Using the right values for the variables, function gives a completely wrong answer

Hello,
I'm afraid of this being a rather simple solution, but I am out of ideas here.. I've been setting up a calcuation in matlab, and I spotted a function giving a completely wrong answer despite of all variables being correct:
clear all; close all;
%%
% Constants used for this calculation
g=9.81; % [m/s2] gravitational constant
rho_w=1000; % [kg/m3] water density
nu=10^-6; % [m2/s] water kinematic viscosity
k=0.4; % [-] Von Karman constant
% Additionally considering the following:
T=6; % [s] wave period
H=1.2; % [m] wave height
D=20; % [m] water depth
%% Problem 2.1
% Calculating wavenumber, wavelength, and celerity at this depth:
syms k
omega = 2*pi/T
omega = 1.0472
Linear = omega^2 == g*k*tanh(k*D)
Linear = 
k_initial = omega^2/g % using deep water wave number was an initial condition
k_initial = 0.1118
sol_k=vpasolve(Linear,k,k_initial)
sol_k = 
0.11413692045541882681977686977527
L=2*pi/sol_k
L = 
My issue is specifically with the very last line here. When checking each variable up to that point one by one, it is clear that pi=3.1416, sol_k=0.1141369, which should give 2*3.1416/0.1141369=55.05. Instead matlab is telling me it's 17.523.
I've been sitting here for some time and I am confused on how this can even be possible here, considering it is a simple division.

1 Comment

"Instead matlab is telling me it's 17.523. "
No, MATLAB is telling you that it is 17.523* pi:
If you want a purely numeric value without pi, then use DOUBLE(L)

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 Accepted Answer

You forgot that 17.523 is multiplied by pi in the calculation of L.

4 Comments

Oh gosh.. before posting a had a bad feeling this might end like this. Thank you, sir!
A quick followup - is there a reason why should it add a pi at the very end instead of multiplying with it, when asked? I had three more lines after this, where pi is used, but the results there are given multiplied by it instead.
For this reason, I didn't even think of checking at the very end of the long result number.
Use
sol_k = double(vpasolve(Linear,k,k_initial))
instead of
sol_k = vpasolve(Linear,k,k_initial)
Then you won't have trouble with long numbers and pi's.
is there a reason why should it add a pi at the very end instead of multiplying with it, when asked?
You told MATLAB to perform the calculations symbolically so it gave you a symbolic answer rather than returning a double precision approximation to that symbolic value. As others have said, if you want to see the double precision approximation call double on the result or you could use sympref to change how MATLAB displays the symbolic expression.
two = sym(2)
two = 
2
root2pi = sqrt(two)*pi
root2pi = 
sympref('FloatingPointOutput', true); % The default is false
root2pi_num = sqrt(two)*pi
root2pi_num = 
4.4429
d = double(root2pi)
d = 4.4429

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R2022b

Asked:

on 3 Dec 2022

Commented:

on 4 Dec 2022

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