MATLAB: Index in postition 1 is invalid. Array indices must be positive integers or logical values.

3d plotsfor loopmatrix array

Trying to write a loop to fill in the values of a 3d array of target coordinates and then one matrix that has the third variable as theta, and the other as velocity, but keep getting this error. eventually wanting to plot these.

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Your Hmin is negative. Your y starts at Hmin so your y starts negative. You use y as a subscript, so that means that you are using a negative value as a subscript.

You need to decide whether you are creating a formula, or an array.

A formula can potentially accept any inputs, including fractional or infinite, but it does not define an ordering: for formula F, f(x, y) is in no way "beside" f(x, y+1) since you might also have f(x, y+0.01) and so on.

An array on the other hand restricts subscripts to be positive integers. With an array, you can talk about the "first" and "second" entry and so on (ordering) but it does not in itself describe how the "first" or "second" correspond to anything in the real or mathematical world. You might have other information to know that, for example, A(5,1) corresponds to x = 17 and y = -2, and that A(5,2) corresponds to y = -1.99, but that is not "baked in" to the array: MATLAB just knows "first", "second" and so on.

What do you do? Well, in the special case that your values are mathematically spaced 1 unit apart, then even if they do not start at an integer, you can use A(x-firstX+1, y-firstY+1) to map from mathematical values x, y to index values. I would, however, recommend round() of the expressions if firstX or firstY are not integers.

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MATLAB: Error: index must be a positive integer or logical.

You are confusing formula and indexing.

In MATLAB, except under some circumstances involving symbolic variables, every time you have () appearing after a variable on the left hand side of an assignment statement, you are doing indexing. Bs(0) is an attempt to index a variable named Bs at offset 0. That is not permitted in MATLAB: offsets must be positive integers, so 1, 2, 3, ...

In particular assigning to Bs(0) does not create a formula for Bs saying that when formula Bs is evaluated at value 0, that a particular value is to be result.

MATLAB has two basic types of objects: 1) variables that can be indexed at positive integer locations (or by logical values), and 2) formula that can be evaluated at arbitrary values (that need not even be real-valued.)

Formulas in MATLAB are called functions. When you can write them in the form of single expressions, you can create them as "handles" to "anonymous functions" and you can assign the handles to variables. For example,

 F = @(x) x.^2 - 1

creates an anonymous function with formula x -> x^2-1, and assigns the handle of that function to F. After that, F(-7.2) and F(0) and F(pi) would all be valid invocations of the function handle to compute the value of the formula.

In MATLAB, you cannot define a piece by piece the way you might abstractly write y(-1) -> 5, y(0) -> -2, y(pi) -> 11, y(pi+1E-50) = -4 . There is no way in MATLAB (without writing a bunch of code yourself) to make a statement along the line of

 %this section is not valid in MATLAB
 Y(-1) = 5;
 Y(0) = -2;
 Y(pi) = 11;
 Y(pi+1E-50) = -4;

to try to define a formula for Y. There are ways you can express computing this, such as

 Y = @(x) (x==-1).*5 + (x==0).*-2 + (x==pi).*11 + (x==pi+1e-50).*-4
 
 Y = piecewise(x==-1, 5, x==0, -2, x==pi, 11, x==pi+1e-50, -4)   %requires symbolic toolbox

Your line of code

 Bs=zeros(1,10);

is creating Bs as a variable that holds 10 (double precision) values and is initialized to all 0. Variables can be indexed at positive integer locations, so Bs(1) = -(imag(yo)); would be valid, but this is not a formula.

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