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Conical Mirror Deformation (Re: Design & Construction of 20" ball scope)

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#1 ckh

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Posted 14 April 2015 - 10:56 AM

In another thread, Pierre Lemay is describing the Design and construction of his 20" ball scope. The 20" mirror is conical (f/3.9) and is centrally supported through a hole in the center. When vertical, on the test stand, the mirror is essentially parabolic. However, when pointed at the zenith, the mirror sags slightly toward the edge because the support is small ring under and just outside of the hole.

 

In post #22 of that thread, Pierre describes and presents the results of a finite element analysis (FEA) done on this mirror. The deformation at the very edge (which is only 1/2" thick) was calculated to be approximately 1/3 wave! Pierre would like to know how the shape changes due to this deformation (although he has been informed that it should still be very close to a parabola).

This thread attempts to answer the question quantitatively.

 

Please beware that I cannot guarantee that these results are correct. It would be great if someone could verify the computations or find errors!

 

We will assume that the mirror's central support on the test stand works very well and that there is little in the way of flexure because the mirror is supported at CoG on the stiffest plane and the outer parts of the conical shape are thin thus light while the inner parts are thick thus stiff.

 

The question is just how parabolic is the amount of deformation in the horizontal position (since a since parabola subtracted from another is still a parabola with slightly longer focus).

 

Since the mirror is parabolic (when the mirror is vertical) the height of the surface above the vertex (which is virtual because of the hole) at any radius r is

 

height = -r^2/8f + sagitta

 

If the FEA deformation were parabolic, then it must follow a quadratic formula:

 

deformation = a*r^2 + b*r + c

 

where a, b, c are some very small constants representing a parabola with an extremely long focal length.

In the case of parabolic deformation, the full formula for the surface height when the mirror is horizontal is:

 

deformed height = height - deformation = -r^2/8f + sagitta - (a*r^2 + b*r + c)

 

or rearranging terms:

 

= (-1/8*f - a) r^2 + (-b)*r + (sagitta - c )

 

which demonstrates that if the deformation is parabolic, the mirror surface will still be parabolic because this formula is a quadratic.

 

So, we want to know how much the FEA deformation deviates from an ideal parabolic deformation:

 

deformation = a*r^2 + b*r  + c

 

We must chose values for a, b and c. This means picking an ideal parabolic deformation for a "best focus" parabola. One choice for "best focus" is a parabola that coincides with the FEA deformations at the outer radius of zones 1, 9 and say 6. (There is a unique vertical parabola through 3 non-collinear points.)

 

A better method could be used to find a best fit. One could use the least squares method to weigh the deviations according to the area of each zone. A brute-force, iterative program could determine such a best fit for a, b and c. However, just to get a rough idea of the deviation from a perfect parabola, this guess at "best focus" is probably acceptable. A fully optimized best-focus parabola might give somewhat better results (less deviation) than will be derived by matching a parabola exactly to just 3 of the zones.

 

To be continued in the next post...



#2 ckh

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Posted 14 April 2015 - 05:34 PM

To find the ideal parabolic deformation, we want these equations to hold for a, b, and c at the chosen zones:
 
Zone     FEA Deformation   = a*r^2      + b*r              + c

1           1.98E-05              = a*13064 + b*114.30     + c
6           1.19E-04              = a*42329 + b*(-205.74)  + c **
9           17.8E-04              = a*64516 + b*254.00     + c

 

** I have used (-205.74) for r in zone 6 instead of 205.74 in order to ensure that the parabola has it's vertex near the center the mirror. Points on both sides of the mirror are needed to ensure the parabola derived is not shifted laterally in contradiction to the rotational symmetry of the mirror.  So I chose zone 6 to be -205.74mm which is a point on opposite side of the mirror from the zone 1 & 9 points.

 

Edit: The following paragraph is wrong concerning the vertex position. In order to get the parabola on axis, b must be taken as 0.  To use all the FEA deformation points in finding an on-axis, best-focus parabola, I'll try to write an optimizing computer program over the next few days. See post #5.

 

"I made the above mistake the first time through. The correction doubled the final deviation from a parabola (since a shifted parabola can fit just one side somewhat better). The parabola generated with points on both sides will turn out have it's vertex 2.0E-5mm off center.  When I used 3 points on just one side of the mirror, the displacement of the vertex from center was 8.0E-5mm. Note that the deformation parabola is extremely shallow so a very small displacement of the vertex has a minute effect on the surface height. This issue would go away in a more optimal parabola derived to best fit all zones weighted by area on both sides."

 

Since these are three linear equations with three unknowns, we can solve for a, b and c. Not trusting myself to do the calculations without errors, I found an online solver for linear equations which produced:

 

Edit: This solution produces a deformation parabola which is not precisely on the same axis as the undeformed mirror surface. The constant b must be constrained to 0 to assure the parabola is on the same axis.

a =  3,13E-09
b = -2.35E-08
c = -1.84E-05

 

Substituting these values into:

 

deformation = a*r^2 + b*r  + c

 

we can calculate perfectly parabolic deformations and compare them with the FEA deformations.

 

To be continued in the next post...


Edited by ckh, 15 April 2015 - 04:11 PM.


#3 ckh

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Posted 14 April 2015 - 06:42 PM

Double post. Still learning... :p


Edited by ckh, 14 April 2015 - 06:46 PM.


#4 ckh

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Posted 14 April 2015 - 06:42 PM

Here are the results:

 

Edit: These results are to be revised with a more accurate and optimal, on-axis solution. See post #5.

 

Attached File  Conical Mirror Deformation.xls   17.5KB   18 downloads
 
Again, please beware that I cannot guarantee that these results are correct. It would be great if someone could verify or find errors!
 
Below and attached is an Excel spreadsheet that lists the outer radius of each zone (column C) and the FEA deformation (column D in mm and in column E in waves of green light) from Pierre's post #22.

 

Column F is the chosen idea parabolic deformation in nm. Column G is the deviation of the FEA deformation from perfectly parabolic deformation in nm, H is the deviation in waves of green light and column I is the fractional deviation in hundredths of a wave.
 
If these results are correct, they are remarkable. Recall that the edge of the mirror sags from the edge of the central hole by nearly 1/3 wave, but the shape is still parabolic to 1/50 wave! Perhaps conical mirrors want to sag into a new parabola?
 
(I vaguely recall reading that a beam with a triangular profile fixed at the side deforms into a parabola, but I need to do some research to see what the conditions actually where.)
 
The formulas used are in the columns of the Excel spreadsheet below and attached . The linear equations solver I used is called Math Lessons, Calculators and Homework Help. (Maybe Excel can do that too?) I had a bit of trouble with this solver. It did not want to do divisions that result in very small numbers (it gave a warning and produced 0.0 for a and b). However, the online calculator showed the step by step solution process so I was able to do the final divisions on a calculator.
 

med_gallery_240847_5047_141259.jpg


Edited by ckh, 15 April 2015 - 04:14 PM.


#5 ckh

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Posted 15 April 2015 - 10:30 AM

I guess all this math is pretty boring... :fingertap: But, I'm still interested in the issue so I'll persist. :crazy:

 

After a night's sleep, I've found some errors in my interpretation of the quadratic formula for an ideal parabolic deformation:

 

deformation = a*r^2 + b*r + c

 

It is a parabola, however my mistake was in the interpretation of it's placement laterally from the axis. Alternatively, the equation for ideal parabolic deformation can be written this way:

 

deformation = a*(r - d)^2 + h

 

where d and h are constants and the point (d, h) is the vertex of the parabola. Constants d and h can be written in terms of the original constants b and c:

 

lateral displacement of vertex = d = -b/2*a                   (mirror horizontal)

height of vertex                      = h = c - b^2/4*a           (mirror horizontal)

 

Because we are assuming that (when vertical) the mirror surface is a centered parabola and since we know that the deformation of the mirror in the horizontal position has rotational symmetry, we want the vertex to be on the axis of the mirror. Hence we want d to be 0 which is the same as constraining b to be 0 in the original equation, in which case h = c. So the equation we should be using for the best-focus, parabolic deformation is instead just:

 

deformation = a*r^2 + c

 

This is the equation for an on-axis parabola and c is simply the vertical displacement of vertex (when the mirror is horizontal). That c may not be 0 is not a problem. It just means the focal point is shifted by c which will be a very small amount.

 

Since this equation constrains the parabola to the mirror axis, we now have only two unknowns, a and c. We can match a and c exactly to only 2 of our 10 given FEA deformation points.

 

At this point I feel more work is needed to obtain the optically best possible parabola by using an area-weighted, least-squares iterative solution. I plan to work on a small program to compute the best-focus parabola and try to embed this program into the spreadsheet. The program will effectively use all of the FEA deformation points symmetrically on both sides of the mirror. The parabola will be precisely on axis and best-fitted to all 10 FEA points instead of an exact fit to just 3 points as before. Also, I will try to make it work for a mirror with or without a hole.

 

The result will hopefully be a spreadsheet the matches the shape of any rotationally-symmetric deformed parabolic mirror to a best-focus parabola (given that the FEA deformations for the mirror are already available) and computes the deformed surface deviation and an area-weighted deviation from a perfect parabola.

 

This may take a while. To be continued...


Edited by ckh, 15 April 2015 - 04:23 PM.


#6 wh48gs

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Posted 18 April 2015 - 07:38 PM

In another thread, Pierre Lemay is describing the Design and construction of his 20" ball scope. The 20" mirror is conical (f/3.9) and is centrally supported through a hole in the center. When vertical, on the test stand, the mirror is essentially parabolic. However, when pointed at the zenith, the mirror sags slightly toward the edge because the support is small ring under and just outside of the hole.

 

In post #22 of that thread, Pierre describes and presents the results of a finite element analysis (FEA) done on this mirror. The deformation at the very edge (which is only 1/2" thick) was calculated to be approximately 1/3 wave! Pierre would like to know how the shape changes due to this deformation (although he has been informed that it should still be very close to a parabola).

This thread attempts to answer the question quantitatively.

The easiest way is to use the conic sagitta expansion series. The first three terms are:

 

S= d^2/2R +(1+K)d^4/8R^3 + (1+K)^2*d^6/16R^5

 

where "d" is the aperture radius, K the conic and R the r.o.c. In this case d=10 and R=156. For K=0 the second two terms give the difference in sagitta between sphere and parabola, with the differential in sagitta depth in between these two shapes being proportional to the conic. The second term gives 15.2 waves for 550nm wavelength, and the third 0.0312 waves, thus can be neglected. Since sag toward the edge makes mirror shallower, i.e. slightly hyperbolic (approximately), the corresponding conic is -1-(0.33/15.2)=-1.022. Since the best focus error is 1/2 of the sphere vs. parabola differential, spherical aberration introduced is 0.022*7.6=0.167, or about 1/6 wave p-v (overcorrection).

 

Vla



#7 GJJim

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Posted 18 April 2015 - 08:29 PM

Thank you for the analysis and links. I'm interested in this problem because one of my projects involves rebuilding a Meade 14" Lx200 SCT. This scope has a conical profile mirror with a central support as you describe. My goal is to fix the mirror (to eliminate flop when imaging) and try and duplicate the support system used by Meade for their now defunct RCX scopes. Meade Described the RCX as having a "bedded" primary mirror support where the back of the mirror rested on some kind of flexible material in addition to the central support at the through hole.

 

Oddly enough, no one seems to know what material Meade used behind the RCX mirrors as a support pad, or how they managed to "float" these heavy pieces of glass without deformation.


Edited by GJJim, 18 April 2015 - 08:29 PM.


#8 Pierre Lemay

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Posted 20 April 2015 - 08:26 AM

The easiest way is to use the conic sagitta expansion series. The first three terms are:

 

S= d^2/2R +(1+K)d^4/8R^3 + (1+K)^2*d^6/16R^5

 

where "d" is the aperture radius, K the conic and R the r.o.c. In this case d=10 and R=156. For K=0 the second two terms give the difference in sagitta between sphere and parabola, with the differential in sagitta depth in between these two shapes being proportional to the conic. The second term gives 15.2 waves for 550nm wavelength, and the third 0.0312 waves, thus can be neglected. Since sag toward the edge makes mirror shallower, i.e. slightly hyperbolic (approximately), the corresponding conic is -1-(0.33/15.2)=-1.022. Since the best focus error is 1/2 of the sphere vs. parabola differential, spherical aberration introduced is 0.022*7.6=0.167, or about 1/6 wave p-v (overcorrection).

 

Vla

Vla,

Thank you for your reply. To see how things fit, I computed the calculation using the sagitta expansion series you mention for all nine zones defined in the FEA analysis diagram. Here is what I got:

 

FEA according to Vla small.jpg

As you point out, the worst p-v spherical aberration (SA) caused by this deformation of the glass happens at the edge and, according to the equation you propose, corresponds to about 1/6 wave.  I don’t know if I’m doing this correctly but I applied this impact on SA for all the other points, as indicated in the table. It shows it is exactly half the mechanical deformation of the glass, which I find strange. Then, I computed the RMS error over the entire surface by taking into account the proportional surface of each zone and their contribution to the overall SA calculation, obtaining an RMS of 1/29 wavelength of light.

 

 

On the other hand, in Post #47 of this thread http://www.cloudynig...nch-ball-scope/ CKH comes up with an entirely different result after fitting an optimum parabola matched to the true shape of the deformed glass. He comes up with a p-v of 1/60 wavelength, which would make the overall RMS error tiny. Although remarkably small, this value is in line, and of similar order of magnitude, with what Robert Houdart calculated a few years ago for Mike Jones’ 24 inch x 3 inch thick (center) x 1 inch thick (edge) blank (see here). Houdart came up with 1/14 wave p-v, after refocus, for that blank (1/55 wave RMS). I was expecting my 20 inch to be less sensitive to static deformation (smaller, lighter, thinner edge).

 

 

Could you elaborate further why you think the sagitta expansion series maybe used to compute the new shape of the glass after deformation? That equation does not seem to take into account the true shape of the glass on the overall surface.

 

Thank you.



#9 wh48gs

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Posted 20 April 2015 - 10:32 PM

Pierre,

 

I assumed that vertex radius doesn't change due to the sag, in which case the surface would be best approximated by a hyperboloid. If that would be the case, there is no refocusing, and the best fit paraboloid is simply the one that focuses at the mid focus (0.707 zone focus) of the hyperboloid. The wavefront error at this (best) focus is 1/2 of the surface error (since surface error doubles in the wavefront at the paraxial focus, but the best focus error is four times smaller than the paraxial focus error).

 

As the second term in the expansion series implies, the hyperboloid vs. paraboloid sagitta differential changes with the 4th power of the zonal radius. That is not the case with the numbers the software gave. They imply deformation changing very closely with the square of the zonal radius, which implies that the deformed surface is practically paraboloid, only of somewhat longer vertex radius (i.e. shallower). If so, that would imply entirely negligible error due to the sag. It may not be so, and the deformed surface may not even be conic, but the worst case scenario seems to be hyperboloid, and if we assume that the software is at least reasonably close, the error due to surface sag should be negligibly small.

 

Vla

Attached Thumbnails

  • csag.PNG


#10 ckh

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Posted 21 April 2015 - 05:09 AM

He comes up with a p-v of 1/60 wavelength, which would make the overall RMS error tiny.

 

 

Pierre,

 

I corrected that in a subsequent post. It was misstated. What I really meant was that the maximum deviation from a parabola in the FEA results was 1/60 wave. Actually the PV was twice because there are two deviations of 1/60 wave, but they happen to deviate in different directions. Probably if those deviations were in the same direct PV would look at lot better.

 

I put little stock in PV, especially when you look at a match as good as that. RMS deviation (over the area of the mirror might be more a more useful and indicate a better match. I did not compute that but I could add it to my spread sheet if you like.

 

 

In another thread, Pierre Lemay is describing the Design and construction of his 20" ball scope. The 20" mirror is conical (f/3.9) and is centrally supported through a hole in the center. When vertical, on the test stand, the mirror is essentially parabolic. However, when pointed at the zenith, the mirror sags slightly toward the edge because the support is small ring under and just outside of the hole.

 

In post #22 of that thread, Pierre describes and presents the results of a finite element analysis (FEA) done on this mirror. The deformation at the very edge (which is only 1/2" thick) was calculated to be approximately 1/3 wave! Pierre would like to know how the shape changes due to this deformation (although he has been informed that it should still be very close to a parabola).

This thread attempts to answer the question quantitatively.

The easiest way is to use the conic sagitta expansion series. The first three terms are:

 

S= d^2/2R +(1+K)d^4/8R^3 + (1+K)^2*d^6/16R^5

 

where "d" is the aperture radius, K the conic and R the r.o.c. In this case d=10 and R=156. For K=0 the second two terms give the difference in sagitta between sphere and parabola, with the differential in sagitta depth in between these two shapes being proportional to the conic. The second term gives 15.2 waves for 550nm wavelength, and the third 0.0312 waves, thus can be neglected. Since sag toward the edge makes mirror shallower, i.e. slightly hyperbolic (approximately), the corresponding conic is -1-(0.33/15.2)=-1.022. Since the best focus error is 1/2 of the sphere vs. parabola differential, spherical aberration introduced is 0.022*7.6=0.167, or about 1/6 wave p-v (overcorrection).

 

Vla

 

 

Hi,

 

I sorry, I'm not sure what you suggesting here. Can you explain what you are trying to do?

 

I posted my results in Pierre's thread and just got back to this one.

 

Terms higher than d^2 are not of interest to meet my goal. To explain briefly, Pierre has the results of an FEA analysis of his mounted mirror facing the zenith. Those results tell him how much the mirror should actually sag at some 10 different zones. The question was, how much does this sagging change the shape of the mirror from a perfect parabola.. So we are simple finding a best-matching parabola to the curve that represents the sag from the FEA.

 

All I did was make a first guess by solving for the unique vertical parabola that has the same sag as 3 of the zones in the FEA data.  This solution was for constants a, b in c in the parabolic (quadratic) equation:

 

h = ar^2 +br + c

 

were r is the radius and h is the sag at that radius.

 

This 3-zone solution was a good match, but the parabola was off center because the b term was not 0. So I made it 0. Then I looked at the resulting graph and manually tuned the constant c a little. This gave me a parabola that is a very good match to the actual FEA sag. I did not have to tune 'a'. The solution for 'a' from just 3 zones just happen pass right through most of the data after the c adjustment.


Edited by ckh, 21 April 2015 - 05:30 AM.


#11 Pierre Lemay

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Posted 21 April 2015 - 12:08 PM

 

The easiest way is to use the conic sagitta expansion series. The first three terms are:

 

S= d^2/2R +(1+K)d^4/8R^3 + (1+K)^2*d^6/16R^5

 

where "d" is the aperture radius, K the conic and R the r.o.c. In this case d=10 and R=156. For K=0 the second two terms give the difference in sagitta between sphere and parabola, with the differential in sagitta depth in between these two shapes being proportional to the conic. The second term gives 15.2 waves for 550nm wavelength, and the third 0.0312 waves, thus can be neglected. Since sag toward the edge makes mirror shallower, i.e. slightly hyperbolic (approximately), the corresponding conic is -1-(0.33/15.2)=-1.022. Since the best focus error is 1/2 of the sphere vs. parabola differential, spherical aberration introduced is 0.022*7.6=0.167, or about 1/6 wave p-v (overcorrection).

 

Vla

 

 

Hi,

 

I sorry, I'm not sure what you suggesting here. Can you explain what you are trying to do?

 

I posted my results in Pierre's thread and just got back to this one.

 

Terms higher than d^2 are not of interest to meet my goal. To explain briefly, Pierre has the results of an FEA analysis of his mounted mirror facing the zenith. Those results tell him how much the mirror should actually sag at some 10 different zones. The question was, how much does this sagging change the shape of the mirror from a perfect parabola.. So we are simple finding a best-matching parabola to the curve that represents the sag from the FEA.

 

All I did was make a first guess by solving for the unique vertical parabola that has the same sag as 3 of the zones in the FEA data.  This solution was for constants a, b in c in the parabolic (quadratic) equation:

 

h = ar^2 +br + c

 

were r is the radius and h is the sag at that radius.

 

This 3-zone solution was a good match, but the parabola was off center because the b term was not 0. So I made it 0. Then I looked at the resulting graph and manually tuned the constant c a little. This gave me a parabola that is a very good match to the actual FEA sag. I did not have to tune 'a'. The solution for 'a' from just 3 zones just happen pass right through most of the data after the c adjustment.

 

Ckh, I think the additional thoughts provided by Vla in post #9 (see above) explain what is assumptions were when he wrote that and what they are now. From my understanding, his post #9 seems in very close agreement with the calculations you provided in my other thread (post # 47). He is finding the maximum deviation from an ideal, new, parabola that matches the deformed surface does not depart by more than 4.5 nm and 7.5 nm peak-to-valley. In the other thread you had computed about 9 nm maximum deviation and 18 nm p-v.

 

However please note that I corrected the positions of some of  the zone limits (X-Zonal), which had some slight errors on my first diagrams (especially near the center). With the corrected zonal position, you should have the same results as Vla.

 

Thank you to both of you for having computed this.  A 7 nm p-v departure from an ideal parabola is insignificant. It clearly demonstrates that the conical mirror blanks, as Ckh has pointed out, are truly perfect glass shapes for mirror blanks, even for diameters up to 24 inches (and possibly larger). Whatever parabolic shape the mirror has when manufactured will naturally droop to another, equally usable parabola, when subjected to earth's gravity. Only a slight re-focus will be required.

 

What happens to the deformation of the glass if the conical mirror blank proportions are varied? That is a good question. A full modelling study would have to be undertaken, subjecting different glass shapes (thin edge vs thick edge; big central hole vs small hole; thick central glass vs thin; large flat on back vs small; etc.) to FEA analysis followed by hand fitting to an ideal parabola. I will let someone else do that but it would be a worthwhile study.



#12 wh48gs

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Posted 25 April 2015 - 07:12 PM

Pierre,

Thank you for answering for me. Should be noted that Houdart's analysis indicates the presence of both deformation components, one significantly larger in magnitude that can be compensated for by refocusing, and the other, smaller, that results in wavefront error. Describing surface deformation with Zernikes (w/o tilt and defocus) gave that it consisted nearly entirely of the primary (17.7nm) and secondary spherical (3.7nm) term Since these (Zernike coefficients) equal the RMS wavefront error, they give the corresponding P-V WFE if multiplied by sqrt(11.25) and sqrt(28), respectively (note that the secondary spherical as a Zernike term represents pure secondary spherical balanced - i.e. minimized - with primary spherical and defocus). The P-V wavefront errors are in the 3:1 proportion, respectively, with the combined deviation given with their sum, as shown on the graph. This combined deviation is nearly identical in form to the pure secondary spherical, hence the two terms were used by the software to reconstruct this deviation form (since pure secondary spherical does not have its Zernike term).

I don't have experience with surface Zernikes, but from what I see the way it is set up is to give surface deviation as the deviation from reference sphere used by interferometer, where the sphere is effectively the perfect surface. This way, even if it is not the actual surface, surface error simply doubles in the wavefront. The actual form of surface deviation can be assessed from the form of deviation. In this case, since the form is pure secondary spherical, which is at the best focus described with z^6-1.5z^4+0.6z^2 ("z" being the zonal height), and at the paraxial focus simply as z^6, with the deviation at the paraxial focus being the directly doubled (actual) surface deviation. In other words, the non-correctable surface deviation due to the sag in this case increases with the 6th power of the radius. The 165nm surface p-v error doubles to 330nm p-v WF error at the paraxial focus, with the best focus error, smaller 2.33 times, being 142nm, or little over 1/4 wave p-v of pure secondary spherical (with the p-v to RMS ratio of about 3.6, that comes to 39nm wavefront RMS).

One thing that doesn't add up in Houdart's numbers is the total surface RMS of 12nm. It cannot be smaller than the largest term, since the RMS error equals Zernike coefficient (as absolute values), and the total RMS error is obtained as a square root of the sum of squared individual coefficients (so the sign doesn't matter). In this case, the primary spherical is 17.7 and secondary 3.7, the total for these two is 18.1nm (that is a bit less than 19.7nm obtained from summing up the two terms' p-v, probably for not using the exact values). That is still 0.065 wave RMS of (pure) secondary spherical on the wavefront, comparable to 0.22 waves p-v of primary spherical. Not negligible.

 

Vla

 

sagB.PNG


Edited by wh48gs, 25 April 2015 - 07:19 PM.


#13 ckh

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Posted 26 April 2015 - 07:12 PM

What happens to the deformation of the glass if the conical mirror blank proportions are varied? That is a good question. A full modelling study would have to be undertaken, subjecting different glass shapes (thin edge vs thick edge; big central hole vs small hole; thick central glass vs thin; large flat on back vs small; etc.) to FEA analysis followed by hand fitting to an ideal parabola. I will let someone else do that but it would be a worthwhile study.

 

 

I wish I had access to FEA tools and knew how to use them. The need comes up all the time. Such tools would be invaluable to ATMs in designing their telescopes. No guessing about adequacy before buying things.



#14 ckh

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Posted 26 April 2015 - 07:29 PM

Hi wh48gs,

 

Your analysis is much more sophisticated than my ballpark shot at it. I've heard of these polynomials for matching curves but really don't anything beyond that. I was a math major long ago, but I studied mostly abstract math (to BA level) rather than applied math (I think abstract math is easier  :lol: ).

 

I appreciate your explanation of the method nevertheless.

 

With that, I am withdrawing from my attempt to code an optimizer for parabola fitting and an Excel calculator. Besides, now we basically know that the conical mirror works very well and further optimization would require some rounds of FEA.

 

Carl



#15 ckh

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Posted 26 April 2015 - 07:29 PM

Hi wh48gs,

 

Your analysis is much more sophisticated than my ballpark shot at it. I've heard of these polynomials for matching curves but really don't know anything beyond that. I was a math major long ago, but I studied mostly abstract math (to BA level) rather than applied math (I think abstract math is easier  :lol: ).

 

I appreciate your explanation of the method nevertheless.

 

With that, I am withdrawing from my attempt to code an optimizer for parabola fitting and an Excel calculator. Besides, now we basically know that the conical mirror works very well and further optimization would require more rounds of FEA which are perhaps unwarranted.

 

Carl


Edited by ckh, 26 April 2015 - 07:32 PM.


#16 abberation

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Posted 27 April 2015 - 01:12 AM

The perfect time for more FEA is when there is uncertainty.

 

If I was deciding on a traditional cylinder blank with 27 point flotation cell vs. a conical blank with central support for a 20+ inch mirror, I would want to know which one simulates better (if the simulation quality is equal).



#17 wh48gs

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Posted 27 April 2015 - 07:06 PM

Hi Carl,

 

It is hard to make optical analysis outside of the optics context. Practically, in order to arrive at the proper interpretation, one would have to re-invent the part of optical theory related to it - not as easy as it sounds ;)  As for the effect of mirror gravity sag, we have two conflicting simulations, one implying no appreciable effect, and the other having it rather significant. As the last poster says, key is in the accuracy of the (sag) simulation.

 

Vla
 



#18 Pierre Lemay

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Posted 27 April 2015 - 07:28 PM

Hi Carl,

As for the effect of mirror gravity sag, we have two conflicting simulations, one implying no appreciable effect, and the other having it rather significant. As the last poster says, key is in the accuracy of the (sag) simulation.

 

Vla

Vla, could you elaborate? I read over and over what you wrote in post #12 but I'm having a hard time understanding the math. When you write "we have two conflicting simulations", which "two" are you refering to? I was under the impression that the FEA results I have for my 20 inch conical mirror fits a new parabola (after sag) that is very good. What am I missing? Also, we know from the experience of several owners of medium sized conical mirrors (12 to 24 inch) that the image is not degraded to the point of uselessness. 

 

Thank you in advance for your patience. 



#19 brave_ulysses

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Posted 27 April 2015 - 08:03 PM

you might have a look here:

 

http://www.caelinux.com/CMS/

 

 

good luck!

 

 

 

 

What happens to the deformation of the glass if the conical mirror blank proportions are varied? That is a good question. A full modelling study would have to be undertaken, subjecting different glass shapes (thin edge vs thick edge; big central hole vs small hole; thick central glass vs thin; large flat on back vs small; etc.) to FEA analysis followed by hand fitting to an ideal parabola. I will let someone else do that but it would be a worthwhile study.

 

 

I wish I had access to FEA tools and knew how to use them. The need comes up all the time. Such tools would be invaluable to ATMs in designing their telescopes. No guessing about adequacy before buying things.

 



#20 wh48gs

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Posted 27 April 2015 - 09:04 PM

When you write "we have two conflicting simulations", which "two" are you refering to? I was under the impression that the FEA results I have for my 20 inch conical mirror fits a new parabola (after sag) that is very good. What am I missing? Also, we know from the experience of several owners of medium sized conical mirrors (12 to 24 inch) that the image is not degraded to the point of uselessness.

 

Pierre,

 

The first is the one which gives practically parabolic deformation, which requires the figure to "open up" all the way to the center, i.e. to have vertex radius increased as a part of the gravity-induced sag. The second one is based on Houdart's Zernike terms assigned to the (deformed) surface, which are - with tilt and defocus removed - practically limited to 17.7nm primary spherical and 3.7nm secondary spherical term (at the end of the Mike Jones' mirror thread). Even if we go with Houdart's combined 12.1nm total - which does not fit with the coefficient values, since it is a square root of the sum of squared individual values - we still get 0.044 wave RMS wavefront error, corresponding to 0.15 wave p-v of primary spherical (if the deformation is proportional to the 4th power of zonal deformation; about as much p-v of pure secondary spherical, if the deformation rate is proportional to the 6th power of zonal radius). The effect depends on the mirror figure w/o gravity effect. It will subtract from undercorrection (which paraboloids generally tend to go), and add to overcorrection. If it is, for instance, 0.15 waves p-v undercorrected, the gravity effect would make it near perfect. But if it is 0.15 waves overcorrected, figure under gravity becomes 0.3 waves overcorrected. But even with that magnitude of wavefront error, a 20+" mirror is quite useful (on most locations this error would have been dwarfed by seeing and thermals).

 

What may be hard to follow is not the math (there is hardly any), but the interpretation of Zernikes. As you probably know, they are used to reconstruct surface shape by a sum of any needed number of aberration forms, in varying magnitudes. Houdart comes up with practically only two Zernikes, mentioned above, as defining the form of surface deviation (after refocusing). To find out what is that form, we merely need to sum up the two surfaces, as shown on the plot (I should have put "primary spherical" for the 4th order surface, and "balanced secondary" for the 6th order surface, since the latter is not a pure 6th order but, as mentioned, also contains 4th and 2nd order terms). When we sum them up, they give a surface nearly identical to that described by pure secondary spherical at the best focus location, described by z^6-0.9z^2 ("z" being the zonal height). It is highly unlikely that this is the actual surface deformation (the black plot), since it would imply that the surface under gravity pops out up to about 74% radius, and then turns down, as if falling off a cliff toward the edge. More likely, it is the deviation from perfect reference sphere acting as perfect surface in the interferometric (software) setup. If so, it implies that the actual deviation has the form of pure secondary spherical, described by z^6 at the paraxial focus. That would make the paraxial p-v 2.33 times greater than the best focus p-v. But what counts is the best focus deviation, and it really doesn't matter whether the actual surface has 2.33 times larger max deviation proportional to z^6, or 2.33 times smaller deviation proportional to z^6-0.9z^2 - the wavefront consequence is identical.

 

Vla




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