why do I get the wrong angle
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why do i keep getting a negatie value for angle
1 Kommentar
Torsten
am 14 Jun. 2022
Why did you delete your question ?
Antworten (1)
David Hill
am 14 Jun. 2022
Bearbeitet: David Hill
am 14 Jun. 2022
mo=2.5;
y=1.4;
angle=10;
fun=@(a)((2*(1+((y-1)/2)*mo^2*(sind(a)).^2))./(((y+1)*mo^2*sind(a).*cosd(a))))-tand(a-angle);
A=fzero(fun,32);
3 Kommentare
Sarah Alhabbas
am 14 Jun. 2022
David Hill
am 14 Jun. 2022
You could also plot and see that there are two solutions between (0 - 90).
mo=2.5;
y=1.4;
angle=10;
fun=@(a)((2*(1+((y-1)/2)*mo^2*(sind(a)).^2))./(((y+1)*mo^2*sind(a).*cosd(a))))-tand(a-angle);
x=3:.1:87;
plot(x,fun(x));
David Hill
am 14 Jun. 2022
Bearbeitet: David Hill
am 14 Jun. 2022
When you graph the function you see solutions should be around 30 and 80. fzero requires an initial guess. 32 is not an exact solution.
mo=2.5;
y=1.4;
angle=10;
fun=@(a)((2*(1+((y-1)/2)*mo^2*(sind(a)).^2))./(((y+1)*mo^2*sind(a).*cosd(a))))-tand(a-angle);
A=fzero(fun,30)
mo=2.5;
y=1.4;
angle=10;
fun=@(a)((2*(1+((y-1)/2)*mo^2*(sind(a)).^2))./(((y+1)*mo^2*sind(a).*cosd(a))))-tand(a-angle);
A=fzero(fun,80)
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