arc question about implementing angle using linspace

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]

what is your question? could you rephrase it?

[Adam Danz]

"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).

Check out madhan ravi's answer to convert deg<==>rad. Here's a summary:

• deg = rad * 180/pi
• deg = rad2deg(rad)
• rad = deg * pi/180
• rad = deg2rad(deg)

Answer 1

[madhan ravi]

if you are looking to convert radians to degrees use the function rad2deg() function

[Amal Fennich]

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 :

circr = @(radius,rad_ang,p0) [radius*cos(rad_ang)+0.5; radius*sin(rad_ang)-2.335]; % Circle Function For Angles In Radians % Circle Function For Angles In Radians

circd = @(radius,deg_ang) [radius*cosd(deg_ang); radius*sind(deg_ang)]; % Circle Function For Angles In Degrees

N = 40; % Number Of Points In Complete Circle

r_angl = linspace(pi/4, 3*pi/4, N); % Angle Defining Arc Segment (radians)

radius = 2.35; % Arc Radius

xy_r = circr(radius,r_angl); % Matrix (2xN) Of (x,y) Coordinates

figure(1)

plot(xy_r(1,:), xy_r(2,:)) % Draw An Arc

axis([-1.25*radius 1.25*radius 0 1.25*radius]) % Set Axis Limits

axis equal