Binoviewer  Comparison Page

This table below is a direct ray trace of how a particular clear aperture and optical path will interact with a set of eyepieces.  All 1.25" units below from 22mm's to 28mm's are represented as having an optical path of 115mm's.  This gives them a drastic benefit of the doubt since most are considerably higher.  115mm's is set very close to the AP which is a little lower by a mm or two.  In the columns below it is broken down into three different f/ratios.  These f/ratios cover many important applications and represent a best case scenario to a worst case scenario.  F/11.5 is a typical f/ratio for an SCT when using a binoviewer.  F/7 is using the binoviewer and SCT with a focal length reducer (FLR) in place giving roughly a 40% reduction off of the f/11.5.  F/5.2 would represent a worst case scenario when a telescope (non SCT) is used without an optical corrector  which is a rare occurrence.  So all of the examples below are on uncorrected systems.  A focal length reducer does not represent an optical corrector.

 Here are a few important facts to understand about the comparisons below.  To see how your clear aperture and optical path would compare to another simply divide the square area of the unit of the lesser square mm size into the larger to see the percentage difference or vice versa to see the percentage of the larger unit.   Examples:  45mm unit at f/11.5 has a square area of 660.2.  The 28mm clear aperture has a square area of 254.3.  660.2 divided by 254.3 = 260% or 2.60x greater fully illuminated area.  Now using a 26mm clear aperture unit which  has a square area of 200.9.  660.2 divided by 200.9 = 328% or 3.28x greater fully illuminated area.  So you are able to see the percentage of fully illuminated area in this way.   Now to find the square area that represents the mm size diameter simply follow this formula:  pi times  radius squared.   What this means is take the diameter number say 29mm's for 45mm unit divide it in half to get the radius, multiply the radius by itself and then multiply it by 3.14 which is pi.   EX.     29 divided by 2 = 14.5 x 14.5 x 3.14 = 660.2   This would represent the square area of 29mm's in diameter.  The rest is simple ray tracing.  If anybody asks I will be glad to explain how to do a complete ray tracing.  Ray tracing is simple and easy and anyone could verify that everything said on this page is 100% true.

Now that we have determined the fully illuminated area consider the fact that the mostly illuminated area is at the same percentage and extends out considerably further than the fully illuminated area.  So when you get to the worst case scenario of f/5.2 you will notice that the percentages stay roughly the same except in the case of the 28mm and the 40mm.  This seems to be a sweet spot for this particular optical path and clear aperture.    But when comparing  45mm unit  to the 40mm and 28mm it has 324% larger fully illuminated area.  You sure can't get that from looking at the mm size can you.  I am sure when we were all puzzled the first time we found out that we should not have bought the 60mm refractor at close to the same price at close to the same price as the 4.5" Newtonian.  The department store didn't tell you that the difference between the light gathering of these two scopes was not 1.9x as the diameter would suggest  but 3.6x because of the larger square area.  So don't let the simplistic numbers fool you regarding diameter of illuminated area.