how do I find the x-axis value from my yline intercept on my acceleration curve
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I have an acceleration curve and i need to find the 0-60mph (0-26.8224m/s) Stuck on how to get there, currently only have a yline that shows where 26.8224m/s along the y-axis. Please help! Here's my code and a picture of the graph:
figure(6)
P1 = plot(tout,yout(:,1));
xlabel('Time (s)')
ylabel('Velocity (m/s)')
grid on
set(P1,'linewidth',2)
title('Acceleration Curve')
yline(26.8224)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1527291/image.jpeg)
2 Kommentare
Dyuman Joshi
am 2 Nov. 2023
@Sam, what is the relation between velocity and time? Do you have an expression/equation?
Akzeptierte Antwort
Voss
am 1 Nov. 2023
% some data:
tout = (0:60).';
yout = 10*log(tout+1);
y_val = 26.8224;
% your plot:
P1 = plot(tout,yout(:,1),'linewidth',2);
xlabel('Time (s)')
ylabel('Velocity (m/s)')
grid on
title('Acceleration Curve')
yline(y_val)
% find the t value where y is y_val:
t0 = interp1(yout(:,1),tout,y_val)
% plot a vertical red line to that point on the curve:
hold on
plot([t0,t0],[0,y_val],'--r')
4 Kommentare
Voss
am 3 Nov. 2023
Linear interpolation gives a point on the curve MATLAB plotted. If there's a non-linear function underlying that curve, use an interpolation method suitable to that function in your judgment.
Dyuman Joshi
am 3 Nov. 2023
But you chose a non-linear function, then why did you use linear interpolation?
Also, the point obtained by linear interpolation is not correct -
syms t
eqn = 10*log(t+1) - 26.8224 == 0;
t = solve(eqn, t)
vpa(t)
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Colormaps 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!