Re: Reflecting a ray off a line

"Dieter Hansen" <dieter.hansen@[EMAIL PROTECTED]
> wrote in message 
news:4749c664$1@[EMAIL PROTECTED]
> Richard schrieb:
>>,,,
>>
>> There are two s's in the above.
>
> Well I didn't really look thru all the math you did (sorry), since I 
> already provided you with a solution (which should work, if I didn't
miss 
> something in the hurry), so you could check it against your results.
>
>>
>> Also, I presume that .x or .y just means only put an x or y coordinate 
>> into
>> the equation where there is a .x or a .y., not an ordered pair. Rich.
>
>
> Yes .x means the x component of the vector and .y the y component.
>
> Dieter

Hi.

Well, I've got someone else checking the math, so that I can be sure about

(6) - which is to get s. Certainly you have given everything  to be able
to 
compute the intersection point (P(s). As long as we have Ro + t Rdir = Po
+ 
s (P1 - Po) we can begin to work things out.

The next step is to figure out the math to get Rdir in terms of x and y,
for 
the next line segment (ray), the ray that is reflecting off the first
line. 
I'll give it some thought.

You have already been very helpful, thanks a lot.

Richard