integration with limits using an array
7 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I am trying to integrate an equation and it works when I do not have limits but when I add them it does not work.
u_wake is a 1x56 array, rho is a constant and 0.07 and 0.13 are the limit boundaries.
This works when I use this:
Drag = rho*trapz(u_wake.*(21.15-u_wake));
But not when I do this:
Drag = rho*trapz(u_wake.*(21.15-u_wake),0.07,0.13);
3 commentaires
the cyclist
le 6 Oct 2022
It's unclear to me why you can't just upload u_wake, which is just a 1x56 vector.
Also, that is not the complete error message, which I am guessing looks something more like ...
Error using matlab.internal.math.getdimarg
Dimension argument must be a positive integer scalar within indexing range.
Error in trapz>getDimArg (line 90)
dim = matlab.internal.math.getdimarg(dim);
Error in trapz (line 36)
dim = min(ndims(y)+1, getDimArg(dim));
Error in wake_scan_sample (line 35)
Drag = rho*trapz(u_wake.*(21.15-u_wake),0.07,0.13);
Réponses (1)
the cyclist
le 6 Oct 2022
Looking at the documentation for trapz, it doesn't seem like the syntax you are trying to use is valid.
2 commentaires
the cyclist
le 6 Oct 2022
I assume that u_wake is the "y" value of the integral. Are 0.07 and 0.13 the endpoints of the "x" value? Are the values of u_wake at evenly spaced points along x? Then
Drag = rho*trapz(linspace(0.07,0.13,length(u_wake)),u_wake.*(21.15-u_wake));
should calculate what you want. For example,
% I just made up some u_wake and rho data
u_wake = sort(rand(1,56));
rho = 1;
Drag = rho*trapz(linspace(0.07,0.13,length(u_wake)),u_wake.*(21.15-u_wake))
Voir également
Catégories
En savoir plus sur Numerical Integration and Differentiation dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!