Do I understand viewing angles for gain calculations? - AVS Forum
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post #1 of 14 Old 01-24-2002, 09:48 PM - Thread Starter
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I am doing some calculations for projector position, viewing area, and screen image uniformity characteristics. I just want to confirm I have the definitions right.

I have seen gain curves of gain vs. viewing angle. As I understand it the viewing angle is the angle in between the line that represents the one 'mirror reflection' path back to the projector and the line to any particular point on the screen. In the limits a mirror would have a delta function gain curve and an ideally dispersive surface would be a flat line gain curve.

So do I have this correct?
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post #2 of 14 Old 01-25-2002, 08:30 AM
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Daryle,

The viewing angle is the angle between the line of main reflection (which may be standard i=r or retro-reflective, i.e. back to the source) and the line from reflection point to viewer.

You are quite right in that a perfect mirror will have a Dirac delta function gain curve (infinity for alpha=0 and 0 otherwise, but still integrates to 1!) and that an ideally diffusive surface would have a constant gain.
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post #3 of 14 Old 01-25-2002, 05:10 PM - Thread Starter
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I believe your definition is slightly ambiguous. It could comfirm mine or suggest the converse (reverse?). To rephrase and improve the semantics of late last night:

Viewing Angle A (My original understanding and one interpretation of yours): The angle between the specular reflection vector of a projector ray to the viewing point and the vector from the viewed point to the viewing point.

Viewing Angle B (Alternate interpretation of yours): The angle between the specular reflection vector of a projector ray to the viewed point and the vector from the viewed point to the viewing point.

Gain Curve: The curve that represents the relative gain with respect to viewing angle. It is a Dirac Delta function for a surface with ideal specular reflection characteristics (ie. mirror) and a constant for a surface with ideal diffuse reflection characteristics.

Attached is a picture illustrating the difference between the two viewing angle definitions (2d side view for simplicity). Now I think neither is 'incorrect' but one is the standard that the graphs use. I assume it is the first since I would thing the curves are measured with a directional sensor at the viewing point. Can you (or anyone else) confirm that definition A is indeed the correct one?
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post #4 of 14 Old 01-26-2002, 03:56 AM
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Nice drawing. Viewing angle is often used to describe different angles depending on the situation. I agree this may produce some problems. In this case, relating to screen gain and the resulting brightness uniformity, it's definition B.
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post #5 of 14 Old 01-27-2002, 09:06 AM - Thread Starter
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I ran some numbers for the initial setup I am considering (8'x6' screen, ceiling mounted projector 17' back, fairly central/optimal viewing position about 14' back). With definition 'B' this leads to viewing angle extremes of ~41 degrees; with definition 'A' it is ~23 degrees.

Now looking at the gain curves for a, say fairly flat dispersion/lowgain, Greyhawk; definition 'B' suggests nearly a factor of 2 gain variation over the screen even for the reasonably ideal setup. Intuitivly this seems large (ie. wrong). Can someone else, perhaps even Mr. Stewart, confirm how the gain measurements are made and what the definition is? Of course, perhaps a factor of 2 gain variation is better than it sounds.
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post #6 of 14 Old 01-27-2002, 10:06 AM
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If you have an interest in screen simulations and ray-tracing, perhaps you should take a look at this thread:

http://www.avsforum.com/avs-vb/showt...hreadid=104630
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post #7 of 14 Old 01-27-2002, 03:11 PM - Thread Starter
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Nice pictures. Not quite sure why you used raytracing since I believe there are analytical vector solutions to all those cases. Anyhow here are my lower res analytical plots for the two definitions. I could send you the Excel sheet if you are interested.
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post #8 of 14 Old 01-27-2002, 03:13 PM - Thread Starter
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And here is the second one. BTW how the hell do you post multiple images inline?
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post #9 of 14 Old 01-27-2002, 11:58 PM
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There may well be analytical solutions, at least for some of the cases presented. A numerical formulation will give you more flexibility and Matlab is really very much specialized for numerical simulations. If properly implemented, calculation times are short, so I would say the pros outweigh the cons.

BTW, the 41 degree angle you refer to above seems very high.
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post #10 of 14 Old 01-28-2002, 12:07 AM
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One other thing, did you compensate for gamma in your pics? As the calculated results are screen values, the gamma has to be taken off the calculated results before producing a picture.

You can use the IMG button to include pics in your posts.
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post #11 of 14 Old 01-28-2002, 09:29 PM - Thread Starter
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Here are another couple of quick sketches of the projections showing how you get ~41 degrees viewing angle in a lower corner of the screen using definition B.

In terms of my intensity maps I applied no explicit gamma correction. What I did was get the gain map from the viewing angle vs. gain curve, normalize it to the peak gain, and then apply approximate gray scale mapping from 1=255 on down.

I guess I am assuming things like half the gain results in half the perceived brightness and that 255/2 is half the perceived brightness of 255.

(PS I see that you want inline images you have to host them off site so I will just attach them)
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post #12 of 14 Old 01-28-2002, 09:30 PM - Thread Starter
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And the second one:
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post #13 of 14 Old 01-29-2002, 02:17 AM
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The maximum angle in your pics is 30 degrees.

As the screen calculated brightness values should be displayed as is, you have to get rid of the gamma that is added by your display. Suitable gamma values may be around 2.2-2.4 and as you have to invert the process, you should correct your calculated greyscale GS as GScorrected = GS^(1/gamma) for a reasonably consistent presentation of the greyscale.
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post #14 of 14 Old 01-29-2002, 12:23 PM - Thread Starter
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Those are 2D projections (top and side views) of the 3D situation for illustration. The projected angles are as noted but perhaps I should have left them out since they confuse the issue somewhat. The proper 3D angle is calculated as follows (using the viewing point of the lower left corner of the screen as the origin):

Vector to viewing position: (14,4,2.5)
Vector of specular reflection: (17,-4,-6)

Plug into angle between vector formula:

cos(theta)=(v1.v2)/(||v1||*||v2||)

Gives you theta=40.64 degrees

I will try your suggested gamma correction and see how it changes the looks of things.
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