how can i use a "while" loop to automatically increment omega?

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0 个投票

theta=120
sign=(cosd (theta)/abs (cosd (theta)))
u=.8
w=(2*pi/5)
R=1
c=4.05/12
H=18/12
y=H-(R*sind(theta/2))
Md=50/32.2
Id=(.5*Md*R^2)
Mp=75/32.2
Mb=30/32.2
alpha=0
u=.8
L=(6/12*R)
g=32.2
ax=(-R*w^2)*cosd(theta)
ab=(-R*w^2*cosd(theta))-(R*w^2*sind(theta))
ay=(-R*w^2)*sind(theta)
AA=[1 0 1 0 0 0 0 0 0; 0 1 0 1 0 0 0 0 0; 0 0 -R*sind(theta) R*cosd(theta) 0 0 0 0 0; 0 0 0 0 1 0 1 0 0; 0 0 0 0 0 1 0 1 0; 0 0 -1 0 -1 0 0 0 0; 0 0 -c 3*R -c -3*R 0 0 0; 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0]
CC=[0 Md*g Id*alpha 0 Md*g Id*alpha Mp*ax Mp*ay+Mp*g Mb*g]
xx=CC*AA
if (0<theta) & (theta<180), x=C*a
else xx=CC*AA
end
w = 2*pi/5;  
theta = 0:2:360

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Emily Gobreski 2016-6-17

This is the email I have from the professor. It might explain the process a little better:

Modify your program to have it solve the kinetoics problem for a full 360 degree rotation of the drive disk at a specified angular velocity omega (start with 2*pi/5 if you'd like). Use a theta increment of 2 degrees)

Increase omega until you find that NB <0 somewhere during the contact phase (e.g., box slides free of the push link). You can choose to manually adjust omega and re-run your program several times OR perhaps use a while loop to automatically increment omega.

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Answer 1

Star Strider 2016-6-17

在 MATLAB Online 中打开

1 个投票

I imagine by ‘omega’, you actually intend:

w=(2*pi/5);

A while loop is best used if you want to increment it and then test to see if a specific value meets a specific criterion, and stop the loop when it does.

If you just want to increment it over a range without testing, use a for loop instead.

I experimented with vectorising your code by making ‘omega’ a vector and using element-wise operations where I could. It works for some calculations, but it makes the ‘xx’ calculation impossible. A loop is likely the only way to go.

I could not run this because ‘C’ and ‘a’ are not defined. Correct that, then see if this works for you:

theta=120;
sign=(cosd (theta)/abs (cosd (theta)));
u=.8;
wv = linspace(0, (2*pi/5), 10);                          % Vector Of Values For ‘w’
R=1;
c=4.05/12;
H=18/12;
y=H-(R*sind(theta/2));
Md=50/32.2;
Id=(.5*Md*R^2);
Mp=75/32.2;
Mb=30/32.2;
alpha=0;
u=.8;
L=(6/12*R);
g=32.2;
for k1 = 1:length(wv)
    w = wv(k1);
    ax = (-R*w.^2)*cosd(theta);                             % Modified For Element-Wise Operations
    ab = (-R*w.^2*cosd(theta))-(R*w.^2*sind(theta));        % Modified For Element-Wise Operations
    ay = (-R*w.^2)*sind(theta);                             % Modified For Element-Wise Operations
      AA=[1 0 1 0 0 0 0 0 0; 0 1 0 1 0 0 0 0 0; 0 0 -R*sind(theta) R*cosd(theta) 0 0 0 0 0; 0 0 0 0 1 0 1 0 0; 0 0 0 0 0 1 0 1 0; 0 0 -1 0 -1 0 0 0 0; 0 0 -c 3*R -c -3*R 0 0 0; 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0];
      CC=[0 Md*g Id*alpha 0 Md*g Id*alpha Mp*ax Mp*ay+Mp*g Mb*g];
      xx=CC*AA;
      if (0<theta) & (theta<180), x{k1} = C*a;
      else xx{k1} = CC*AA
      end
  end

I left ‘x’ and ‘xx’ as cell arrays because I don’t know how large they are.

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Emily Gobreski 2016-6-17

It works! You're the best! Thank you so much!!!!!

Star Strider 2016-6-17

As always, my pleasure!

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