superposition of two vectors

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Haleh on 5 May 2013
Hi Everybody,
I am trying to superpose two protein structures using Kabsch and absor functions but in both cases I do not get correct transformed structures. I input the vectors 3by N as stated in the description but the output in both cases is a bunch of coordinates gathered around one coordinate. I mean all the numbers resemble each other at least in first 2 digits. It is like they are all sitting on top of each other. I can not use pdbsuperpose since my coordinates are not in pdb format so any idea please on how to do the superposition of the rigid body translation and rotation?
here are my data 5 x 3 data set :
A = 0.0669166284993940 0.0302935157371087 0.185882093887639 0.0240147296251836 0.00925539980246008 0.0925160692954149 -0.0317431774275438 -0.0147916036102607 -0.00436453950606019 -0.0337673308633457 0.00697062096721668 -0.0660928572116686 -0.0893274016664437 -0.0244575537488091 -0.166298369404389
I am trying to superpose these numbers on to B which is :
B= -4.30300000000000 22.4930000000000 11.0710000000000 -1.09900000000000 20.5260000000000 10.5110000000000 1.65600000000000 18.9540000000000 12.4680000000000 5.21800000000000 19.4570000000000 11.4000000000000 8.04100000000000 17.1650000000000 12.4880000000000
by using absor (A' , B'). Thank you again.
Matt J on 5 May 2013
Edited: Matt J on 5 May 2013
In your 3xN data sets, are the columns arranged in corresponding order? In other words, is A(:,i) supposed to map to B(:,i)? Give us a 3x5 truncation of each of data set, so we can get an idea of the quality of the data.

Matt J on 6 May 2013
Edited: Matt J on 6 May 2013
so any idea please on how to do the superposition of the rigid body translation and rotation?
If the 2 data sets differ only by a rotation/translation, then the separation between corresponding pairs of points should be the same. We can easily see from the example below, however, that this is not the case. Conclusion - you have bad data or a bad model for the data or the data points are not ordered the same in both A and B.
>> norm(A(1,:)-A(2,:))
ans =
0.1487
>> norm(B(1,:)-B(2,:))
ans =
29.3488