"Index exceeds matrix dimensions" Error. solution please.

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h b 2016-11-4

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https://ww2.mathworks.cn/matlabcentral/answers/310590-index-exceeds-matrix-dimensions-error-solution-please

评论： h b 2016-11-11

x1=60;x2=40; y1=40; y2=60;
 d = sqrt((x2-x1)^2+(y2-y1)^2); 
  a = atan2(-(x2-x1),(y2-y1)); 
  b = asin(d/2/r); 
 c = linspace(a-b,a+b); 
  e = sqrt(r^2-d^2/4); 
  x = (x1+x2)/2-e*cos(a)+r*cos(c); 
  y = (y1+y2)/2-e*sin(a)+r*sin(c);
  axis([0 150 0 150])
 hold on
 plot(x,y,'r'); 
axis([0 150 0 150])
h1 = plot(x(1),y(1),'bs','MarkerSize',7,'MarkerFaceColor','r'); 
for n = 1:numel(t)
set(h1, 'XData', x(n), 'YData', y(n));   %HERE
drawnow %// refresh figure
end

The error is displayed in the highlighted line.

suggest solution. (what should be the value of t here)

Here, the program runs even without specifying the value of t. how is it possible?

Can the speed of the pointer be controlled (in terms of frequency/velocity) by the user given input (from dialog box)?

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Jan 2016-11-4

@h b: Please format your code using the "{} code" button. It is easier to help you, when the code is readable.

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Answer 1

Walter Roberson 2016-11-4

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https://ww2.mathworks.cn/matlabcentral/answers/310590-index-exceeds-matrix-dimensions-error-solution-please#answer_241913

You have an old t variable sitting around or passed in,and it is longer than your x is.

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h b 2016-11-7

编辑：Walter Roberson 2016-11-7

在 MATLAB Online 中打开

thanks. yes i had got the solution.

Could you rectify the error in the below mentioned code as the pointer stops without trailing the entire path.

    t=linspace(0,40,180)
  x3=linspace(54,54,180);% left
  y=linspace(0,40,180);
  a1=54;a2=44; b1=40; b2=60;
   d = sqrt((a2-a1)^2+(b2-b1)^2); % Distance between points
    a = atan2(-(a2-a1),(b2-b1)); % Perpendicular bisector angle
    b = asin(d/2/r); % Half arc angle
   c = linspace(a-b,a+b); % Arc angle range
    e = sqrt(r^2-d^2/4); % Distance, center to midpoint
    q = (a1+a2)/2-e*cos(a)+r*cos(c); % Cartesian coords. of arc
    z = (b1+b2)/2-e*sin(a)+r*sin(c);
    nt = numel(t);
  q(end+1:nt) = nan;
  z(end+1:nt) = nan;
    xa=linspace(44,0,180);
  ya=linspace(60,60,180);
  x111=[ x3 q xa ]
  y111=[ y z ya ]
  axis([0 150 0 150]);
  hold on;
h1 = plot(x111(1),y111(1),'bs','MarkerSize',7,'MarkerFaceColor','r'); %// plot initial position of object 1
for n = 1:numel(t)
set(h1, 'XData', x111(n), 'YData', y111(n)); %// update position of object 2
drawnow %// refresh figure
end

Walter Roberson 2016-11-11

Yes, it is possible to divide your output figure into cells of desired dimension. It is also a lot of work.

Easier is to calibrate your output rate to the virtual distance to be traveled.

See attached.

The first parameter to the function is r, the radius, which you never defined in your posting so I decided it must be something you wanted to adjust.

The second parameter is v, the velocity. The total distance to be traveled is about 120, so approximately 120/v will give the number of seconds the output will take to plot.

An easier version of the code would be to say that the distance between adjacent points is "close enough" to equal, and so to divide the calculated time by the number of points and advance equally. It turns out that the overhead of doing that gets fairly high for higher velocities (shorter wait times.) And because you are traveling on a curve, you should not really be calling the distances to be equal.

The version of the code I attach calculates the distances between adjacent points by Euclidean formula, totals the distances, uses the velocity to calculate times. Then, to allow the plot to proceed at constant linear velocity, it divides the time up into a number of chunks and figures out where on the curve (which sample) one would be at after that time. Then it plots with the trail increasing one chunk at a time.

Walter Roberson 2016-11-11

plot111.m

Forgot, sorry.

h b 2016-11-11

Thank you so much.

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