can anyone provide a code for converting 2d point into 3d cordinate system
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praveen rai
el 13 de Sept. de 2017
Comentada: Rik
el 15 de Sept. de 2017
i want to map the 2d point into 3d cordinate usinf direct linear transformation algo(DLT) or using homography I am not sure how to do or which method shd i do I want the code with brief xplanation for eg i have (x1,y1),(x2,y2)...x and y denote cordinate and with ref to d same point i want in object space i.e (x,y,z) coordinate
6 comentarios
Rik
el 15 de Sept. de 2017
You just got upgraded from unintelligible to a case of doit4me. Read those links I gave you and then follow my suggestion from the second comment:
"You should at the very least provide the image you are working on, because that is a very difficult problem, which may not be possible with your data. I have little experience with computer vision, but usually it uses two cameras side by side or a single camera whose movements are known."
I don't know much about computer vision, but if I were you I would start Googling examples until you have something that is close to working. As you know the name from a specific algorithm, I suspect you are in some educational program or course. If that's the case, ask your teacher/instructor for help. It seems you need more than what you can reasonably expect to get here.
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John D'Errico
el 13 de Sept. de 2017
Sure. Trivial to do, in fact.
xy2xyz = @(x,y) [x,y,1];
For any value of x and y, this creates a new set of coordinates in the R^3 (x,y,z) domain.
It is a rather boring transformation, but it satisfies the requirements. I suppose you could have done something more exciting. Perhaps...
xy2xyz = @(x,y) [x,y,randn()];
or
xy2xyz = @(x,y) [x,y,x+2*y];
The latter can be written as
xy2xyz = @(x,y) [x,y]*[1 0 1;0 1 2];
Any of infinitely many transformations will exist. What you have not said is what mapping is appropriate.
2 comentarios
John D'Errico
el 13 de Sept. de 2017
But only you know what you want to do. So only you can write the code.
Más respuestas (1)
Image Analyst
el 13 de Sept. de 2017
I don't exactly know what you want, but perhaps you're thinking of colorcloud().
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