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I am working with 3d point data. When I checked the data I realized that there is some error on my data and need to do some kind of rotational rectification because the points which should be lie on a vertical line don’t lie but lie on a oblique line and also points which should lie on a vertical plane do lie on a bit oblique plane to the vertical.

So, what I did was, I have fitted the two planes by carefully selecting points lie on two nearby vertical planes (also by avoiding outliers). Then, I got an intersection line, in 3D, (the red line in figure) made by two vertical planes which I have fitted. If I get the directional vector of that line 3d, then it was (-0.04, -0.01, 0.9). As this line should be truly vertical I think these values should be equal to (0, 0, 1). Now, I want to apply rotational corrections along each axis to be come to this line parallel to (0, 0, 1).

enter image description here

I am bit confused how can I compute my rotational angles with these information. (I think I have enough information to compute correction angles.)

As I am thinking that my question is not clear, so that I am explaining the question again here under updates.-

Updates:

I have a point data set; Pi(xi, yi, zi). I want to find coordinate errors of this data set. However, I don’t have reference data for that, so that I found two vertical planes, perpendicular to each other, (for example building walls) by fitting planar fitting algorithm. Though these two planes must be truly vertical, it appears like it has small deviation from vertically. Then, I intersect these two planes and get a 3D line. In reality, this line should be truly vertical so vector of the line should be (0, 0, 1), however the detected line vector is (-0.04, -0.01, 0.99). I guess this is because of axis rotation error occurred at the time of data capturing. So what I really want to find rotation matrix (rotation angles of X,Y, Z axes i.e alpha, beta, gama) to be applied on to my points. I think I should use the following equation for this; enter image description here

So, please help me to find my rotational correction angles along X, Y and Z axes. Thank you.

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why dont you just round up numerically as you did in your example. –  al-Hwarizmi Jul 15 '13 at 19:37
    
@al-Hwarizmi: sorry, I did not get what you said. I have found the rotational equation codes. it need rotation angles. However, finally I want to get new coordinates of whole points and then I hope all come to true positions. Me, I am poor in mathematics and could you please show me the way to do this. thank you. –  slinga Jul 15 '13 at 20:53
    
yes but if I understood correctly, the differences $\Delta x$, $\Delta y$, $\Delta z$ for correction are likely equivalent to the round up of your coordinates namely $\Delta x=+0.04$, $\Delta y=+0.01$, $\Delta z=+0.1$. corrections can be done in each direction, cartesian, what is equivalent to rotation, polar. –  al-Hwarizmi Jul 15 '13 at 20:59

1 Answer 1

up vote 1 down vote accepted

If I understood your question correctly, the differences $\Delta x$, $\Delta y$, $\Delta z$ for correction are likely equivalent to the round up of your coordinates namely: $$\Delta x=+0.04$$ $$\Delta y=+0.01$$ $$\Delta z=+0.1$$ If you must have the correction in spherical coordinates, further you may take the general transformation rules for line element (approximating small differences $\Delta$) you have the Jacobi matrix: $$\frac{\Delta(x,y,z)}{\Delta(r,\theta,\varphi)}\simeq J=\begin{pmatrix} \sin\theta\cos\varphi &r\cos\theta\cos\varphi&-r\sin\theta\sin\varphi\\ \sin\theta \sin\varphi&r\cos\theta\sin\varphi&r\sin\theta\cos\varphi\\ \cos\theta&-r\sin\theta&0 \end{pmatrix}$$ and: $$(\Delta x,\Delta y,\Delta z)^T=J\cdot(\Delta r,\Delta \theta,\Delta \varphi)^T$$

See also here>>>; The system of equation must be solved. You may also need to take into consideration:

$$r=\sqrt{x^2+y^2+z^2}$$

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thank you very much for the explanation. I will refer all and try to understand. If I need further clarification on something, then when I ask it (in here), I think you may again response me. till many thanks. –  slinga Jul 15 '13 at 23:02
1  
you are absolutely welcome. –  al-Hwarizmi Jul 16 '13 at 6:18
    
thanks a lot.... –  slinga Jul 16 '13 at 9:39
    
Thank you very much for your clarification, but I feel that this is not what I really need. I will again explain in more detail to be clear it. Please refer the updates which I posted and help me to solve my problem. thanks again. –  slinga Jul 19 '13 at 21:06
    
could you please help me to solve my problem. still i am struggling and suffering a lot. –  slinga Jul 20 '13 at 19:04

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