Finding Corresponding X Value for Y value
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Collin McKenzie
am 21 Feb. 2022
Bearbeitet: Les Beckham
am 21 Feb. 2022
Looking to find the corresponding X value for the maximum force P in my force vs. displacement graph.
P=zeros(1,750);
theta=zeros(1,750);
L=zeros(1,750);
F=zeros(1,750);
W=10;
g=9806;
for x = 1:750
theta(x)=atand(400/x);
L(x)=sqrt((x^2)+(400^2));
F(x)=W*g*cosd(theta(x))*500;
P(x)=F(x)/L(x);
end
plot(1:750,P)
xlabel('x in mm')
ylabel('P in kg/mm/s^2')
Pmax= max(P);
fprintf('The maximum P value in kg/mm/s^2 is: %0.2f \n',Pmax);
0 Kommentare
Akzeptierte Antwort
Les Beckham
am 21 Feb. 2022
Bearbeitet: Les Beckham
am 21 Feb. 2022
If you ask for two outputs from the max() function you can find the index of your peak:
P=zeros(1,750);
theta=zeros(1,750);
L=zeros(1,750);
F=zeros(1,750);
W=10;
g=9806;
for x = 1:750
theta(x)=atand(400/x);
L(x)=sqrt((x^2)+(400^2));
F(x)=W*g*cosd(theta(x))*500;
P(x)=F(x)/L(x);
end
plot(1:750,P)
xlabel('x in mm')
ylabel('P in kg/mm/s^2')
[Pmax, index] = max(P);
fprintf('The maximum P value in kg/mm/s^2 is: %0.2f at x = %d mm\n', Pmax, index);
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Structures finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!