RE: Table lookup & interpolation [follow-up]

[Ted Harding]

On 19-Mar-03 Ted Harding wrote:
> For this case, log(p(h)) is quite close to a linear function of h
> (there is slight curvature). Possibly, linear interpolation of
> log(p) between points may be adequate. If not, quadratic interpolation
> using three consecutive points almost certainly is adequate.

I just checked: using quadratic interpolation of log(p), the
following compares p(h) with phat(h) where the points used for
interpolation are (h-1000,p(h-1000)), (h+1000,p(h+1000)),
(h+2000,p(h+2000)). The octave commands were:

y=log(p);c=[];
for i=2:29,
x=h(i);
h1=h(i-1);h2=h(i+1);h3=h(i+2);
y1=y(i-1);y2=y(i+1);y3=y(i+2);
phat=exp((x-h2)*(x-h3)*y1/((h1-h2)*(h1-h3))+ ...
         (x-h1)*(x-h3)*y2/((h2-h1)*(h2-h3))+ ...
         (x-h1)*(x-h2)*y3/((h3-h1)*(h3-h2)));
c=[c;[p(i) phat]];endfor;c

Now the agreement between p(h) and phat(h) is very close indeed.

    p(h)   phat(h)
  2040.90  2040.85
  1967.70  1967.75
  1896.70  1896.65
  1827.70  1827.68
  1760.80  1760.89
  1696.00  1695.96
  1633.00  1632.91
  1571.90  1572.09
  1512.90  1512.76
  1455.40  1455.45
  1399.80  1399.79
  1345.90  1345.89
  1293.70  1293.72
  1243.20  1243.19
  1194.30  1194.36
  1147.00  1146.89
  1101.10  1101.22
  1056.90  1056.82
  1014.00  1014.06
   972.60   972.56
   932.50   932.52
   893.80   893.79
   856.40   856.46
   820.30   820.19
   785.30   785.39
   751.70   751.66
   719.20   719.26
   687.90   687.82

Ted.


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E-Mail: (Ted Harding) <address@hidden>
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Date: 19-Mar-03                                       Time: 09:34:11
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