What resolution are you looking for?  Would an 8 bit result do?  If so
with a little cleverness you may be able to use 16 bit integer math.  My
company measures ships.  We do it in 24 bit integer millimeters,
anything from a rowboat to a supertanker!  We don't use a PIC, we use a
network of ten PICs.

Sherpa Doug

> -----Original Message-----
> From: Tom Mariner [mailto:tmariner@OPTONLINE.NET]
> Sent: Friday, July 06, 2001 8:33 AM
> To: PICLIST@MITVMA.MIT.EDU
> Subject: Re: [PIC]: Scaling of a curve
>
>
> Hello Andrew,
>
> Your numbers fit a third order polynomial quite well and if
> your problem was
> to add a value to each point, you would be in business (just
> modify the "a"
> term). Frankly, the only way I know that will work in all cases is to:
>
> a) represent your curve by an equation (in this case a polymonial)
> b) get back to your original numbers by reversing the equation.
> c) modify the y intercepts in the desired manner.
> d) re calculate the polynomial coefficients.
>
> We are doing this in a Pic 17C756A. You can use 24 bit float for the
> polynomial, but will need 32 bits for the solver.
>
> I guess I got to ask the most important questions of how big
> a Pic? and how
> much time do you have? (Both real time and coding time)
>
> On the other hand, you could use whatever technique you choose for you
> original curve and then scale the y values after they are computed.
>
> Tom Mariner
>
> > -----Original Message-----
> > From: pic microcontroller discussion list
> > [mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Drew Vassallo
> > Sent: Sunday, July 01, 2001 09:54 AM
> > To: PICLIST@MITVMA.MIT.EDU
> > Subject: [PIC]: Scaling of a curve
> >
> >
> > Math geniuses of PICLIST:
> >
> > I have a series of 40 points that represent a curve.  It's
> > not a linear
> > curve.  In fact, it doesn't necessarily even have a positive slope
> > throughout.  What I want to do is to create a scaling that
> is variable
> > depending on the position on the curve.  It's probably best
> to give an
> > example:
> >
> > point 40 = 100 (dec.)
> > point 30 = 80
> > point 20 = 75
> > point 10 = 70
> > point 1 = 35
> >
> > Let's say at point 30, we want to add an additional 10% to
> > the value.  So, I
> > want to add an additional 10% also at all the other points (3
> > or 4 at point
> > 1 and 10 at point 40).  This will, of course, change the
> > characteristics of
> > the curve as a whole, but it is necessary in this application.
> >
> > Any quick ways?  My initial idea was to take the percentage
> and try to
> > multiply it out, but I'm not sure how this would work over
> the entire
> > percentage range (i.e., I could be adding as little as 1% or
> > as much as
> > 200%) and I would probably need some sort of floating point
> > math for any
> > sort of precision at the lower values.
> >
> > My other idea was just to take the endpoints, find the ratio
> > between them,
> > and linearly scale the intermediate points, even though
> > that's not entirely
> > accurate, but it's better than nothing.
> >
> > Thanks,
> > --Andrew
> > _________________________________________________________________
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