# Rate of change of heading in terms of normal acceleration and air speed

Let's say an aircraft is traveling with speed $$|\mathbf v|$$ with respect to an inertial frame. The aircraft is able to execute turns by producing an acceleration $$\mathbf a$$ that is normal to $$\mathbf v$$ at all times.

For simplicity we will assume the aircraft is flying at level altitude above a flat surface (the $$x-y$$ plane) and that all motion is in the plane of constant altitude ($$z$$ constant).

If we denote the aircraft's heading (which we will define as the angle $$\mathbf v$$ makes with the x-axis) as $$\theta$$, is it correct to say that the heading rate magnitude $$|\dot\theta|$$ is: $$|\dot\theta| = \frac{|\mathbf a|}{|\mathbf v|}$$ This definition of heading assumes that the wind axis and body axis are co-incidental, i.e. zero angle of attack and zero sideslip angle. This an approximation, but not a terrible one for a simple analysis.

Justification:

If the aircraft is executing a turn by applying normal acceleration $$\mathbf a$$, the turn radius $$R$$ is given by: $$R = \frac{|\mathbf v|^2}{|\mathbf a|}.$$ The magnitude of the attitude rate $$\dot\theta$$ is the same as the angular velocity of the aircraft about the center of curvature of the flight path, therefore: $$R |\dot\theta| = |\mathbf v|,$$ which, after substituting, yields $$|\dot\theta| = |\mathbf a|/|\mathbf v|$$.

Is this train of reasoning correct? If not, what assumptions are incorrect?

• Everything seems right, I don’t think there are any problems with your reasoning, and the result looks correct. Is there any reason why you would think it’s not right? – Hugo V Oct 10 '18 at 0:10
• I think there is some terminology confusion en.m.wikipedia.org/wiki/Heading_(navigation) heading is where the plane points. – Emil Oct 10 '18 at 1:57
• @Emil-- this is the intended use. For purposes of this question, we will assume the velocity vector is co-incident with the aircraft nose direction. I'll edit the question to clarify. – Robert L. Oct 10 '18 at 13:11