[loggeo]

I have been looking for some time now to get an ambient rejecting light screen. Of course most of them, if not all, have limitations regarding throw ratio. I can see for example that there is a minimum of 1.5 throw ratio for dnp and SI screens.
My system currently consists of a jvc dla rs57, a Isco IIIL and a 2.35 screen (9.8 feet wide). Throw distance is about 14 feet.
When projecting without using A-lens, I get a throw ratio of about 1.92. After moving the A-lens in front of the projector I get a ratio of about 1.43.
This maybe stupid but which result should I take under consideration to avoid hot spotting etc?
Projector is natively throwing a picture of 7.3 feet wide. Should I consider projector+Alens as one system throwing at 9.8 feet wide?
Unfortunately moving the projector is not an option.
Thank you in advance.

 

[loggeo]

Anyone?

 

[stanger89]

It would be best to talk to the screen manufacturer, but in this case, I suspect you'd want to use the lens in place (full screen width) throw ratio, since the recommendation is related to the angle of light hitting the screen and the gain falloff.

 

[GetGray]

There is only one "TR". TR = (distance to screen/screen height*0.5625)

When screen companies give advise including TR, that's the number they are talking about. The 16:9 geometry does not change if you scretch the beam to fill more screen. the 16:9 portion, including the angles relative to the PJ remain unchanged. If it hotspots at the TR, becasue of the TR, then spreading it out won't change anything except you may not notice it as much due to decreased lumens per unit area.

Hope that helps clear it up.

 

[stanger89]

Interesting, so hotspotting/falloff doesn't get worse with the wider screen? I though gain was generally a function of angle (between you, the screen, and the light), and with a wider screen you have a greater angle, akin to a 16:9 screen with the larger width.

 

[GetGray]

Yes, it may. But if you ar eusing a mfgr's specs, typically a minimum TR recommended, then you need to use the TR based on 16:9. If oyu meet that minimum, then the expansion won't make it worse.

 

[loggeo]

There is only one "TR". TR = (distance to screen/screen height*0.5625)

When screen companies give advise including TR, that's the number they are talking about. The 16:9 geometry does not change if you scretch the beam to fill more screen. the 16:9 portion, including the angles relative to the PJ remain unchanged. If it hotspots at the TR, becasue of the TR, then spreading it out won't change anything except you may not notice it as much due to decreased lumens per unit area.

Hope that helps clear it up.

Yes this helps a lot. Thank you! Some manufacturers give a minimum throw ratio and others give a minimum throw distance though. I was wrong about SI. They mention a minimum throw distance of 1.5 X Image Width. How does the A-lens affect all that? My projector throws an image 1.9 X image width, before adding the A-lens to the equation. Then the number falls to 1.44! Which number should I consider being closer to reality? This is very important to me since I would like to buy the biggest possible screen.

 

[GetGray]

Yes this helps a lot. Thank you! Some manufacturers give a minimum throw ratio and others give a minimum throw distance though. I was wrong about SI. They mention a minimum throw distance of 1.5 X Image Width. How does the A-lens affect all that? My projector throws an image 1.9 X image width, before adding the A-lens to the equation. Then the number falls to 1.44! Which number should I consider being closer to reality? This is very important to me since I would like to buy the biggest possible screen.

You are making this too hard

Your PJ thows an image that is 1.0 the image width. It throws the image width, period.

Did you use the formula I gave above to arrive at a TR of 1.9?

If you did then your 1.9 TR is over the SI spec'ed minimum of 1.5. Farther (= higher TR number) the better, but you are beyond their minimum.

The bigger screen you select, the lower the TR. Using some simple algebra and my formula:

TR= Dist/height * .5625
For a 1.5 TR minimum (max screen size), at 14' (168")

Height = Dist * .5625 /TR => 168 * .5625/1.5 = 63"

Height (max for 1.5 TR) = 63". @ 2.35 that's 63 x 148.

DO NOT just buy the biggest screen you can. The projector needs to be considered or you won't have enough to light it up.

 

[stanger89]

You are making this too hard

Your PJ thows an image that is 1.0 the image width. It throws the image width, period.

Did you use the formula I gave above to arrive at a TR of 1.9?

If you did then your 1.9 TR is over the SI spec'ed minimum of 1.5. Farther (= higher TR number) the better, but you are beyond their minimum.

The bigger screen you select, the lower the TR. Using sme simple algebra and m yformula:

TR= Dist/height * .5625

Let me ask you this, what's the difference, as far as the screen goes (hotspotting, brightness uniformity) between a 120" wide scope screen and a 120" wide 16:9 screen? I would think across the areas that are common to the two, there would be no difference, yet your throw calculations come out very different.

Just trying to understand, I get how your throw formula works for throw vs anamorphic lens spec, since that's always relative to the size without the lens in place, but as far as screen performance (hotspotting, etc) it seems like whether a lens is in place or not should be irrelevant.

 

[loggeo]

I understand that and thank you. Probably my English are not that good (I am from Greece). I was saying that SI doesn't suggest a minimum TR but a minimum throw distance (1.5 image width they say) which is something different. When I'm saying 1.9 I am referring to throw distance (1.9 image width). I hope this makes sense.

 

[GetGray]

Let me ask you this, what's the difference, as far as the screen goes (hotspotting, brightness uniformity) between a 120" wide scope screen and a 120" wide 16:9 screen? I would think across the areas that are common to the two, there would be no difference, yet your throw calculations come out very different.

Just trying to understand, I get how your throw formula works for throw vs anamorphic lens spec, since that's always relative to the size without the lens in place, but as far as screen performance (hotspotting, etc) it seems like whether a lens is in place or not should be irrelevant.

This is a constant width example. What changes is the geometry of the PJ at a set throw distance. In your example, the "middle" of the respective 2 16x9 screens is in a different place. The middle is typically where the hot spot occurs. In your example, middle of the larger screen is lower, thus the reflective angles have changed. Don't misunderstand me, I agree you can't completely discount it. I'm just explaining how to use the Mfgr number if that's given. In the case of SI, apparently they spec distance, not TR, which is reasonable. And aspect neutral.