Sorry, I re-red in one of my own old post that Angular velocity is RBD
only..
Le 04/07/2014 11:33, olivier jeannel a écrit :
Hi Raffael,
Out of curiosity, do you have an example of how you work with angular
velocity ? Never tried that before.
For example right now, I've built a tornado mostly based on
CrossProduct. It's superb, but let's say the client want's it to turn
faster, with a multiplied CrossProduct the particle will go away from
the "center" (unless I add a vector to suck it back to the center
(Centripetal) ).
With an angularVelocity, do I have the possibility to just make it
spin faster ?
Le 04/07/2014 03:29, Raffaele Fragapane a écrit :
You introduced a magnitude to the velocity which will over time
amount to further distance, but reasoned the problem out without that
variable.
It would work if the velocity was infinitely small.
Multiplying that vector you pump into the velocity by an extremely
small number should reduce the slingout, until you hit precision
limits and you end up with an immobile object because the velocity
drops to 0, but for the right combination of numbers it might stay
subpixel for long enough to be "artistically" the result you want.
If you needed it for a practical application you would of course be
better off rotating the vector around y instead, affecting angular
velocity instead of linear won't compromise the system and will give
you the result you want.
On Fri, Jul 4, 2014 at 4:21 AM, olivier jeannel
<olivier.jean...@noos.fr <mailto:olivier.jean...@noos.fr>> wrote:
Ah, thank's for clarifying this. Makes perfect sense.
Le 03/07/2014 11:34, pete...@skynet.be <mailto:pete...@skynet.be>
a écrit :
this does make sense to me, if I think of it as a rocket
orbiting a planet.
at each moment in time the rocket is pushing itself forward with
a linear force (the vector) - so it will tend to move from where
it is to where the force is telling it to go – in a straight
line, tangent to the circle you are after – but it already has
it’s current speed, so you don’t end up exactly where you are
pointing but a bit further out - leaving the circle a bit. The
next moment in time you are correcting with the new tangent
vector – so you are approximately following the circle.
if you want to get the perfect circle, you will need to add
another force, pulling towards the centre. ( check on
centripetal force: http://en.wikipedia.org/wiki/Centripetal_force )
in ice: subtract the pointposition from the center of the circle
and multiply by scalar to finetune – add this vector to the one
you have
In the example of the orbiting rocket I guess that would be gravity.
*From:* olivier jeannel <mailto:olivier.jean...@noos.fr>
*Sent:* Wednesday, July 02, 2014 10:00 PM
*To:* softimage@listproc.autodesk.com
<mailto:softimage@listproc.autodesk.com>
*Subject:* Running in circle, The CrossProduct question
Hi gang,
with my partner we were discussing crossproduct "theory" and I'm
not sure what to believe or think.
I was persuaded that the result of a Cross Product of a
PointPosition (x,y,z) and a vector 0,1,0 plugged in a the
PointVelocity, would give a particle orbiting around 0,0,0
describing a perfect circle.
In fact, not exactly.
with simulation substep 1 I get this :
with simulation substep 10 I get this (but it travels much slower) :
So my question is : Is this a problem of approximation from the
or the computer, and then the mathematical nature of cross
product is able to "describe" a circle.
or is this a normal behaviour, considering that the cross
product vector is pushing in straight line a particle and that
it could never "describe" a circle.
--
Our users will know fear and cower before our software! Ship it! Ship
it and let them flee like the dogs they are!