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Absolute Orientation - Horn's method

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Absolute Orientation - Horn's method


Matt J (view profile)


23 Dec 2009 (Updated )

Solves weighted absolute orientation problem using Horn's quaternion-based method.

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 ABSOR - a tool for solving the absolute orientation problem using Horn's
 quaternion-based method, that is, for finding the rotation, translation, and
 optionally also the scaling, that best maps one collection of point coordinates
 to another in a least squares sense. The function works for both 2D and 3D
 coordinates, and also gives the option of weighting the coordinates non-uniformly.
 The code avoids for-loops to maximize speed.
 As input data, one has
   A: a 2xN or 3xN matrix whos columns are the coordinates of N source points.
   B: a 2xN or 3xN matrix whos columns are the coordinates of N target points.
 The syntax
 solves the unweighted/unscaled registration problem
            min. sum_i ||R*A(:,i) + t - B(:,i)||^2
 for unknown rotation matrix R and unknown translation vector t.
 This is a special case of the more general problem
            min. sum_i w(i)*||s*R*A(:,i) + t - B(:,i)||^2
 where s>=0 is an unknown global scale factor, to be estimated along with R and t,
 and w is a user-supplied length N vector of weights. One can include either
 s or w or both in the problem formulation using the syntax,
 with parameter/value pair options
   'doScale' - Boolean flag. If TRUE, the global scale factor, s, is included.
   'weights' - the length N vector of weights, w. Default, no weighting.
  regParams: structure output with estimated registration parameters,
      regParams.R: The estimated rotation matrix, R
      regParams.t: The estimated translation vector, t
      regParams.s: The estimated scale factor (set to 1 if doScale=false).
      regParams.M: Homogenous coordinate transform matrix [s*R,t;[0 0 ... 1]].
      For 3D problems, the structure includes
         regParams.q: A unit quaternion [q0 qx qy qz] corresponding to R and
                    signed to satisfy max(q)=max(abs(q))>0
      For 2D problems, it includes
         regParams.theta: the counter-clockwise rotation angle about the
                    2D origin
   Bfit: The rotation, translation, and scaling (as applicable) of A that
         best matches B.
  ErrorStats: structure output with error statistics. In particular,
              defining err(i)=sqrt(w(i))*norm( Bfit(:,i)-B(:,i) ),
              it contains
       ErrorStats.errlsq = norm(err)
       ErrorStats.errmax = max(err)


Absolute Orientation inspired this file.

MATLAB release MATLAB 7.9 (R2009b)
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Comments and Ratings (29)
31 Jul 2015 RyanG

RyanG (view profile)

Hi Matt,
I understand now.
Thank you!

Comment only
30 Jul 2015 Matt J

Matt J (view profile)

Hi Ryan,

M is the optimum transformation operator that best maps A to B.

As for Bfit, my explanation in my previous comment was in error. Bfit(:,i) is what you get if you apply the optimum transformation M to the original data A(:,i). It is the best fit to B(:,i). If you are not interested in fitting the B data, there is no need to call Bfit.

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30 Jul 2015 RyanG

RyanG (view profile)

Hi Matt,
Thank you again for the great script and helpful response. I assumed that the output of M was already the best fit solution. Is this not the case? Do I need to call 'Bfit' when I call absorb for this to be applied?


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25 Jul 2015 Matt J

Matt J (view profile)

Hi Ryan,

The general way to transform an arbitrary point P is s*R*P+t, though s will simply be 1 when you call absor without the doScale option. To perform the transformation in homogeneous coordinates, you can do M*[P;1] padding P with a 1, as you mentioned.

If you call absor with the Bfit output, this transformation will be applied for you internally on all B(:,i), sparing you the effort of doing so yourself manually.

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24 Jul 2015 RyanG

RyanG (view profile)

Hi Matt,
A follow up to my earlier question, what is the use and purpose of the Bfit? Should I be using Bfit to map points A to B versus the M transformation matrix?

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24 Jul 2015 RyanG

RyanG (view profile)

Hi Matt,

Thank you for supplying this useful script. I had a question regarding applying the resultant M transformation matrix. Below you state that to map your A to B points you can perform the following: Pnew=M*P. With P being A basically.

But that it requires homogeneous coordinates. I padded the 4th row of my data with 1's so that the above matrix multiplication could be performed. What is the fourth row? Are 1's required there so that when scaling is applied that the matrix dimensions match?

Thank you,

10 Jul 2015 Matt J

Matt J (view profile)

Hi Yingjuan,

I'm not sure you can do it if you mean to consider rotation only (no translation). If you do,


then the code will give you the best combination of rotation matrix and translation vector simultaneously matching all A to the corresponding B. It cannot constrain the transformation to be a rotation only.

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10 Jul 2015 Yingjuan Wu

Hi Matt,

Thanks for your code. I have 15 A vectors (A1 to A15) and 15 correspondign B vectors (B1 to B15)in 3D. Now I want to figure out if these vectors have the same rotation angle (or rotation matrix). For example, A1 to B1 a has rotation angle theta, I want to know if A2 to B2 or A3 to B3 has the same rotation angle. Could I use your code to solve this problem? Thanks!

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09 Jun 2015 Matt J

Matt J (view profile)

Hi Alexander,
No, the algorithm does not handle refelections. Only rotations, translations, and (optionally) scalings.

Comment only
09 Jun 2015 Alexander Kraskov

Hi Matt,

I probably should read the original paper but though it would be quicker to ask you.

It seems that your code (and maybe method) does not treat correctly reflection cases. Is it correct?

Here is an example my example is
17.6607 53.2300 69.2854
14.5424 31.3154 50.5354
28.9079 5.8403 55.3454
43.7231 3.0544 78.9054
48.0822 25.4468 98.7854
34.5058 50.4406 92.8754
19.4431 37.4269 83.5354
-24.0000 24.3000 72.3000
-5.4000 2.5000 78.2000
-10.3000 -25.4000 66.5000
-34.0000 -31.3000 51.7000
-55.4000 -8.4000 42.2000
-48.2000 19.3000 56.0000
-38.5000 7.8000 72.4000

It produces wrong (big error) transformation matrix.


-0.7865 -0.3739 -0.4916 40.6255
-0.5476 0.7903 0.2751 -29.5564
0.2856 0.4855 -0.8262 102.4473
0 0 0 1.0000

>> ErrorStats

ErrorStats =

errlsq: 20.7210
errmax: 18.1723

If I reflect the first coordinate of A (multiply by
-1 0 0
0 1 0
0 0 1)

than I get more reasonable transformation matrix

-0.0127 0.0020 -0.9999 44.3410
0.1158 0.9933 0.0005 -27.5493
0.9932 -0.1158 -0.0129 96.5021
0 0 0 1.0000
ErrorStats =

errlsq: 3.9922
errmax: 2.9032

My question is whether it is possible to deal with it in a automatic way?


Comment only
15 Apr 2015 Matt J

Matt J (view profile)

Hi Fox,

I'm not sure if you've been through the help documentation or not, but the input syntax is described there and in the Description section above, in particular where it says

A: a 2xN or 3xN matrix whos columns are the coordinates of N source points.
B: a 2xN or 3xN matrix whos columns are the coordinates of N target points.

So, put your points to be matched in the columns of matrices A and B. They must be in correspondng order, i.e., B(:,j) must be the rototranslation of A(:,j).

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13 Apr 2015 Fox

Fox (view profile)

I'd just like to know how to write the co-ordinates out. How would I input the 3 sets of A values and 3 sets of B values?


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12 Apr 2015 Fox

Fox (view profile)

Thanks for the quick response. I've just sent you a message.

I've just realised this might not work for what I require.

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12 Apr 2015 Matt J

Matt J (view profile)

You haven't shown your call to absor(), so we don't know precisely what you're doing. However, your data example,

A = [1,2,3]'
B = [2,1,2]'

only contains 1 point in each set. You need at least 3 points in each set to do a sensible registration. They are to be placed in the columns of A,B in corresponding order.

Comment only
11 Apr 2015 Fox

Fox (view profile)

I'm pretty new to this type of coding, could someone tell me where to put my two sets of 3D co-ordinates?

If I put for example:
A = [1,2,3]'
B = [2,1,2]'

It says that A and B are used in the function, so they get changed to - in the first line of the code.

It does run, however doesn't give me any values. Just gives like [3x1 double matrix] in writing.


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06 Nov 2014 ZZZZZ

ZZZZZ (view profile)

23 Jun 2014 Zohar Bar-Yehuda  
14 Apr 2014 Matt J

Matt J (view profile)

The points must be placed into the columns of a and b, not the rows.

Comment only
14 Apr 2014 Steve

Steve (view profile)

One question about the translation vector (regParams.t):
I have the following 3 points before rotation:
A1 = (400, 400, 200,)
A2 = (600, 600, 200,)
A3 = (2000, 2000, 2000)
and after rotation:
B1 = (400, 400, 200,)
B2 = (600, 600, 200,)
B3 = (2000, 2000, 2001)
so all point are the same, just B3z is 2001 instead of 2000. Then the computed translation vector using your code is: (162, -76,4001)
What doesn't make sense to me? Because just one single point changed it's z-value from 2000 to 2001 what is just a difference of 1...?
Nevertheless, the result seems to be right, because:
b(:,1) = regParams.R * a(:,1) + regParams.t
is true...

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14 Apr 2014 Steve

Steve (view profile)

Thanks for this very nice tool!

14 Nov 2013 Cong

Cong (view profile)

14 Nov 2013 Matt J

Matt J (view profile)

Hi Cong,

If you're asking how to apply the rotation matrix R to rotate a point P=[x;y;z], you would just use matrix multiplication Pnew=R*P.

Note, however, that ABSOR calculates a translation vector, t, as well (and sometimes also a scaling if you select the doScale option). The full transformation would be Pnew=R*P+t.

Finally, if you are working in homogeneous coordinates P=[x;y;z;1], ABSOR also returns a 4x4 total transformation matrix M. In that case, the transformation can then be done Pnew=M*P.

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14 Nov 2013 Cong

Cong (view profile)

Thank you to provide this very useful toolbox. I have a question: if I obtain the rotation matrix from your method, then how can I get the coordinates of any point in the new coordinate system (target).

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18 Apr 2013 monica

monica (view profile)

Thank you very much.

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16 Apr 2013 Matt J

Matt J (view profile)

Hi Monica. You can find the derivation of the scaling factor in the original paper by Horn, Section 2D

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16 Apr 2013 monica

monica (view profile)

Thank you Matt for this submission. I want to know how you got scaling factor sss=summ(right.*(R*left))/summ(left.*left). I know t+RSp-q=0 but i am not able to figure out the formula you have used for getting scaling factor. Could you please elaborate.

06 Jun 2012 Georg Wiora

Very nice work!

10 May 2012 Matt J

Matt J (view profile)

Thanks for the feedback, Georg. I'm not entirely sure why the case of 3 points has been giving you trouble, though. It is true that alternative SVD-based methods and orthogonal matrix methods had to be modified to handle coplanar point data, but I always understood that to be a fairly stable solution.

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10 May 2012 Georg Stillfried

great work! (seems to work fine and also allows matching of three points)

25 Dec 2009 1.1

1. Important typo fix in help doc. Model is s*R*A+t not s*R*A-t
2. Added version for earlier MATLAB versions lacking bsxfun

22 Jan 2010 1.2

Added absorientParams. See README.txt for details

09 Apr 2010 1.3

Added tools for purely 2D registration

29 Sep 2010 1.4

*The many previous files in the distribution have been consolidated into a single file absor.m.

*New capability:weighted least squares registration.

*Fixed minor bug in the 2D registration routines, occurring when rotation angle was 0.

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