How can I write a loop to evaluate theta at every two degrees from 0-360?

4 visualizzazioni (ultimi 30 giorni)
n=(360/Dth)+1
theta=0
for I=1:2:n
R=12
w=(2*pi/5)
Md=50
Thetarads= theta*pi/180
Id=(.5*Md*R^2)
Mp=75
Mb=30
alpha=0
u=.8
L=6*R
theta=0
Dth=2
thetarads=theta*pi/180
rx=(R)*cos(theta)-L*cos(180)-L
ry=(R*sin(theta))-(L*sin(180))-(R*sin(theta))
vx=(-R)*(w)*sin(theta)
vy=(R)*(w)*cos(theta)
alpha=0
ax=(-R)*alpha*sin(theta)-(R*(w^2)*cos(theta))
ay=(R)*alpha*cos(theta)-(R*w^2*sin(theta))
theta=theta+Dth
end
  2 Commenti
Roger Stafford
Roger Stafford il 11 Giu 2016
There are a number of errors or questionable items in your code:
1) Inside the for-loop you set theta = 0 so that it will never change.
2) You write sin(theta) and sin(180), and theta and (I suspect) 180 are in degrees, whereas ‘sin’ uses radian measure.
3) In the line
ry=(R*sin(theta))-(L*sin(180))-(R*sin(theta))
the R*sin(theta) terms cancel each other.
4) You set alpha = 0, so why use it in the lines for ax and ay?
5) Why have you defined Md, Id, Mp, Mb, u. They are never used.
Emily Gobreski
Emily Gobreski il 11 Giu 2016
Modificato: Emily Gobreski il 11 Giu 2016
thank you so much! I have no idea what I am doing. I really appreciate your time and help! what do i need to do to make sure theta changes?

Accedi per commentare.

Risposte (1)

Steven Lord
Steven Lord il 11 Giu 2016
You probably want simply to use the degree-based trig functions like sind, cosd, etc. instead of sin and cos.
  2 Commenti
Emily Gobreski
Emily Gobreski il 11 Giu 2016
thank you! how can i write it in radians and still account for the degree change?
Chad Greene
Chad Greene il 11 Giu 2016
Steven's suggesting an alternative function. If you're working in degrees, you can use sind(theta) or you can use sin(theta*pi/180). They will both give the same result.

Accedi per commentare.

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by