The best way I know to show what is involved in making one of these telescopes is to lead you through the design process. I am currently building an 8" f/12.
The paraboloidal primary mirror should have a focal length of 96 inches, but the grinding and polishing of the primary is already done and I have missed my target focal length by two inches. It is only 94 inches so we are now talking f/11.75. With a primary r.o.c. of 188 inches, the auxiliary mirrors need a curve of 75.2 inches radius. This is the value I will shoot for while grinding.
Starting with equation 3, and selecting the proper value of C from the table, we divide by 11.75, and assign a primary tilt of 3.04 degrees (slightly rounded up). At a radius ratio of .4, k has a value of 1.75. Using equation 4, we find the secondary tilt will be 5.32 degrees. Equation 5 yields a tertiary tilt of 2.28 degrees. The focal plane tilt (Eqn 6) works out to 9.12 degrees, very close to the exact value of 9.07 degrees derived from raytracing. We shall find later that focal plane tilt may be reduced somewhat if we are willing to accept the introduction of a small amount of astigmatism into the field.
Next, we may easily calculate the spaces between surfaces with equations 7, 8, and 9. We mustn't forget to assign a negative focal length to the convex secondary mirror.
The design so far looks like this:
| Mirror | Radius | Figure | Tilt | Dist. to next surface |
| 1 | 188 | Para | 3.04 | 56.4 |
| 2 | -75.2 | Sph | 5.32 | 75.2 |
| 3 | 75.2 | Sph | 2.28 | 37.6 |
Our first questions at this point are, "What are the mirror diameters, especially the diagonal, and where does the diagonal mirror go?". Equations 1 and 2, for mirror diameters, are for the clear field and the diagonal mirror was not included there. The design field has a one-half magnitude loss at the edge and cannot be used in these equations.
The diameter of the clear field which results in the prescribed loss at the edge of the design field can be found in the accompanying table. I have chosen to design for a 1 inch diameter field. Smaller telescopes suffer if this value is adhered to rigorously. Therefore, I downsize the design field to 3/4 inch for a 6 inch telescope, and to 1/2 inch for a 4.25 incher. Without this concession, these sizes wouldn't have enough free cone except at higher tilts, or at the longer f-ratios.
| Pri. size | Design field diam. | Clear field diam. |
| 4.25" | 1/2" | .28" |
| 6" | 3/4" | .45" |
| 8" | 1" | .62" |
| 10" | 1" | .50" |
| 12.5" | 1" | .36" |
| 16" | 1" | .18" |
These values of clear field should be considered minimums. Larger telescopes (above 8 inches), or ones with long f-ratios may well be capable of a larger clear field (up to as much as the design field), and still yield enough free cone for the focuser.
Right now, let's see what the minimum required mirror diameters are in the 8 inch f/11.75 telescope. Using a clear field diameter of .62 from the table in equations 1 and 2, the answers are:
Because an oversized secondary mirror may be offset sideways (it is spherical, after all), and because there is plenty of space for an oversized tertiary, I will be using standard 4.25" and 5" blanks.
We now need to know the required size of the diagonal mirror, its proper placement, and the available free cone. The easiest way to determine these is from a full-scale drawing. Such a drawing is necessary anyway for actual construction of the telescope.
Curator: Hartmut Frommert
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