Translate vector to a center point

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Alessandro Ruda
Alessandro Ruda el 3 de Mzo. de 2022
Comentada: Jan el 4 de Mzo. de 2022
Dear MatLab comunity,
I have an array of few thousands xyz coordinates of three points in a molecule: A,B and C.
I created two array of vectors:
AB = [(A(:,1)-B(:,1)), (A(:,2)-B(:,2)), (A(:,3)-B(:,3))];
BC = [(C(:,1)-B(:,1)), (C(:,2)-B(:,2)), (C(:,3)-B(:,3))];
The two vectors share one of the point (B). Now I want to translate the two vector - keeping of course the same orientation and magnitude - so that the origin is the center of mass , which in my case is:
COM = [3.9721615314483643 -8.11227798461914 -0.34564414620399475]
How can I do that?
Thank you in advance!
/alex

Respuestas (1)

Jan
Jan el 3 de Mzo. de 2022
At first simplify
AB = [(A(:,1)-B(:,1)), (A(:,2)-B(:,2)), (A(:,3)-B(:,3))];
BC = [(C(:,1)-B(:,1)), (C(:,2)-B(:,2)), (C(:,3)-B(:,3))];
to:
AB = A - B;
BC = C - B;
Now I'm not sure if I understand, what you are asking for. Vectors can be seen as directions or location. Moving them while keepinmg the direction is not meaning ful. This works with lines, but not with vectors, which can be move in space freely.
But maybe all you want to do is:
AB - COM
BC - COM
  2 comentarios
Alessandro Ruda
Alessandro Ruda el 3 de Mzo. de 2022
Okay, not really. I haven't explained clearly.
So I have a molecule and I calculated the center of mass and the principal axis of inertia and these are:
COM = [3.9721615314483643 -8.11227798461914 -0.34564414620399475]
X_ax = [0.43753358721733093 0.719929575920105 -0.5387632250785828]
Y_ax = [0.5724487900733948 0.2390475869178772 0.7843204736709595]
Z_ax = [0.6934455633163452 -0.6515809297561646 -0.3075314462184906]
Then I have these array of points A(x,y,z), B(x,y,z) and C(x,y,z). Each of these array has few thousands of coordinates.
Now I want first to generate the vectors AB and BC and then calculate the angle between these and each of the principal axis separately.
I hope I have explained it clearly, what would be the best way to do it?
/alex
Jan
Jan el 4 de Mzo. de 2022
Sorry, I do not get it.

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