How to make successive for loops?!

Question

Mariam Sheha 2016-4-1

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编辑： Mariam Sheha 2016-4-8

Ass_2_solu.m

Hey everybody;

i want to make 3 successive loops that the next loop start from the end of the previous ....can any one help me through that?!

code is attached

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Answer 1

Torsten 2016-4-8

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If you compare with my code, you'll notice that the use of i2 and i3 in these lines is not correct.

          v2(i2)=70.*w1;
          u2(i2)=70.*w12;
          x2(i2)= y1(i+1); 
          y2(i2)=y1(i+1); 
          v2(i3)=30.*w2;
          u2(i3)=30.*w22;
          x2(i3)=y1(i2+1); 
          y2(i3)=y1(i2+1);

Furthermore, I never introduced v2,u2,x2,y2.

Best wishes

Torsten.

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Mariam Sheha 2016-4-8

编辑：Mariam Sheha 2016-4-8

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Yes i got your point and i fixed out but the problem is that once i entered this lines in the loop it takes to long time (doesn't stop till i force close the program...) and if i remove it out of the loop, it warning out asking (what is i2?))

Noting that: i must introduce these parameters to work as an initial values for the second loop...

          v2(i2)=50.*w1;
          u2(i2)=50.*w12;
          x2(i2)= y1(i); 
          y2(i2)=y1(i);

Thanks a lot for your continuous help

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Answer 2

Image Analyst 2016-4-1

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It's basically like this. Adapt as needed:

for k1 = 1 : 4
  k1 % Echo to command window
end
for k2 = (k1+1): (k1 + 5);
  k2 % Echo to command window
end
for k3 = (k2+1): (k2 + 2);
  k3 % Echo to command window
end

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Mariam Sheha 2016-4-1

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Thanks a lot, would you please be patient to and help me...

i do the code as following but it doesn't work... can you tell me what is the problem

if true
  % code

% variable values w(1)= 45; % unit in degree u(1)= 100*cosd(w); v(1)= 100*sind(w); x(1)=0; y(1)=0;

% using Euler method;

% first step size h=0.005; t1=1000; % unit in sec w1= 45; % unit in degree u1(1)= 100*cosd(w1); v1(1)= 100*sind(w1); x1(1)=0; y1(1)=0;

        for i=1:t1;
          % position of the ball
          % y-axi
          y1(i+1)=y1(i)+(h*v1(i));
          %x-axis
          x1(i+1)=x1(i)+(h*u1(i));
          % air resistance
          z1(i)= ((50*R.^2)/m) *(sqrt(u1(i).^2 + v1(i).^2));
          % velocity in x and y
          v1(i+1)=v1(i)-(h*(g+(z1(i))*v1(i)));
          u1(i+1)=u1(i)-(h*z1(i)*(u1(i)));
          if y1(i+1)<0
            break
          end
     %%second solution     
     for i2=i+1:1000;
          v2(i2)=0.7*v1(i+1);
          u2(i2)=0.7*u1(i+1);
          x2(i2)= y1(i+1); 
          y2(i2)=y1(i+1); 
           % y-axi
          y2(i2+1)=y2(i2)+(h*v2(i2));
          %x-axis
          x2(i2+1)=x2(i2)+(h*u2(i2));
          % air resistance
          z2(i2)= ((50*R.^2)/m) *(sqrt(u2(i2).^2 + v2(i2).^2));
          % velocity in x and y
          v2(i2+1)=v1(i2)-(h*(g+(z2(i2))*v2(i2)));
          u2(i2+1)=u1(i2)-(h*z2(i2)*(u2(i2)));
        end 
        end
  end