Trying to implement a code for the algorithm described in this paper I found something not very clear to me that leads me to misunderstand the whole concept.

To calculate the vector $\vec{b_{3d}}$ the paper (page 3) suggests to use the following equation:

enter image description here

To me it look like a normal vectorial equation where you divide one vector by its normalized value. Something very trivial.

But I was wondering that the variables $e_{x}, e_{v}$ are previously defined as a vector of the error of the variables $x, v$ as: $ e_{x} = x-x_{d} \quad e_{v} = v-v_{d} $ at the beginning of the paper where $x,v$ are vectors: $ x \in \mathbb{R}^{3} $ and $ v \in \mathbb{R}^{3} $

So now I have the following questions:

  1. I suppose that the equation above is not useful to get $\vec{b_{3d}}$ if I treat $x,v$ as vectors. But in this case I cannot do nothing since in the paper doesn't explain which vector element should I consider to calculate $\vec{b_{3d}}$;
  2. In the equation compares the term $e_{3}$ which is the unity vector of the $Z$ axis. Maybe the authors wanted to point out that you should consider the gravity along Z, but why should I put a unity vector in an equation, if it is suppose to have all term as scalar?;

Anyway... good day.

  • $\begingroup$ I don't know about anyone else, but your picture of the equation isn't displaying on my page and I've tried refreshing a couple of times. $\endgroup$ Commented Dec 28, 2014 at 9:31
  • $\begingroup$ Look at equation nr. 12 on this paper $\endgroup$
    – Dave
    Commented Dec 28, 2014 at 9:41

1 Answer 1


the expression is a vector equation as you initially thought. ex,ev,e3,xd and hence xd'' are also vectors. 2. mg is a scalar and the force acts in -e3(-Z) direction.Force is a vector. From a first glance it seems like you find b3d corresponding to a the xd'' acceleration.Also b3d doesnt seem to be vertically downward (as its name suggests) but in a direction depending on centre of mass vs rotor position.Maybe thats why its b3d and not just b3

Even if you dont have x,v it appears like you know ex,ev (based on geometry?)


You must log in to answer this question.

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