Finding Corresponding X Value for Y value

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);

 采纳的回答

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);
The maximum P value in kg/mm/s^2 is: 61287.50 at x = 400 mm

更多回答(0 个)

类别

帮助中心File Exchange 中查找有关 Deep Learning Toolbox 的更多信息

产品

版本

R2020b

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by