Filter Size to Avoid Vignetting

[russellhq]

I want to work out what size of filter I can get away with without excess vignetting but I've no idea

I've sketched up my equipment arrangement with dimensions but not sure where to go from there:

Would I be able to get away with a 1.25" (31.7mm), or maybe 36mm or have to settle with a 2" (50.8mm)? Anyone know the math behind this?

[Lonestar70]

Hi Russel,

If this is for use with yor QHY8L camera, which has a 28.4mm active sensor diagonal dimension, then a 1.25" (31.75mm)  filter should be plenty big enough... assuming this is the clear optical diameter of the filter window (inside the mounting ring).

Otherwise a 36mm filter would be adequate.

Best regards.

Sandy.

[russellhq]

... assuming this is the clear optical diameter of the filter window (inside the mounting ring)...

Very good point. I had a search about and it looks like for Astrodon filters at least, the clear optical diameter is 25.5mm

http://www.astrodon.com/Orphan/astrodon_1.25_filter_cells/

[russellhq]

Well, I think I've gotten close to the answer with the help of the following website:

http://www.handprint.com/ASTRO/ae2.html

For simplicity I've ignored the reducer and assumed a refractor of equivalent focal length (510mm instead of 600mm).

Based on the following diagrams and the above web page:

I first used the following formula to work out the field angle to produce an image height equivalent to half the sensor diagonal:

tan(α) = h' / ƒo

tan(α) = 14.2 / 510

tan(α) = 0.0278

I then used the following formula to work out the filter diameter that would give no vignetting at the solved field angle above. Note that No is the f-ratio and I calculated w assuming the filter was at the furthest possible point from the sensor based on the arrangement in the first post.

z = (ƒo–w).[tan(α)/±0.5] + (w/No)

z = (510–33.7).[0.0278/±0.5] + (33.7/6.375)

z = (476.3).[±0.0556] + (5.286)

z = 31.8mm

Therefore, I'd need a clear aperture of 31.8mm. Just too much for a 1.25" filter

I must say, I've not studied optics since high school, so this took most of the afternoon to work through. I hope I've picked it up correctly!

[Lonestar70]

Hi Russell,

Your calcs look about right... I did a quick similar calc but used 14mm for image height and a filter displacement of 27.2mm (centre of filter wheel)... this came out at 30.72mm clear aperture being required.

However, since the clear aperture of the 1.25" filters is only 25.5mm it rather a non-starter.

A 36mm filter looks likely to be the smallest practical size... if your filter wheel can take them.

Best regards.

Sandy.

[russellhq]

Thanks Sandy, glad it checks out!

From the same site, it suggests:

At each obstruction, for any angle greater than φ, the amount of the light reduction increases in approximate proportion to the ratio cosine(φ)/pi. Vignetting effects can be calculated separately and are linearly subtractive.

Using this approximation, if I have the 1.25" filter in place but need an angle of 1.592deg for full illumination, the amount of light reduction would be cos(1.592)/3.142 = 0.318. So roughly a 30% reduction in light at the edge of the sensor. Would that make sense?

[Lonestar70]

Hi Russell,

Given your required angle of 1.592deg your assumption is correct.

How did you arrive at 1.592deg?

I am not entirely sure our original calcs are correct, since quickly reading through the article, I believe 'w' should include the length of the focusser draw tube... as this would be the last obstruction aperture before the focal plane ignoring any filters.

(3) The third limitation is the internal diameter of the last smallest obstruction before the focal surface, which may be the last tube baffle (in a refractor), the cylindrical or conical primary mirror baffle (in a Cassegrain), the opening in the visual back, or the internal diameter of the focuser drawtube (symbolized in the diagram above as opening z in relation to distance w):

tan(φ₃) = ±0.5[z – (w/N_o)] / (ƒ_o–w).

The ID of the draw tube would limit the max angle possible before the drawtube introduced Vignetting. 

I must admit I am very rusty on all this stuff now... not done much on optics since college some 50 years ago.

What do you think.

Best regards.

Sandy.

[russellhq]

Hi Sandy, I worked out the angle based on a field angle required to produce an image height equivalent to the sensor size:

tan(α) = h'/ƒo

tan(α) = 14.2 / 510

tan(α) = 0.0278

α = tan-1(0.0278)

α = 1.592

I worked out w as the distance of the filter from the sensor. I assumed that would be the last obstruction but maybe that's not correct. I'm working on that at the moment. The end of the draw tube is about 260mm from the sensor.

[Lonestar70]

LOL,

I worked out the angle based on a field angle required to produce an image height equivalent to the sensor size:

tan(α) = h'/ƒo

tan(α) = 14.2 / 510

tan(α) = 0.0278

α = tan-1(0.0278)

α = 1.592

Looking at the original calcs I can see it now... silly old me...

Told you I was getting rusty... I think maybe I have siezed up  

Keep happy.

Sandy.

[Lonestar70]

Hi Russell,

I found this in part 3 of the same article: -

If the field stop diameter and objective focal length are known, then the TFOV can be calculated as:

TFOV = 57.3·(D_fs/ƒ_o) degrees.

This formula can also be used to calculate the maximum true field of view possible with an eyepiece barrel or a visual back of a certain diameter. For example, in a 254 mm (10") ƒ/8 telescope and a 1.25" (27 mm) eyepiece barrel, the maximum true field of view is:

TFOV = 27/2032 x 57.3° = 0.76° x 60 = 46 arcminutes.

Using a 2" (46 mm) eyepiece barrel, the maximum true field of view is:

TFOV = 46/2032 x 57.3° = 1.30° x 60 = 78 arcminutes.

Substituting 510mm Focal length @f6.37 for the ED 80 with it's 0.85x ff/fr and using 46mm for the focusser tube dia... 

TFOV = (46/3248.7) x 57.3 = 0.8113 degrees x 60 = 48.678 arc minutes.

This would appear to be the maximum for the ED80.

How does this fit with your calcs?... it would seem to limit the image size somewhat.

Best Regards.

Sandy.

[russellhq]

Thanks Sandy.

I think it's time for another sketch (apologies for the crudeness!)

This is a sketch of the telescope without the reducer in place. This shows where the draw tube gate is in relation to the objective back.

Using these dimensions I can work out the maximum field angle before vignetting occurs by using: tan(φ3) = ±0.5[z – (w/No)] / (ƒo–w)

This gives tan(φ3) = ±0.0286

Then going back to equation: h' = tan(α)·ƒo; This gives h' = 17.2mm (or 34.4mm image diameter)

Now when the reducer is in the system, the draw tube is adjusted to the following position:

This gives tan(φ3) = ±0.0250; Note I used fo of 600mm as the light from the objective has not reached the reducer yet so this assumption made sense to me.

But if I use fo = 510mm in equation h' = tan(α)·ƒo then: h' = 12.75mm (or 34.4mm image diameter)

Now, I don't know whether these last 2 assumptions add up or not, I'll probably need to read a bit more.

[russellhq]

I went back to the drawing board today as I wasn't happy with the calcs saying I wouldn't get vignetting with my current set-up when my observations show that I do.

I had that Eureka moment when I realised I need to illuminate a larger image that is then reduced to the sensor size!

As my measured reduction is 0.80x, I would therefore need to illuminate an image 35.5mm to give a reduced image of 28.4mm!

Using the same equations as before, I found out that the draw tube was actually vignetting the image when the reducer was attached!

I calculated the vignetting effect to be 32% reduction at the edge of the sensor. This correlated well with the observed value of 35% from the master flat.

[Lonestar70]

Hi Russell,

That all makes sense and is not a million miles from the relationship between the 34.4mm iage dia. and the 25.5 image dia figures calculated in the previous post. 

A couple of things about your tables etc below....

The 285mm figure for (w) in the above includes the 55mm back focus distance required AFTER the FF/FR image plane... surely a distance of 230mm would be the correct figure since it would be the last point where the F7.5 is actually correct and would be the point where you would need to project the 35.5mm dia. image prior to reduction.

Or am I missing something?

Also... if you have a measured 0.8x reduction would this not make the overall focal length 480mm rather than the 510mm figure obtained with the claimed 0.85x reduction.

The other minor difference in the tables above is the value for (z) in the last column.

I believe the correct value to use is the INTERNAL DIA of the draw tube (this being 51.1mm as measured on my ED80 scope)  rather than the OD. 

I have not calculated what any of these differences would make as I am not entirely sure they are all real... other than the last one, which could possibly account for the difference between the latest theoretical value for vignetting and your actual measured value.

Given that it is now proven that the draw tube is vignetting the final image It would seem that a 36mm clear aperture filter would not increase this.

Very interesting problem this one and I feel it deserves more study to be sure of the exact methods to employ.

Best regards.

Sandy.

[russellhq]

Thanks Sandy.

I measured the ID of the tube as 53.7mm but I can see a step further down the tube that I can't measure with my ruler. So let's go with your 51.1mm figure, that's more conservative.

The way I derived the figure for w was:

I assumed the focus tube was extending 105mm into the telescope as per configuration when reducer is attached.

But when assuming where the focal point was, I use the objective's focal length of 600mm. I imagined that regardless whether the reduced was in the light path or not, the vignetting at the focus tube entrance would be unchanged. Hope that makes sense.

Also, I reviewed the drawing after reading your post and noticed a mistake with a couple of the dimensions. I've attached an updated drawing:

I'm now working on an assumption that if the focal tube is vignetting the image, then the image size on the sensor I should use to check for vignetting caused by the filter should, instead of being the sensor diagonal, should be the fully illuminated image height instead. i.e. 23.2mm from calcs vs 28.4mm.

Update calc table:

New calc shows the min filter diameter as 21mm. Quite a bit smaller than before!

[Lonestar70]

Hi Russell,

But when assuming where the focal point was, I use the objective's focal length of 600mm. I imagined that regardless whether the reduced was in the light path or not, the vignetting at the focus tube entrance would be unchanged. Hope that makes sense.

 Yes that is how I see it.

I'm now working on an assumption that if the focal tube is vignetting the image, then the image size on the sensor I should use to check for vignetting caused by the filter should, instead of being the sensor diagonal, should be the fully illuminated image height instead. i.e. 23.2mm from calcs vs 28.4mm.

 I also agree with this, since the 28.4mm image would include the vignetting from the draw tube.

New calc shows the min filter diameter as 21mm. Quite a bit smaller than before!

I am having difficulty agreeing with this one... In a converging image field between reducer and sensor, with the final fully illuminated reduced image size being 23.2mm... how can the filter be smaller than the finished image...without causing severe vignetting?... at a filter distance between filter and sensor of 32.7mm (approx 59% of the reducer back focus distance nearer to the reducer) I figure it must be at least 27mm clear dia. (assuming a fully illuminated non-reduced image dia of 29-30mm dia).

Final query... in your modified diagram of the ED80 with reducer you show the focal length as being 575mm...with the SW 0.85x reduction this would be 510mm... Hence I am a little puzzled by your figure

Or is this not showing the focal length?

Best regards.

Sandy.

[russellhq]

Ah, you're correct, it can't be less than the fully illuminated height! I think I got my algebra wrong somewhere. I re-calculated the number to be 28.6mm which is closer to your 27mm.

The 575mm at the bottom of the drawing is the physical distance from sensor to objective. It was getting confusing for me to draw focal lengths so I just used the measurements

I've been working on the vignetting a bit more an realise that my initial measurement of 65% remaining light was incorrect. I had used the max and min pixel values from the master flat, but any variance would throw that out!

So I measured again, this time using a 100x100 pixel area average in the centre and 4 corners. I then averaged the 4 corner results. I did this for each channel. This gave a new reduction of only 90% remaining light.

Next, to estimate the % remaining light, I assumed it was directly proportional to the % of unobstructed light in the marginal rays. I used the following sketch and calcs to work out that the remaining light would be 97% which is a bit more that the measurement. Maybe there is another obstruction somewhere, but that check will have to wait until tomorrow. Or maybe the yellow circle is actually an ellipse? I'll have to think about that one a bit more

http://mathworld.wolfram.com/Circle-CircleIntersection.html

[Lonestar70]

Hi Russell,

Yes, 28.6mm dia makes much more sense to my small brain... I was beginning to think I had lost the plot somewhere along the line. 

I came up with the 27mm figure by drawing the light cone at a seperation of 55mm, using the 2 fully illuminated image diameters at opposite ends, and then measured the width at 32.7mm from the final image plane... rough and ready, I agree, since it did not allow for edge attenuation or refraction in the filter etc.

Light loss.

This is an area which is certainly outside my knowledge comfort zone to a large degree, however, I do know that filters have a specific bandwidth and that frequencies towards the outer limits of the band, in either direction will have more attenuation... in exactly the same way as an RF filter would attenuate more on the flanks of the passband (albeit for slightly different reasons).

Blue light for example has an accepted wavelength bandwidth range of 490nm - 425nm, however, at the extremes of these wavelengths the attenuation could be as much as 90% or more... filter manufacturers tend to pick a certain centre wavelength, say 460nm, and then specify the bandwidth of the filter at the wavelengths which give FWHM transmittance (50%)... which could be e.g. 470nm - 450nm. 

Of coarse that does not mean you can ignore the transmittance outside the specified passband... which will/may also include a small percentage of the other colours... it makes for interesting calculations.

I would think, therefore, that you would certainly need to take into account the transmittance of the particular filter for the different wavelengths making up the pass band.

Also... no filter will pass 100% of the available input light within the specified bandwidth, and the peak transmittance may not be exactly in the centre of the bandwidth.

Perhaps this is your missing link? in the equations.

This is an interesting document I have on filter characteristics: -

wp-inside-filters.pdf

How did you measure the levels on your pictures... did you use a software package for this or have you got access to a spectrum analyser of some type?

I am certainly enjoying these discussions and hope I can learn a lot more from them... as an electronics engineer (retired) I am more familiar with the RF spectrum which is somewhat more predictable... light, on the other hand, can appear to do crazy things   

Best regards.

Sandy.

 

[russellhq]

Thanks Sandy, though I have even less idea about filters! I'm not sure if my drawing is clear, I've not drawing any filters. what I've shown is the entrance to the draw tube as a blue circle and the light cone from the marginal rays as a yellow circle. You can see the top of the light cone getting "cut off" by the draw tube. Hope that makes sense.

With regards to measurement, I took the master flat file I created for the camera and used a free program called Fitswork4. It allows you to draw a selection on the image and measure the mean pixel value. So I simply draw a square 100x100 pixels in the centre and four corners and noted the mean pixel value.

[Lonestar70]

Hi Russell,

Thanks Sandy, though I have even less idea about filters! I'm not sure if my drawing is clear, I've not drawing any filters. what I've shown is the entrance to the draw tube as a blue circle and the light cone from the marginal rays as a yellow circle. You can see the top of the light cone getting "cut off" by the draw tube. Hope that makes sense.

Yes, that makes sense, however... correct me if I am wrong... your actual measurements are from images obtained after the remaining calculated light has passed through your RGB filters; which could easily account for the difference.

This is getting complicated     

I will look up that software as it sounds usefull.

Best regards.

Sandy.