3 meters away and focal length
of primary and secondary known and corrector
that can be adjusted out by redesigning the tube.

Paul
.

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