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In an assignment A(I) = B, the number of elements in B and I must be the same.

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Hi, I know this question must have been asked a million times before but I can't find a problem quite like mine. I am running a loop to predict a trajectory where I calculate time t1 from an equation to give a constant. However this t1 varies with every iteration of the program, hence t1(i).
I then want to use this to get t(i) from
t(i)=0:1:(t1(i)-2);
However it is spitting out the statement as in the title. Can someone suggest a way around this?
Thanks.

Respuesta aceptada

Walter Roberson
Walter Roberson el 26 de Abr. de 2012
The expression 0:1:(t1(i)-2) creates a row vector. You cannot store a row vector into the space of a single numeric array element t(i) .
You could consider using cell arrays:
t{i} = 0:1:t1(i)-2;

Más respuestas (2)

Honglei Chen
Honglei Chen el 26 de Abr. de 2012
I don't quite know what your intention is, but the left side is a scalar and the right side in general is a vector, could be scalar, could be empty, so an error is thrown.

Daniel Shub
Daniel Shub el 26 de Abr. de 2012
You need to step back and think about the error. What is the size of
t(i)
From inspection and a little bit of MATLAB knowledge you should realize it will be the same size as i. Since you are likely looping over i, i will be a scalar (1x1) so t(i) will have 1 element independent from the value of i. In other words, on every iteration t(i) will have 1 element.
Okay now look at
0:1:(t1(i)-2)
How big do you think that is going to be? Do you think it is going to have exactly 1 element on every iteration? Do you see where the error is yet? If not lets consider
0:1:n
where n is equal to (t1(i)-2). This obviously has n+1 elements. So for your assignment to work (i.e., the number of element in B and I to be the same) n must be equal to 0, and in turn t1(i)-2 must be zero and t1(i) must be -2. Are you initializing t1 to be a vector of -2's?
  4 comentarios
Stuart
Stuart el 26 de Abr. de 2012
Sorry I'm not being clear.
A projectile takes a different time to cover a ceratin distance depending on its velocity in the x and y directions.
I am running 5 iterations with 5 different angles of launch and so the times will be different.
I then use these times to calculate a lot of other things.
Does that make more sense?
I was expecting variables like 0:1:2, 0:1:2.4, 0:1:2.45 etc
Daniel Shub
Daniel Shub el 26 de Abr. de 2012
While that is not going to work because will all be the same array 0:1:2.

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