Oh, sorry  my mistake

Not theta=x/y in my code but theta=atan(x/y)


2016년 9월 5일 월요일 오후 9시 22분 49초 UTC+9, Jean-Paul Pelteret 님의 말:
>
> Hi Kyusik,
>
> r is calculated by "r=sqrt(x^2+y^2) ", and theta is calculated by 
>> "theta=x/y" 
>
>
> At the very least, your definition of theta in the (radial) coordinate 
> system is incorrect. See the wikipedia page 
> <https://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Cartesian_coordinates>
>  for 
> information as how to convert from Cartesian coordinates to radial 
> coordinates.
>
> Regards,
> J-P
>
> On Monday, September 5, 2016 at 2:04:20 PM UTC+2, hank...@gmail.com 
> <javascript:> wrote:
>>
>>  Hello. 
>>
>> I'm studying step20
>>
>> I tried to change the Eq in step20 a little.
>>
>> So, I change the element of matrix Inverse K
>>
>> What I changed is like... (in 3D)
>>
>> K_Inv[0][0]=2*r/(cos(theta)*cos(theta)), 
>> K_Inv[1][1]=2*r/(sin(theta)*sin(theta)), K_Inv[2][2]=r
>>
>> r is calculated by "r=sqrt(x^2+y^2) ", and theta is calculated by 
>> "theta=x/y"
>>
>> But, as you can expect that ,on the points where |x|<0.0000001 or 
>> |y|<0.000001,  cos(theta) or sin(theta) is almost zero. So, It seems that 
>> It causes the singularity.
>>
>> So, At first I tried to change the above 2 element in K_Inv like this...
>>
>> K_Inv[0][0]=2*r/(cos(theta)*cos(theta)+del)
>>  K_Inv[1][1]=2*r/(sin(theta)*sin(theta)+del),
>> (where del is very small value)
>>
>> So, the calculation is converged, But the result is not good. Because, on 
>> the plot, The azimuthal symmetry is broken. even though The Eq I try to 
>> calculate(Grad-Shaf Eq) is independent of theta.
>>
>> I attached the result what I made.(What I'm trying to see is pressure 
>> that is noted by p, I don't need to see u,v or w )
>>
>> Is there simple way for me to avoid this singularity error?
>>
>> Please  help me....
>>
>> Best Regards.
>>
>> Kyusik. 
>>
>

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