Home Science and Industry Astronomy Contact Us

 

CALCULATING THE SIZE OF YOUR DIAGONAL
and
The Impact of Diagonal Surface Errors

Over the years I have seen many complex formulas for computing the size of a Newtonian diagonal. Granted, some of these take into consideration the fact that the diagonal will be offset, as is necessary with very fast systems. But even then, I have found them to be at times needlessly complex and not particularly intuitive. While my method does not take into consideration offsetting the diagonal, I think it is fairly intuitive and easy to remember and understand and suitable for Newtonians of f/5 and longer. 

Begin by determining the length of the cone of light from the diagonal to the focal plane. Divide that by the total focal length. That gives you the percentage of the total cone of light cut off by the diagonal. It also gives you the percentage of the diameter of the primary mirror at the point where the diagonal is. Take that percentile number and multiply that by the diameter of the mirror. You now have the diameter of the minimum diagonal size for on axis rays. An actual diagonal calculation for a 6 inch, f/8 mirror might be as follows: The distance from the diagonal to the focal plane is measured to be 7". 7" divided by 48" equals 0.15. 0.15 times 6" equals 0.9".

But, you want a reasonable unvignetted field for the observing off axis objects. For the present, let us assume that the physical diameter of the field area at the eyepiece is 1/3". In other words, the image from the primary mirror forms a circle of light 1/3" in diameter. Simply add that 1/3" area to the on axis diagonal size you just calculated. This is the minor axis size of the diagonal you want. 

Yes, I understand that there are some very slight changes in the diameter of the off axis cone of light that causes the size of the diagonal calculated by the above method to be very slightly overstated, but the overstating is extremely tiny and of absolutely no consequence. The following drawing graphically demonstrates the geometry involved. The red rays represent axial image rays and the blue and green rays represent off axis image rays.

As to how the size of the hypothetical 1/3" image disk used above was calculated one has to first decide how large an unvignetted field is desired. Typically, the full moon or 1/2 degree is taken as reasonable for wide-field viewing. For planetary work where a diagonal of the smallest reasonable size is desired a field of 1/4 or even 1/8 degree is sufficient. To calculate a 1/2 degree unvignetted field take the focal length of the objective times 2 and multiply that times pi. Divide that by 360 degrees. You now have the size of a 1 degree image circle in inches. You want 1/2 degree of that circle so divide the one degree amount by 2. An actual example is as follows: 48" focal length times 2 equals 96" times 3.1416 equals 301.59 divided by 360 equals 0.84" divided by 2 equals 0.42".

Based upon or prior calculation for a diagonal of 0.9" we add the 0.42" for a diagonal minor axis size of 1.32". This results in a central obstructions of 22 percent. Sizing for planetary use at 1/4 degree we arrive at a diagonal minor axis size of 1.11" or 18.5 percent central obstruction. Realistically, any diagonal of less than 20% obstruction is of no material consequence

To practically calculate the length of the cone from the diagonal to the final focus, go to your focuser, insert an eyepiece and place the focuser at 1/2 racked out position. Stick a pencil up the bottom of the focuser until the eraser comes into focus. Note the length of the pencil from the bottom of the inside of the tube to the eraser. (Do this with several eyepieces.) Add that amount to one-half the inside diameter of the tube and you have the distance from the diagonal to the principal focus. 

I'm often asked about the impact of errors in the diagonal on the final image. Some theories have been circulating to the effect that since the diagonal is very close to the image plane errors in the diagonal do not materially affect the image to the same degree as errors in the primary mirror. Actually, the diagonal appears to be as important as the primary mirror with regard to surface quality and must be made well. The problem is that has the cone of light is contracting and everything is scaling down in proportion so that while there may not be a long distance for the light to travel after it strikes the diagonal imperfections are impacting proportionally.

To calculate the impact of errors on the final wavefront one method is to take the error of the primary mirror plus the error of the secondary mirror (if its surface error its times 2) times the cosine of the angle of incidence of the diagonal, in this case 45 degrees or .7071. So, assuming a 1/10 wave primary mirror and a 1/10 wave secondary mirror (wave front accuracy, not surface) we have the following: .1 + .1 * .7071) = .171 or Lambda/5.8. But some people think this is much too severe an application of arithmetic. I like the product of the complement method (which I developed independently) but it may only be foolishness. It goes like this: assume you have a .1 wave mirror and .1 wave diagonal. .9 is the complement of .1 or 1 - .1 = .9 so the entire formula is 1- (.9 * .9) * .7071 = .134 or Lambda/7.44 It looks nicer. Then there is the root sum squared method. This is valid for systems having many elements. This method is be as follows: ((.1^2 + .1^2)^.5) * .7071 = .141 or Lambda/7.07. Confusing?