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I'm currently trying to calculate the Zero Moment Point for a game I develop, however im terrible at physics and therefore have trouble calculating the “rate of angular momentum at the center of mass”.

As I understood the angular momentum is defined as:

$\qquad \qquad L =$ distance from the center of rotation $\times$ mass $\times$ velocity

The mass of my object is $10$ kg.

It's velocity is $(3, 4, -1)$.

However, I don't get how to calculate the distance from the center of rotation, since my object has no fixed point its rotating around.

I'm referencing the following article, where I want to calculate HG: Zero moment point.

Thanks for any help in advance!

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  • $\begingroup$ $\sum_i{m_i r_i}=0$ with respect to CM.so you need to find the coordinate of cm by those equations. $\endgroup$ – baponkar Oct 12 at 15:13
  • $\begingroup$ Zero Moment Point for a game ? Can you give more details on your game $\endgroup$ – Eli Oct 12 at 19:42
  • $\begingroup$ Okay thanks, the coordinate of my center of mass is (3, 2, 3) but i still cant figure out how to get it's angular velocity momentum from it. $\endgroup$ – Tim von Känel Oct 12 at 19:45
  • $\begingroup$ Im trying to code a game with ragdolls fighting each others. I need the zero momentum point to calculate the optimal position of their feet so they can walk and stand. $\endgroup$ – Tim von Känel Oct 12 at 19:46
  • $\begingroup$ So lets say i have a leg a body and an arm as the characters components. First i calculate the center of mass of the whole character. Then i calculate the distance of each bodypart to the COM of the whole character. After that i calculate the angular momentum of each bodypart by multiplying the distance * it's mass * it's velocity. Am i right until this step? $\endgroup$ – Tim von Känel Oct 12 at 19:53
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A particle of mass $m$ located at $\vec{r} = \pmatrix{x & y & z}$ from the origin, and having velocity $\vec{v} = \pmatrix{ vx & vy & vz} $ has the following properties

  • Linear momentum $$\vec{p} = m \vec{v} = \pmatrix{m\, vx \\ m\, vy \\ m\, vz} $$

  • Angular momentum about the origin $$ \vec{L} = \vec{r} \times \vec{p} = \pmatrix{m ( vz\, y-vy\, z) \\ m (vx\,z - vz\,x) \\ m( vy\,x -vx\,y )} $$

    Where $\times$ is the vector cross product.

I think you are asking about finding the line of action of the reaction forces from an equipollent system of forces $\vec{F} = \pmatrix{Fx & Fy & Fz}$ and moments $\vec{M} = \pmatrix{Mx & My & Mz}$ at the location $\vec{r}$.

The is found with the calculation

$$ \vec{r}_{\rm zero moment} = \vec{r} + \frac{ \vec{F} \times \vec{M}} { \| \vec{F} \|^2} = \frac{1}{Fx^2+Fy^2+Fz^2} \pmatrix{Fy Mz-FzMy \\ Fz Mx - Fx Mz \\ Fx My - Fy Mx} $$

Anyway, I strongly suggest you do some reading on vectors and cross products in order to understand the math of mechanics.

My favorite way of calculating a cross product in via the cross-product matrix

$$ \vec{a} \times \vec{b} = \pmatrix{0 & -a_z & a_y\\ a_z & 0 & -a_x \\ -a_y & a_x & 0 } \pmatrix{b_x \\ b_y \\ b_z} $$

Use the shorthand notation $[\vec{a}\times]$ to denote the 3×3 skew symmetric matrix show above that multiplies $\vec{b}$. The above is concisely written as a matrix-vector multiplication which is easily computed

$$ \vec{a} \times \vec{b} = [\vec{a} \times ] \vec{b} $$

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  • $\begingroup$ Thanks! I will look into it tomorrow and tell you if it worked :) $\endgroup$ – Tim von Känel Oct 13 at 0:55
  • $\begingroup$ Can i use (0, 0, 0) as the origin or should i use the objects Center of mass? $\endgroup$ – Tim von Känel Oct 13 at 1:04
  • $\begingroup$ It depends on what you are doing. The origin is a fixed point you measure things at, and the center of mass is a moving point you are interested in rigid body mechanics since it simplifies the equations of motion. $\endgroup$ – ja72 Oct 13 at 2:50
  • $\begingroup$ I know at least computed a vector which seems to be the zmp. It looks correct, but it is probably complete garbage :)) Thank you really much! $\endgroup$ – Tim von Känel Oct 13 at 3:14
  • $\begingroup$ No, it is not garbage and if you want a little deeper understanding read on the subject of screw theory as it applies to robotics. $\endgroup$ – ja72 Oct 13 at 3:26

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