let weight(kgs)/100 =x
then: fflow(wt) = 0.0118x^2 - 0.3454x + 343.43
let alt (fl)=a
then: fflow(alt)= 0.0012a^2 - 0.9261a + 355.21
What you've presumably done, compressor stall, is make quadratic approximations to fflow(x,a) in the two different dimensions of x and a. In doing so, you must have used an operating point in the other variable. i.e. your first equation must hold for a particular altitude a0, and your second must hold for a particular weight x0.
i.e.
fflow(x, a0) = 0.0118x^2 - 0.3454x + 343.43
fflow(x0,a)= 0.0012a^2 - 0.9261a + 355.21
fflow(x0, a0) must be the same calculated by either formula.
If your operating point is at a0 = 0, the remainder is fairly easy -- your x0 is about 49.5, to make fflow(x0, a0) = 355.21 both ways. (You clearly haven't used x0 = 0, as aircraft always weigh something!)
If that presumption is correct, then you can just add the correction coefficients for a about a0 into the first equation:
fflow(x, a) = 0.0118x^2 - 0.3454x + 343.43 + 0.0012a^2 - 0.9261a
That then comes up with the same value, 355.21 as both equations for fflow(49.5, 0).
If you've used an operating point other than a0=0, you probably want to go back and try again -- or at least tell us what it is! In general, you need to expand in differences around your operating point:
fflow(x, a0) = p*(x-x0)^2 + q* (x-x0) + r
fflow(x0, a) = s*(a-a0)^2 + t* (a-a0) + r
So then:
fflow(x, a) = p*(x-x0)^2 + q*(x-x0) +s*(a-a0)^2 + t*(a-a0) + r
Note that fflow(x0, a0) = r in all cases. If you felt really brave you could throw in a cross term (x-x0)*(a-a0), but there may be little correlation.
Worth bearing in mind that because you've ignored higher orders in (x-x0) and (a-a0), the approximation starts to break down as you move away from the operating point.
HTH
We crossed:
Anyway the first lot of data is at 5000kgs.
Ie. x0 = 50. Not a bad guess then!
