Retroreflective Screen = Curved Screen - Page 2 - AVS Forum
Forum Jump: 
 
Thread Tools
Old 01-27-2002, 04:26 AM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
snaggs,

You will get a very good brightness uniformity and low color shift with a ceiling mounted PJ and retro-reflective screen. What you will not get is high brightness since the main part of the light is reflected back to the PJ. Tilting the screen will not make any difference for a retro-reflective screen as the light will still be reflected back to the source.


MrWigggles,

1. I have not checked your calculated angles, but they seem correct - I am sure there is no conflict here. Illuminance is inversely proportional to distance squared, as I am sure you already know. The 2.5X and 1.5X PJ positions have differing left edge to right edge distance ratios. Accounting for this effect, the results shift quite a lot.

2. Just looking at the differences in angles is not enough as the absolute angles also have an influence depending on the behavior of the gain curve.

3. I believe the gain curve I am using was in fact measured for the Hi-power, but as I have lost track of its origins, I am not completely certain. Any gain curve reasonably close will do, however.

Summing up, I am quite confident that my calculated results are correct and the differences depend on my algorithm including several additional steps.
Iceman is offline  
Sponsored Links
Advertisement
 
Old 01-27-2002, 06:46 AM
KBK
AVS Special Member
 
KBK's Avatar
 
Join Date: Apr 2000
Location: The Wilds Of Canada
Posts: 7,112
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 11
The simulations are fine for getting close to the realities presented by the 'rated' materials we work with. The real problem as we see it,(as we see it! hah! as we see it!...never mind...) is that the screens in use are actually combinational characteristics of differing considerations. They stray too far from the simplistic theoretical ideal -in real terms. The differences can be minor, but are real, and are there.

A very handy simulation tool nonetheless.

Ken Hotte

"Never forget that only dead fish swim with the stream." -- Malcolm Muggeridge.
KBK is offline  
Old 01-27-2002, 07:30 AM
Advanced Member
 
snaggs's Avatar
 
Join Date: Feb 2001
Location: Perth, Australia
Posts: 594
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
Ok, well I've got my best friend roped in (woodworking expert), and we're going to be building a super DELUXE version of Don Stewarts TORUS. Now I just need more info on high-gain screeen materials in this context....... and that information is thin on the ground.

I particullarly interested in the silver high gain surfaces in this application. But due to their unsuitability in normal flat screens, there has been no discussion on information on them. How does the Sony VPS-100HG1, Vutec, and Draper Da-Curve achieve gains of 12-14 ?
snaggs is offline  
Old 01-27-2002, 11:21 AM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
KBK,

There is definitely more to a screen surface than its gain properties. Still, as long as we are only dealing with easily defined properties such as brightness, brightness uniformity and color shift, simulations will be quite accurate.
Iceman is offline  
Old 01-27-2002, 09:22 PM - Thread Starter
AVS Special Member
 
MrWigggles's Avatar
 
Join Date: Aug 2000
Location: Houston, TX, USA
Posts: 6,240
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
Let me make it even simpler...

Asumming the projector has the capability to be zoomed from 1.5X to 2.5X, to the viewer located .5X screenwidths to the right standing at 2.0X away:

1. The LEFT side of the image is BRIGHTER with 1.5X than 2.5X.

2. And, the RIGHT side of the image is DARKER with 1.5X than 2.5X.

Therefore given 1 and 2, the 2.5X location has better left to right uniformity than the 1.5X

Q.E.D.

I am not saying how much better because I don't know that's not important yet. Either 1 or 2 has to be wrong for my conclusion not to be correct. The only way that I can see that being possible is if the screen isn't behaving in a true retroreflective manner (i.e. peak intensity is NOT observed at an angle where the light is going directly back to its source.)

-Mr. Wigggles

The Mothership is now boarding.
MrWigggles is offline  
Old 01-28-2002, 12:44 AM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
MrWigggles,

I feel I am sounding like the proverbial parrot here. For the third and final time: You are only considering the angles and are getting incorrect results from oversimplifying the problem. You HAVE TO include the distances as well to calculate the illuminance at the viewing position. If you do this, you will see that for 1.5X, the longer distance to the left edge cancels some of the brightness increase from the smaller angle. The opposite is true for 2.5X. This makes the 1.5X have better brightness uniformity.

If you do not believe me, please do the FULL math yourself and calculate the illuminance from each screen edge at the viewing position. This would include the appropriate edge angles, distances and gain curve. One can quite easily do this by hand in 2D even if 3D will of course be even better.
Iceman is offline  
Old 01-28-2002, 02:43 AM - Thread Starter
AVS Special Member
 
MrWigggles's Avatar
 
Join Date: Aug 2000
Location: Houston, TX, USA
Posts: 6,240
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
Quote:
Originally posted by Iceman
MrWigggles,

You HAVE TO include the distances as well to calculate the illuminance at the viewing position. If you do this, you will see that for 1.5X, the longer distance to the left edge cancels some of the brightness increase from the smaller angle. The opposite is true for 2.5X. This makes the 1.5X have better brightness uniformity.

If you do not believe me, please do the FULL math yourself and calculate the illuminance from each screen edge at the viewing position. This would include the appropriate edge angles, distances and gain curve. One can quite easily do this by hand in 2D even if 3D will of course be even better.
The illumination intensity of a plain does NOT change with distance. If you used a spot meter to measure your screen intensity, you would get the same reading from 10 feet as 100 feet etc.

Your simulation is obviously treating the 20,000 points as individual radiating points each with intensity that decays at a 1/D^2 relationship. This is fine but...

What your analysis doesn't take into effect is that the left side of the screen is SMALLER when the viewer is on the right side. Therefore the point charges are more densely concentrated.

Your BR answer needs to be corrected by an aproximate factor of (sqrt(5)/2)^2 or 1.25 which would make the new BR for 1.5X go from 1.33 to 1.67 and the BR from the 2.5X location go from 1.53 to 1.22. At least this is the best I can do from your current calculations which include your own 2.8 screen mathmatical model.

This is the same reason why your matte screen had a BR of 1.15 which should by definition have a perfect 1.00 BR (at least with a one gun digital projector, A 3 gun CRT can change things concievably in pratice). Your 1.15 comes from the distance difference of the center of the screen versus the corners squared or (sqrt(3^+1^2+.5625^2)/3)^2 = 1.146. That factor shouldn't be there when everything is all said and done. In fact all of your BR's from your previous reflective flat screen analysis should be divided out by that amount.

Good luck you almost have it.

-Mr. Wigggles

Ps. Q.E.D.

The Mothership is now boarding.
MrWigggles is offline  
Old 01-28-2002, 05:27 AM
AVS Special Member
 
wireburn's Avatar
 
Join Date: Feb 2000
Location: Muskegon, Michigan
Posts: 1,355
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
Quote:
Originally posted by Iceman

wireburn,

??? This would make the screen reflect quite a lot more than 100% of the incident light. If I am not misinformed, the Hi-Power goes below unity gain somewhere around 30 degrees off-axis.
Of course, that is correct. I should have said that the gain is consistent outside the optimum viewing cone and behaves similarly to matte white. Here is the link to the article on the Dalite website: http://www.dalite.com/educational_ma...ils&issueid=50

-Mike

I didn't do it
Nobody saw me do it,
You can't prove a thing

An Haiku by Bart Simpson
wireburn is offline  
Old 01-28-2002, 05:38 AM
AVS Special Member
 
wireburn's Avatar
 
Join Date: Feb 2000
Location: Muskegon, Michigan
Posts: 1,355
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
And here is some more interesting reading about the Hi-Power:

http://www.dalite.com/educational_ma...ils&issueid=29

-Mike

I didn't do it
Nobody saw me do it,
You can't prove a thing

An Haiku by Bart Simpson
wireburn is offline  
Old 01-28-2002, 07:27 AM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
MrWigggles,

Ahh, the good ol’ point source / area source debate once again. This discussion somehow always pops up in many differing fields of physics. I will try my best to straighten out the concepts.



Quote:
The illumination intensity of a plain does NOT change with distance.
This is not really relevant here but, granted - true – if the area in question (plane) can be considered infinitely large. This is approximately the case if the distance is much smaller than the area dimensions. If this is the case you are in the so called near-field and total contribution (integrated over the area) intensity will be constant with varying distance. But this is not true in this case as the distance to the screen is even larger than the screen dimensions (and, much more importantly, a lot larger than the screen segments we are considering). Thus, we are in the near to far-field transition region and the statement above is not applicable even when considering total contribution illumination. Note that my ray-tracing algorithm will also produce approximately constant total screen contribution illumination for different distances close to the source or, in other words, as long as the source dimensions are large in comparison to the distance. This does not mean that the contribution from each screen segment is equal, however.



Quote:
If you used a spot meter to measure your screen intensity, you would get the same reading from 10 feet as 100 feet etc.
Only if you had one large muther of a screen –1000x1000 ft :D. This may well hold true for the planetarium where you work, but will very definitely not be true for any regular size screen, and the following explains why:

When we are far from a matte white screen, say 10 times screen width, it can be approximated as being a point source radiating into a half-sphere. So let’s consider radiation from a point source. Intensity is defined as units/m2 or units/ft2 and thus falls by distance squared. Total radiated power will be the same for all distances, what we are doing is expanding the area of our “control sphereâ€. In fact, if the intensity did not fall off this way we would increase the power output of the source just by increasing the distance. I hope there is no controversy here.

It is perfectly permissible that, without violating any physical principles, divide a large area into small segments (approximating point-sources) and consider the contribution from each source. This type of modeling is used extensively in many fields. Still no controversy, I am sure.

What my ray-traced results relate to is this situation:
1. Perfectly evenly illuminated screen. For example, if the projector has a light output of 1000 lumen, each screen segment is considered a source with a light output of 1000/20000 = 0.05 lumen. All segments are naturally equi-distant.
2. The illuminance [lux], [lumen/m2] or [foot-lamberts] from each screen segment is calculated assuming point source radiation, while also accounting for the gain characteristics.
3. The viewer is considered to be looking directly at all screen segments simultaneously (or in practice, scanning the screen area), otherwise there would be even more fall-off for off-center segments (cosine function).



Quote:
What your analysis doesn't take into effect is that the left side of the screen is SMALLER when the viewer is on the right side. Therefore the point charges are more densely concentrated.
The point charges or point sources have nothing to do with viewing position as they are only PJ-related, see 1. above.



Now let’s build an example to illustrate my point:
Construct an imaginary array with the same dimensions as the screen we have been discussing in our previous posts. Cover it with 20000 50 W bulbs. Place yourself with a spot-meter 1.5X array width away, level with the right array edge. Turn off all bulbs except for the column on the very far left and measure the illumination. Now do the same but with the right edge bulbs active. The measured illuminances WILL differ.



Quote:
This is the same reason why your matte screen had a BR of 1.15 which should by definition have a perfect 1.00 BR (at least with a one gun digital projector, A 3 gun CRT can change things concievably in pratice).
You are comparing apples to oranges here. Brightness Ratio (BR) relates to the combined effects of distance and gain. It is perfectly possible for a perfect matte white material to have a constant gain of one over all angles while a particular setup with that material produces a BR > 1. See my bulb example above, the imaginary array is equivalent to a perfect flat matte white screen.



Quote:
Ps. Q.E.D.
:D, Thanks for an interesting discussion.
Iceman is offline  
Old 01-28-2002, 07:49 AM
KBK
AVS Special Member
 
KBK's Avatar
 
Join Date: Apr 2000
Location: The Wilds Of Canada
Posts: 7,112
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 11
The acoustical analogy is fully worked out. in the form of planar speakers vs. point source. I am well familiar with these straightforward situations. it is interesting how the numbers are immediately transferable. It immediately becomes obvious why this is so.. it is the same math.

Ken Hotte

"Never forget that only dead fish swim with the stream." -- Malcolm Muggeridge.
KBK is offline  
Old 01-28-2002, 08:40 AM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
KBK,

Very true indeed. In fact, acoustical modeling is a bit trickier as you also often have to keep track of the phase angles. It is principally the same, though. The acoustical near-field may be very complex with reactive sound-fields not contributing to the radiated power. To further complicate matters, the radiation impedance of the air itself will in some cases make analysis even messier.
Iceman is offline  
Old 01-28-2002, 01:02 PM - Thread Starter
AVS Special Member
 
MrWigggles's Avatar
 
Join Date: Aug 2000
Location: Houston, TX, USA
Posts: 6,240
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
ICEMAN,

You are wrong on this one.

Intensity is X per cm^2. In your matte white screen example a square cm is in the corner of the screen is not a square cm to the eye at the viewing location. It has an area of .8726. It will have .8726 less brightness than the square cm in the center of the screen. But its intensity = .8726X/.8726 = 1X. Same as the center square.

Quote:
Now let’s build an example to illustrate my point:
Construct an imaginary array with the same dimensions as the screen we have been discussing in our previous posts. Cover it with 20000 50 W bulbs. Place yourself with a spot-meter 1.5X array width away, level with the right array edge. Turn off all bulbs except for the column on the very far left and measure the illumination. Now do the same but with the right edge bulbs active. The measured illuminances WILL differ.
Huh? with a spot meter or light meter? To a light meter turning off the bulbs closer will lower the reading, but to a spot meter (or intensity meter) the difference will be the same - the bulbs closer will be brighter but fewer of them will be in the circle of the spot meter. The bulbs further away are dimmer but will have more in the circle. To make it clearer, here are some examples for you:

1. Take a cluster of four bulbs 2m away and a single bulb 1m away. Which has greater intensity from the viewing location the cluster of four or the one? Both will have the same intensity. The combination of the four bulbs will not appear dimmer.

2. Here's one that you can do at home easilly. Look at a 4 ft or so flourescent tube from a distance. The light should look very uniform from one side to the other. Now put your head very close to one of the ends. Did the end further away change in intensity?
No.

3. The same can be done with a good flat computer monitor. Look at the monitor from a distance is perfectly uniform from corner to corner. Now get close to one side and look at the other side. Did the intensity change of the other side, no.

4. Ever measure a TV's ft-L with a spot meter? It will measure the same from any distance as long as the spot meters circle stays within the viewing area.

If you don't like those examples, do me a favor, build me a matte white screen that has a perfect uniformity or BR. Your current matte white screen model will have a greater unifomity (or smaller BR) the further you are viewing it. That is wrong, period. Tell anyone that a matte white screen has different uniformity from different viewing angles/positions and they will laugh at you.

The world isn't flat it is round. A flat 16:9 rectangle doesn't appear as 16:9 rectangle to the eye. The left and right ends are smaller than the center. Our mind might make it rectangular but that's not how it enters the eye.

Maybe something is lost in the terms brightness and intensity that we are using here. I have from the very beginning assumed that BR = 1/uniformity. Maybe your trying for BR to be something else completely. I would hate for all of your work to go to waste do to a conceptual error.

consistant intensity = perfect uniformity.

-Mr. Wigggles

The Mothership is now boarding.
MrWigggles is offline  
Old 01-28-2002, 04:16 PM
Senior Member
 
Iceman's Avatar
 
Join Date: Oct 2000
Location: Stockholm, Sweden
Posts: 439
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
MrWigggles,

You are right. My line of reasoning is entirely correct - except for one thing: lenses. The eye does not react to true intensity because of the lens. The lens will make equal light output objects with differing distances appear as having differing sizes - not differing brightnesses. An object 10 times farther away will be perceived as having a 100 times smaller area, not 100 times lower brightness. A spot-meter is intended to emulate the human eye and will of course present values that are consistent with the optical characteristics of the eye.
Iceman is offline  
Old 01-28-2002, 04:32 PM - Thread Starter
AVS Special Member
 
MrWigggles's Avatar
 
Join Date: Aug 2000
Location: Houston, TX, USA
Posts: 6,240
Mentioned: 0 Post(s)
Tagged: 0 Thread(s)
Quoted: 0 Post(s)
Liked: 10
Did I mention how much I liked your use of system gamma of 2.4 in your example images? (That is a ball park guess from the pixel value of your images) That is a much better choice than say 2.2 which is not representative of the CRT monitors that people use as their displays.

Anyway, happy viewing! ... that's what it all about.

-Mr. Wigggles

The Mothership is now boarding.
MrWigggles is offline  
 
Thread Tools


Forum Jump: 

Posting Rules  
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off