Just beware that you need a data type which is large enough to store
x^2 to do this. You might be better doing c*x first and then
multiplying by x again. That minimizes the range of values which your
variables need to store.

Sean


On Thu, Oct 30, 2008 at 6:48 AM, Matthew Rhys-Roberts <matt@nu-ins.com> wrote:
> Hi Gerhard
>
> I figured out that
> x^2/309023 is approx. equal to
> (x^2 / 2^16) / 4.72, which is approx. equal to
> [(x^2 / 2^16) * 13885] / 65536 !!
>
> The graph gives a nice enough fit in OpenOffice Calc, so it should be OK
> when the PIC spits it out.  :)
>
> Thanks for your encouragement.
>
> Best regards,
> Matt
>
>
> Gerhard Fiedler wrote:
>>>> I'm calculating a quadratic expression, y=a+bx+cx^2, using PIC18
>>>> assembler, as simply as possible in binary.
>>>>
>>>> Calculating bx is easy;
>>>> b=-0.0407;
>>>> 0.0407x == 2667x/65536. (just make a 32bit product, then chop off the
>>>> bottom 16 bits).
>>>>
>>>> However:
>>>> c = 3.236e-6, or 1/309023.
>>>> My range of x^2 is between 171,793,449 and 3,310,820,528.
>>>> 171,793,449/309023 is about 556; 3,310,820,528/309023 is about 10714.
>>>>
>>>> Now my brain hurts. How can I calculate cx^2 as easily as I did bx?
>>>>
>> Try using the same approach... one thing that's missing here is how precise
>> you need it. Based on this, you look for a multiplier/dividend pair that
>> approximates 1/309023. Depending on your precision needs, you can chop off
>> low-order bits from the squared x before multiplying and dividing, and thus
>> be able to fit it into 32-bit math.
>>
>> Gerhard
>>
>>
>
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