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seif eldin Ahmed on 10 Jan 2021
Answered: Alan Stevens on 10 Jan 2021
%calculating velocity
V3 = .1; % velocity of the piston in m/sec
%global c
th2 = pi/6; % initial value for th2
th3 = pi/6; %initial value for th3
% c= 28.8:0.2:52.6;
C = 28.8:0.2:52.6;
for ii = 1:length(C)
% c = C(ii);
th3(ii) = acos((-1425 - (C(ii)).^2)./(110.*C(ii)));
th2(ii) = acos((C(ii).^2 - 4625)./(-4400));
M = [-40*sin(th2(ii)) C(ii)*sin(th3(ii)); 40*cos(th2(ii)) -C(ii)*cos(th3(ii))];
N = [V3*cos(th3(ii)); V3*sin(th3(ii))];
W(:,ii) = M\N;
end
W2 = W(1,:);
W3 = W(2,:);
figure;
subplot(2,1,1); plot(th2,W2,'r','linewidth',2);grid
subplot(2,1,2); plot(th2,W3,'r','linewidth',2);grid
%Acceleration calculation
V3 = .1; % acceleration of the piston in m/sec
%V3dot = V3 we substituite with V3 not V3dot
%global c
th2 = pi/6; % initial value for th2
th3 = pi/6; %initial value for th3
% c= 28.8:0.2:52.6;
C = 28.8:0.2:52.6;
for ii = 1:length(C)
% c = C(ii);
th3(ii) = acos((-1425 - (C(ii)).^2)./(110.*C(ii)));
th2(ii) = acos((C(ii).^2 - 4625)./(-4400));
M = [40*sin(th2(ii)) -C(ii)*sin(th3(ii)); 40*cos(th2(ii)) -C(ii)*cos(th3(ii))];
N = [-V3*cos(th3(ii))-40*W2(ii)^2*cos(th2(ii))+C(ii)*W3(ii)^2*cos(th3(ii))+V3*W3(ii)*sin(th3(ii)); V3*sin(th3(ii))+V3*W3(ii)*cos(th3(ii))-C(ii)*W3(ii)^2*sin(th3)+40*W2(ii)^2*sin(th2(ii))];
A(:,ii) = M\N;
end
A2 = A(1,:);
A3 = A(2,:);
figure;
subplot(2,1,1); plot(th2,A2,'r','linewidth',2);grid
subplot(2,1,2); plot(th2,A3,'r','linewidth',2);grid
seif eldin Ahmed on 10 Jan 2021

Alan Stevens on 10 Jan 2021
You need
N = [-V3*cos(th3(ii))-40*W2(ii).^2*cos(th2(ii))+C(ii)*W3(ii).^2*cos(th3(ii))+V3*W3(ii)*sin(th3(ii)); V3*sin(th3(ii))+V3*W3(ii)*cos(th3(ii))-C(ii)*W3(ii).^2*sin(th3(ii))+40*W2(ii).^2*sin(th2(ii))];
If you compare this with your expression you will see that you have th3 instead of th3(ii) in a couple of places.

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