Author Topic: Math behind Projection / Rectification  (Read 14689 times)

Offline freak42

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on: January 13, 2008, 03:58:00 AM

can someone shed some light how one can calculate the 2d coordinates of a (ir-)point after one did the 4-point synchronisation with a projection plane?
I looked at the source of Warper.cs but the non-genius I am I couldn't quite figure out what's going on.
Btw I'm trying to code the same screen-tracker just with a webcam and java (poor man's version, i guess)
Any help or links are appreciated.


ps I moved this message to this forum, I guess I posted it in the wrong forum first. sorry.
« Last Edit: January 13, 2008, 03:59:35 AM by freak42 »

Offline romlel

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Reply #1 on: January 13, 2008, 04:23:23 AM

A matrix is used to transform the coordinates.
By reading the source code, you should be able to find how the matrix is "fed" with the different values given during the initialisation of the wrapper. Once your matrix build, you will have to multiply it with the tracked 2D coordinates to get the "rectified" values.

Johnny do not use a specialized class for it and do the math "by hand", but Java should have a matrix class

Hope this helps

Offline ujs

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Reply #2 on: February 04, 2008, 04:24:53 PM

you can save yourself the work to do that all manually in Java.
Just can use the class PerspectiveTransform from JAI.

The method getQuadToQuad creates a PerspectiveTransform. Then you just have to use one of the transform-methods to convert from one perspective to the other.


Java WiimoteWhiteboard for Mac OS X, Windows and Linux.

Offline Helza

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Reply #3 on: February 04, 2008, 04:47:05 PM
The WarpedX & WarpedY coordinated that this produce are they related to the current screensize?
I have to transform the warpedx/y values to an value between 0..1, currently i do this by dividing the warpedx/y by my screenwidth & height, is that correct?

Offline JTurtle

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Reply #4 on: February 06, 2008, 05:07:38 PM
maybe you shoud look at my wii project page - i've got a little explanation behind my maths there...

My Wii Projectpage:

Offline Helza

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Reply #5 on: February 07, 2008, 01:46:50 AM
Thanks, my math is horrible, but after a couple hours i figured out the problem. Turned out to be bad calibration data and a switch of screenwidth/height i used.. together it made the results rather... funny.. :)

Offline chema

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Reply #6 on: September 16, 2009, 01:35:09 PM
Hi all,

that's an old post, but didn't find any newer.

So, when calibrating we take 4 points, but the perspective transform matrix is just 3x3. What do you do with the 4th point?? It's there any point discarded or is used to rectificate the calibration??

By the way, how do you do the smoothing?? Do you take the linear predicted point calculated with the 7 last points??

Thanks you for your help


Offline shakespeare1212

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Reply #7 on: November 05, 2009, 02:24:33 AM
Wow, you guys are bright. I hope I'm not intruding too much but I think that this is a decent place to put in my two cents.

I was wanting to help teachers an implementers do a good job of placing the wii-remote to get the best tracking resolution.  We are given the viewing angle of the Wii camera to be 45 degrees. So, I let Turbo Cad do my calculations, i.e. I drew an accurate sketch of a 45 degree angle at varrying distances from various screen sizes. Can one of you gentlemen work out the trigonometry and come up with a formula that translates a desired or given screen size into an optitmal distance that the Wiimote should be from the center of the screen, given that we will place on a normal coming straight out from the center of the screen. Are you with me?

Anyway here is a table of my results from Turbocad. Please someone tell me if I'm on the right track here. Do these numbers make sense?

Width of your screen             Opitmal distance on a normal (straight out from the screen)
from left to right                    from the screen center for the Wiimote

32   inches                               38.93        inches
52   inches                               62.78        inches
102 inches                               123.185    inches
200 inches                               240.3469  inches

This data was just based on the demsion tools in TurboCad. I would be happy to email my turbo cad file ( or a JPEG) to anyone who is interested, but I would prefer to use trig. and develop a simple formula.

OK,... I thought a little harder about it....
The formula I have is: Optimal distance from the screen = (Sceen width (left to right) / 2)/ tan 22.5 degrees.

Bam!! I check it against the table. It works!!

James Mclain
Owner and Creator of
Whalebone IR and Software

or view and buy our products on ebay:   Search for IR pen or products by shakespeare1212.