Scope/camera calculations

[matt]

I'm in the process of working out a new scope and ccd camera combination.

I'm looking at the Intes Micro MN86 for planetary imaging.

I'd like to know what formula or considerations I'd need to make to arrive at the ideal ccd camera to use with it?

I know a lot of it comes down to the size of the pixels on the chip and the camera's resolution etc

But what about the interplay between the scope and the camera? Is there any way of calculating which scopes work best with which cameras?

Many thanks

[Dennis]

Hi Matt

I note that this is an 8” F/6 Maksutov –Newtonian with a native focal length of 1200mm. This is a short focal length for purely planetary and lunar imaging.

On the C9.25 with a native focal length of 2350mm, using a x2.5 Powermate you are working at an effective focal length of 2350mm x 2.5 = 5875mm.

With the MN86 native fl of 1200mm, you will need a x5 Powermate to give you efl of 6000mm.

If I were looking for a specialist planetary OTA, I would be looking at a native focal length of at least 2000mm. This would allow you to use your collection of x2.5 and x4 Powermates more effectively?

Another thing to check is the back focus of these scopes. I assume you will be using a ccd camera / webcam with a colour filter wheel and these require a fair amount of back focus.

If you run out of back focus, it means that you have wound your focuser in, all the way of it’s inside travel and your camera or filter wheel has hit the OTA – there is no more inward travel left yet you still haven’t reached focus.

Cheers

Dennis

[gts055]

Hi Matt,
A program which you may find useful, called CCD Calc, can be downloaded free at
http://www.newastro.com/newastro/book_new/camera_app.asp
Mark

[74tuc]

Hello Matt,


In answer to your 'scope / ccd matching the sums are simple - all you need is a calculator to work it out. If you assume square pixels then:

1. The basic formula:

Calculate the FoV as seen by one pixel - also called the Instantaneous FoV (IFoV)

IFov(Radian) = Pixel size (M) / Scope Focal length(M)

the above gives an angle in Radians.

eg. Pixel size = 5.8E-06 M (5.8 micro metres)
Scope Focal length = 1.1M

IFoV = 5.8E-06 / 1.1 = 5.27E-06 Radians or 5.27 micro-radians.

To convert this to degrees note that a circle(360 degrees) has 2*Pi Radians
and further there are 3600" in a degree (Pi = 355/113)

so

IFoV (arc sec.) = (3600*Pi/180)*IFov(Radians)

or IFov(arc sec.) = 206.26*IFov (Radians)

In the example

IFoV (arc sec.) = 206.3*(Pixel size (M) / Scope Focal length(M))

or IFoV(arc sec.) = 1.09"

What this means is that each pixel of the ccd "sees" 1.1" of sky.

Now if you multiply the number of pixels in the ccd by the IFoV you get total Fov so

in the example the IFoV was calculated at 1.1" and if your ccd is 640x480 pixels then the ccd sees a FoV of (1.1*640) X (1.1*480) arc secs.

or 704 x 528 arc sec. or 12 x 9 arc minutes. Adding barlows scales these numbers so if you used various barlows in the above 'scope and ccd you get for example:

Barlow FoV

x1 704" x 528"
x2 352 x 264
x2.5 282 x 211
x5 120 x 105

These calcs will give you an idea as to what the camera sees. If Jupiter is 20" in diameter (guess) then with a x5 barlow it would occupy a square 90 x 90 pixels or about 14% of the width of the ccd may need to go to a longer focal length in this case.

Hope this is of help.

Jerry.:)

[gbeal]

Hi Matt,
as we both know the MN86 is a fine scope.
Without wanting to suggest Dennis isn't on the mark her (cos really he is), let me add that I currently lunar and planetary image with a newt of 1250mm focal length. I simply use a 5x Powermate, or sometimes eyepiece projection. I did this as well when I had the MN106. You will not find this an issue.
He is also right with the "in-focus" issues. These can be negated somewhat by careful focuser choice, and also (as I did with my MN76) careful adjustment of the secondary and primary. I was able to bring a DSLR to focus without any major structural changes.