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Catseye vs. Howie Glatter and Blug??

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#201 Vic Menard

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Posted 05 December 2008 - 07:54 PM

...Although this is probably just another academic point, considering the calculations Jim presented earlier on this page, it does seem to me that the dissappearing compromises the 2x-4x-6x-8x accuracy claim of the AC at least a little. If that didn't happen its accuracy superiority to a cheshire would be more convincing. I'd buy one, anyway.

Using a carefully decollimated primary, the focuser axial error is magnified 2X, no compromises. Reducing the primary mirror after correcting the focuser axial error, the error can be read to 1/8 of the autocollimator pupil diameter, or 0.015-inch. I can usually read a Cheshire alignment to about 0.01-inch, and maybe a little closer with a good Barlowed laser. For high magnification performance, my f/4 primary mirror axial alignment needs to be in the vicinity of 0.014-inch, which is pretty close to the resolution of the autocollimator and well within the resolution of my "Barlowed" (1mm aperture stop) laser. Since I use both tools, I'm always satisfied with my alignment accuracy.

The real power of the autocollimator lies in its ability to show persistent axial errors, magnified as much as 8X, when a less reliable tool is used (i.e., an "economy" laser), or when the user's alignment skill isn't quite up to speed compared to his performance expectations. With more reliable tools (including the user's skill, experience, and knowledge), the autocollimator's accuracy may not be necessary for final axial alignment, but it can still be used for redundant verification.

#202 CatseyeMan

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Posted 05 December 2008 - 08:43 PM

Don, the 1-way discussion is not to solve the parallax issue but rather to solve the disappearance act of the last 3 images. The peep hole is the reason why the last 3 images disappear under "perfect" collimation. The 1-way mirror will allow all stacked images to remain – at least in theory.
Jason


Technically it's only the last "2" images that disappear. The 2nd (upright) image remains (slightly offset from the 1st image) even after the last 2 are eclipsed out of sight. This offset diminishes with continued reduction in axial alignment error until perfect axial alignment is achieved and image 2 is coincidental with image 1 or as Nils would say.. there is actually no image 2 at all (because the triangle is not a parallel-ray reflector like the mirror)

#203 CatseyeMan

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Posted 05 December 2008 - 10:11 PM

Jim, I was referring to reflections at the Primary end -- not at the AC end. When all images are stacked, it means the last rays (for images “2”, “3”, and “4”) have to emanate from the primary center where the triangle sticker is located.
With a peep hole, that last 3 images are invisible therefore the above will not come to play. With a peep hole, the above will reduce the intensity of the last 3 images significantly – perhaps to nil.
Jason


in developing your Excel ray tracer, I think you may have a misconception as to how the 3rd and 4th images are formed. Image 3 is a "real" image formed by rays coming from ALL points on the parobolic Primary generated from the "virtual" image of the center spot 1 fl behind the AC mirror or 2 focal lengths away from the Primary center or at its ROC. By definition, any object (or in this case, an "image") placed at the ROC generates an inverted image of the same size at the ROC. The caveat is that when there is sufficient axial error to prevent the "real" image rays from going through the AC pupil (bundled into a point at the focal plane) the AC mirror effectively reflects the "real" image rays back to the Primary surface where it can be seen as (the inverted) image 3. We see image 4 as image 3's virtual reflection (located 1 fl behind the Primary) as a result of image 3's reflection being returned by the AC mirror.

See diagrams and animations at: Concave Mirror Ray Diagram Tutorial

#204 Jason D

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Posted 06 December 2008 - 12:26 AM

Jim, thanks for the posts. After reading your last two posts, I realized what I was doing wrong. I modeled the center spot as one point which is too simplistic. I improved the modeling and now I can see your point about how image “2” persists while images “3” and “4” disappear.

See attachment. I raised the AC above the focal plane by a small amount for clarity.

The left figure represents image “2” which persists because “vertical” reflected lights will bounce down then on to the pupil hole -- visible

The middle figure represents image “3” which will end up being truncated by the pupil hole – invisible – unless 1-way mirror is used.

The right figure represents image “4” which will also end up being truncated by the pupil hole – invisible – unless 1-way mirror is used

Few more interesting observations:

1- Image “3” will disappear before image “4"
2- As I continued to move the AC above the focal plane, image “4” eventually became visible but magnified
3- As I continued to move the AC higher above the focal plane, image “3” also became visible but not too magnified

Jim, in your explanation, did you assume the AC located at the focal plane?
According to my math, ROC for a paraboloid is not fixed in space (ROC=(2*FL)+((X^2)/(4*FL)) but I suppose for small angles (small X) we can treat the mirror surface as a sphere which has a fixed ROC (ROC=2*FL).

FYI. The way I implemented the ray tracer is by taking care of the math of bouncing a light ray between a flat mirror and a parabola – I did not factor in the ROC formula or any complex optical formula.

Jason

Attached Thumbnails

  • 2789227-ac_act8.JPG


#205 CatseyeMan

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Posted 06 December 2008 - 09:58 AM

... I raised the AC above the focal plane by a small amount for clarity.


Once you move the AC mirror off the FP, things get busy and bizzare in a hurry. Only at or very near the FP are the visible reflections reduced to 4 and are the same "size" (at exactly the FP).

.... Few more interesting observations:
Jim, in your explanation, did you assume the AC located at the focal plane?


Yes, the simulations have been with AC mirror "at" the FP which is the practical application.

According to my math, ROC for a paraboloid is not fixed in space (ROC=(2*FL)+((X^2)/(4*FL)) but I suppose for small angles (small X) we can treat the mirror surface as a sphere which has a fixed ROC (ROC=2*FL).


In the narrow angular window of AC functionality, yes.

If you really want to see a kaleidoscope of reflections, get yourself a 1" extension tube for your focuser and pop in the AC. When you rack the focuser up and down through it's range you might see something like this animation. The scenario illustrated is a 2.75" AC mirror placed directly over and perpendicular to an 8"/f2 mirror. The animation shows the reflection effects as the mirror is moved from the FP to a height 2" above the FP and back. The same effect happens going below the FP as well.

Posted Image

Posted Image

Larger AVI Version

#206 hudson_yak

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Posted 06 December 2008 - 10:06 AM

I realized what I was doing wrong. I modeled the center spot as one point which is too simplistic


Yeah, that had me puzzled a bit in your last set of images.

FYI. The way I implemented the ray tracer is by taking care of the math of bouncing a light ray between a flat mirror and a parabola – I did not factor in the ROC formula or any complex optical formula.



I'd think that's all the math you need. On the other hand, I believe it might be helpful if your traces extended up to the ROC region, and also below the primary mirror to show where virtual images would be formed, as in the very helpful link Jim posted.

Just in general, I have to confess though I do get some understanding from these diagrams you're creating (and appreciate you posting them), they are pretty terse and it takes time. Anything you can do to embellish them a bit would broaden their appeal I imagine. I'd also like the horizontal and vertical scaling to match, though I understand that would require much bigger images to see the details.

Mike

#207 Starman1

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Posted 06 December 2008 - 10:09 AM

Few more interesting observations:

1- Image 3 will disappear before image 4
2- As I continued to move the AC above the focal plane, image 4 eventually became visible but magnified
3- As I continued to move the AC higher above the focal plane, image 3 also became visible but not too magnified


How far above the focal plane? The reason I ask is that the focal plane of my scope is close to the tube, and I typically use the autocollimator with the focuser racked out to nearly its maximum outward extent, or probably about 1.5" farther out.
I do this because the AC fits so snugly in my focuser that the pressure of the focuser bearings on the outside of the drawtube is sufficient to make the AC difficult to insert. So I got in the habit of racking the focuser out to insert the tool, and it didn't dawn on me that this might make a difference in what appeared in the AC.
It might explain what I noticed when I was using a triangle on the mirror and it might explain what I see when the images are completely stacked.

I typically do my first collimation in daylight with the sun actually hitting the center of the mirror (note: the telescope is NOT pointed at the sun--I have a truss tube and the centermark is illuminated by sunlight at an angle). This makes all 4 images in the AC very visible and even bright. I use a reflective centermark on the mirror--even the 4th image is easily seen.
When I was using a triangle, I noticed the upside down image would dim significantly, but not disappear behind the upright images. I could still see a Star of David image even when all the images' central perforations were lined up and the pupil was centered. When I first was told that the 3rd and 4th images would disappear when the stack was exact, I thought "What am I doing wrong?" Because the 3rd and 4th images didn't disappear, they just became quite dim.
Now I use a circular "donut" centermark, and when I am collimated exactly--all 4 images stacked, the pupil dead center, and the cheshire showing perfect primary collimation--the edges of the stack appear "fuzzy", as if I've stacked 4 images of unequal size. I attributed that to my older vision, which cannot simultaneously focus on all 4 images because they are seen at differing distances away from the eye.

But if the 3rd and 4th images don't disappear when the AC is above the focal plane, that could explain what I saw with the triangles, and what I see with the circles.

Now if the AC has to be, say, 25cm behind the focal plane in order for that to hold true, then it is my vision, and the non-disappearing triangles just indicated a less-than-perfect collimation.

Any comments?

NOTE: JUST SAW JIM'S POST ABOVE, AND NOW I BEGIN TO UNDERSTAND ABOUT THE NECESSITY OF POSITIONING THE AC NEAR THE FOCAL PLANE. THANK YOU. :bow:

#208 Howie Glatter

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Posted 06 December 2008 - 12:04 PM

After viewing the larger version, I think I want an autocollimator and an extension tube. I may never use the telescope for watching the sky again.

H.

#209 CatseyeMan

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Posted 06 December 2008 - 12:21 PM

After viewing the larger version, I think I want an autocollimator and an extension tube. I may never use the telescope for watching the sky again.

H.


Hi Howie - LOL!

Certainly could be an entertaining diversion on cloudy nights especially if you're inclined to smoke those funny cigarettes. ;)

It was fun to animate this effect and is definately :cool: to watch.

#210 Jason D

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Posted 06 December 2008 - 12:54 PM

Jim, I hypnotized my kids my having them stare at your last animation :lol:
I guess the autocollimator can be used to hypnotized mischievous kids during school star parties :question:

Jason

#211 Jason D

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Posted 06 December 2008 - 12:57 PM

the edges of the stack appear "fuzzy", as if I've stacked 4 images of unequal size

Don, my guess is that not being exactly at the focal plane and the coma affect of the Primary parabolid could contribue to the fuzziness.
Jason

#212 Jason D

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Posted 06 December 2008 - 02:25 PM

I realized what I was doing wrong. I modeled the center spot as one point which is too simplistic


Yeah, that had me puzzled a bit in your last set of images.


Mike, using a simple model works most of the time. By "simple", I mean shrinking the whole center triangle to one point as I have illustrated in the following post
http://www.cloudynig...sb/5/o/all/vc/1
However, when considering the scenario when the AC and optical axes coincide, then the above simple model breaks down. In the this case, we need to trace both edges of the triangle.


I believe it might be helpful if your traces extended up to the ROC region, and also below the primary mirror to show where virtual images would be formed, as in the very helpful link Jim posted.


I am not sure if the following is correct
{{Since the AC is located at the focal plane then all images will not undergo any magnification and since all images have the same perceived sizes then we can fairly assume all images are located at a distance equals to the focal length in front or behind the AC.}}

As far as the relative perceived locations of the 4 images, they are included in many of my illustrations as shown in the same link above (the right most figure)

Just in general, I have to confess though I do get some understanding from these diagrams you're creating (and appreciate you posting them), they are pretty terse and it takes time. Anything you can do to embellish them a bit would broaden their appeal I imagine. I'd also like the horizontal and vertical scaling to match, though I understand that would require much bigger images to see the details.


I am open to suggestions.
Unfortunately, shrinking the Y axis has the adverse affect of distorting right angles.

AC images were confusing to me until I started playing with the ray tracer then things started making more sense. By shifting/rotating the AC axis while tracing light rays for each of the 4 images, you develop a better feel of how these images behave; especially, around interesting scenarios such as CDP and BOW TIE.

Jason

#213 Jason D

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Posted 06 December 2008 - 02:45 PM

in developing your Excel ray tracer, I think you may have a misconception as to how the 3rd and 4th images are formed. Image 3 is a "real" image formed by rays coming from ALL points on the parobolic Primary generated from the "virtual" image of the center spot 1 fl behind the AC mirror or 2 focal lengths away from the Primary center or at its ROC. By definition, any object (or in this case, an "image") placed at the ROC generates an inverted image of the same size at the ROC. The caveat is that when there is sufficient axial error to prevent the "real" image rays from going through the AC pupil (bundled into a point at the focal plane) the AC mirror effectively reflects the "real" image rays back to the Primary surface where it can be seen as (the inverted) image 3. We see image 4 as image 3's virtual reflection (located 1 fl behind the Primary) as a result of image 3's reflection being returned by the AC mirror.


Jim, I read the above post again and it made even more sense.
Jason

Attached Thumbnails

  • 2790159-ac_rays.JPG


#214 CatseyeMan

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Posted 06 December 2008 - 03:58 PM

I think this is what you want:

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  • 2790299-AC Reflections - R1.gif


#215 Jason D

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Posted 06 December 2008 - 04:24 PM

Thank you Jim... Of course, "3" and "4" need to have the same orientation.
Jason

#216 hudson_yak

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Posted 06 December 2008 - 05:36 PM

I am open to suggestions.
Unfortunately, shrinking the Y axis has the adverse affect of distorting right angles.

AC images were confusing to me until I started playing with the ray tracer then things started making more sense. By shifting/rotating the AC axis while tracing light rays for each of the 4 images, you develop a better feel of how these images behave; especially, around interesting scenarios such as CDP and BOW TIE.



I guess I was misunderstood. There were some images back on page 9 where the scaling did not match, those were the ones I was commenting about. Later on you did explain that in the fine print. My preference would be that they did match and yes, you do that most of the time. I guess I just remember the ones the most where you didn't, as it jumped out at me that the mirror reflections didn't look right.

As to understanding the diagrams, I'll just say I think they can be pretty tough to figure out unless the viewer is just as inclined as you were to start at the ground level and work your way up. My assumption is that you want to give the viewer a bit faster path than that.

What I still hope to see, and haven't yet, is a full raytraced diagram showing all AC images, real and virtual, stretching from the ROC to behind the primary mirror. One of Nils Olof's pages shows something close to this, though treating the marks as points.

Mike

#217 Jason D

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Posted 06 December 2008 - 11:17 PM

Mike, how about these diagrams

Image "1" is the real thing
Image "2" is formed behind the AC. Given a fixed point on the primary, all random rays emanating from that given point will eventually run through a unique point behind the mirror at the ROC level after few or many iterative reflections between the primary and the AC-- as Jim described it.
Image "3" and Image "4" are the flat mirror reflections of "1" and "2"
The location of the 4 images is fixed in space for a given tilt of the AC. That is, if you slide the pupil hole along the mirror, the location of the 4 images in space will remain constant; however, their relative locations will change since your are observing them from the sliding (variable location) pupil hole.
(Jim correct me if I got it wrong)

For a given AC tilt, here is a BOW TIE scenario

EDIT NOTE: AC Axis and AC mirror are perpendicular but are shown at a different angle in the illustrations due to shrinking the Y axis -- an unfortunate adverse affect. That is, the Y and X axes in the illustration are at different scales.

The figure to the left shows the path of reflected light rays -- there are only 4 images.
The figure to the right shows the light path between the pupil and the different images. In the BOW TIE case, image "2" behind the AC will be reflected/inverted then coincides with image "1". The same goes for images "3" and "4".

Attached Thumbnails

  • 2790981-ac_bow_tie.JPG


#218 Jason D

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Posted 06 December 2008 - 11:18 PM

And for the same tilt, here is a CDP scenario

EDIT NOTE: AC Axis and AC mirror are perpendicular but are shown at a different angle in the illustrations due to shrinking the Y axis -- an unfortunate adverse affect. That is, the Y and X axes in the illustration are at different scales.

The figure to the left shows the path of reflected light rays.
The figure to the right shows the light path between the pupil and the different images. In the DCP case, image "3" behind the AC will be reflected/inverted then coincides with image "1". Image "4" will be reflected/inverted; however, it will end up on the opposite side of image "3".

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  • 2790985-ac_cdp.JPG


#219 Jason D

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Posted 06 December 2008 - 11:53 PM

One thing I noticed is when the pupil hole moves significantly away from the optical axis, either "3" or "4" shift. I do not know if this is due to numerical rounding errors or due to the parabolic shape of the primary.
You can see this effect in the last attachment where the blue line does not run through one of the images.
Jason

#220 Jason D

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Posted 07 December 2008 - 12:03 AM

Jim, shouldn't it be as shown in the attachment?
"2" will overlap with "1" once you run it through the AC mirror and "3" will overlap with "4" for the same reason. When I ran my ray tracer on "1", I ended up with "2".
I have to admit that over the past few days I must have blown up 100s of brain cells thinking about this stuff :)
Jason

Attached Thumbnails

  • 2791055-ac_rays.JPG


#221 Jason D

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Posted 07 December 2008 - 05:12 AM

Another example with the same AC tilt but this time the pupil is located at the focal point

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  • 2791253-ac_focal.JPG


#222 Jason D

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Posted 07 December 2008 - 05:13 AM

And another example. Same tilt but a random pupil location

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  • 2791254-ac_random.JPG


#223 Nils Olof Carlin

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Posted 07 December 2008 - 06:53 AM

The reason the two rotated images (3 and 4) vanish is:
For these to be visible to the observer, first time around they must hit the reflective part of the autocollimator (=miss the pupil), but the next time around, they must hit the pupil to be seen (or else be lost to further reflections, eventually ending up on the spot they came from).

When the reflection of the autocollimator is aligned with its origin, (as I outlined earlier), no light rays can make it past both hurdles - they will be lost at either the first or the second hurdle (at least if the AC is close enough to the focal plane).

The condition for this to happen is when the axis of the AC passes through the COC. This need not imply true collimation - Vic Menard's "bowtie" simulation illustrates this. The visible condition for this is that the first reflected image, or the second image (whichever you prefer - it's the non-rotated one) is stacked with the spot itself (the two rotated images are also stacked, but the pairs are in general not).

Imagine this scenario - you have the bowtie pattern with the true and rotated pairs separated, but very little separation within each. Then you tweak the primary a little - enough to make the first pair stack and the second pair vanish. You might then -falsely- believe that you have true collimation.

Not too likely in practice, perhaps, but if you are not aware of this trap, you just may fall into it.

Nils Olof

#224 CatseyeMan

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Posted 07 December 2008 - 09:52 AM

Jim, shouldn't it be as shown in the attachment?
"2" will overlap with "1" once you run it through the AC mirror and "3" will overlap with "4" for the same reason. When I ran my ray tracer on "1", I ended up with "2".
I have to admit that over the psat few days I must have blown up 100s of brain cells thinking about this stuff :)
Jason


Here is a collage of the progression of reflections as generated in the PovRay virtual telescope scenario. There is both focuser axial eror and Primary axial error present. Each frame going from left to right and down represents the view to the eye after 1 incremental additional reflection is permited in the render. The number associated with each Frame is the number of reflections generating the view. As has already been mentioned, the 4th spot image appears after 13th reflections between the 3 optical components. It's interesting to note that at 15 reflections the background of the A/C is lightened; after that there is no change in the view - the last frame is after 100 reflections. It'a also interesting to note that the inverted section of the Secondary holder appears 1 frame later after 10 reflections than the inverted "3rd" triangle appearing in 9 reflections.

Any 2-dimensional ray tracing scenario must be consistent with the progression shown here. The 2 "upright" images are formed first in a 1-2 sequence and then the 2 inverted images are formed last in a 3-4 sequence.

Attached Thumbnails

  • 2791528-Reflection Collage.gif


#225 CatseyeMan

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Posted 07 December 2008 - 10:16 AM

... Imagine this scenario - you have the bowtie pattern with the true and rotated pairs separated, but very little separation within each. Then you tweak the primary a little - enough to make the first pair stack and the second pair vanish. You might then -falsely- believe that you have true collimation.

Not too likely in practice, perhaps, but if you are not aware of this trap, you just may fall into it.

Nils Olof


The subtle but important clue to avoiding this error is to note the symmetry of the AC pupil reflection relative to the spot perforation. If it is NOT centered, then then the viewer is alerted to this condition of "false" collimation.

Attached Thumbnails

  • 2791582-Vic Menard.png



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