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Re: [ferret_users] @din integration with spatially limit


Yes, You will need the @WEQ transformation along with the indefinat integral transform @IIN

Example 3 here shows how to get salinity on an isotherm


Now, reather than salinity, you want the value of the indefinite integral


let isotherm_26 = temp[Z=@WEQ:26]
let  myvar = var[z=@iin]                           !!! Actually this might be wrong. There is a new transform in the latest Ferret let myvar=var[gz=var@iin] or something similar might be better.
let integrand_26 = myvar * isotherm_26

and so on.

If you computed d_26 earlier and no longer have the temperature field then we nee to create a new variable based on the depth field ad find the zero of the depth-d26
See example 5 for something similar.

let zero_at_26 = z[g=integrand] - d26
let isotherm_26= zero_at_26[z=@WEQ:0]

Oh yeah, You might need to replace missing values of the integrand with zero in case the temperature is >26 all the way to the sea floor.


On 20/08/18 16:23, Satyesh Ghetiya wrote:
Dear all,

  I want to calculate integration for a variable Integrand till depth of one isotherm say 26 degree .

let d26=temperature[z=@loc:26]

 My variable Integrand is having x,y,z and t dimension and d26 is having x,y and t dimension. I want to integrate Integrand  in z dimension from 0 to d26. So I want,

Integration=Integrand[z=0:`d26`@din] at each x,y and t.  From this Ferret gives message/error that "d26 must be a scalar" which is expected. I don't want to use d26 by its each dimension as this is lengthy process. 

Is there any way I can tell Ferret to integrate Integrand at each place till `d26` ?

Hope to get suggestions about this.

G Satyesh

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