[OpenLayers-Users] Point in parrallel with line formula

[Chris Adams]

For the second part, you use this equation first, to find 'u':

http://local.wasp.uwa.edu.au/~pbourke/geometry/pointline/pointline2.gif
(Where (x1,y1)-(x2,y2) is the line, and (x3,y3) is the point)

Then, you use this equation:

x = x1 + u (x2 - x1)
y = y1 + u (y2 - y1)

to calculate the new point (x,y)

Then you have to use some distance equation, to find the distance 
between (x3,y3) and (x,y).

Is your data in Latitude/Longitude? or Spherical Mercator? or something 
else?

newbie wrote:
> Dear Chris,
>               So once I have I have decide whether is +1 or -1 then I must
> apply the other formula. So must I still apply the first formula to decide
> +1 or -1 or can I just go to the second formula. Secondly I still cant
> really see the implementation of the formula. So is there any example
> implementation of this formula cause I dont know to how to get the answer in
> metres? Thank you.
>
> Chris Adams wrote:
>   
>> +1 means to the left
>> -1 means to the right
>>
>> before calling the 'ccw' function, you'll want to order p0 and p1 so 
>> that p0 has the smaller latitude
>>
>> To get the distance on the left or right, use the formula linked from my 
>> second response. It finds the closest point, on the line to the other 
>> point, and then you need to take the distance between the two. If you 
>> use Pythagorean theorem, it will give you distance in whatever 
>> measurement  the projection of your map is in.
>>
>> newbie wrote:
>>     
>>> Dear Chris, I am a bit confuse. So the d will be in what measurement 
>>> is it meter or km ? Then if is + or - what is the difference too. 
>>> Thank you.
>>>
>>>     Chris Adams wrote:
>>>     Here's a C++ function (Should be easy to convert to JavaScript) /*
>>>     Taken from Robert Sedgewick, Algorithms in C++ */ /* returns
>>>     whether, in traveling from the first to the second to the third
>>>     point, we turn counterclockwise (+1) or not (-1) */ int ccw( Point
>>>     p0, Point p1, Point p2 ) { int dx1, dx2, dy1, dy2; dx1 = p1.x -
>>>     p0.x; dy1 = p1.y - p0.y; dx2 = p2.x - p0.x; dy2 = p2.y - p0.y; if
>>>     (dx1*dy2 > dy1*dx2) return +1; if (dx1*dy2 < dy1*dx2) return -1;
>>>     if ((dx1*dx2 < 0) || (dy1*dy2 < 0)) return -1; if ((dx1*dx1 +
>>>     dy1*dy1) < (dx2*dx2 + dy2*dy2)) return +1; return 0; } >From this
>>>     post on GameDev.net
>>>     http://www.gamedev.net/community/forums/topic.asp?topic_id=457450
>>>     Note it will also return 0 if the point is precisely on the line.
>>>     The deals with turn directions, so it depends on the ordering of
>>>     the points on the line. To get around this, you'll want to swap p0
>>>     and p1, if if p1 is below p0. (If p1's latitude is smaller than
>>>     p0's) newbie wrote: > Dear All, > Just say I got a pair of lat and
>>>     long p1 and p2 and I draw a line > based on both the points. So
>>>     now I want to know if a particular point in the > parallel range
>>>     of say 50m either on left or right sides of the line. How can > I
>>>     decide this what is the formula please ? >
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>>>
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