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?
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.
[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.