Adobe After Effects Forum
Real world dimensions After Effects
Real world dimensions After Effects
by Sarah Leitner on Apr 14, 2018 at 4:50:16 pm

I'm making a 3D animation for an event. At the event, two cameras will be filming a man while he is making a presentation. The audience will see nothing on stage next to the presenter in real life. But on large screens behind him, the feed from the two cameras will be playing and they will insert our 3D model next to the speaker (where the red cube is). We have to render two versions of our animation that match those camera angles so it appears as if the hotel animation is next to him from those two angles.

I know the distance the cameras will be from the stage and also their angles (one will be dead on and the other will be off to the right and slightly angled). How do I translate those real-world distances/angles into After Effects so I can figure out where to place my cameras in my composition?

Camera 1:

Camera 2:

Re: Real world dimensions After Effects
by Mark Whitney on Apr 14, 2018 at 6:25:00 pm

Are you doing this live? If so that adds a whole other element & don't think AE is your solution as it's not a real time tool.

Among other considerations, it really depends on what your 3D app considers as "Real World" dimensions. I'm not finding my notes at the moment but I seem to recall that the 3D program I prefer, Lightwave, 100 AE pixels equals 1 meter. LW has a tool as part of it's feature set however to do data exchange between it & AE. Other, but not all 3D apps have something similar.

If your cameras are locked off, then it doesn't matter; simply go by eyball. If not, then I'd take the footage into AE, use it's tools to create a 3D camera, then export that data out to LW. That would give you camera movements & focal length. Use your footage as a background guide to line up your 3D object & it should lock & stay in place to the camera moves.

Re: Real world dimensions After Effects
by Dave LaRonde on Apr 14, 2018 at 7:16:55 pm

Moreover, AE knows nothing of real world dimensions. It is up to you to calculate dimensions in terms of pixels... which is the only unit AE understands.

I suspect you could get close enough by eyeballing it.

Dave LaRonde
Promotion Producer
KGAN (CBS) & KFXA (Fox) Cedar Rapids, IA

Re: Real world dimensions After Effects
by Steve Bentley on Apr 14, 2018 at 9:47:33 pm

In a 3D package everything is relative - you can enter all those dimensions you have and you will be bang on (assuming the data is correct). Export the camera from the 3D package along with nulls for object position and height and you can match those markers in AE when you import all that data.
But if you want to use a 3D model inside AE, you can't say to the model, be six feet high. As Dave said you have to say "be 359 somethings high" (AE can't natively use 3d models, there is always a plug in, even a built in one, and that may not translate to pixels).
But it can do renders of models - so once its rendered in a 3D package all the hard stuff is done.
But often eyeballing it is better. Lens distortion can screw up lots of measurements because AE won't automatically take the distortion into account.

And Mark may have said this - AE isn't real time so it's not the best tool if you are live keying a model into a stage shot in front of an audience (it wasn't totally clear if this was the case). And you might have a killer graphics card that can get close to real time but any type of system activity can interrupt that (like autosave).

Re: Real world dimensions After Effects
by Walter Soyka on Apr 16, 2018 at 10:19:36 am

[Sarah Leitner] "How do I translate those real-world distances/angles into After Effects so I can figure out where to place my cameras in my composition?"

You have to determine the relationship between physical measurement and pixels.

Let's assume in the Camera 1 view that you're working in HD (1920x1080) and that the 6' cube is 640 px. That relationship tells us what we need to know to convert back and forth:
640px / 6ft = 106.67 px/ft

To position a camera 151'7" away,
151.583 ft * 106.67 px/ft = ~16168px

As the others note, there are other physical and optical factors at play -- but this should get you close.

Walter Soyka
Designer & Mad Scientist at Keen Live [link]
Motion Graphics, Widescreen Events, Presentation Design, and Consulting
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