Why are the symbolic units not evaluated by a square root?
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clear, clc
u = symunit;
d = .025*u.m; %Automatically convert to meters
l = .6*u.m;
A = pi*d^2/4;
g = 9.81*u.m/u.s^2;
E = 207e9*u.Pa;
I = pi*d^4/64;
lambda = 76.5e3*u.N/u.m^3;
omega_n = vpa((pi/l)^2*(g*E*I/(A*lambda))^(1/2),4)
When I plug this in, the units stay the same inside the square root.
Why aren't they evaluated?
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Accepted Answer
Star Strider
on 12 Jan 2020
It appears that the square root is still expressed as a square root, so the units remain the same within it:
omega_n =
27.42*(1037.0*(([Pa]*[m]^6)/([N]*[s]^2)))^(1/2)*(1/[m]^2)
Doing this:
[omega_n,omega_n_u] = separateUnits(omega_n)
produces this:
omega_n =
882.80792016869907152094706300452
omega_n_u =
1*((1/[N]^(1/2)*[Pa]^(1/2)*[m])/[s])
So the units convert correctly. It is necessary to specifically request the conversion.
3 Comments
Star Strider
on 12 Jan 2020
My pleasure.
The reason it did not work the first time is that the units will not convert until you request that they be converted. This likely makes things easier for the Symbolic Math Toolbos, so that it does not have to convert and then re-convert interim results, converting them only when requested at the end.
More Answers (1)
David Goodmanson
on 13 Jan 2020
Edited: David Goodmanson
on 13 Jan 2020
Hi Lucas,
try your original code but with the last two lines
omega_n = (pi/l)^2*(g*E*I/(A*lambda))^(1/2)
omega_n = vpa(simplify(omega_n),4)
omega_n = 882.8*(1/[s])
Then you don't have to do any conversions yourself, e.g Pa to SI.
3 Comments
Paul
on 15 Mar 2024
Maybe I'm missing something, but I don't unerstand what the issue is. One extra call to simplify seems to solve the problem. What is the downside of that?
u = symunit;
d = .025*u.m; %Automatically convert to meters
l = .6*u.m;
A = pi*d^2/4;
g = 9.81*u.m/u.s^2;
E = 207e9*u.Pa;
I = pi*d^4/64;
lambda = 76.5e3*u.N/u.m^3;
omega_n = vpa((pi/l)^2*(g*E*I/(A*lambda))^(1/2),4)
omega_n = simplify(omega_n)
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