Can you look to previous velocity for a clue to direction?
You might consider using a default direction algorithm based upon
position or other vertex/particle attributes?
David Farmer
email@hidden
On May 30, 2007, at 8:12 AM, Lorenzo wrote:
Hi,
I have to find the vector "pv" perpendicular to an arbitrary vector
"tv".
The perpendicular vector "pv" has to be parallel to the XZ (floor)
plane.
How to do it?
I have successfully found a generic perpendicular with the cross
product
between tv and a random vector rv. Of course the result is a vector
not
always laying on the XZ plane so pv[1] != 0.
So I have found this trick. Firstly I set rv = tv. Then I set rv[1]
= 0 or
rv[1] += 12345; So tv and pv lay on the same vertical plane.
So I find pv[1], that is pv is parallel to the plane XZ. Good.
The problem is that this trick doesn't work for any tv. E.g. when
tv is
{0, 1, 0} it doesn't work.
Best Regards
--
Lorenzo
email: email@hidden
_______________________________________________
Do not post admin requests to the list. They will be ignored.
Mac-opengl mailing list (email@hidden)
Help/Unsubscribe/Update your Subscription:
http://lists.apple.com/mailman/options/mac-opengl/email@hidden
This email sent to email@hidden
_______________________________________________
Do not post admin requests to the list. They will be ignored.
Mac-opengl mailing list (email@hidden)
Help/Unsubscribe/Update your Subscription:
http://lists.apple.com/mailman/options/mac-opengl/email@hidden
