Does anyone know the math on the relationship of the need for a Paracorr and magnification. In other words are there diminishing returns when the exit pupil gets down to a certain size?
It's fairly easy, actually.
It depends solely on apparent field, not magnification.
I'll explain:
Coma has a linear size that increases linearly from zero at the center of the field to a certain size at a certain distance from the center of the field.
I.e., coma's size, in linear terms, is twice as great at 20mm off-axis as it is at 10mm off-axis.
Coma also has an apparent size. Since the star is not a mathematical point, it has a size, and the apparent size grows larger with increased distance from center.
I.e. a star image will appear to be twice as long, in a radial direction, 20mm off-axis as at 10mm off-axis. The Apparent size is what we see with the eye in the field of view.
Take two 20mm eyepieces, one of 50°, and one of 100°.
The 100° eyepiece extends twice as far from center in the field of the scope, so the apparent size of coma at the edge will be twice as large as in the 50° eyepiece.
As you can see, the larger the apparent field, the larger is both linear size of coma and the apparent size of coma.
Now, double the magnification of the 100° eyepiece--make it a 10mm eyepiece.
The field is now 1/2 as wide, so the star images at the edge have 1/2 the linear size and are identical to the 50° eyepiece at the 20mm focal length.
Ah, but now the magnification is literally twice the magnification of the 20mm eyepieces, so the comatic star image at the edge, though 1/2 as large linearly,
is twice as large apparently. Twice times 1/2 = 1. So the apparent size of coma at the edge of the field in the 10mm 100° is exactly the same as it was in the 20mm 100° eyepiece.
And the math works out to carry this all the way up. Essentially, the visibility of coma to the eye is related to apparent field, not magnification. 100° eyepieces, at all powers,
display coma to be apparently twice the size of 50° eyepieces. I can see coma in my 3.7mm 110° eyepiece quite easily at f/5 when I do not use a coma corrector.
Is there an exception to the rule? Yes. When the field stop of the eyepiece is small enough that the linear size of coma is hidden in the Airy disc.
No 100/110° eyepiece has a field stop that small, but some small field ortho and monocentric eyepieces do.
So why doesn't every observer see coma at all magnifications?
--they use narrower apparent field eyepieces at high power. That reduces the visibility of coma.
--they are looking at primarily tiny targets at high power, in the center of the field. If coma is present at the edge, who would notice? You're not viewing there.
--the outer edges of the comatic star image are a lot fainter than the inner point. If the sky is light polluted, or the scope is small, the outer comatic part of the star image might not be visible
--they are looking at bright targets that reduce night vision so the faint outer parts of comatic stars are less visible.
But, in essence, the visibility of coma has a formula:
The larger the apparent field, the more visible coma is, in a simple ratio: the ratio of the wider field to the narrow field = the increased visibility of coma.
In our example, 100°/50° = 2x the apparent size of coma at the edge. We can't necessarily say it means it will be twice as visible.
