Finding Aerodynamic Center of an Airfoil: Assuming that the normal force $N$ and the moment $M_\frac{c}{4}$ acts at the quarter cord, the aerodynamic center lies somewhere behind this point. First let's assume that the aerodynamic center lies at $x_{AC}$ distance from the leading edge as shown in Figure 1 below.
Figure 1: Airfoil Diagram
Nomenclature
$c$: Chord Length
$N$: Normal Force
$M_{c/4}$: Moment about the quarter chord
$M_{AC}$: Moment about the aerodynamic center
$q_\infty$: Dynamic Pressure
$S$: Wing Planform Area
$C_M$: Coefficient of Moment
$C_L$: Coefficient of Lift
$\alpha$: Angle of Attack
Now let's sum the moment about the assumed aerodynamic center assuming counterclockwise direction is positive.
\[M_{AC}= M_{c/4} + (x_{AC}-\frac{c}{4}) \vec{N} \tag1\]
Divide all moment terms by the $q_\infty S c$.
\[\frac{M_{AC}}{q_\infty S c} = \frac{M_{c/4}}{q_\infty S c} + \frac{(x_{AC}-\frac{c}{4}) \vec{N}}{q_\infty S c}\tag2\]
By dividing all the moment terms by $q_\infty S c$, we get the coefficients.
\[C_{M_{AC}} = C_{M_{(c/4)}} + C_L \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag3\]
Differentiate the term with respect to $\alpha$. The reason for differentiating with respect to $\alpha$ is that the coefficient of moment about the aerodynamic center is invariant to the angle of attack.This makes it a constant; therefore, taking the derivative of the coefficients with respect to $\alpha$ we can set the above equation equal to zero and solve for the Aerodynamic Center.
\[\frac{dC_{M_{AC}}}{d\alpha} = \frac{dC_{M_{(c/4)}}}{d\alpha} + \frac{dC_L}{d\alpha} \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag4\]
Rewrite Equation 4:
\[0 = \frac{dC_{M_{(c/4)}}}{d\alpha} + \frac{dC_L}{d\alpha} \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag5\]
Divide by $\frac{dC_L}{d\alpha}$.
\[0 = \frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} + \frac{\Big[\frac{dC_L}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]}\Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag6\]
Rewrite Equation 6:
\[0 = \frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} + \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag7\]
Now, assume that the chord is airfoil is normalized. This lets us assume that the chord of the airfoil is equal to 1. We can now replace the $c/4$ in Equation 7 with $0.25$ and solve for $x_{AC}$.
\[x_{AC} = 0.25-\frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} \tag8\]
$N$: Normal Force
$M_{c/4}$: Moment about the quarter chord
$M_{AC}$: Moment about the aerodynamic center
$q_\infty$: Dynamic Pressure
$S$: Wing Planform Area
$C_M$: Coefficient of Moment
$C_L$: Coefficient of Lift
$\alpha$: Angle of Attack
Now let's sum the moment about the assumed aerodynamic center assuming counterclockwise direction is positive.
\[M_{AC}= M_{c/4} + (x_{AC}-\frac{c}{4}) \vec{N} \tag1\]
Divide all moment terms by the $q_\infty S c$.
\[\frac{M_{AC}}{q_\infty S c} = \frac{M_{c/4}}{q_\infty S c} + \frac{(x_{AC}-\frac{c}{4}) \vec{N}}{q_\infty S c}\tag2\]
By dividing all the moment terms by $q_\infty S c$, we get the coefficients.
\[C_{M_{AC}} = C_{M_{(c/4)}} + C_L \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag3\]
Differentiate the term with respect to $\alpha$. The reason for differentiating with respect to $\alpha$ is that the coefficient of moment about the aerodynamic center is invariant to the angle of attack.This makes it a constant; therefore, taking the derivative of the coefficients with respect to $\alpha$ we can set the above equation equal to zero and solve for the Aerodynamic Center.
\[\frac{dC_{M_{AC}}}{d\alpha} = \frac{dC_{M_{(c/4)}}}{d\alpha} + \frac{dC_L}{d\alpha} \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag4\]
Rewrite Equation 4:
\[0 = \frac{dC_{M_{(c/4)}}}{d\alpha} + \frac{dC_L}{d\alpha} \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag5\]
Divide by $\frac{dC_L}{d\alpha}$.
\[0 = \frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} + \frac{\Big[\frac{dC_L}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]}\Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag6\]
Rewrite Equation 6:
\[0 = \frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} + \Bigg[\frac{x_{AC}-\frac{c}{4}}{c}\Bigg] \tag7\]
Now, assume that the chord is airfoil is normalized. This lets us assume that the chord of the airfoil is equal to 1. We can now replace the $c/4$ in Equation 7 with $0.25$ and solve for $x_{AC}$.
\[x_{AC} = 0.25-\frac{\Big[\frac{dC_{M_{(c/4)}}}{d\alpha}\Big]}{\Big[\frac{dC_L}{d\alpha}\Big]} \tag8\]
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