To center the fully illuminated area (in the focal plane) on the optical axis, it is necessary that the center of the secondary is not placed on the intersection of the optical axes of primary and focuser, but slightly shifted towards the primary and away from the focuser.
Though some people argue that only scopes which have the offset applied are able to generate the best possible images, I personaly do not know if this is true or, when true, if the difference is visibile. The only case where I know it is needed is in scopes used for imaging. When calculating the the size of the secondary, the size of the fully illuminated area is the driving factor. Ofcourse the smallest possible size for the secondary is used. But when the fully illuminated area is NOT centered on the optical axis then it becomes near to impossible to position the imaging device correctly. Other factors can be tought of, but in the end they are mostly a matter of aesthetics (assuming the telescope is not used for scientific work).
It could be argued that it is not necessary to apply the offset, since the effect could be compensated for by tilting the primary. However, then the "on-axis" picture would not be on the optical axis of the primary and thus have a considerable aberration. (Especially in fast systems, where the offset is necessary).
If the secondary offset is applied your collimation methode may have to this into account. The 'circle within circles' methode will no longer provide accurate collimation.
Note : All formula's presented are valid only for secondaries placed at 45 degrees w.r.t the optical axis of the primary.
The figure below shows a primary with a secondary at 45 degrees (w.r.t the optical axis of the primary). Also shown in red and blue are the outer rays of light from an object on the optical axis, located at infinity, reflected by the primary. The yellow lines show the outer rays of the light cone for the fully illuminated field. The focuser would be placed above the secondary. The f-number is about f/1.75 so the effect of the offset is much clearer as for the more common values.
A secondary for the optical system below would be calculated to have the size A-B (the circle with full illumination would then have a radius of "k"). It is clear to see, that if this secondary would be geometrically centered on the optical axis of the primary it would cause a decentering of the area that receives 100% illumination.
In the figure below, the line AB represents the correct position of the secondary. It can be seen that the geometrical center of the secondary is shifted away from the focuser, and also towards the primary. It is important to realize that the surface of the secondary still intersects with the interception point of the optical axes of the focuser and primary. One could visualize this as sliding the secondary along the line F1 in the figure below.
To confuse things further, the offset can be given in two value's. A value for the displacement of the center (CD) of the secondary (i.e. the amount of movement along the F1 line) or as a value for the 'horizontal' and 'vertical' displacement (HD and VD). The relation between these values is shown in this formula :
In the figure above :
The primary is symbolized by the pink curved line on the left hand side.The green line shows the diameter of the area that receives 100% illumination.
The focus of the primary is at "f".
The depth of the sagitta of the primary is "S"
The radius of the 100% illuminated area is "k"
The radius of the primary is "r"
The size of the secondary is sqrt(2) * (Xb - Xa)
The function F1 : f(y) = x
The function F2 (upper yellow line) : f(y) = (Ym + r) - (Ym + k) / Xf
The function F3 (lower yellow line) : f(y) = (Ym - r) + (Ym - k) / Xf
The offset is : HV = (Xb - Xa) / 2 - (Xb - Xm)
The value Xf = f - S, the focal length minus the depth of the sagitta.
The value Xm = h - S, distance between primary and secondary on the optical axis minus the depth of the sagitta
It can be shown from the above figure that the displacement (offset) can be calculated with the following formula :
The above formula is exact. However, if it seems to complex, just throw out the "sagitta" and maybe the "radius of fully illuminated area". The calculation becomes much simpler, and the error you make will be very small and (IMHO) not observable. On the other hand, since the invention of computers and the pocket calculator ... :-)
If you don't want the 100% illuminated area to be centered, then you can also substitute another value for "k". For example the area wich will be illuminated for at least 70%. An interesting (theoretical) point is that you cannot center the entire vignetting envellope.
Refer to "fully illuminated area" for a formula on how to determine the size of this area from the size (and placement) of the secondary.
Examples :
All units in [mm] (1 millimeter is about 1/25 of an inch), all values cut-off after the second decimal place (no rounding). The 'Delta' values refer to the difference w.r.t. the leftmost HD colum.
Primary Diameter : 300 mm (12 inch)
Focal plane to optical axis : 200 mm
Radius of fully illuminated area : 15 mm
f/# HD HD Delta HD Delta HD Delta
Assume k=0 Assume s=0 Assume s=k=0
2 14,64 13,78 0,85 14,21 0,42 13,33 1,30
3 6,98 5,79 1,19 6,90 0,08 5,71 1,27
4 4,30 3,19 1,10 4,27 0,02 3,17 1,12
5 3,00 2,03 0,97 2,99 0,01 2,02 0,98
6 2,26 1,40 0,86 2,26 ---- 1,39 0,87
7 1,80 1,02 0,77 1,79 ---- 1,02 0,77
Primary Diameter : 150 mm (6 inch)
Focal plane to optical axis : 125 mm
Radius of fully illuminated area : 15 mm
f/# HD HD Delta HD Delta HD Delta
Assume k=0 Assume s=0 Assume s=k=0
2 8,56 8.61 0,05 8,33 0,22 8,33 0,22
3 4,34 3,62 0,72 4,29 0,04 3,57 0,77
4 2,79 1,99 0,79 2,77 0,01 1,98 0,80
5 2,02 1,26 0,75 2,01 ---- 1,26 0,75
6 1,56 0,87 0,68 1,56 ---- 0,87 0,69
7 1,27 0,64 0,62 1,26 ---- 0.64 0,63