# arc question about implementing angle using linspace

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Amal Fennich on 17 Mar 2020
Edited: Adam Danz on 19 Mar 2020
r_angl = linspace(pi/4, 3*pi/4, N);
what does this mean? because I have an angle of 3.433 degree and I do not know how to implement it here ! any help ?

madhan ravi on 17 Mar 2020
what is your question? could you rephrase it?
Adam Danz on 19 Mar 2020
"what does this mean?"
Check out the linspace page in the documentation. That line is basically creating N points between (1/4)pi (45 deg) and (3/4)pi (135 deg).
• deg = rad * 180/pi
• rad = deg * pi/180

madhan ravi on 17 Mar 2020
if you are looking to convert radians to degrees use the function rad2deg() function

#### 1 Comment

Amal Fennich on 17 Mar 2020
I have to draw an arc in Matlab which is going to represent half of airfoil .
the slope of the arc is dy/dx=(69.722-x^2+x-0.25)^-1/2(-x+1/2)
theta =3.433 degree
R=2.35
center is (0.5,-2.335)
someone has posted this code for arc and I have made slite changes on it but I do not know if I have resepted the requirements that I have :
N = 40; % Number Of Points In Complete Circle
r_angl = linspace(pi/4, 3*pi/4, N); % Angle Defining Arc Segment (radians)
xy_r = circr(radius,r_angl); % Matrix (2xN) Of (x,y) Coordinates
figure(1)
plot(xy_r(1,:), xy_r(2,:)) % Draw An Arc