SCT/Newt/central obstruction debate [split from 16" Dob]

[jase (Jason)]

It is impossible to evaluate the contrast reduction caused by central obstruction by simply looking at a Strehl ratio. It is also impossible to evaluate the performance of any telescope with a refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet) with the simple knowledge of the Stehl Ratio for one wavelength (usually green), which is sometimes given by telescope manufacturers.

Harold Suiter - http://www.willbell.com/tm/tm5.htm developed another ratio (EER), which takes into account central obstructions.

Encircled Energy Ratio - The EER is the ratio between the total energy inside a given angular radius from a mirror's diffraction pattern to the total energy inside the errorless circular aperture's diffraction pattern.

The point I'm making is that the SR alone can't deliver the final verdict.

[mickoking]

Originally Posted by bird

If you want the best performing scope, and don't mind getting your hands dirty with some tinkering and maintenance then the newt is very hard to beat.

Give me a newt over a SCT any day of the week

[Starkler (Geoff)]

I recall getting into a lot of trouble for saying similar things in the past

[jase (Jason)]

Perhaps to be politically correct, the perfect telescope has yet to be invented. All optical designs have strengths and weaknesses. The best telescope is one you use.

[74tuc]

Hi Jase,

1. Re:
"It is also impossible to evaluate the performance of any telescope with a
refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet)
with the simple knowledge of the Stehl Ratio for one wavelength (usually green), "

Please tell us what you understand by "Strehl Ratio" and further why you believe
the quoted statement is true?


2. Re:
"Encircled Energy Ratio - The EER is the ratio between the
total energy inside a given angular radius from a mirror's
diffraction pattern to the total energy inside the errorless
circular aperture's diffraction pattern."

You have just given us a definition of Strehl Ratio called by a different name.

The above statement may be read as:

"Strehl Ratio - The SR is the ratio between the
total energy inside a given angular radius from a mirror's
diffraction pattern to the total energy inside the errorless
circular aperture's diffraction pattern."

where: "given angular radius" = Radius of the Airy Disc.

I leave it to you to figure out why Strehl Ratios in a system are multiplied.
When you find the answer you will discover that the first statement is not true.

Regards,

Jerry.

[74tuc]

Suiter's definition of EER (SR) is in the spatial domain in terms of energy in the Airy disc.

SR may also be described in the frequency domain as "The amplitude of the 0 frequency term in the MTF"

The use of SR is always accompanied with a caution on its use - the optical system must be Aplanatic.

BTW the use of corrector plates can introduce an insidious chromatic aberration due to oblique (off axis) rays and the inherent spherical aberration, this aberration goes under the grand name of "Oblique spherro-chromatic Aberration"

Cheers,

Jerry.

[jase (Jason)]

Originally Posted by 74tuc

1. Re:
"It is also impossible to evaluate the performance of any telescope with a
refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet)
with the simple knowledge of the Stehl Ratio for one wavelength (usually green), "

Please tell us what you understand by "Strehl Ratio" and further why you believe
the quoted statement is true?

The Strehl Ratio put bluntly, is the measurement between reality and perfection. More specifically, it is the ratio of the light intensity at the peak of the diffraction pattern of an aberrated image to that at the peak of an aberration free image. Correct or have I missed something

A refracting element will refract wavelengths with varying performance. Some scopes are optimised for red, blue etc. Note: I have not mentioned anything about reflecting telescopes only refracting.
Potentially, it could be possible to have an achromat with the same Strehl ratio for green light wavelength as other designs (such as a high-end APO). This does not indicate the achromat is a better performer. Different wavelengths with different refractions. Some interferometers test at 550nm such as those used by Obsession telescopes (now I'm mentioning reflecting telescopes), so the red wavelength is used.

http://www.obsessiontelescopes.com/o...Interferometry

As a side note: Quoting from their website, there appears to also be descrepancies in the Strehl testing, so how can one be certain of the manufacturers specifications...
""Assumed" Strehl Ratio
There's a big difference in the Strehl ratios calculated only by zonal Foucault testing and interferometric testing. The Foucault test measures only a few points across one diameter of the mirror to generate a 2 dimensional cross section of the mirror's wavefront. There isn't enough data in this representation to accurately calculate an RMS or Strehl value. Some opticians take that cross section and assume that the mirror has perfect rotational symmetry. Because of this false assumption a Strehl based on Foucault test data is always over optimistic."

Originally Posted by 74tuc

2. Re:
"Encircled Energy Ratio - The EER is the ratio between the
total energy inside a given angular radius from a mirror's
diffraction pattern to the total energy inside the errorless
circular aperture's diffraction pattern."

You have just given us a definition of Strehl Ratio called by a different name.

The above statement may be read as:

"Strehl Ratio - The SR is the ratio between the
total energy inside a given angular radius from a mirror's
diffraction pattern to the total energy inside the errorless
circular aperture's diffraction pattern."

where: "given angular radius" = Radius of the Airy Disc.

I recall reading something about EER and IIRC a long time ago in one of the ATM books. Let me try dig up some info. It could have potentially been the combination of both to obtain the central obstruction affects.

Originally Posted by 74tuc

I leave it to you to figure out why Strehl Ratios in a system are multiplied.
When you find the answer you will discover that the first statement is not true.

Please give me an example. I'm assuming you are specifically talking about RMS values?

I'm learning lots here...

[jase (Jason)]

Hi Jerry,

Originally Posted by jase

The Strehl Ratio put bluntly, is the measurement between

I recall reading something about EER and IIRC a long time ago in one of the ATM books. Let me try dig up some info. It could have potentially been the combination of both to obtain the central obstruction affects.

I checked out the ol'telescope making book ("hand me down" with missing pages) for calculation of the SR associated with optical obstructions. Originally was thinking apodisation, but realised I got confused.

Central obstruction effect on intensity distribution of the diffraction pattern can be expressed in very simple terms. Obstruction of the relative size υ in units of the aperture diameter affects peak diffraction intensity I normalized to 1 as given by this simple relation:
I=(1-u2)2


The change in normalized peak intensity - given by dI = 1-I = 2(1-u2)u2 - represents the relative amount of energy transferred from the disc to the rings area. This attribute is consistent with the effect of wavefront aberrations, where the relative drop in peak intensity - for relatively small wavefront errors - practically equals relative amount of energy transferred to the rings area. The similarity extends to the consequential contrast drop-off for the range of spatial frequencies bellow ~0.5 which is the range of resolvable low-contrast details. In other words, for this range of spatial frequencies, the peak intensity I resulting from the central obstruction is comparable to the Strehl ratio for wavefront aberrations. They both indicate the relative amount of energy transferred to the rings area - the main factor determining contrast transfer at low- to mid-frequencies of the MTF.



Thus, with the RMS wavefront error w in terms of the Strehl ration "S" being ω=0.24(-logS)1/2, direct relation can be established between the relative linear size u of central obstruction and the similar in effect RMS wavefront error ώ (in units of the wavelength) for the low-contrast detail performance as:
ώ~0.24[-log(1-u2)2]1/2


For u=0.33, this gives ώ~0.076, or slightly worse than 1/4 wave P-V of spherical aberration level. Comparison with the effect of spherical aberration is most appropriate, due to both effects causing radially symmetric intensity distribution, with the predominant change being brightening of the first bright ring.

I'm still toying with these formulas to see the effects, but suspect I'm missing some values. If you have any great sites with more information on the above, I'd be interested in checking them out.

Cheers

[Geoff45 (Geoff)]

What about a refractor of equal size to the Dob or SCT?

[jase (Jason)]

Hmmm. Tough call.
Assuming both telescopes have the same focal length and focal ratio it would be a close call. I'd need to research this topic further. I have seen on other forums that it is has generated some debate amongst telescope enthusiasts.

I'm also assuming that you have sufficient funds for a refractor of equal aperture to a newt or dob. Cost differences per aperture inch is considerable.

Changing the topic...
I'd like to reference a comment by master optician Roland Christen (owner of astro-physics the manufacturer of what many regard as the best refractors money can buy with a Strehl ratio of .99)

Reference: http://geogdata.csun.edu/~voltaire/r...O_testing.html

"Manufacturers use interferometers to compare the wavefront errors of a finished optic against some reference standard. In the case of the interferometer, it is a reference sphere of known high quality which is used to form interference fringes with the optic under test. When testing a mirror, it would not matter what wavelength was used, since mirrors are totally achromatic. In the case of lenses, it matters greatly what wavelength is used, since there is typically only one point in the wavelength range where the lens was nulled or figured by the optician. Testing at another wavelength almost always results in a lower wavelength rating."

My earlier statement of:
"It is also impossible to evaluate the performance of any telescope with a refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet) with the simple knowledge of the Stehl Ratio for one wavelength (usually green), which is sometimes given by telescope manufacturers."
Is indeed true.

For those interested in more of Roland's optical wisdom and fantastic imagery please check the site out.
http://geogdata.csun.edu/~voltaire/roland/index.html

Regards

[74tuc]

Hi Jase,

I'd like to call time out and get back (for the time being) to the thread.
Sejanus must be thoroughly confused with all this
.
Question and Answer

The question posed was in the comparison of two types of telescope.
The answer to the question lies in the treatment of the telescope as an optical system - not just a primary mirror or lens.

In general:
To compare the telescopes optically we need to take into account the performance of each the optical element in the telescope then "combine" the performance of each of these elements such that we get a measure of the performance of the telescope as a whole. The overall performance of each 'scope is then compared.

The metric chosen here is the system Strehl ratio. As we have seen the Strehl ratio tells us the proportion of light reaching the Airy disk.

By definition:
SR = (Intensity at centre of PSF of the aberrated optic)
/(Intensity at centre of PSF of a diffraction limited optic)

PSF = Point spread function ie. the response of an optical system to a point image
cf. Impulse response in the time domain.

What happens next?

The psf of each element comprising the telescope must be known.
The PSF carries all the information about the optical element - to make life easy let us talk about one colour only.

I now try to describe in words the application of "Linear Systems Theory" to this telescope system.

(Why this theory may be applied to these systems is a long story that
starts with the fundamental question as to why we get diffraction in the
first place!!)

The light leaving the optical element is the light entering the
element modified by the psf of the element itself. This "modification" is
actually a mathematical operation called convolution (*). What may be said is that the function representing the incoming light (iin(x,y) is convolved with the psf(x,Y) to produce the output light function (io(x,y)) or

io(x,y) = iin(x,y)*psf(x,y)

For a system with a number of psf's in a line (eg. Telescope)
i(x,y) is the intensity profile. ii = input, io = output
io(x,y) = iin *(psf1*psf2*psf3)

Where psf1, psf2 etc are the individual psf's
and psft = psf1*psf2*psf3 ... psft = system psf

The equation: io(x,y) = iin(x,y)*psft(x,y) is very computationally intensive to implement - there must be an easier way (?) and there is!

We invoke an animal called the fourier transform - you may take the fourier transform of the above equation so:

if F{io(x,y)} = Io(wx,wy) ... notice change from i to I

note:x and y are spacial (eg mm) and wx and wy are frequency (cycles per mm)

In the time domain x and y would be seconds and wx and wy would be Hertz.
Io(wx,wy) = F{iin(x,y)*psft(x,y)}

or

Io(wx,wy) = Iin(x,y) . F{psft(x,y)} Where "." = multiply

note: We have changed the complex operation of convolution
to the simple operation of multiplication

Next, to simplify things by invoking the property of radial symmetry
we will remove one of the variables

So: Io(wx) = Iin(x) . F{psft(x)}

or write the above as:
Io = Iin . F{psft} ... Nice and simple!

Now remember that: psft = psf1*psf2*psf3

and

F{psf1*psf2*psf3} = F{psf1} . F{psf2} . F{psf3}

Now F{psf} is a complex function ie. It has amplitude and phase for
this exercise we will use the amplitude of F{psf} or |F{psf}|

In plain english:
|F{psf}| is the Modulus of the Fourier Transform
of the Optical Transfer Function and this is called the
Modulation Transfer Function (MTF)

Io = Iin . MTF1 . MTF2 . MTF3

or to extend the above to a 'scope with n components

Io = Iin.MTF1.MTF2.MTF3 ---- MTFn-1 . MTFn

MTF in optics is the same as frequency response in audio systems.

If we used the zero frequency term only then

I0o = Iin0o.S01 . S02 . S03 ----- S0 = Strehl ratio (SR)
I0o = Iin0o . SR1 . SR2 . Sr3 -----

So evaluating the component SR's in a telescope and multiplying them
gives us the overall SR which is a good measure on the performance of the scope.

The above sets the argument behind my original post. We have talked about the telescope as a system.
The next stage would be to discuss the points you raised but this is outside this thread.

It could be discussed as a separate thread but I am sure it would bore most people. The points would would address the tie between Strehl ratio and the phase front of the energy and how the optics affect this phase front. Once we understand this one could look at the atmosphere and the issue
of image de-convolution - notice many programmes have de-convolution algorithms and we have been talking about convolution - do you have a strange feeling that if we go down that path we will be going in the opposite
direction to the one we've been taking!!

Regards,

Jerry.

[74tuc]

Hi Jase

Re:
My earlier statement of:
"It is also impossible to evaluate the performance of any telescope with a refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet) with the simple knowledge of the Stehl Ratio for one wavelength (usually green), which is sometimes given by telescope manufacturers."
Is indeed true"

Please Show us the proof that this is true yours or by someone else.

Thanks.

Jerry.

[jase (Jason)]

Hi Jerry,
Thanks for the definition. This is now understood. Took me a while to get the flow. So the MTF can be defined as the magnitude of the Fourier transform result. It is the response of an imaging system to an infinitesimal point or line of light. Correct eerr perhaps I need to read your previous post again.

Agree, our conversations have become off topic. Apologies to all who are bored or scared of this thread.


Thanks again. Perhaps we can brain meld another time

[jase (Jason)]

Originally Posted by 74tuc

Hi Jase

Re:
My earlier statement of:
"It is also impossible to evaluate the performance of any telescope with a refracting element (Schmidt corrector, Maksutov corrector, doublet or triplet) with the simple knowledge of the Stehl Ratio for one wavelength (usually green), which is sometimes given by telescope manufacturers."
Is indeed true"

Please Show us the proof that this is true yours or by someone else.

Thanks.

Jerry.

This is based on what I have read. Don't own a refractor, nor a interferometer to valid these results.

"Manufacturers use interferometers to compare the wavefront errors of a finished optic against some reference standard. In the case of the interferometer, it is a reference sphere of known high quality which is used to form interference fringes with the optic under test. When testing a mirror, it would not matter what wavelength was used, since mirrors are totally achromatic. In the case of lenses, it matters greatly what wavelength is used, since there is typically only one point in the wavelength range where the lens was nulled or figured by the optician. Testing at another wavelength almost always results in a lower wavelength rating."

http://geogdata.csun.edu/~voltaire/roland/APO_testing.html

If you can get your hands on a interferometer with reference lens, I'll get some refractors. I'm sure there will be plenty of IIS members who wouldn't mind their optics tested.

[Fox]

Originally Posted by sejanus

hey guys out of curiousity, does a dobsonian of equal size to a SCT absolutely kill it optically? is the difference in brightness really big?

I'll add my 2 cents... A long time ago, after I bought my brand new 8 inch SCT, I took it to my friends place who had a home made (excellent) 8 inch Newt. In brief, the Newt just left the SCT dead in its tracks. Side by side, Centaurus A (Caldwell 77) and its dark lane were plainy visible in the Newt, but simply invisible in the SCT, not even with averted vision. OK, we had to use different eyepieces to achieve similar magnification, but in our minds there was no question the Newt just had so much more brilliance and contrast - it was such a disheartening experience since the SCT was brand new. Light gathering and resolution were OK, so planets, star clusters and globulars were pretty good in the SCT (IMHO). But when contrast became more critical, I reckon the SCT just seemed to fall apart.

Maybe at the time I just had a bad one, and maybe todays SCTs are generally better with better coatings etc. who knows, I have to say that modern SCTs look so inviting with Goto mounts. But I eventually replaced the SCT with my Genesis and never looked back. Fox.

[74tuc]

No Problems Jase. I find this stuff most interesting it is complex and the fine detail bores most people.

Yeah I agree about the Newtonian. I was at a viewing night when I looked through this chap's Newtonian (dob) - 8" F10 I think. The mirror was hand made in 1957 and it had a cardboard tube and hand painted; using orthoscopic eyepieces - this was one of the more impressive 'scopes I've seen through so much so I want to make one myself - I'm still looking for one of those amazing mirrors. This 'scope was excellent where it counted!

Jerry