Suppose a flying missile's velocity vector has components $u, v, w$, as illustrated below.


From my knowledge of flight dynamics, the angle of attack is

$$\alpha = \arctan \left(\frac{w}{u}\right)$$

where $u, w$ are time-varying. Then,

$$\frac{d\alpha}{dt} = \frac{1}{1+\frac{w^2}{u^2}} \left(\frac{w}{u}\right)'= \frac{w'u - wu'}{u^2+w^2}$$

Some books indicate

$$\frac{d\alpha}{dt} = q + \frac{Z}{mu}$$

where $q$ is the second component of angular velocity, $Z$ is the third component of the total force, and $m$ is the mass of missile. How can these two formulas be the same?


4 Answers 4


They are two equations describing two different states.

The first equation is the momentary state of $w$ and $u$ without regards as to what causes variation of $w$ and $u$.

The second equation is the change in angle of attack as an inertial response to a force or moment acting on the missile.

  • $\begingroup$ I get the idea but how could we derive the second equation explicitly? $\endgroup$
    – Dat
    Commented May 5, 2019 at 15:42

Assuming that there isn't an error in the books you've read:

Both formulas are true. It's true that

$$\frac{d\alpha}{dt} = \frac{w'u - wu'}{u^2 + w^2},$$

and it's also true that

$$\frac{d\alpha}{dt}= q + \frac{Z}{mu}.$$

From this, we can conclude that

$$\frac{w'u - wu'}{u^2 + w^2} = q + \frac{Z}{mu}.$$

Both of these two formulas describe the way that the angle of attack changes, but they describe it in different ways.

I'm not sure exactly how the $q + Z/mu$ formula was derived, but it seems to be related to the fact that angle of attack is related to pitch and the flight path. Specifically, for flight with zero bank angle, the angle of attack is the pitch angle minus the climb angle. It looks like the $q$ term comes from the rate of change of pitch, and the $Z/mu$ term comes from the rate of change of climb angle.

  • $\begingroup$ Yes, I believe they are both true but how can we formulate from one to another? I don't understand how to derive formula 2 $\endgroup$
    – Dat
    Commented May 4, 2019 at 0:35
  • $\begingroup$ @Dat I added another paragraph describing my best guess about the $q + Z/mu$ formula. $\endgroup$ Commented May 4, 2019 at 17:43
  • $\begingroup$ I see, I understood the q term but still can't get the Z/mu term. The second formula is from en.wikipedia.org/wiki/Flight_dynamics_(fixed-wing_aircraft) $\endgroup$
    – Dat
    Commented May 5, 2019 at 12:46

First we can use the definitions of w, u, alpha, and beta to define

$$ \frac{w}{u} = \tan\alpha $$

$$ u = Vcos(\alpha)cos(\beta) $$

Now lets look closer at the first equation you provided using the definitions above:

$$\frac{d\alpha}{dt} = \frac{1}{1+\frac{w^2}{u^2}} \left(\frac{w'}{u}-\frac{u'w}{u^2}\right) = \frac{1}{1+\tan^2\alpha} \left(\frac{w'}{V\cos\alpha\cos\beta}-\frac{u'\tan\alpha}{V\cos\alpha\cos\beta}\right) $$

Which can be simplified further with trig identities to

$$\frac{d\alpha}{dt} = \frac{\cos(\alpha)w'-\sin(\alpha )u'}{V\cos\beta} $$

We can see that we're on the right track here, using first-order approximations, w'/V is approximately q and u'/m is approximately Z (by Newton's second law). From the sines and cosines we can tell that pitch rate on its own is only a valid approximation while alpha is small.

Why The Term For Z?

Why is force included here? Because angle of attack and pitch are measured in different frames and a change in airspeed could cause you to overestimate or underestimate the new angle of attack. This isn't novel or particular to the case of missiles, as a similar equation appears in Real-Time Aerodynamic Parameter Estimation without Air Flow Angle Measurements by Eugene Morelli, and I've seen similar kinematic equations used in avionics.

Let's draw what happens graphically: enter image description here

From the image we see that while flight path is changing, pitch angle changes alone won't account for all the differences. We also have to account for a change in flight path angle relative to the airmass, which can be approximated using the component of flight path change that's perpendicular to the flight path. If we assume angle of attack is small or that Z is perpendicular to the flight path instead of to the body, this rotation equals arctan(Z_w/ (m*V)).

Further derivation with fewer approximations

Now the piece omitted in other answers. The kinematic equations of motion as shwon in Morelli, et al:

$$ w' = q u-p v + g*a_z+g\cos\phi\cos\theta $$

$$ u' = r v - q w + g a_x + g \sin\theta $$

Plugging in the definitions for u,v, and w as function of V, alpha, and beta into these equations and then inserting those into our formula for the derivative of alpha gives (see Morelli, et al. for more information)

$$\frac{d\alpha}{dt} = q + p\sin\beta + \frac{g}{V}\left(-\sin\alpha\left(a_x-sin\theta\right)+\cos\alpha\left(a_z+\cos\phi\cos\theta\right)\right) $$

Note that even though we used calculus and trig identities, not kinematics to get this equation, it matches the expected rotation matrix for angular rates between the body frame and the wind frame, using this example, or also this question:

$$ \begin{bmatrix} \delta \phi \\ \delta \alpha \\ \delta \beta \end{bmatrix}= M_{a,b2w} \begin{bmatrix} \delta p \\ \delta q \\ \delta r \end{bmatrix} $$

$$ M_{a,b2w} = \begin{bmatrix} \cos\alpha\cos\beta & \sin\beta & 0 \\ -\cos\alpha\sin\beta & \cos\beta & 0 \\ \sin\alpha & 0 & 1 \\ \end{bmatrix} $$

We can convert those pesky accelerations (well, pesky if you're not trying to measure them, then they're really handy) and include gravity in the total force:

$$ T_{f,z}/m = g \left(-\sin\alpha\left(a_x-sin\theta\right)+\cos\alpha\left(a_z+\cos\phi\cos\theta\right)\right) $$

And we get:

$$\frac{d\alpha}{dt} = q - p\sin\beta + \frac{T_{f,z}}{m V} $$

So with some generous assumptions that sideslip and roll rates are small so p*sin(beta)≈0, that Z includes gravity already, and that Z is mostly aligned perpendicular to the incoming wind, this equals your original formula, we've shown that

$$\frac{d\alpha}{dt} \approx q + \frac{T_{f,z}}{m V} $$

Variables used

p,q,r = roll, pitch and yaw rates in the body frame

ϕ,α,β = roll, angle of attack, and sideslip angles

u,v,w = Components of aircraft velocity along the body longitudinal, lateral, and normal directions

ϕ,θ = Roll and pitch in the inertial frame

V = Total velocity, vector sum of u,v,w

T_fz = Total force perpendicular to the flight path, pointing in opposite direction of angle of attack

m = Aircraft mass


  1. Eugene A. Morelli, Real-Time Aerodynamic Parameter Estimation without Air Flow Angle Measurements

I think the equations are not referred to the same axys.

The first equation refers to the angle of attack, the angle between both components of the velocity refered to the body axys (x along the body), using a non-inertial frame (as moves with the missile) . Well, actually following some comments, is not actually the angle of attack, it will be the angle of attack in the specific case of v=0.

The second equations looks to be referring to the rotation of the missile referred to an inertial frame of an external observer (for example, Earth), so actually, it looks like the actual movement of the body frame refered to that observer in a specific rotation component .

Imagine an example, that I am able to provide the right thurst to the missile in such a way that I keep u, w and q constant rotating in a circle.

In this case, the angle of attack will be constant and its increase 0 with time but the second alpha will be increasing/decrasing with a constant increase/decrease, also this change will be linked to the circle rotation.

I think you can also make the math and relate both equations (you will need more components!) taking care of inertial forces.

Final comment, as you are introducing as well a second term in teh second alpha equation, it really looks that it is an specific case in the movement of missile.


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