integration with limits using an array
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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);
댓글 수: 3
the cyclist
2022년 10월 6일
편집: the cyclist
2022년 10월 6일
It would be helpful if you uploaded the data, instead of just describing it. You can use the paper clip icon in the INSERT section of the toolbar.
Also, you should be more specific than saying, "it does not work". Does the code give an error message? (If so, what is the complete error message?) Or, do you just get a result you do not expect?
James Keiser
2022년 10월 6일
편집: James Keiser
2022년 10월 6일
the cyclist
2022년 10월 6일
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);
답변 (1개)
the cyclist
2022년 10월 6일
0 개 추천
Looking at the documentation for trapz, it doesn't seem like the syntax you are trying to use is valid.
댓글 수: 2
James Keiser
2022년 10월 6일
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))
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