**comp.graphics.algorithms**

## Subject: **Re: How to get angle and speed to reach a movable point**

Hans,

I'm trying to do real example using your algorithm and I got stuck. Let

me show you what I have so far.

Let's suppose that A is located at (2,4) and its speed is 5 (let's say

meters per sec)

B is located at (12,16), its speed is 3 and it's moving 30=BA east.

If we have:

(B'^2-A'_max^2)*t^2 + 2*t*(B-A).B' + (B-A)^2 =3D 0

We replace our data in that equation:

( 3^2 - 5^2 ) * t^2 + 2*t*(B-A)*3 + (B-A)^2 =3D 0;

-16^2 + 6*t*(B-A) + (B-A)^2 =3D 0;

t =3D -6*(B-A) +- sqrt [ (6*(B-A))^2 + 64* (B-A)^2 ] / -32

At this moment, I don't know how to continue. Since A and B are vectors

I thought that (B-A) would be another vector V where V.x =3D10 and V.y

=3D12

How should I get t from there? and one more, where can I specify the

angle of B?

Thanks and sorry for this probably dumb question :(

Reply

View All Messages in

**comp.graphics.algorithms**

path:

How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>

Replies:

Re: How to get angle and speed to reach a movable point

Re: How to get angle and speed to reach a movable point

Copyright © 2006 WatermarkFactory.com. All Rights Reserved.