Baker Reflector Corrector help
#51
Posted 14 February 2013  04:28 AM
John

#52
Posted 14 February 2013  06:51 AM
#53
Posted 14 February 2013  08:08 AM
If suitable equipment is about it would be cheaper to get them to machine the corrector and a lot simpler. It might even work out at long IR.
John

#54
Posted 15 February 2013  08:50 AM
Problem though. In some respects it's not a schmidt plate as Vla pointed out. Worse still in some ways Baker seems to have put the "neutral zone" at 70% to minimise the amount of glass that has to be removed. In quotes as naughty me I assumed something. Mainly that it was desirable. The neutral zone in terms of the total sag zone doesn't seem to exist.
Looking at the figures more closely the aspheric depths from a rad don't have a neutral zone either other than if the power of the asphere overcomes the sphere. Might be possible with some other set of curves. At inch intervals they are
0.0,0.01,0.02,0.05,0.08,0.09,0.12,0.11,0.08,0.00
in thou's as they used to be called. The total sags show ever increasing depths indicating a sphere with a sag of 0.174 thou but he suggests starting from 0.2 so some of the form may project past the 0.174 sag rad. More likely margin for error on an analogue spherometer. These are for a 18in dia corrector. The scope has an 17in aperture corrector, 18in o/d. These needn't tie up as he expects the final figure to be put in by hand anyway. Basically it's a ve asphere on a positive sphere. He also gives figures for a 40in corrector for an 100in FL mirror. The total sag does drop of near the outer rim of these figures from around a 15in zone height.
If the system is scaled for focal length first for a smaller faster mirror than he uses it's possible to finish up with a slight reduction in F ratio rather than an increase in speed. It's much easier to play with without getting severe astigmatism problems.
In days of old when knighst were bold and having plates coated was rather difficult it seems people put slight concaves on the none aspherised side of the corrector to avoid problems with reflections.
John

#55
Posted 15 February 2013  11:05 AM
Let r = radius of the aspheric aperture
n = power,
Cn = nth order aspheric coefficient, and
S = the desired scale factor
A general aspheric deformation term is Cn (r^n). n can be both even and odd integers.
Scaling the system or particular aspheric surface by S scales the aspheric coefficient as
C'n = Cn * S^(1n)
The scaling routine in OSLO should already do this properly. Have you tried using it?
Mike
#56
Posted 16 February 2013  05:57 AM
Scaling the ^4 powers etc would be even more tedious as olso wont scale the plates power on it's own. MKV has posted a plate design with the neutral zone in the correct place and in principle that could be scaled to suit any scope that needs one while keeping the neutral zone in the right place.
Mentioned as an aside really as I'm fairly sure the Baker doesn't use a schmidt type plate.
John

#57
Posted 16 February 2013  07:28 AM
My point was to show that 86.6% NZ corrector gives better correction, as you said, but the 70.71% may be easier to make.
The 0.707 zone is certainly easier to make, being less than half as deep as 0.866 zone profile. The profile dept is also directly proportional to the wavefront error of spherochromatism, so it also has less than half the spherochromatism. The only "advantage" of the 0.866 neutral zone is that it focuses at the smallest blur focus, thus produces the smallest geometric blur. Many sources (including Schroeder and Mahajan) state that it minimizes spherochromatism based on this criterion, but the criterion is flawed. We don't see the geometric blur, we see diffracted energy, and it is the wavefront error that determines its distribution.
It is easy to check: place neutral zone at either location and look at the wavefront error in the same unoptimized wavelength. The 0.866 zone will always have it more than twice larger.
Vla
#58
Posted 16 February 2013  09:17 AM
Mike
#59
Posted 16 February 2013  11:38 AM
I posted comparative results three 200mm f/3 Schmidt cameras and they show that the geometrical blur is the smallest with the NZ at 86.6%, but it also confirms Vla's assertion that the wavefront error and chromatic aberration will be the smallest at the 70.71% NZ location.Vla, this is not consistent with (1) refereed, published articles and (2) my own analysis. I do respect your analytical abilities, though, so maybe this would be a good subject for a new thread, as this is sort of OT and buried too far down in this one.
Is it significant (i.e. perceptible)? I think not, certainly not photographically or interferometrically. I think the PSFs make that perfectly clear. A thing to remember is that geometric raytrace analysis of the image blur (spot diagram) does not show accurate energy distribution because it neglects diffraction effects.
Mladen
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#60
Posted 16 February 2013  01:13 PM
I would guess that the 86.6% wins out in that respect.
John

#61
Posted 16 February 2013  03:47 PM
Mike
#62
Posted 16 February 2013  07:50 PM
Yes, they are axial. This is a Schmidt camera. Its images are pretty much unchanged even 3 degrees off axis.I take it those are axial MKV? If as Vla suggests the profiles are distinctly different they will have differing degrees of colour correction off axis.
I would guess that the 86.6% wins out in that respect.
John
#63
Posted 17 February 2013  12:45 AM
I do mention that many (including the respectful ones) sources erroneously state that 0.866 neutral zone minimizes Schmidt spherochromatism. And, like I said, it is easy to confirm with raytrace that the wavefront error for nonoptimized wavelengths is more than twice smaller with 0.707 neutral zone (it is also directly implied by the diffraction theory, which places best s.a. focus midway between the paraxial and marginal, i.e. coinciding with 0.707 zone focus)
That may change if other refractive elements are present. For instance, this particular system has a field lens made of common glasses that produces both longitudinal and lateral color error. They cannot be both minimized at the same time with the field lens alone: when the lateral color is corrected, red and blue foci are separated, and vice versa. Increase in the optical power of the corrector by heightening the neutral zone can bring the red and blue together, without affecting (corrected) lateral color.
Vla
#64
Posted 18 February 2013  05:23 AM
On Vla's point he mentions that the "bulge" in the centre is part of the colour correction even in just blue light. Optimisation routines tend to move that way too increasing it significantly.
I'm beginning to suspect my best bet is to try and optimise it at the original size with new glasses as that way I should gain when the size is reduced. The problem with what I have at the moment is that there is no real margin for error. The biggest problem is working on the lens to the exclusion of the corrector. The paraxial stuff isn't accurate enough.
John

#65
Posted 18 February 2013  05:39 AM
John

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#66
Posted 18 February 2013  07:46 AM
...chromatic aberration will be the smallest at the 70.71% NZ location.
Is it significant (i.e. perceptible)? I think not, certainly not photographically or interferometrically.
At 200mm f/3 it probably isn't, although it is always better to have less of the aberrations, than more, in the context of minimizing all error sources. But at f/2 it becomes significant, as illustrated by polychromatic MTF (typical CCD spectral sensitivity).
Vla
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#67
Posted 18 February 2013  08:23 AM
This is Oslo's standard asphere formulae. Most is pretty obvious apart from the value of Cc it uses as it seems that can't be changed. Is the value zero?
Since the Schmidt profile combines two wavefront deformations, defocus and primary spherical, and it is the former that defines the first term, being a parabolic function proportional to the zonal height squared, the corresponding conic is 1, and the first term comes to (r^2)/2R, where "r" is the zonal height defined in polar coordinates (i.e. in the pupil) as (x^2+y^2)^0.5, and curvature is 1/R, R being the vertex r.o.c. of the corrector, i.e. of the applied defocus wavefront deformation.
The OSLO formula is identical to one posted by Mike, and I caught myself in not seeing the obvious, which is that if the specific Schmidt profile formula uses parabolic (defocus) term, as it does, it can be also expressed using the general conic sagitta relation. It isn't revealing what the actual profile parameters are, but is formally correct.
Vla
#68
Posted 18 February 2013  12:42 PM
You're probably more familiar with the specific sagitta expressions for a sphere (using the same xy coordinate notion as OSLO): z(s) = R  sqrt(R²  r²), or, as Vla mentions, the parabola sagitta z(p) = r²/2R, where R is vertex ROC.
In the case of Schmidt plates, the cc = 1 (parabolic) because that's the only conic that has zero axial sphericla aberration (i.e. cc + 1 = 0) over the aperture. When cc = 1, then, as Vla says, the whole first term defualts simply to the familiar r²/2R.
In ATM3, the profile (z values) of the Schmidt corrector is given algebraically, avoiding the deformation constants, by what is called a biquadratic parabola equation. Remember though that this equation is a 4th order formula and incorporates only the first two terms in the expansion series,
z = (r^4  kr^2y^2)/[4(n1)R^3], where y = zonal height, and R = ROC of the mirror you're trying to correct, and n is the refractvie index of the glass used.
One more thing, OSLO can correct a perfect sphere (cc + 1 = 1), by using nothing but deformation constants, as shown below. Setu up a sphericla mirror, D = 200, R = 1200, leave conic = 0 and enter the deformation constants listed below. You get the same results as if you hand entered 1 for conic.
The OSLO formula however does not give you the vertex radius, which is not the same as the ROC of the mirror. And the vertex radius is the tgeoretical radius at the very cnete rof the corrector plate to which the plate is deformed.
Also, notice that the OSLO equation does not account for refractive index of the glass used. It is therefore much better to use the alternate equation that does: z = Ay^2 + B y^4 + Cy^6...etc. The term A = [(y/r)²D²]/[8(n1)R³]. Once you have A, the vertex radius Rv = 1/2A. The 4th terms is even simpler. For anything faster than f/3 you're probably fine with these two terms. Thus B = 1/[4(n1)R³]. Your total z, then is simply Ay^2 + By^4, where y = zone along the semidiameter of the corrector.
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#69
Posted 18 February 2013  04:20 PM
This is interestingly about halfway between the 70% and 86.6% zones. The study I did years ago that gave NZ's more in the lowtomid 80's was encircled energybased, which does not appear to be the best metric for image optimization of wideband linearresponse sensors like CCD's.
Just thought you'd find this interesting, and worthy of further study.
Mike
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#70
Posted 18 February 2013  04:22 PM
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#71
Posted 18 February 2013  05:01 PM
Mike
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#72
Posted 18 February 2013  06:22 PM
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#73
Posted 18 February 2013  06:26 PM
#74
Posted 18 February 2013  06:48 PM
Vla's point about the base curve of the corrector allows the SA curves of any 2 colours to coincide. As I have it at the moment bringing red and green in throws blue way out.
John

#75
Posted 18 February 2013  07:08 PM
And what, specifically are you referring to about "today's Schmidts"? Can you post up prescription files or patent numbers so that we might discuss them analytically and quantitatively, or are you just guessing it might be true?
Mike