Calculate Focal Point

[uminded]

When I put my webcam in my focusing tube I get a well magnified image. Is there a way to calculate what magnification my webcam is getting?

Also, I will be making a crayford focuser soon, how do I figure out the length of tube I will need??

[Merlin66]

You don't say which type of scope you have....

The "magnified" field of view (FOV) you get in the webcam (or any camera) is dependent on the focal length of the system.

have a look at CCDCalc

http://www.newastro.com/book_new/camera_app.html

(Roughly a webcam "shows" a similar FOV to a 6mm eyepiece.)

Plug in details of your scope/ camera and you'll have the answers.

If the scope you're adding the Crayford to is already functional, the easiest way is to image the Moon on a card and measure distances.

Hope this helps.

[D4N]

Magnification doesn't really exist in imaging terms, the scopes focal length and the chip size determine the field of view.

If you are able to capture an image with a good number of stars in it then you could plate solve it to find the exact field of view.

There are also online calculators for this provided you know the size of the sensor.

Otherwise you could capture something like the moon and work out how big it was at the time to infer your FOV.

Once you have this you can then work out the FOV of your different EP's so you have a general comparison.

TSED70Q, iOptron Smart EQ pro, ASI-120MM, Finepix S5 pro.

[uminded]

According to CCD Calc my image scale is 0.69 arcsec/px and FOV of 5.5x7.3 arcmin with the webcam.

I can measure how far out of the tube my 20mm EP requires to focus and do the same with the webcam. What EP is my webcam comparable to in mm?

[ronin]

Your webcam is not compariable to an eye piece.

The scope will create an image of the object that is in millimeters, not as a magnification.

If the object is say 3 degrees across (Andromeda) and the focal length of the scope is 1000mm then the image formed on the chip of the webcam is 54.2mm

As a good approximation the image created by a scope is:

Image size = Tan(object size in degrees) x Scope Focal Length.

So an 800mm scope pointed at the moon creates an image of 6.9mm (say 7mm)

It does not magnify it simply creates an image of a determined size.

Concerning the Crayford the camera needs to be positioned at the prime focus of the objective so whatever length is required to accomplish that.

Magnification is really only relevant when you have an eyepiece and an eyeball in the optical path, with a webcam in then you have neither an eyepiece nor an eyeball. an easy way of putting it is: Visual is not equal to imaging. Magnification is a visual term not an imaging one.

The data:

According to CCD Calc my image scale is 0.69 arcsec/px and FOV of 5.5x7.3 arcmin with the webcam.

is simply data that may be useful, change scope or webcam and it is irrelevant. One thing is not to get bogged down in numbers, I have read several things on the arcsec/pixel value, and these range from recommended values of 1 to 10 seem common, unless of course it is a planet when for whatever reason 0.5 and such like seem OK. So now it is 0.5 to 10, all are recommended by someone somewhere and lets face it not all can be correct.

With 5.5x7.3 arcmin FOV are we talking SCT/Mak?

Seems kind of narrow.

[uminded]

Thanks for the info, I am quite new to optics in general.

My 114/900mm scope will be used for the moon and planets. I have a WF 50mm EP for deep sky stuff but the FOV is pretty small at F7.9

Prime focus on my scope should be 900mm from the center of the mirror, reflected off the secondary and be somewhere in the focusing tube just outside the scope's tube, correct? I am going to center dot the primary this weekend so I might grab some string and measure it out...

On average for a F8 scope how much EP travel does a crawford have? They all look really low profile.