Replicating a video projector
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Hello!
I feel like I asked this question in the past and got an answer on the old cineversity, but I couldn't find it...I'm trying to create a setup that allows me to replicate a video projector's throw ration and lens shift to calculate a projector placement in a space that I have a 3d model for.
This is the exact projector I'm trying to replicate.
https://www.optomausa.com/product-details/zu720tst#specificationsI can't remember what the procedure was
Thanks for any reminders! -
Hi AlexC,
Yes, we had some exchange on that, but I also know each render engine might require different handling. Is it for Standard or Redshift?
All the best
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Redshift is what I am using mostly now (except for very quick sketches).
thanks Sassi!
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Hi AlexC,
For RS, there is currently a little detour needed. The shift Y moves in the opposite direction. At the same time, the Camera Map node is 100% useless here.
I have attached the workaround here. It simply uses a second camera as translation to get the rendering matching the initial camera for the Shift.
A trapezoid would be the result if you pointed the projector from any point other than perpendicular to the projection surface. This is compensated with these projectors typically.
With Redshift, we do not have that option. If you leave the projector perpendicular to the screen, even way off the axis of the screen, the two offset parameters will give you the option to get a proper rectangular (or square) image on the projection surface.
Use the camera on top, not the one "fixed," to set the parameters.
Sorry for the extra complexity here.
https://stcineversityprod02.blob.core.windows.net/$web/Cineversity_Forum_Support/2023_PROJECTS_DRS/20230309_CV4_2023_drs_23_RScm_01.zip
Render to see correct results, the editor preview is wrong, and the problem is reported.
All the best
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Thanks Sassi!
So, out of curiosity, how did you derive the camera parameters to match the 0.75 throw ratio of the projector?
That's the thing that always confuses me....Thanks!
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Hi AlexC,
I had a square image set up. But the ratio can be set up in the Material Tag> Tag> Film Aspect.
If you have any trouble, please share a low res image that you like to use with the aspect ratio. It can be just white.
Cheers
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Hello Sassi
Maybe it is easier with a project file!
So in this project you will see an RS camera (called "Projector") that needs to cover a wall which has a slight obstruction in front of it.
I know the projetor is .75 throw ratio, and the wall is 6.7 meters wide, so according to my math, the projector needs to be at about 5 meters away from the wall to cover the whole surface
However there is that "obstruction" wall, and so I need to look at where to place the projector in the X axis to make sure I can clear this obstruction and cover the whole wall with my projected content.
https://www.dropbox.com/s/8vd1rwvau1fwuym/Lobby_3_Simplified_MaxSimple_2.c4d?dl=0
Does that make sense?Thanks!!
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Hi AlexC,
Please have a look at the files below:
https://stcineversityprod02.blob.core.windows.net/$web/Cineversity_Forum_Support/2023_PROJECTS_DRS/20230310_CV4_2023_drs_23_TXcm_01.zipThis needs a render preview, and if any changes happen, refresh.
File 01 explores your camera height, assuming model 1.25 is used (there are three models.) The manual tells me to be around 8m away to cover that wall.
The manual is pretty quiet about what the "slide-width" is. The lens at its widest point (model 1.25) is given as 11.11mm, which means nothing if the image creation size is not as well provided; like cameras and sensors. Smaller sensors use only a portion of the lens, hence the idea to call that a sensor crop.
So I tried to explore with the given distance width data from the manual what image element this one has. If all the data given is correct, the width is 9.02 mm. Yes, I'm aware that the laser doesn't physically have this Image size anywhere visible, but how much it uses the lens (field of view wise) creates this virtually.The Projector has a limitation in shift, and if both vertical and horizontal are maxed out, you have a "vignetting".
So, file 01 will have problems with the steel beam and the shift limitation.
This brings me to file 02.
The Projector needs to be much lower and around 8+ meters away. I have placed a 180cm figure to show how much space above is given.
This option maxes out the shift and tries to be as equal in brightness as possible.Please test all the data, and if possible, compare my data with the manual. I can't guarantee that my assumptions about what the data of the manual gives are correct. I had not even a single image. Hence I assume you use the 4:3 aspect ratio in the footage.
All the best
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Ah wow, thank you for going into all this detail Sassi
I see what you are saying about the beam and the height...I will be installing this projector tomorrow morning, so this is extremely valuable information, thank you so much! -
Thanks for the reply, AlexC.,
If you have any "stand" like a C-Stand, a Microphone stand, or anything that can extend to that height where the project needs to be, would be a great help. Set up the top part where you want the projector's lens to be. From that point, the projection wall must be fully visible. To have that stand there helps to get fast to the final position. (They certainly have a ladder that might work, but typically that is moved when mounting
If my exploration is not off, then a 45º is the horizontal field of view, which is easy to create by folding a piece of paper (square, fold diagonal); this will be the fastest guide to see. To see means here that you can see the projected wall ultimately.Many iPhone cinematography apps allow you to work as a director's viewfinder; with a full-frame camera, a standard 50mm landscape mode will give you a safe estimation (a 40mm, like the Sigma Art or Canon's Pancake lens, is closer, full frame as well.)
My best wishes for your project.
