Vignetting Page

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QuickCam and Unconventional Imaging Astronomy Group -


> It was taking me a while to figure how I could work it out. The 8mm hole > extends another 2mm past the face to give support to the neck so it is > probably a bit worse then you have calculated.

I assumed 27.4mm from the CCD surface to the point where the 8mm hole breaks out into the larger diameter of the adapter, so the numbers should be close. In reality, this is only the f-ratio required to get all of the on-axis light to one pixel in the center of the detector.

> Maybe that is why the neck was 9mm in the original?

The problem didn't present itself in the original, because the lens, and therefore the entire light cone, was entirely contained within the tube. We decided to replace a 5mm diameter lens with a 200mm diameter lens, and created the problem all by ourselves. If you have an 8mm diameter aperture 27.4mm from the CCD, you will get a fully illuminated on-axis circle of 4mm diameter at f/6.91.

> I got to think about this on axis/off axis stuff a bit more yet.

The best way to think of on and off axis is by position on the sensor. Assuming you are in focus, there is one pixel in the array which is on-axis, and all others are off-axis to varying degrees. How much? It depends on your actual focal ratio. An example, based on our f/6.91 limit for a QC grayscale (and a scope of f/6.91) would be 6.5 x 5 minutes of arc. The corners will be illuminated at something like 63% (approximation) of the value at the center. As the light cone moves off axis at all, it no longer coincides with the 8mm hole. The amount of falloff is that amount of light blocked by the adapter.

To calculate variations on the problem, you need to divide the distance from the CCD to the aperture by the difference between the aperture diameter and the CCD diagonal dimension.

f = distance/(Daperture - Dillumination)

where: f = f-ratio distance is the distance from the CCD to the aperture Daperture is the diameter of the aperture Dillumination is the diameter of the fully illuminated circle (the CCD diagonal)

To calculate the aperture required for a given f-ratio:

Daperture = (distance / f) + Dillumination

To calculate the fully illuminated circle for a given aperture and spacing:

Dillumination = (distance / f) - Daperture

For an example of a fully illuminated circle in 3-D, look at:

There are some sample flat fields and some wire-frame graphs which clearly show a fully illuminated circle (the flat topped peak in the wire frame), and the effect of vignetting on off axis light. They are good examples, because they show that the vignetting aperture is not so far away or so small as to limit the fully illuminated circle to just on-axis light, but allow some off-axis angles to be fully illuminated as well. The example f-ratios I gave in my original post would not have the flat fully illuminated circle in the center, but would instead look like a hill, with a peak in the center. My example is made with a Meade LX-200 at f/3.3 with a 25mm filter approximately 25mm in front of the CCD. The physical size of the fully illuminated circle (on- and off-axis) is smaller than the size of the TC-255 in the QC. The images are reduced in size. The TC-255 has approximately 1/4 the surface area of the KAF-0401 CCD which was used to get these flat frames.

If you want a real-world example, take an on-axis image of an out of focus star. You should get a nice round donut. Now move your scope off-axis, by moving the star to one edge of the detector. The donut is clipped by the obstructing aperture, and is no longer round. This of course assume that you have an obstructing aperture to test with.


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