How does Anamorphic Art work mathematically?
February 1, 2009 12:29 PM Subscribe
There's an artist who creates anamorphic paintings on sidewalk pavement,
which when viewed from a certain perspective appear as a separate, embedded 3D scene. How does this illusion work, mathematically speaking?
Also, what other kinds of interesting or peculiar mathematical problems are similar? I'm interested in researching this topic and would like to explore related tangents. The more technical the answer the better, so please do feel free to delve into linear algebra, differential forms or topology in your answers.
Also, what other kinds of interesting or peculiar mathematical problems are similar? I'm interested in researching this topic and would like to explore related tangents. The more technical the answer the better, so please do feel free to delve into linear algebra, differential forms or topology in your answers.
I can't help with the math, but modern artistic perspective was "discovered" by Brunelleschi in 1415 in Florence. Nice video clip by James Burke here about that innovation...
posted by wfrgms at 12:40 PM on February 1, 2009
posted by wfrgms at 12:40 PM on February 1, 2009
I've long had a suspicion that the artists who do this (there are more than one) start by mounting a slide projector on a tripod and projecting the image (or a line drawing of it) onto the ground. They then use their chalk or paint and trace out what's being projected.
Once they're finished, the slide projector would be removed, and a camera would be mounted on the same tripod in the same location.
I don't see that advanced math is inherently necessary to solve this problem. The projected image from the slide projector would automatically produce all the right distortion to make the scene look right for the camera.
posted by Chocolate Pickle at 1:07 PM on February 1, 2009
Once they're finished, the slide projector would be removed, and a camera would be mounted on the same tripod in the same location.
I don't see that advanced math is inherently necessary to solve this problem. The projected image from the slide projector would automatically produce all the right distortion to make the scene look right for the camera.
posted by Chocolate Pickle at 1:07 PM on February 1, 2009
Best answer: OK, imagine that, in a cartesian coordinate system, your head is at 0,0,0. You're looking straight ahead, along the z axis.
There's a canvas right in front of you, it's 2m away. There are 4 dots forming a 20cm by 20cm square at the center of the canvas. So in the coordinate system we're talking about, the dots are at (0.1,0.1,2.0), (0.1,-0.1,2.0), (-0.1,0.1,2.0), and (-0.1,-0.1,2.0).
OK, now you tilt your head downwards by 45 degrees. Except when you do, a strange thing happens - the canvas is a magic advertising canvas which is always at the centre of the viewer's view.
So that downward rotation by 45 degrees is the same as taking the points' coordinates, and rotating them by 45 degrees around the z axis. So now the 4 points are at (0.1000,-1.3435,1.4849), (0.1000,-1.4849,1.3435), (-0.1000,-1.3435,1.4849), and (-0.1000,-1.4849,1.3435).
You wonder to yourself, if you painted 4 spots on the pavement so that, to your eye, they looked the same as those 4 points on the canvas, where would those 4 spots be? You know light travels in straight lines, and you know the pavement is at Y=-2 (i.e. 2m below your head, you happen to be pretty tall).
In other words, each spot on the pavement would be on a straight line which ran from the pavement, through the point on the canvas, and to your eye.
So you take your rotated points, you work out a scale factor that will extend the line so Y=-2, which is simple enough as it means newX = oldX * (-2/oldY), newY = oldY * (-2/oldY) = -2, newZ = oldZ * (-2/oldY).
So the spots on the ground would be at (0.1489,-2.0000,2.2105), (0.1347,-2.0000,1.8095), (-0.1489,-2.0000,2.2105), and (-0.1347,-2.0000,1.8095)
Now, I flew overhead in a helicopter and looked down, I would see you looking down at the pavement, and the pavement would have 4 spots on. One would 14.8cm to your right and 2.21m ahead of you, one would be 13.4cm to your right and 1.81m ahead of you; and so on.
What I've just explained is basically some simple ray tracing if you want to learn more.
As people have observed in the past, just projecting the thing is probably easier.
posted by Mike1024 at 1:40 PM on February 1, 2009 [3 favorites]
There's a canvas right in front of you, it's 2m away. There are 4 dots forming a 20cm by 20cm square at the center of the canvas. So in the coordinate system we're talking about, the dots are at (0.1,0.1,2.0), (0.1,-0.1,2.0), (-0.1,0.1,2.0), and (-0.1,-0.1,2.0).
OK, now you tilt your head downwards by 45 degrees. Except when you do, a strange thing happens - the canvas is a magic advertising canvas which is always at the centre of the viewer's view.
So that downward rotation by 45 degrees is the same as taking the points' coordinates, and rotating them by 45 degrees around the z axis. So now the 4 points are at (0.1000,-1.3435,1.4849), (0.1000,-1.4849,1.3435), (-0.1000,-1.3435,1.4849), and (-0.1000,-1.4849,1.3435).
You wonder to yourself, if you painted 4 spots on the pavement so that, to your eye, they looked the same as those 4 points on the canvas, where would those 4 spots be? You know light travels in straight lines, and you know the pavement is at Y=-2 (i.e. 2m below your head, you happen to be pretty tall).
In other words, each spot on the pavement would be on a straight line which ran from the pavement, through the point on the canvas, and to your eye.
So you take your rotated points, you work out a scale factor that will extend the line so Y=-2, which is simple enough as it means newX = oldX * (-2/oldY), newY = oldY * (-2/oldY) = -2, newZ = oldZ * (-2/oldY).
So the spots on the ground would be at (0.1489,-2.0000,2.2105), (0.1347,-2.0000,1.8095), (-0.1489,-2.0000,2.2105), and (-0.1347,-2.0000,1.8095)
Now, I flew overhead in a helicopter and looked down, I would see you looking down at the pavement, and the pavement would have 4 spots on. One would 14.8cm to your right and 2.21m ahead of you, one would be 13.4cm to your right and 1.81m ahead of you; and so on.
What I've just explained is basically some simple ray tracing if you want to learn more.
As people have observed in the past, just projecting the thing is probably easier.
posted by Mike1024 at 1:40 PM on February 1, 2009 [3 favorites]
If you really wanted to be tricky, you'd go to the anticipated site, put up your tripod and use your camera to take a picture of the raw pavement. And you'd spray-paint a spot under the tripod where a plumbob touched down, so you could find it again.
You'd take the picture and with photoshop add to it the image you want ultimately to paint, and then turn that into a slide. Return to the site, find the spot again and set up the tripod in the same location, use the slide projector to project the new image onto the pavement, and then paint what the projector tells you to paint. That way you could fit your image into the existing pavement texture (e.g. bricks, manhole covers).
Still doesn't require any math, though; everything related to that is being done for you automatically by the slide projector.
posted by Chocolate Pickle at 5:05 PM on February 1, 2009
You'd take the picture and with photoshop add to it the image you want ultimately to paint, and then turn that into a slide. Return to the site, find the spot again and set up the tripod in the same location, use the slide projector to project the new image onto the pavement, and then paint what the projector tells you to paint. That way you could fit your image into the existing pavement texture (e.g. bricks, manhole covers).
Still doesn't require any math, though; everything related to that is being done for you automatically by the slide projector.
posted by Chocolate Pickle at 5:05 PM on February 1, 2009
This thread is closed to new comments.
posted by niles at 12:38 PM on February 1, 2009