**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

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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 =>

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