Hi Jim, Dawn -
No, I don't think Jim is correct. If a reference pressure is used, then you
must use potential temperature theta instead of temperature. It does not make
sense to calculate density at 0 pressure but with temperature at depth (where
the pressure affects the temperature). Instead, bring the water parcel
adiabatically to the surface by finding theta, then use your reference
pressure:
let theta=theta_fo(salt,temp,z[gz=temp],0) ! potential temperature theta
let sigma_theta=rho_un(salt,theta,0)-1000
See the documentation for the THETA_FO function. For sigma-2, use 2000
instead of zero in the above.
But I think Dawn is asking something different. Instead of z, she wants to
use pressure. Frankly, I don't know the answer to her question ("the results
are obviously wrong" doesn't give me much of a clue). However, I think she is
on the right track, because I believe the 3rd argument to THETA_FO really
should be pressure, not depth (the documentation is unclear on this point). I
suggest looking at the reference given under the THETA_FO function (Bryden
1973). Then tell us so the documentation can be fixed if necessary.
Billy K
On Jan 22, 2007, at 9:59 AM, James Orr wrote:
Dawn,
Your pressure in the 3rd argument of the rho_un function should be the
reference pressure:
0 for sigma_0
2000 for sigma_2
Cheers,
Jim
On Mon, 22 Jan 2007, ferret ocean wrote:
Dear ferreters,
I want to calculate the density with pressure instead of depth (more
accuate way), but when I get the pressure with the following scipt to
convert the depth to pressure,
let x0 = sin(y[g=v]*3.1415926/180);
let c1 = 5.92E-3 + x0*x0 * 5.25E-3;
let pres = ((1-c1)-(((1-c1)*(1-c1))-(8.84E-6 * z[g=v]))^0.5) / 4.42E-6;
then let rho=rho_un(salt,temp,pres)-1000;
But the results are obviously wrong. Do you have any suggestions?
Dawn