I have a quadratic equation (for a power curve) which produces a value for efficiency at a fixed number of Watts:
(1) $e = E_0 + E_1 x + E_0 x^2$
I have a second quadratic equation which should work for any value of $x$ to produce the same value for efficiency, using a reference nominal value for $x$ and $e$, $x_{nom}$ and $e_{nom}$ respectively:
$e = [E_0 + E_1 (x/x_{nom}) + E_2(x/x_{nom})^2]e_{nom}$
So where $x = 3400$ Watts in (1):
$e = 0.642388 -1.62\cdot 10^{-4} x + 2.26\cdot 10^{-8} x^2 = 0.353$
What I need to do is make so that using (2), where:
$x_{nom} = 3400$
$e_{nom} = 0.353$
$x = 3400$
produces a value for $e$ of $0.353$ but I don't know what to do to the coefficients of the equation to make this true.
So my first question is is there sufficient information here to do what I'd like? And second, if so what do I need to do to the coefficients to achieve the expected result?
I am not sure if I properly understood; so, forgive me if I am off-topic.
Let us start with $$e = E_0 + E_1\, x + E_2\, x^2=E_0+E_1\, x_{0}\Big(\frac{x}{x_{0}}\Big)+E_2 \,x_{0}^2\, \Big(\frac{x}{x_{0}}\Big)^2$$ and $$e_0=E_0+E_1\, x_{0}+E_2 \,x_{0}^2 $$ So $$e=\frac{E_0+E_1\, x_{0}\Big(\frac{x}{x_{0}}\Big)+E_2 \,x_{0}^2\, \Big(\frac{x}{x_{0}}\Big)^2}{E_0+E_1\, x_{0}+E_2 \,x_{0}^2}e_0$$
Is this what you wanted to see?