I've been learning how to do dead-reckoning using just an ordinary IMU chip (a 3-axis accelerometer and a 3-axis gyroscope). In a number of textbooks I've read on aircraft navigation, the motion of a navigating object is often summed up in four equations. Aside from the last equation - (3.17) rotational dynamics - where we need to separately calculate the inertial matrix in CAD, the other three equations requires only the input data registered by the accelerometers or gyroscopes, or integration of the values.

I have so far no problem understanding the derivation for each equation or knowing what they physically mean. The trouble I had is I couldn't relate the rotational dynamics equation (3.17) of an object to its navigation.

Navigation is a matter of tracking the position and orientation of an object and perhaps also its acceleration and velocity at a specific point in space. The first three equations (3.14) (3.15) (3.16) will suffice:

  • Given an initial condition and gyro data, we can numerically update the Euler angles in (3.16) and determine the object's orientation
  • With data measured by accelerometers and gyros in the body-frame, and if we know our externally applied force, we can find out the object's velocity in (3.15)
  • Since we already know our orientation and body's velocity, we can calculate our inertial velocity in (3.14). Integrating that, we get the object's inertial position

Up to this point I would have everything I need for navigation and the last equation (3.17) seems out of place here. I've already had my orientation per unit time - what's the point of knowing the angular acceleration around the bodily axis? Calculating all the different moments of an object doesn't seem to have any meaning to navigation of an object either. So why does that equation (3.17) matter in navigation?

enter image description here

I start to suspect it could be my entirely lack of basic knowledge in aeronautics or aviations that gives rise to this question. I don't see any mathematical difference in mounting an IMU on a banana or on an actual aircraft. The IMU provides me with accelerations and angular rates that I can calculate where the object navigates in what orientation. Navigation care less for the inertia or moments of the object. Even if I need angular acceleration for whatever reason, wouldn't it be much easier (if not more accurate) to differentiate the gyros data directly?

The only situation I can think of that makes inertia and rotational acceleration matter in navigation is when the object collides into something else where a transfer of momentum occurs. But that's just my wild imagination and am unsure if that makes any sense.

  • $\begingroup$ Will your IMU be affected by body rates? $\endgroup$
    – Jim
    Jan 4, 2023 at 3:43
  • $\begingroup$ I’m a little rusty here but generally in spherical motions, you have to account for the Euler term AND the Coriolis term. Except at the equator, a great circle line always has a northern or southern component, irrespective of the direction of motion. That northern or southern component, if performed at a constant altitude, mean I am moving inward or outward from the vertical axis of my motion, and I am going to have to account for that in my navigation solution. I believe that that’s 3.17, it’s where you account for the conservation of angular momentum. $\endgroup$
    – Max R
    Jan 4, 2023 at 20:02
  • $\begingroup$ What do p/q/r represent in the equations? looks like something related to angular velocity? Thanks to the Dzhanibekhov effect an object with constant angular momentum might not have constant angular velocity though I don't know if that's what this equation means. $\endgroup$ Jan 4, 2023 at 23:51
  • $\begingroup$ @Jim If you're referring to spinning of an object, then no. I only care for the orientation and position (and it's derivatives) of the IMU. If by body rates you mean linear and angular velocity ... well that's what the IMU is measuring. Otherwise I'm not sure if I understand your question. $\endgroup$
    – KMC
    Jan 5, 2023 at 4:08
  • $\begingroup$ @Max R, I believe the equations are in their most general forms. Acceleration $\mathbf{a}(t) = \left(\ddot{r}-r\dot{\theta}^2\right)\hat{\mathbf{e}}_r + \left(r\ddot{\theta}+2\dot{r}\dot{\theta}\right)\hat{\mathbf{e}}_{\theta}$ encapsulates both radial and tangential terms (Coriolis acceleration being the last term). But this fraction of acceleration shouldn't have anything to do with navigation. It really is just a secondary abstraction to be calculated from the data in hand (e.g. find out the Coriolis force etc.). After all, each numerical data is as linear as it can be per unit time. $\endgroup$
    – KMC
    Jan 5, 2023 at 4:14

1 Answer 1


As you noticed, you do not need this information for navigation per se, and you are right. If you can measure the rotational rates $p$, $q$ and $r$ directly, then you do not need to worry about updating the rotational rates.

However there are a couple of reasons why you might be interested in this equation anyways.

  1. This completes the set of equations. These are generally called the equations of motions of an aircraft and are central for flight dynamics. Therefore this set of equations describes how an aircraft moves/rotates/behaves in the air.

  2. For real-world navigation you also might use this formula for example inside a data-fusion scheme. No sensor is perfect, therefore directly using the rotational rates for your navigation solution might lead to undesirable results. For this reason, often data fusion approaches like a Kalman Filter are used which fuse sensor data and model data. For the model data part of this filter you need a mathematical model, which is exactly posed by the rotational dynamics. (Probably related: this question)

  3. Last but not least, these equations are used for simulation purposes. If you want to numerically simulate an aircraft for whatever purpose, you absolutely need these equations. Simulations are of course central for a whole myrade of applications such as flight simulators (e.g for testing your navigation filter), general flight dynamics analysis, controller synthesis, and so on.

From the context you provide, I would speculate that the author wanted to give the complete set of equations.

Perhaps as a side-note regarding this set of equations: These are flat earth approximations (not related to the conspiracy theory :D). Therefore not really suited for navigation purposes if you want to navigate further then lets say 30-50km. After this distance, the error accumulates and gives distance errors in the 10s of meters. In applications like short term path planing like landing planing or short-term path following, these are fine however.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .