Effective angle of attack of a wing

Question

Is the angle of attack of a rectangular wing influenced by the dihedral ($\Gamma$) of the wing? If e.g. a wing exists with $\Gamma$ = 0$^{o}$ it has angle of attack $\alpha$. If the dihedral is set at $\Gamma$ = 89$^o$ (fictional example, no practical value) I suspect that the effective angle of attack should be:

$$ \alpha_{eff} = F(\Gamma) \cdot \alpha $$

Such that $\alpha_{eff}$ is close to zero and with $F(\Gamma)$ being a factor dependent on the dihedral angle. Is there such a relation between angle of attack and dihedral?

Edit:

The answers given did sound logical, but after fiddling around with simple 3D drawings I still can't understand that dihedral has no effect on the angle of attack.

Here's a zero degree dihedral flat plate wing at 0$^{o}$, 20$^{o}$ and 40$^{o}$ angle of attack.

Here's a wing with an 80$^{o}$ dihedral and again 0$^{o}$, 20$^{o}$ and 40$^{o}$ angle of attack.

I understand that the lift vector tilts with the dihedral, but in the last picture I also clearly see that the wing does not have an angle of attack of 40$^{o}$ w.r.t. the free stream. Sure the middle part does (as it is unnaffected by the dihedral) but the left and right part certainly don't. Imagine if the diherdal would be 90$^{o}$ then it would be just a vertical plate tilted back with no angle of attack.

So my question is still: how does the dihedral influence the angle of attack.

Answer 1

The angle of attack will not change. However, the lift vector will change direction. Thus, at $\Gamma=89^\circ$, the wings' lift vectors will be almost at odds with one another, but they will still have the same angle of attack as they did when $\Gamma=0^\circ$. The purpose of the dihedral (as you probably know) is to change the angle of attack when horizontal forces are present when lift is not opposite the effective weight vector. Dihedral prevents spiraling by improving the lift of the lower, up-wind wing relative to the higher, down-wind wing, thereby opposing the roll of the aircraft.

Answer 2

The answer is in the page you linked to. Dihedral does not affect the angle of attack until the aircraft has a yaw angle w.r.t. the direction of motion.

Answer 3

Dihedral does affect the angle of attack, but this effect is only significant at high dihedral angles.

I will try to explain this by using normal vectors in a xyz-axis system.
The positive x-direction is pointed downstream parallel to the flow, the y-axis is pointed to the right when facing the flow. This means that the z-axis points upwards.

$w$ is the normal vector of the incoming flow in positive x-direction $$ w = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right] $$

$f$ is the normal vector of the flat plate. When the plate has no incidence angle w.r.t. the flow f is:

$$ f = \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right] $$

To compute the angle of attack the following procedure is followed:

Normalisation of $f$ and $w$. Normalised vectors $f_n$ and $w_n$ are the result.

Then the angle of these two vectors is:

$$ \theta = \arccos \left( {{f_n} \cdot {w_n}} \right) $$

And the effective angle of attack $\alpha$ is: $$ \alpha = 90 - \theta $$

Now I use transformation matrices on $f$ to give it a pitch $\alpha$ and dihedral $\Gamma$. First dihedral, so rotation about the x-axis with an angle $\Gamma$ gives the following rotation matrix:

$$ \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos \left( \Gamma \right)}&{ - \sin \left( \Gamma \right)}\\ 0&{\sin \left( \Gamma \right)}&{\cos \left( \Gamma \right)} \end{array}} \right] = {\Gamma _{rot}} $$

After applying the dihedral a pitch angle is applied. This is equal to rotation about the y-axis:

$$ \left[ {\begin{array}{*{20}{c}} {\cos \left( \alpha \right)}&0&{\sin \left( \alpha \right)}\\ 0&1&0\\ { - \sin \left( \alpha \right)}&0&{\cos \left( \alpha \right)} \end{array}} \right] = {\alpha _{rot}} $$

Working this out with a dihedral of 45 degrees and a pitch of 30 degrees:

Dihedral: $$ f' = \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos \left( {45} \right)}&{ - \sin \left( {45} \right)}\\ 0&{\sin \left( {45} \right)}&{\cos \left( {45} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ { - 0.70711}\\ {0.70711} \end{array}} \right] $$

Angle of attack:

$$ f'' = \left[ {\begin{array}{*{20}{c}} 0\\ { - 0.70711}\\ {0.70711} \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} {\cos \left( {30} \right)}&0&{\sin \left( {30} \right)}\\ 0&1&0\\ { - \sin \left( {30} \right)}&0&{\cos \left( {30} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.35355}\\ { - 0.70711}\\ {0.61237} \end{array}} \right] $$

Both $f''$ and $w$ are already normalised, so the (effective) angle of attack is:

$$ \alpha = {90^o} - \arccos \left( {\left[ {\begin{array}{*{20}{c}} {0.35355}\\ { - 0.70711}\\ {0.61237} \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right]} \right) = {90^o} - 69.295 \approx {21^o} $$

So the effective angle of attack is smaller than the pitch angle given to the plate due to the dihedral