It 'appears' that in most flights the horizontal stabilizer is set for positive angle of attack.

So does this produce an 'up force' (lifting force) on tail thus making the aircraft nose-heavy?

Or does the horizontal stablizer presents itself to the relative air at a different angle of attack than the wing does (which creates down force on tail)?

  • 2
    $\begingroup$ it "appears" from where? also, what is not clear from your previous question? $\endgroup$
    – Federico
    Commented Apr 29, 2016 at 10:45
  • $\begingroup$ Appears in the sense when I see that there is a positive angle of incidence. I thought means it is set for positive aoa. Earlier question assumed that horizontal stab is fixed at 0 angle of attack and by using trimming its angle of attack is varied. $\endgroup$ Commented Apr 29, 2016 at 10:50
  • 3
    $\begingroup$ The AoA is between the airflow and the chord. But the airflow rear of the wing is not horizontal when the aircraft moves straight and level. Maybe the pitch is positive. $\endgroup$
    – mins
    Commented Apr 29, 2016 at 10:51

1 Answer 1


You are right, the horizontal tail of a conventional airplane appears to have a higher incidence, but the actual angle of attack is smaller than that of the wing.

The wing, flying ahead of the tail, produces downwash, so the flow at the tail location has a distinct downward component. The downwash angle can be calculated from the lift coefficient and the geometry of the aircraft: To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$

Lift $L$ is then the impulse change which is caused by the wing and equal to weight. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$

$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$ with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$ A typical airliner cruise lift coefficient is 0.4, and a typical aspect ratio is around 8: This results in a downwash angle of nearly 2° if the lift distribution over span is elliptical. In reality, it is more triangular-shaped, so the downwash angle is larger near the center of the aircraft. Note that the engine nacelles of the DC-9 and the MD-80 range are tilted 3° up to align them with the local flow.

The resulting angle of attack is lower by those 3°, and if the angle of attack difference between wing and tail is less than that, the tail surface will appear angled upward. To achieve static stability, the tail will have to fly at a slightly lower angle of attack than the wing.

  • $\begingroup$ so If I understand it correctly, air meets the tail at lower angle of attack when the horizontal stabilizer is placed at a positive angle of incidence? (thus producing less lift on tail surface) $\endgroup$ Commented Apr 30, 2016 at 18:02
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    $\begingroup$ @user2927392 The angle of attack is just the difference between flow direction and wing/stabilizer incidence. Increasing the incidence increases the angle of attack. The point is that behind the wing the local flow direction is different than ahead of it. This makes it appear as if the angle of attack at the stabilizer is higher when in fact it is lower. $\endgroup$ Commented Apr 30, 2016 at 18:36

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