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Videowriter, issues with making a video

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Lissa
Lissa el 3 de En. de 2021
Respondida: Walter Roberson el 5 de En. de 2021
Hello.
We have some troubles with making a video by using the function VideoWriter. Are we not referring to it correctly? Thanks!
This is our code of the function:
function [AnimationFigure] = SimulationVraag5Function(TDnumber,condition,trialnumber,tibialength,x_knee,y_knee)
% SimulationVraag5Function
% SimulationVraag5Function is a function that will create a plot to visualize
% the pendulum test in a stick figure
% Input arguments
% TDnumber: fill in number of subject, f.e. for TD3 fill in '3'
% condition: fill in '1' if condition is HR, fill in '0' if condition is MR
% trialnumber: fill in number of trial HR/MR, f.e. for MR5 fill in '5'
% tibialength: fill in length of tibia of the subject [in cm]
% x_knee: fill in x-co of the knee (we considered the knee to be a static
% point, the tibia was able to move during the pendulum test) [in cm] (fill in a value that fits in coordinate system of -70 by 70)
% y_knee: fill in y-co of the knee [in cm] (fill in a value that fits in coordinate system of -70 by 70)
% Output arguments
% AnimationFigure: an animation figure of the knee during pendulum test will be made, we considered other segments to be static at a chosen coordinate by us
SimulationVraag5Video = VideoWriter('SimulationVraag5Video.avi');
open(SimulationVraag5Video);
% IMPORT DATA (for this exercise we need IK data)
startpath ='/MATLAB Drive/Assignment_V2/MBEXAMEN/EXAMEN/Data';
Datapath = [startpath '/Data'];
addpath '/MATLAB Drive/Assignment_V2/MBEXAMEN/EXAMEN'
% right/left subjects HR
if condition == 1 % this is the first main loop, condition = 1 stands for HR trials
for i = TDnumber % TDnumber was filled in (see function), we will retrieve all the files for specific TDnumber
dataFolder = [startpath '/TD' num2str(i)];
dataFile = fullfile(dataFolder, '/IK/HR*.mot');
pathwayIK_HR = dir(dataFile);
% loop left subjects HR
if i == 3 | i == 5 | i == 7 | i == 9 | i == 14 % making a distinction between left/right leg tested subjects, 3/5/7/9/14 are left leg tested
x_hip = x_knee+30 % since the left leg is tested, we want the x-coordinates of the other joints (hip, shoulder, head) to be positive (will be to the right of knee x-co), so we add a number to the x-co of the knee (to get a relative visual representation)
y_hip = y_knee+10 % we want the y-coordinates of the other joints (hip, shoulder, head) to be positive (will be above knee y-co)
x_shoulder = x_knee+35;
y_shoulder = y_knee+50;
x_midhead = x_knee+35;
y_midhead = y_knee+60;
dataFolder = [startpath '/TD' num2str(i)];
for k = trialnumber % getting the files for the specific HR trial
% data HR
IKHRdata = importdata(fullfile(pathwayIK_HR(k).folder,pathwayIK_HR(k).name));
DataHRlinks = IKHRdata.data(:,strcmp('knee_angle_l',IKHRdata.colheaders)); % get the knee_angle_l because left leg testings
TimeHRlinks = IKHRdata.data(:,strcmp('time',IKHRdata.colheaders));
kneeangle = IKHRdata.data(:,strcmp('knee_angle_l',IKHRdata.colheaders))
for a = 1:100:length(kneeangle) % we chose steps of 30 between the rows, otherwise the running would take a long time
% computing the lengths of the triangle: femur length (length between knee and hip) and length between hip and ankle
lengthbetweenkneehip = sqrt(((x_knee-x_hip)^2) + (y_knee-y_hip)^2) % computing distance between two coordinates (co hip and co knee are known)
lengthbetweenanklehip = sqrt(tibialength^2 + lengthbetweenkneehip^2 - (2*tibialength * lengthbetweenkneehip * cos(kneeangle(a)))) % law of cosines in a triangle
lengthbetweenanklehipsquared = lengthbetweenanklehip^2 % to get rid of the root square on the other side of the equation
% vgl1 = (xhip-xankle)^2 + (yhip-yankle)^2 - lengthbetweenanklehipsquared
lengthbetweenkneeankle = tibialength % this is known (requested in the funtion as input argument)
lengthbetweenkneeanklesquared = tibialength^2 % to get rid of the root square on the other side of the equation
% vgl2 = (xknee-xankle)^2 + (yknee-yankle)^2 - lengthbetweenkneeanklesquared
syms x_ankle y_ankle
E = [((x_hip-x_ankle)^2 + (y_hip-y_ankle)^2 - lengthbetweenanklehipsquared == 0), ((x_knee-x_ankle)^2 + (y_knee-y_ankle)^2 - lengthbetweenkneeanklesquared == 0)]; % two equations are set to zero, two equations so the two unknown variables of x_ankle and y_ankle will be solved
S = solve(E,x_ankle,y_ankle)
x_ankle = S.x_ankle % there are two possible values for x_ankle
y_ankle = S.y_ankle
x(1) = x_knee
x(2) = x_ankle(1)
y(1) = y_knee
y(2) = y_ankle(1)
xlim([-100 100]); % we chose the width of the plot, do not fill in coordinates that are outside of the plot size
ylim([-100 100]);
hold on; % hold on so all the plots will be created
plot(x_hip,y_hip,'o'); % plot the hip coordinates as a circle
plot([x_hip x_knee], [y_hip y_knee]) % plotting a line, we connected the hip co with the knee co
% we had to write the x and y co seperatly in a vector
text(x_hip,y_hip,'H') % plotting a text next to the circle of the hip coordinate
plot(x_shoulder,y_shoulder,'o'); % same procedure as explained above
plot([x_shoulder x_hip],[y_shoulder y_hip])
text(x_shoulder,y_shoulder,'S')
plot([x_shoulder x_midhead],[y_shoulder y_midhead])
plot(x_midhead,y_midhead,'o','linewidth',20)
text(x(1),y(1),'K')
text(x(2),y(2),'E')
grid on;
plot(x,y) % plotting the line between the requested knee coordinate (see function)
% and the ankle angle computed above
comet(x,y) % with this function the curve will be followed, so it looks like a drawing
frame = getframe(gcf);
writeVideo(SimulationVraag5Video,frame);
end
end
else % loop for right subjects HR
x_hip = x_knee-30 % we subtract a number from the x-knee coordinate so that the joints will be on the left side of the x-knee coordinate (because right leg tested)
y_hip = y_knee+10
x_shoulder = x_knee-35;
y_shoulder = y_knee+50;
x_midhead = x_knee-35;
y_midhead = y_knee+60;
for k = trialnumber
IKHRdata = importdata(fullfile(pathwayIK_HR(k).folder,pathwayIK_HR(k).name));
DataHRlinks = IKHRdata.data(:,strcmp('knee_angle_r',IKHRdata.colheaders)); % get the knee_angle_r because right leg testings
TimeHRlinks = IKHRdata.data(:,strcmp('time',IKHRdata.colheaders));
kneeangle = IKHRdata.data(:,strcmp('knee_angle_r',IKHRdata.colheaders))
end
for a = 1:100:length(kneeangle)
% computing the lengths of the triangle: femur length (length between knee and hip) and length between hip and ankle
lengthbetweenkneehip = sqrt(((x_knee-x_hip)^2) + (y_knee-y_hip)^2) % computing distance between two coordinates (co hip and co knee are known)
lengthbetweenanklehip = sqrt(tibialength^2 + lengthbetweenkneehip^2 - (2*tibialength * lengthbetweenkneehip * cos(kneeangle(a)))) % law of cosines in a triangle
lengthbetweenanklehipsquared = lengthbetweenanklehip^2 % to get rid of the root square on the other side of the equation
% vgl1 = (xhip-xankle)^2 + (yhip-yankle)^2 - lengthbetweenanklehipsquared
lengthbetweenkneeankle = tibialength % this is known (requested in the funtion as input argument)
lengthbetweenkneeanklesquared = tibialength^2 % to get rid of the root square on the other side of the equation
% vgl2 = (xknee-xankle)^2 + (yknee-yankle)^2 - lengthbetweenkneeanklesquared
syms x_ankle y_ankle
E = [((x_hip-x_ankle)^2 + (y_hip-y_ankle)^2 - lengthbetweenanklehipsquared == 0), ((x_knee-x_ankle)^2 + (y_knee-y_ankle)^2 - lengthbetweenkneeanklesquared == 0)]; % two equations are set to zero, two equations so the two unknown variables of x_ankle and y_ankle will be solved
S = solve(E,x_ankle,y_ankle)
x_ankle = S.x_ankle % there are two possible values for x_ankle
y_ankle = S.y_ankle
x(1) = x_knee
x(2) = x_ankle(1)
y(1) = y_knee
y(2) = y_ankle(1)
xlim([-100 100]);
ylim([-100 100]);
hold on;
plot(x_hip,y_hip,'o');
plot([x_hip x_knee], [y_hip y_knee])
text(x_hip,y_hip,'H')
plot(x_shoulder,y_shoulder,'o');
plot([x_shoulder x_hip],[y_shoulder y_hip])
text(x_shoulder,y_shoulder,'S')
plot([x_shoulder x_midhead],[y_shoulder y_midhead])
plot(x_midhead,y_midhead,'o','linewidth',20)
plot(x,y);
text(x(1),y(1),'K')
text(x(2),y(2),'E')
grid on;
comet(x,y);
frame = getframe(gcf);
writeVideo(SimulationVraag5Video,frame);
end
end
end
end
if condition == 0 % MR trials
for i = TDnumber
dataFolder = [startpath '/TD' num2str(i)];
dataFile = fullfile(dataFolder, '/IK/MR*.mot');
pathwayIK_MR = dir(dataFile);
% for loop for left subjects MR trials
if i == 3 | i == 5 | i == 7 | i == 9 | i == 14
x_hip = x_knee+30
y_hip = y_knee+10
x_shoulder = x_knee+35;
y_shoulder = y_knee+50;
x_midhead = x_knee+35;
y_midhead = y_knee+60;
for k = trialnumber
IKMRdata = importdata(fullfile(pathwayIK_MR(k).folder,pathwayIK_MR(k).name));
DataMRlinks = IKMRdata.data(:,strcmp('knee_angle_l',IKMRdata.colheaders));
TimeMRlinks = IKMRdata.data(:,strcmp('time',IKMRdata.colheaders));
kneeangle = IKMRdata.data(:,strcmp('knee_angle_l',IKMRdata.colheaders));
end
for a = 1:30:length(kneeangle)
% computing the lengths of the triangle: femur length (length between knee and hip) and length between hip and ankle
lengthbetweenkneehip = sqrt(((x_knee-x_hip)^2) + (y_knee-y_hip)^2) % computing distance between two coordinates (co hip and co knee are known)
lengthbetweenanklehip = sqrt(tibialength^2 + lengthbetweenkneehip^2 - (2*tibialength * lengthbetweenkneehip * cos(kneeangle(a)))) % law of cosines in a triangle
lengthbetweenanklehipsquared = lengthbetweenanklehip^2 % to get rid of the root square on the other side of the equation
% vgl1 = (xhip-xankle)^2 + (yhip-yankle)^2 - lengthbetweenanklehipsquared
lengthbetweenkneeankle = tibialength % this is known (requested in the funtion as input argument)
lengthbetweenkneeanklesquared = tibialength^2 % to get rid of the root square on the other side of the equation
% vgl2 = (xknee-xankle)^2 + (yknee-yankle)^2 - lengthbetweenkneeanklesquared
syms x_ankle y_ankle
E = [((x_hip-x_ankle)^2 + (y_hip-y_ankle)^2 - lengthbetweenanklehipsquared == 0), ((x_knee-x_ankle)^2 + (y_knee-y_ankle)^2 - lengthbetweenkneeanklesquared == 0)]; % two equations are set to zero, two equations so the two unknown variables of x_ankle and y_ankle will be solved
S = solve(E,x_ankle,y_ankle)
x_ankle = S.x_ankle % there are two possible values for x_ankle
y_ankle = S.y_ankle
x(1) = x_knee
x(2) = x_ankle(1)
y(1) = y_knee
y(2) = y_ankle(1)
xlim([-100 100]);
ylim([-100 100]);
hold on;
plott = plot(x,y);
grid on;
comet(x,y);
frame = getframe(gcf);
writeVideo(SimulationVraag5Video,frame);
end
else % right subjects MR trials
x_hip = x_knee-30
y_hip = y_knee+10
x_shoulder = x_knee-35;
y_shoulder = y_knee+50;
x_midhead = x_knee-35;
y_midhead = y_knee+60;
for k = trialnumber
IKMRdata = importdata(fullfile(pathwayIK_MR(k).folder,pathwayIK_MR(k).name));
DataMRlinks = IKMRdata.data(:,strcmp('knee_angle_r',IKMRdata.colheaders));
TimeMRlinks = IKMRdata.data(:,strcmp('time',IKMRdata.colheaders));
kneeangle = IKMRdata.data(:,strcmp('knee_angle_r',IKMRdata.colheaders));
end
for a = 1:30:length(kneeangle)
% computing the lengths of the triangle: femur length (length between knee and hip) and length between hip and ankle
lengthbetweenkneehip = sqrt(((x_knee-x_hip)^2) + (y_knee-y_hip)^2) % computing distance between two coordinates (co hip and co knee are known)
lengthbetweenanklehip = sqrt(tibialength^2 + lengthbetweenkneehip^2 - (2*tibialength * lengthbetweenkneehip * cos(kneeangle(a)))) % law of cosines in a triangle
lengthbetweenanklehipsquared = lengthbetweenanklehip^2 % to get rid of the root square on the other side of the equation
% vgl1 = (xhip-xankle)^2 + (yhip-yankle)^2 - lengthbetweenanklehipsquared
lengthbetweenkneeankle = tibialength % this is known (requested in the funtion as input argument)
lengthbetweenkneeanklesquared = tibialength^2 % to get rid of the root square on the other side of the equation
% vgl2 = (xknee-xankle)^2 + (yknee-yankle)^2 - lengthbetweenkneeanklesquared
syms x_ankle y_ankle
E = [((x_hip-x_ankle)^2 + (y_hip-y_ankle)^2 - lengthbetweenanklehipsquared == 0), ((x_knee-x_ankle)^2 + (y_knee-y_ankle)^2 - lengthbetweenkneeanklesquared == 0)]; % two equations are set to zero, two equations so the two unknown variables of x_ankle and y_ankle will be solved
S = solve(E,x_ankle,y_ankle)
x_ankle = S.x_ankle % there are two possible values for x_ankle
y_ankle = S.y_ankle
x(1) = x_knee
x(2) = x_ankle(1)
y(1) = y_knee
y(2) = y_ankle(1)
xlim([-100 100]);
ylim([-100 100]);
hold on;
plott = plot(x,y);
grid on;
comet(x,y);
frame = getframe(gcf);
writeVideo(SimulationVraag5Video,frame);
end
end
end
end
close(SimulationVraag5Video);
this is the code of our other script:
SimulationVraag5Function(3,1,1,30,0,0)
  1 comentario
Lissa
Lissa el 3 de En. de 2021
This is the error that we get:
Error using VideoWriter/writeVideo (line 368)
Frame must be 448 by 336
Error in SimulationVraag5Function (line 95)
writeVideo(SimulationVraag5Video,frame);
Error in SimulationVraag5 (line 4)
SimulationVraag5Function(3,1,1,30,0,0)
It does makes a file for the video in the current folder.

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Respuestas (1)

Walter Roberson
Walter Roberson el 5 de En. de 2021
https://www.mathworks.com/matlabcentral/answers/329710-how-to-save-a-figure-s-content-as-image#comment_1241428 explains

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