Group maths on for loop

I am trying to solve this

omegas=0:1:10;
RR=40:70;
d=90;
for i=1:length(omegas)
    for c=1:length(RR)
        gg=d+(omegas.*RR);
    end
end

for every value of omegas, I want to calculate gg. The idea is that gg should be a 31 by 11 matrix. Means that each omegas is calculated with all values of RR

[Stephen23]

DARLINGTON ETAJE 's "Answer" moved here:

Guys...nice try...it's still not working...maybe because my question excluded something. see the real code below

Thet=5:10005;
Theta=Thet';
R=15;
r=5;
epsilon=100;
delta=(R-r)*(epsilon/100);
for omega_b1=1:9
   for t=1:10001
        x_b1(omega_b1,t)=(delta.*cos(t'.*omega_b1))+(r.*cos(t'.*Theta));
        y_b1(omega_b1,t)=(delta.*sin(t'.*omega_b1))-(r.*sin(t'.*Theta));
        Vb_x1(omega_b1,t)=- (r.*Theta).*sin(t'.*Theta) - ((((R - r)*epsilon).*omega_b1).*sin(t'.*omega_b1))/100;
        Vb_y1(omega_b1,t)=- (r.*Theta).*cos(t'.*Theta) + ((((R - r)*epsilon).*omega_b1).*cos(t'.*omega_b1))/100;
        ab_x1(omega_b1,t)=- (r.*(Theta.^2)).*cos(t'.*Theta) - ((((R - r)*epsilon).*(omega_b1^2)).*cos(t'.*omega_b1))/100;
        ab_y1(omega_b1,t)=(r.*(Theta.^2)).*sin(t'.*Theta) -((((R - r)*epsilon).*(omega_b1^2)).*sin(t'.*omega_b1))/100;
   end
end

[Guillaume]

Since Theta is a vector, each expression, such as:

delta.*cos(t'.*omega_b1) + r.*cos(t'.*Theta)

is a vector. This can't be stuffed into the scalar x_b1(omega_b1,t). So, it's not clear what you're trying to calculate. Maybe x1 should be a function of Theta as well? (i.e. 3D)

It's also unclear why t is transposed (t' everywhere), since t is scalar. The transpose doesn't do anything.

Perhaps you started with code which had no loops (where the transpose would have made sense) but if you don't give us the right starting point, it's hard to help you.

[DARLINGTON ETAJE]

I was just trying to show that each value of t has a corresponding Theta. Both are 10001 in length. I just want to get answers for all the variables in the left in the format of 9 by 10001

[DARLINGTON ETAJE]

qq=8:10008;
for m = 1:9
 for n = 1:10001
 A(m, n) = (m+n)+qq;
  end
end

this is an easier way of looking at the problem

[Guillaume]

this is an easier way of looking at the problem

Not really, it's exactly the same issue. qq is a vector, so you still have a vector that you're trying to store in a scalar.

However, I've understood what you're asking from your description. Sometimes, explaining in words is better than broken code.

Answer 1

[Adam]

omegas=0:1:10;
RR=40:70;
d=90;

numOmegas = numel( omegas );
numRR = numel( RR );
gg = zeros( numOmegas, numRR );
for i=1:numOmegas
    for c=1:numRR
        gg(i,c) =d+(omegas(i).*RR(c));
    end
end

You could do it perfectly fine without a loop at all though too probably.

[DARLINGTON ETAJE]

please the codes with the real problem below

Answer 2

[Guillaume]

The vectorised version of what you intended with your loop (but didn't achieve):

R=15;
r=5;
epsilon=100;
delta=(R-r)*(epsilon/100);
Theta = 5:10005;        %a ROW vector
omega_b1 = (1:9)';      %a COLUMN vector. The two vectors must be along different dimensions

x_b1 = delta*cos(omega_b1) + r*cos(Theta);   %automatic expansion of compatible arrays
y_b1 = delta*sin(omega_b1) - r*sin(Theta);
Vb_x1 = -r*Theta.*sin(Theta) - (R - r)*epsilon*omega_b1.*sin(omega_b1)/100;
Vb_y1 = -r*Theta.*cos(Theta) + (R - r)*epsilon*omega_b1.*cos(omega_b1)/100;
ab_x1 = -r*Theta.^2.*cos(Theta) - (R - r)*epsilon*omega_b1.^2.*cos(omega_b1)/100;
ab_y1 = r*Theta.^2.*sin(Theta) - (R - r)*epsilon*omega_b1.^2.*sin(omega_b1)/100;

I've removed a lot of unnecessary brackets.

For how this works, see compatible array sizes