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# Perspective Transform using Chessboard

Hey,

I need some help with this problem: I have a camera that takes a picture of something on a horizontal plane with a specific angle. That creates a perspective transform of this "something". And I would like to get this picture as if I would look down from top of it.

What I did already and one thing that I don't know how to do:

1. I placed a chessboard there.
2. I find the corners of the chessboard.
3. ???
4. cvGetPerspectiveTransform
5. cvWarpPerspective

My problem is point 3. I have to find out Source and Destination Points which depend on the corners of the chessboard and the width of the picture that was taken, because they show the transformation. Source is easy: (0,0), (Width, 0), (0,Height) and (Width,Height), because I want the whole picture to be transformed. Destination however is difficult for me. I don't know how to find those points. I want that the whole picture (Not just the part with the chessboard inside) is transformed within a single step. Like in the picture below. I would appreciate any help.

Greetings and my thanks in advance, Phanta

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## Comments

have a look in this sample here in order to see how to obtain the bird's eye view.

( 2015-05-31 10:27:16 -0500 )edit

No, that's not doing what I want. It just transforms the chessboard back to it's bird's eye view, but not the complete picture. I understood the basics that I need to do that and transforming a given part of a picture back to it's bird's-eye view is quite simple.

I just don't know how to calculate the 4 destination points to transform the whole picture back to it's bird's-eye view given a single chessboard that doesn't fill the complete picture

( 2015-05-31 17:31:06 -0500 )edit

## 1 answer

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actually what you are asking it cannot be done to the exact result that you wish. The reason for that is that you are lacking information, and here you can read a nice answer about it. Especially focus on this:

The aspect ratio of the destination rectangle can be chosen arbitrarily. In fact, it is not possible to deduce the "original aspect ratio" of the object in world coordinates, because "this is 2D homography, not 3D".

Thus, choosing arbitrarily destination points I managed for example to obtain the following results which I do not know how close are to your will.

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## Stats

Asked: 2015-05-31 07:20:48 -0500

Seen: 2,441 times

Last updated: Jun 01 '15