interpolation between two curves

Hi, I'd like to do an interpolation to draw three curves between curves of 900 and 1100K. the curves I need are at 950,1000,1050K. thanks,

Answer 1

[Matt J]

INTERP1 should work well here,

 A =[     NaN         900        1100        1500
          -2     -4.4175     -5.1954     -6.1525
          -1     -4.3845      -5.186     -6.1478
       -0.02     -4.0357      -5.186     -6.1525
           0     -2.0603     -5.1624     -6.1525
        0.02     -1.2729     -5.1389     -6.1525
        0.04     -1.0136     -5.0776     -6.1525
        0.06    -0.88162     -4.9408     -6.1525
        0.08     -0.7779     -4.2667     -6.1525
           1    -0.70718     -3.1823     -6.1525
        1.02    -0.69304      -2.923     -6.1289
        1.06    -0.65061     -2.6873     -6.0346
        1.08    -0.64589      -3.088     -5.7753
           2    -0.63175     -2.5694      -5.219
        2.02    -0.61289     -2.5459     -4.9267
        2.06    -0.61289     -2.5223     -4.6674
           3    -0.61289     -2.4846     -4.5495]

T=[900,950,1000,1050,1100,1500];

Anew =[[nan,T]; [A(2:end,1),...
      interp1([900,1100,1500],A(2:end,2:end).',T).']];

Answer 2

[Andrei Bobrov]

Please use griddedInterpolant.

Let A your data:

A =[     NaN         900        1100        1500
          -2     -4.4175     -5.1954     -6.1525
          -1     -4.3845      -5.186     -6.1478
       -0.02     -4.0357      -5.186     -6.1525
           0     -2.0603     -5.1624     -6.1525
        0.02     -1.2729     -5.1389     -6.1525
        0.04     -1.0136     -5.0776     -6.1525
        0.06    -0.88162     -4.9408     -6.1525
        0.08     -0.7779     -4.2667     -6.1525
           1    -0.70718     -3.1823     -6.1525
        1.02    -0.69304      -2.923     -6.1289
        1.06    -0.65061     -2.6873     -6.0346
        1.08    -0.64589      -3.088     -5.7753
           2    -0.63175     -2.5694      -5.219
        2.02    -0.61289     -2.5459     -4.9267
        2.06    -0.61289     -2.5223     -4.6674
           3    -0.61289     -2.4846     -4.5495]

[X,Y] = ndgrid(A(2:end,1),A(1,2:end));
F = griddedInterpolant(X,Y,A(2:end,2:end),'cubic');

use:

x = 0; % log(P)
y = 1000; % T = 1000 K
out = F(x,y)

[rana saleh]

Thank you ur answer. but, when I tried to replace x and Y such that F= giddedInterpolant(0,1000,A(2:end,2:end),'cubic');

it gives me message: The size of the grid point arrays should match the size of the value arrays!!

[rana saleh]

If you can please tell me all the code and also the plot function I'd be grateful

[Andrei Bobrov]

Example (full code):

A =[     NaN         900        1100        1500
          -2     -4.4175     -5.1954     -6.1525
          -1     -4.3845      -5.186     -6.1478
       -0.02     -4.0357      -5.186     -6.1525
           0     -2.0603     -5.1624     -6.1525
        0.02     -1.2729     -5.1389     -6.1525
        0.04     -1.0136     -5.0776     -6.1525
        0.06    -0.88162     -4.9408     -6.1525
        0.08     -0.7779     -4.2667     -6.1525
           1    -0.70718     -3.1823     -6.1525
        1.02    -0.69304      -2.923     -6.1289
        1.06    -0.65061     -2.6873     -6.0346
        1.08    -0.64589      -3.088     -5.7753
           2    -0.63175     -2.5694      -5.219
        2.02    -0.61289     -2.5459     -4.9267
        2.06    -0.61289     -2.5223     -4.6674
           3    -0.61289     -2.4846     -4.5495]

[X,Y] = ndgrid(A(2:end,1),A(1,2:end));
F = griddedInterpolant(X,Y,A(2:end,2:end),'cubic');

log_P = -3:.5:3; % log(P)
T = 900:50:1100;

[P,T_arr] = ndgrid(log_P,T);
out = F(P,T_arr);

plot(log_P(:),out);

[rana saleh]

Thank you so much for your answer. I made a printscreen of applying the code. I found all the curves are straightlines? I don't know what is the wrong?

[Andrei Bobrov]

my typo, corrected

[rana saleh]

Hi, Thank you for your answer , but this is what I found .. Any further help?