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Old 03-05-2019, 11:30 AM
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The_bluester (Paul)
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Technical question on reducer/flatteners

I am sure there is a simple formula but i have not found it so far.

When looking at changng the spacing between a reducer/corrector and image sensor (Or Eyepiece I) experience suggest that the change in position of the focuser to bring the assembly to focus is greater than the change in spacing.

If I am not leading myself astray and seeing an affect that is not there, is it as simple as dividing the change of spacing by the reduction ratio? So for instance if you changed reducer to sensor spacing by 10mm using a 0.7 reducer the change in focuser position to bring it back to focus would be about 14.2mm?

I am fiddling around with my ED72 and trying to work out what bits and pieces I can put where in the image train, being of a technical bent I would prefer to be able to measure things first and have a good idea if they will work rather than just chuck it on and see.
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Old 03-05-2019, 03:14 PM
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Merlin66 (Ken)
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Paul,
You just need to know/ measure the focal length of the reducer.
Once you have this you can use the simple lens equation to find the "back focus" which depends on the distance from the prime focus to the reducer.

1/F = 1/U + 1/V
where F is the fl of the reducer
U is distance from reducer to prime focus
V is the distance from the reducer to the new focus.
and V/U = magnification
HTH
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Old 03-05-2019, 04:13 PM
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Merlin66 (Ken)
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Paul,
The spreadsheet I prepared for Meade/ Celestron reducers used in an f10 SCT may be of some help.
Note this only gives the effective focal length/ f ratio and does not make any comment on the quality and performance of the reducer.
(In spectroscopy we are working on the optical axis and interested in the final efl and f ratio)
Attached Files
File Type: zip reducers_V2a.zip (6.3 KB, 5 views)
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