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The airfoil force-axis transformation expresses a section-plane aerodynamic force in chord-fixed axes or flow-fixed axes. Choose $x$ aft along the airfoil chord, $z$ toward the upper surface, and a flow direction with body components $(\cos \alpha ,\sin \alpha )$. Taking positive lift along $(-\sin \alpha ,\cos \alpha )$ givesHere $A$ and $N$ are the chord-axial aerodynamic force and chord-normal aerodynamic force. The convention for angle and axis direction must be stated before applying the signs. NASA's definitions make drag parallel and lift perpendicular to the flow.^[ref]

The rotation matrix is orthogonal: $R^{T}R=I$. Its inverse is therefore its transpose, givingIt also preserves $A^2+N^2=D^2+L^2$, a useful numerical check on a conversion. The two component pairs need not individually agree.

An original conversion uses measured $D=10$ N and $L=100$ N at $\alpha =10{}^{\circ }$. It gives $A\simeq -7.517$ N and $N\simeq 100.217$ N. Rotating these values back recovers the original flow-axis loads. Setting $A=D$ because the angle appears small would miss the term $L\sin \alpha$, which is large compared with the small drag in this example.

Angle uncertainty can likewise mix the large lift component into inferred drag. Holding body-axis readings fixed,For the example, an angular error of $0.1{}^{\circ }\simeq 0.001745$ radians contributes approximately 0.175 N to inferred drag. This is a coordinate-sensitivity calculation, not a prediction of the aerodynamic change caused by physically rotating the model. Changing its actual incidence can change the underlying load as well as the axes.

The same rotation applies to the lift coefficient, drag coefficient, chord-normal force coefficient and chord-axial force coefficient only when they share the same pressure and area normalization.

The chord-axial force coefficient is the dimensionless chord-axial aerodynamic force:Here $S$ is the stated aerodynamic reference area, $\rho$ the reference mass density and $U$ the reference speed. The chord axis and its positive direction accompany the normalization. A signed coefficient can be negative even when the flow-relative drag coefficient is positive.

An original example uses $\rho =1.2$ kg/m${}^3$, $U=25$ m/s, $S=0.40$ m${}^2$ and aft-directed load $A=9$ N. Then $q=375$ Pa and $C_A=9/(375\cdot 0.40)=0.060$. Reporting that load with area 0.80 m${}^2$ instead gives 0.030. The reference-area scaling of aerodynamic coefficients changes the reported coefficient without changing the balance reading.

For a section, axial load per unit span $A^{\prime }$ is instead normalized as $c_a=A^{\prime }/(qc)$ using airfoil chord $c$. The denominator $qc$ has units N/m, whereas $qS$ has units N. A per-span load divided by a finite area does not produce the same dimensionless quantity.

With common references, the inverse airfoil force-axis transformation isFor stipulated $C_D=0.040$, $C_L=0.60$ and $\alpha =5{}^{\circ }$, this gives $C_A\simeq -0.01245$. The subtraction shows why labeling a body-axis coefficient as drag can change even the sign of the apparent result. NASA defines lift and drag relative to the flow, which is the physical basis for keeping these axes distinct.^[ref]

At fixed area, density and measured force, a small fractional speed error contributes approximately $-2\delta U/U$ to the coefficient's fractional error. Normalization consequently does not replace measurement calibration or an uncertainty assessment of the reference flow.

In the airfoil section plane, the chord-normal aerodynamic force $N$ is the component of an aerodynamic force perpendicular to the airfoil chord. The chord-axial aerodynamic force $A$ points along the chord, chosen positive aft. These body directions differ from the flow-relative lift and drag axes.^[ref]

Choose $z$ toward the upper surface and let the incoming flow direction have body components $(\cos \alpha ,\sin \alpha )$. Resolving the same vector in the two orthonormal bases gives the airfoil force-axis transformation,The rotation matrix preserves the squared Euclidean norm: $A^2+N^2=D^2+L^2$. An original example with $A=20\mathrm{N}$, $N=100\mathrm{N}$ and $\alpha =30{}^{\circ }$ gives $D\simeq 67.32\mathrm{N}$ and $L\simeq 76.60\mathrm{N}$. Calling the 100-newton normal reading lift would confuse two different components. The transformation is kinematic; it does not assert that a particular wing produces those loads at that angle.

For a pressure-only reconstruction, the chord-normal component is $N_p^{\prime }=\int (p_l-p_u)dx$ per unit span when upper and lower surfaces are graphs over the chord. The total $N$ can also include viscous stress. A force balance resolving body axes and a pressure reconstruction therefore need matched axes and a stated load decomposition before they are compared.