Lpc2000

There's no easy way to interpolate non-orthogonal data. You must first 
orthogonate it, at leas in one axis. Then use simple PWL method. I am 
using same procedure for calculating temperature compensation of 
pressure sensors.

Here's simple 2D aproximation algorithm (in matlab). Check SPAPS 
documentation for more details.

function [coefs,fn,err,breakx,breaky]=bivsp(x,y,z,dx,dy,kx,ky)
sp = spap2({kx,ky},[dx dy],{x,y},z);
pp = fn2fm(sp,'pp');
coefs =fnbrk(pp,'coefs');
breaks =fnbrk(pp,'breaks');
l =fnbrk(pp,'l');
k =fnbrk(pp,'order');
d =fnbrk(pp,'dim');
breakx =breaks{:,1};
breaky =breaks{:,2};
err=norm(z-spval(sp,{x,y}))^2;
m = length(breaks);
breaksx=breaks{:,1};
breaksy=breaks{:,2};
v = coefs;
sizev = [d,l.*k];
dd = prod(sizev(1:m));
v=reshape(v,dd*l(m),k(m));
for i=1:1:l(m) %scan all number of x segments
  sizesegx=k(1)*l(1);
  a(:,:,i)=v((i-1)*sizesegx+1:(i-1)*sizesegx+sizesegx,:);
end
v=[];
for i=1:1:l(m) %scan all number of x segments
  for p=1:1:l(1)
    for j=1:1:k(1) %scan all number of y segments
      u(j,:,i,p)=a((j-1)*l(1)+p,:,i);
    end
    v=[v;reshape(fliplr( flipud(u(:,:,i,p)))',k(1)*k(m),1)'];
  end
end
fn=v;



I hope that helps.

Onestone wrote:

> Somebody's doing fuel maps from the look of this, however fuel maps just
> aren't linear the way yours are. You may as well just set the start
> point and the end point thus:-
>              0      3500
> 0         12      61
> 20       17      52
> 40       18      151
> 60       23      198
> 80       44      121
> 100     56       140
>
> Then for example2300(rpm?) at 55kPa can easily be converted as follows:-
>
> for 40kpa, 2300rpm point:
>
> F1 = (151-18) * 2300/3500. Of course you would precalculate most of this
> so:-
> F1 =  133/3500 * 2300. It makes sense to precalulate the increment /rpm,
> ie 133/3500
> F1 =  0.038. In 0:16 (unsigned decimal format) this is equal to 0x09BA
> (2490)
> F1 = 09BA * 2300 =  0x57.6318 (.038 * 2300 = 87.4)
>
> Now repeast this for the second 60(kPa) row where rpm factor = 175/3500
> = 0.05 = 3277 or 0x0CCD
> hence F2 =  0x0073.01CC (0.05 * 2300 = 115)
>
> Now to interpolate the kpa part do (F2-F1)/(60-40)*(55-40)
>
> Again you could speed this up by pre-storing the 60-40 or divide by 20
> as a binary decimal. Which happens to be 0.05 or 0x0CCD in 0:16 format 
> so:-
>
> F2 - F1 =  * 15 =  0x001B.9EB4 * 15 (this orders maintains lowest
> calculation error)  = 0x019E.4C8C
>
> Now divide by 20 or multiply by 0x0.0CCD = 0x0014.B759DC1C => 20.716.
>
> Finally add the 40kPa 2300 rpm value to this, keeping track of the DP =
> 0x006C.1A71DC1C. returning to the land of 32 bits this is equivalent to
> 108.103. Honestly a LOT simpler than it looks. With your tables having a
> hump you would only need to track the sign of results, which is possibly
> simpler than tracking signed arithmetic. Anyhow it comes down very 
> simply:_
>
> For each kPa point store the difference between 0RPM value and 3500RPM
> value divided by 3500 as a binary decimal.
> This gives you 6 values in your scenario, call them K0 - K5, call the
> pressures at these values P'0 - P'5
>
> read RPM (R), read kPa (P)(obviously if I have guessed wrongly
> substitute whatever the axes represent)
> if P exactly matches one of the stored values Kn simply multiply RPM by
> Kn else
> multply R * K(P'n<P) = Ix
> multiply R * K(P'm>P) = Iy
>
> next calculate  Ix + (Iy - Ix) * (P-P'n)/(P'm-P'n)
>
> Might look a bit long winded, but by prestoring a lot of the calculable
> values you can run the calcualtions very fast.
>
> Al
>
>
>
>
> Rolf Bredemeier wrote:
>
> > Hi,
> >
> > i'm searching for an example for 2d linear interpolation.
> > The data is in an array unsigned char [10][400].
> >
> > Here is my "worksheet", may be, it help's to explain for i looking on:
> > http://www.nur-solutions.de/tmp/2dinterpol.gif
> >
> > There is the way to solve  the prob. for x-value 55 and y-value 800 =
> > 59,35
> >
> > I need no floting point, unsigned char is only what i need.
> >
> > Best regards, Rolf
> >
> >
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