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Net momentum for quadrotor

  1. Feb 9, 2016 #1
    While watching lecture at Coursera, i tumbled over this fourmula

    Moment
    [tex]\mathbf{M}=\sum_{i=1}^{4}(\mathbf{F}_i\times \mathbf{r}_i+\mathbf{M}_i)[/tex]
    where F is uplift force from a propeller, and M is drag moment for a propeller.

    But why they add drag moment(torque) like that? Maybe it will differ for central point.

    quadrotor.png
     
  2. jcsd
  3. Feb 9, 2016 #2

    A.T.

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    Try it out by computing the moments of the blade drag forces around different reference points.
     
  4. Feb 9, 2016 #3
    Thanks, it is true for two points on the picture, but can you name the rule? I want to read it whole to understand it better. Thank you.
    two_reference.png
     
  5. Feb 21, 2016 #4
    Could somebody write what are these [itex]M_i[/itex]? Because i still don't understand how it works, i think every moment must be calculated about some point.

    I have found this formula
    [tex]
    \begin{equation}
    \dot{\mathbf{H}}_O = \sum(\mathbf{r}_i \times m_i\dot{\mathbf{v}}_i) = \sum(\mathbf{r}_i \times \mathbf{F}_i + \mathbf{M}_i)
    \end{equation}[/tex]
    but i don't understand from where [itex]\mathbf{M}_i[/itex] came, because i think that [itex]m_i\dot{\mathbf{v}}_i = \mathbf{F}_i[/itex]
     
    Last edited: Feb 21, 2016
  6. Feb 21, 2016 #5

    A.T.

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    Didn't you explain it yourself:
    It's the total moment of all aerodynamic forces on the blades, which are in the plane of the propeller disc.
     
  7. Feb 21, 2016 #6
    Well, yes, but here i want to know from where [itex]M[/itex] came in this general formula [itex](1).[/itex] It is unrelated for quadrotor, just to understand underlying theory.
     
    Last edited: Feb 21, 2016
  8. Feb 21, 2016 #7

    A.T.

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    In general it's just some external moment.
     
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