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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 20177 Location: Richland, Washington State, USA

Posted: Sat Dec 22, 2012 4:33 pm Post subject: 


ichbtm wrote:  Am I correct?
If this is the case, then i see absolutely no difference, in term of passing rays , between the left and the right baffle. 
You are correct, there is no difference in passing rays that should form the image.
But the point of a baffle is mainly to block rays that we do not want. Simply passing all the rays that we do want is easy: put in no baffles at all.
If you now draw some lines in autocad, I think you will find that baffles near the center of the extension are more effective at blocking stray rays, compared to baffles near either end.
Two limiting cases are clear. If you put a baffle exactly at the aperture, then rays that it would have blocked were already blocked by the aperture. If you put a baffle exactly in the sensor plane, then rays that it blocks have already missed the sensor.
The baffle is not effective in either of those positions. Baffles near those positions are not very effective either. For example your right baffle back near the sensor plane will not protect the center of the sensor from seeing those stray reflections around the lens in canonian's pictures. You need to stick the baffle someplace in the middle so that it can block stray rays while letting pass the ones you want.
The very best position is not easy to identify. It depends on where there are surfaces for those stray rays to bounce off, in addition to how well the shape of the baffle matches the boundary of the light cone at any particular position.
For some further discussion of how to design baffles, see the discussion and links HERE and in the surrounding thread. In the post HERE (same thread), Charles Krebs describes a clever device for directly observing what stray rays the sensor can see.
Rik 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Sun Dec 23, 2012 6:11 am Post subject: 


Quote:  But the point of a baffle is mainly to block rays that we do not want. 
yes , but i had to understand the difference betwwen the effect of the circle baffle and the rectangular baffle.
In fact, now i understand that there's two aspects :
1  The baffle has to be adapted to the section of the volume containing the rays hitting the sensor, and this to avoid vignetting and to be the spmallest possible to avoid non necessary lines to the e sensor.
Between the lens and (let's say) the middle of the extension, the best section is a circle. Between the middle and the sensor, it's a rectangle.
2  The baffle has to be positionned so that the maximum of stray rays are blocked. And it's clear that the better position is the middle (for a circle of for a rectangle one).
I was blocking on the first aspect because I was thinking that you wanted to tell me was that : the circle baffle had to be in the middle of the extension whereas the rectangular one could be near the sensor. In fact, it's not what you was saying
I'll do some 3D drawings to represent all that and post them here soon.
thank you rik 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Sun Dec 23, 2012 9:59 am Post subject: 


I'm baffle (like would say canonian )
Here is the drawing of the comparison between circle and rectangle baffles.
The sensor is a 24x36 (dark green rectangle on the right).
The lens aperture is 10mm diameter (bule circle on the left).
The distance between the lens and the sensor is 210mm.
The truncated cone containing all the rays which have to hit the sensor is magenta.
The cone containing this is yellow.
I did 3 sections (1/4 between them). Then i redrawed them on the right with 3 colors :
 in dark green, the surface containing the sensor rays
 in yellowgreen, the surface non blocked by circle baffle
 in magenta, the surface non blocked ba a rectangular baffle
Then I calculated the ratio between these surfaces.
In my previous post i was believing that balance between circle and rectangle baffle was in the middle of the extension, but in fact, I realize it's absolutely not the case now
i'll try to draw the stray rays now. 

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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 20177 Location: Richland, Washington State, USA

Posted: Sun Dec 23, 2012 12:47 pm Post subject: 


Wow! This is an excellent drawing, well beyond my ability to construct.
I too have had to struggle some to understand what is being shown. Let's see if I got it right.
In the figure above the scale, I think that the circles are sized to exactly contain the aperture (on the left), to just touch the sensor (on the right), and to scale linearly at the intermediate points. Similarly, the rectangles are sized to exactly contain the sensor (on the right), to scale linearly at the intermediate points, and to just touch the aperture (on the left). The rectangle touching the aperture is not drawn.
In the figure below the scale, I think that the rectangles and circles are the same as shown above the scale, with their areas calculated.
And I guess that "perfect" must be the shape shaded in green and black. That appears to be constructed as the intersection of the circle and the rectangle. I think that's not exactly correct[*], but certainly it captures the essence of the problem.
I look forward to seeing your diagram of some stray rays.
Rik
[*] The exact shape (I think!) would be the boundary of the convolution of two shapes. One shape would be the aperture, scaled from full size on the left to a single point on the right. The other shape would be the sensor, scaled from full size on the right to a single point on the left. When the aperture is a circle, the boundary of the convolution will not have any sharp corners at the intermediate distances. This contrasts with the simple intersection of a circle and a rectangle, which has 8 sharp corners.
You could closely approximate the perfect shape for any shape aperture by counting rays in a discrete simulation:
Code:  for each point Pa in the aperture
for each point Ps in the sensor
compute point Pb in the plane of the baffle, on the ray from Pa to Ps

The perfect shape is then all those points in the plane of the baffle where no rays pass.
For the special case of a circular aperture and a rectangular sensor, I think you can imagine an NC milling machine cutting away material to leave a rectangle. The inner boundary of the path will be a rectangle to match the sensor, with its size scaled by position of the baffle. The size of the cutting tool is a circle to match the aperture, again with its size scaled by the position of the baffle. Near the aperture, you're cutting a small rectangle with a large tool, so the outer boundary is almost a circle. Moving toward the sensor, the shape turns into a rectangle with quarterround corners whose radii get smaller and smaller until exactly at the sensor the radii drop to zero and you have exactly the shape and size of the sensor. 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Mon Dec 24, 2012 5:37 am Post subject: 


Warning : picture spam bellow
rjlittlefield wrote:  I think that the circles are sized to exactly contain the aperture (on the left), to just touch the sensor (on the right), and to scale linearly at the intermediate points. 
It's that but I didn't sized them, they are the section of the aperture cone at 3 positions on the lenssensor axis (1/4, 1/2 and 3/4). The section plane is perpendicular to the lenssensor axis.
rjlittlefield wrote:  Similarly, the rectangles are sized to exactly contain the sensor (on the right), to scale linearly at the intermediate points, and to just touch the aperture (on the left). The rectangle touching the aperture is not drawn. 
In fact the only rectangle of the left drawing is the sensor. What you see here are rectangles with rounded corners.
This time, what you are seeing is the section of the volume containing all the direct rays between the lens and the sensor. These pseudo rectangles are, too, the result of the intersection with the 3 section planes (at 1/4, 1/2, 3/4). There is no rectangle at the lens level because, at this position, the section of this volume is a circle.
I think that you thought that these rectangles was the representation of an eventual rectangular baffle. This is not the case (for clarity, i didn't drawed this volume).
I realize that the volume of good rays (purple/magenta volume of direct rays between the lens aperture and the sensor  I will say good rays now ), may be hard to apprehend so I will detail it's construction.
First I put the lens aperture, the sensor and the circumsribing circle to the sensor :
The green circle is just a construction object to create the following cone :
This cone is the envelope of 'good rays' hitting the sensor and lost rays which are focused outside the sensor. This cone, too, describes the minimal diameter of an iris inserted on the lenssensor axis. If the diameter of the iris is smaller than the diameter of the cone we will have vignetting.
Now with this cone I can create the volume of 'good rays'. I just need to take my better sharpened razor blade and cut the parts focusing outside of the sensor . For that, for each of the 4 sides of the sensor rectangle, i create a plane which is tangent to the lens aperture circle :
And I cut the external part. The result (after the 4 cut) is the volume that we needed :
Here are other views with sections added to help the visualisation of this volume :
(lens is near us, sensor further)
(sensor is near us, lens is further)
As you can see, as we walk away from the lens, the section is quickly like a rectangle (with rounded corners). It only become a real rectangle when we arrive at the sensor. Beware that, on the drawing, the corners may look like straight oblique lines, but it's not the case. They really are arcs.
This new volume allows us to know the perfect shape of a baffle so that :
 it let pass all 'good rays' to avoid vignetting
 it fits perfectly the volume of 'good rays' to avoid unnecessary opening for stray rays.
The projection of this volume on the section plans is what I've called in my previous post 'perfect' (surface).
rjlittlefield wrote:  The exact shape (I think!) would be the boundary of the convolution of two shapes. One shape would be the aperture, scaled from full size on the left to a single point on the right. The other shape would be the sensor, scaled from full size on the right to a single point on the left. When the aperture is a circle, the boundary of the convolution will not have any sharp corners at the intermediate distances. This contrasts with the simple intersection of a circle and a rectangle, which has 8 sharp corners.
You could closely approximate the perfect shape for any shape aperture by counting rays in a discrete simulation:
Code:
for each point Pa in the aperture
for each point Ps in the sensor
compute point Pb in the plane of the baffle, on the ray from Pa to Ps
The perfect shape is then all those points in the plane of the baffle where no rays pass. 
You're right!! But I Think the result of your computation will gives you the same volume In fact my volume is the union of all your small cones.
I Think I'm right because :
 for each point in my volume I can find a straight line between the lens aperture and the sensor
 I can't find any straight line between the lens aperture and the sensor outside of my volume.
Now the analysing : the right part of my previous post drawing.
In this part I copied the 3 sections that I got at 1/4, 1/2 and 3/4.
For section 3/4 I had this shape :
In Green/Black we find the perfect surface containing all the good rays that hit the sensor.
I want to compare circle baffle and rectangular baffle.
With a circle baffle :
The circle is the shape of a best fitting circle baffle (explained above), it contains 'good rays' (green/dark) and lost rays (in yellow).
For a rectangle baffle, we need to extend the border of the rounded corner perfect surface and we'll get our best fitting rectangle.
Here is the result :
Now we have all what we need to compare the efficience of circle with rectangle.
I just had to measure the surface of each colors.
For example at 3/4 :
 the surface of the perfect surface (green/black) is 212.7mm².
 the surface of lost rays with a circle baffle is the sum of the 4 yellow surfaces (78.4mm²). It's 37% more than the perfect surface.
 the surface of lost rays with a rectangular baffle is the sum of the 4 magenta/purple corners (10.8mm²). It's 5% more than the perfect surface.
The conclusion of these measures is that : between the sensor and 3/4 of the distance to the lens, the rectangle baffle is with no doubt better than the circle one (as there is far less lost rays surface).
What is really instersting is that, by determinig the volume of 'good rays' we are now able to describe the best fit for a circle baffle and a rectangular baffle. I did it experimentaly (with drawings) but now I can see that all that is simple geometry mathematics. And it's possible to calculate easily the following things :
 diameter of the best fit circle baffle in fonction of it's position on the axis
 width and height of best fit rectangular baffle in function of it's position on the axis
 and, really interesting, the position from which it's better to have a circle baffle or a rectangular baffle (comparison of surface).
So far, I have not considered the positionning to avoid stray rays. I just compare circle and rectangle. 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Mon Dec 24, 2012 6:59 am Post subject: 


here are the formulas to determine :
 the circle baffle diameter
 rectangle baffle width
 rectangle baffle height
My studies are too far to be able to resolve the equation to calculate the equilibrium between circle and rectangle
Anyway it's fast enough to find it by playing with the baffle distance param so that circle baffle surface and rectangle baffle surface are nearly equal (example: with the previous case, I find the equilibrium at 185mm (from the sensor).


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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 20177 Location: Richland, Washington State, USA

Posted: Mon Dec 24, 2012 1:58 pm Post subject: 


icmbtm, thank you for the clarifications.
I agree with the spirit of your construction. But I think a few of the details are not quite right yet.
ichbtm wrote:  Now with this cone I can create the volume of 'good rays'. I just need to take my better sharpened razor blade and cut the parts focusing outside of the sensor . For that, for each of the 4 sides of the sensor rectangle, i create a plane which is tangent to the lens aperture circle :
<diagram>
And I cut the external part. The result (after the 4 cut) is the volume that we needed :
<diagram>

It is not enough to make 4 planar cuts. You must make an infinite number of cuts, running the plane all around the circle of the aperture. For all except four of those cuts, the plane will be pivoting on a corner of the sensor.
Quote:  As you can see, as we walk away from the lens, the section is quickly like a rectangle (with rounded corners). It only become a real rectangle when we arrive at the sensor. Beware that, on the drawing, the corners may look like straight oblique lines, but it's not the case. They really are arcs. 
Close, but the corners should be quarter circles. They are not small arcs of a big circle as you have them drawn.
I get it like this:
Quote:  But I Think the result of your computation will gives you the same volume In fact my volume is the union of all your small cones.
I Think I'm right because :
 for each point in my volume I can find a straight line between the lens aperture and the sensor
 I can't find any straight line between the lens aperture and the sensor outside of my volume. 
It is close. But you can see in the corners of the diagrams that there are some small areas between your green shape and mine. This is due to the difference of the construction. You have taken a circular cone and sliced off four sections. If you had sliced off an infinity of sections, you would have gotten the same as mine. The small differences are places where a few rays can get through the aperture and just barely miss the sensor.
Quote:  here are the formulas to determine :
 the circle baffle diameter
 rectangle baffle width
 rectangle baffle height
. 
I agree with your calculations for these numbers. I have extended your spreadsheet to calculate also the area of the perfect shape with quartercircles on the corners. You can download that HERE.
Quote:  Anyway it's fast enough to find it by playing with the baffle distance param so that circle baffle surface and rectangle baffle surface are nearly equal (example: with the previous case, I find the equilibrium at 185mm (from the sensor). 
Yes. It looks like you did this using trial and error by hand. A more accurate result can be obtained automatically using Excel's builtin "Goal Seek" function. I get 185.2502889, with a calculated difference of zero between the circular and rectangular baffles.
Quote:  My studies are too far to be able to resolve the equation to calculate the equilibrium between circle and rectangle 
It is very difficult to do this symbolically. I tried running the equations through the symbolic solver HERE, but it just threw up its hands and said "Can not solve".
Rik 

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canonian
Joined: 31 Aug 2010 Posts: 890 Location: Rotterdam, Netherlands


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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Tue Dec 25, 2012 10:55 am Post subject: 


rjlittlefield wrote:  It is close. But you can see in the corners of the diagrams that there are some small areas between your green shape and mine. This is due to the difference of the construction. You have taken a circular cone and sliced off four sections. If you had sliced off an infinity of sections, you would have gotten the same as mine. The small differences are places where a few rays can get through the aperture and just barely miss the sensor. 
[EDITED]OK I understood my mistake!!!
I had points which was not on a line between the lens and the sensor
A good and easy method of construction would be to do a rectangular tunel with the sensor at one side and a square containing the lens aperture at the other side. Then, to create 4 cones with the lens aperture at one side and the corners of the sensor at the other. Then to cut the exceding corners of the rectangular tunel.
Well done Rik![/EDITED]
rjlittlefield wrote:  I agree with your calculations for these numbers. I have extended your spreadsheet to calculate also the area of the perfect shape with quartercircles on the corners. You can download that HERE. 
Thank you for the complement.
rjlittlefield wrote:  A more accurate result can be obtained automatically using Excel's builtin "Goal Seek" function. 
Thank you again. I didn't knew this excel function and it can be very interesting.
rjlittlefield wrote:  I tried running the equations through the symbolic solver HERE, but it just threw up its hands and said "Can not solve". 
It is reassuring to see that the machine does not beat us over (I've put it in my favs).
canonian wrote:  ichbtm, maybe a coincidence, or by design, maybe to fold them easier, but the diagrams you draw show similarities with how vintage bellows were made, starting small on the side of the lens, a series of bigger rectangular baffles ( if you consider the inner crease of the bellows as a baffle) towards the film/sensor. 
I'm not a specialist but maybe it's because the photographic plate is so large that it didn't made sense to have a constant width bellow...
I can see some advantages too :
 it certainly gains space when folded.
 and, maybe (to be confirmed), this shape is better to avoid stray rays... 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Wed Dec 26, 2012 8:11 am Post subject: 


Here is it
The cones :
And the complete volume :
Corners are nice quatercircles 

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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 20177 Location: Richland, Washington State, USA

Posted: Wed Dec 26, 2012 1:23 pm Post subject: 


The new version looks perfect. I like the new diagrams because they make clear to me what's happening. You've shown the four cones at the limiting corners, and I think it's then clear that the entire area between them must be part of the baffle shape also.
Quote:  I didn't knew this excel function [Goal Seek] and it can be very interesting. 
It's a very powerful feature, not widely known. There is an even more powerful tool called Solver that hardly anybody knows about. Solver is a multivariate optimization package. It's one of the "Addins" that comes bundled with every copy of Excel but does not appear in the menu system until you hunt it down and install it. See lessons #4, #5, and #7 HERE.
Rik 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Wed Dec 26, 2012 2:13 pm Post subject: 


Here is the stray rays simulation
I used a bellow instead of an extension tube as it is easier to simulate the reflections. (In fact I can not even imagine how to simulate the reflexions on an extension tube )
The width of the bellow is 44mm.
The drawing is certainly not very clear but if needed I could detail it.
Under are the sections at 1/4, 1/2 and 3/4.
In yellow you have the ideal circle baffle.
In green the ideal rectangular baffle.
Red areas are areas bringing stray rays (the levels of red show superpositions of different 'source' of stray rays.
The best plot is, like said it Rik at the start ( ) in the middle of the bellow.
I look a bit before and a bit after the middle and the clean zone is degrading in the 2 cases. (I didn't add these sections).
I would notice that it could be possible to use a front baffle and a rear baffle (the rear baffle would block the stray rays that the front one didn't blocked.
For exemple, with a rectangle baffle, the 1/4 and the 3/4 above are complementary and any stray rays is allowed to hit the sensor. (you have to think that at 1/4 the stray ray volume is almost inverted by the reflection).
Sadly, I think the mathematics behind the calculation of the ideal complementary positions is of a very high level 

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ichbtm
Joined: 26 Nov 2012 Posts: 28

Posted: Thu Dec 27, 2012 6:58 am Post subject: 


And here is the spreadsheet calculating the % of stray rays blocked
Not sure yhat there's no errors but it seems to be cohérents with the sections...
Nota : For simplification reasons, I didn't took in account the rounded corners of the stray rays beams. And I the baffle is always the perfect fitting rectangular one.
Bellow are the calculations for the 3 sections at 1/4, 1/2 and 3/4 :
The result of the calculation is on line 16
A link to a google spreadsheat 

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