Home Products Download Order Contacts

comp.graphics.algorithms

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



staryon@gmail.com wrote:

> 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 30ΒΊ east.

> If we have:
> (B'^2-A'_max^2)*t^2 + 2*t*(B-A).B' + (B-A)^2 = 0

> We replace our data in that equation:
> ( 3^2 - 5^2 ) * t^2 + 2*t*(B-A)*3 + (B-A)^2 = 0;
^^^

No. (B-A) and B' are both vectors. The '.' in between indicates a
scalar product, a.k.a. "dot product". That can't be replaced by a
multiplication by the length of B'. You have to construct B' from
length 3 and angle 30 degrees, then carry out the dot product

> Since A and B are vectors


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

Replies:

Copyright © 2006 WatermarkFactory.com. All Rights Reserved.