[PD] bouncing

[Uğur Güney]

# I read the first mail and find an algorithm for the special case of
"moving point particles in  a 2D pentagon boundary". The answer is given as
using pyode script but I am excited about my solution and want to share it
with you :-)
# The easy part of using a rectangular boundary (of which sides are parallel
to the axes) is that, one should only check if x position of a particle is
greater than x_max (coordinate of the right side of the rectangle), if
x<x_min, y<y_min and y>y_max. And if it exceeds the boundary in the next
step of the simulation according to ODE (or any kind of calculation) of the
simulation, then reverse the velocity in that direction (and mirror the
coordinate using the boundary line as the axis of symmetry).
# So, We can try to use this simple idea in pentagon case.
# We can divide the pentagon in five triangles: Put a point in the center
and draw lines to vertexes.
http://www.math.union.edu/~dpvc/courses/1999-00/MTH012-02-WI00/notes/DividedPentagon.jpg<http://www.math.union.edu/%7Edpvc/courses/1999-00/MTH012-02-WI00/notes/DividedPentagon.jpg>
# Call the bottom triangle 0th triangle, turn in counter clockwise direction
and call other triangles as 1st, 2nd...
# The algorithm is as following:
# First determine in which triangle the particle is. One can get this using
an atan2(x,y) like function. atan2 returns the angle between the position of
the particle and the x-axis. We can beforehand calculate the angles between
the sides of triangles and the x-axis. And using this information we can
find in which triangle the particle is. (or maybe y/x is just enough for
determination)
# If it is in the zeroth triangle, all we have to do is to look for the y
position. If it is less than minus of the height of the bottom triangle then
it exceeds the boundary. Apply the reflection like in a rectangular
boundary.
# One can not do this in other triangles (1st, 2nd...). Because their
boundaries are not parallel to axes but they have slopes. So, if we rotate
the coordinate system, until the bottom side of the triangle becomes
parallel to y-axis, e.g. rotate 72 degrees clock wise for 1st triangle, 2*72
degrees for the 2nd..., (360/5=72) we can apply this reflection law. (this
is the trick)
# After applying the reflection one must rotate the coordinate system
counter clock wise.
# Here is the formula of rotating a coordinate system (taken from Arfken,
Mathematical Methods for Physicists). (x,y) is the coordinate in the old
system and (x', y') is the coordinate in the new (rotated) system, a is the
angle of rotation.

x' = x*cos(a) + y*sin(a)
y' = -x*sin(a) + y*cos(a)

# But one don't have to make trigonometric calculations for every rotation,
because a is constant. It is 72 or integer multiples of 72 and one can
calculate sines and cosines beforehand e.g. sin(72) = 0.9511, cos(72)=0.3090,
for CCW rotations, and sin(-72)=-0.9511, cos(-72)=0.3090 for CW rotations.
# Here is the pseudo code:

take_the_simulation_one_step_further()
for_each_particle:
n = the_number_of_triangle_which_includes_the_particle(x,y)
(x',y',vx',vy')=rotate_the_velocity_and_coordinate_of_the_particle(-n*72,x,y,vx,vy)
(x',ynew',vx',vynew')=reflect_the_particle_(if_necessary)_as_if_it_reflects_from_the_bottom_of_a_rectangle(x',y',vx',vy')

(x,y,vx,vy)=rotate_the_velocity_and_coordinate_of_the_particle(n*72,x',ynew',vx',vynew')

# I didn't try this of course, but seems plausible. Thanks for reading :-)
-uğur-


On 10/13/07, marius schebella <marius.schebella at gmail.com> wrote:
>
> hi,
> I need to bounce objects against a border that is not running parallel
> to x or y, but in a gradient angle. actually I need to bounce the
> objects within a pentagon. iSeg2D allows object to cross the border when
> they bounce, there is no x/y-max for gradients.
> I am also looking into frank's pyode script. Is there something in it,
> that can do that?
> marius.
>
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