Dividing a sine function with scalar values - why do i get a complex double vector?
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Hello all,
please have a look on the following code:
if true
% %%ZUWEISUNG NEUER VARIABLEN
Gemessene Daten
delta_v = Delta_V_40_20;
v_ohne_d = V_400x2D200x25GeschwindigkeitAlteFederOhneD0xE4mpfer_Red;
v_mit_d = V_400x2D200x25GeschwindigkeitAlteFederMitD0xE4mpfer_Red;
a_ohne_d = a_400x2D200x25BeschleunigungAlteFederOhneD0xE4mpfer_Red;
a_mit_d = a_400x2D200x25BeschleunigungAlteFederMitD0xE4mpfer_Red;
Berechnungsdaten
m = 16.6; % Masse [kg]
b = 170; % Länge [mm]
c = 30; % Länge [mm]
x2 = 200; % Länge [mm]
x1 = 301; % Länge vom Drepunkt zum Schwerpunkt
BERECHNUNG VERÄNDERLICHE LÄNGE DES DÄMPFERS
a = sqrt(b^2+c^2-2*b*c*cos(Winkel))-b;
BERECHNUNG KRAFT F1, F2, F3
y = asin((c*sin(Winkel))./a);
F1 = m * delta_a;
F2 = F1 * x1/x2;
F3 = F2 ./ sin(y); % y meint den winkel omega end
why is F3 a complex double vector? I´d like to have F3 over the angle y.
Thanks
3 Comments
Accepted Answer
Wayne King
on 10 Oct 2012
Edited: Wayne King
on 10 Oct 2012
The issue is that in your asin(), you have values outside of [-1,1] that is what is causing your output, y, to be complex-valued.
For example:
asin(2)
Real-valued sin() and cos() are bounded by 1, that is not true for complex-valued sin() and cos(), so when you give asin() a value outside of [-1,1], MATLAB correctly returns a complex number.
3 Comments
Walter Roberson
on 8 Mar 2022
MATLAB has no implied multiplication so you need .* between the ) and the (
More Answers (1)
Wayne King
on 10 Oct 2012
Edited: Wayne King
on 10 Oct 2012
I do not see that you show us how you get Winkel in cos(Winkel), so we cannot reproduce the issue, but I suspect that your expression
a = sqrt(b^2+c^2-2*b*c*cos(Winkel))-b;
is producing a complex-valued output because you are taking the sqrt() of a negative number.
Please provide Winkel if that is not the case.
2 Comments
Wayne King
on 10 Oct 2012
Edited: Wayne King
on 10 Oct 2012
You have values that you are feeding asin() that are outside the interval, [-1 ,1], see my answer below. Asking for the inverse sine of a number outside of [-1,1] correctly results in a complex number.
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