[OpenLayers-Users] Point in parrallel with line formula

[newbie]

Dear Chris,
              I am actually using php at my backend is ok I will try to
figure out how to convert your given javascript function. So this function
is sufficient rite to decide in meters ? Thank you.

Chris Adams wrote:
> 
> Here is a JavaScript code which will calculate the distance in meters. The
> points are (lon1, lat1) and (lon2, lat2)
> 
> var R = 6371*1000; // m
> var dLat = (lat2-lat1).toRad();
> var dLon = (lon2-lon1).toRad(); 
> var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
>         Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) * 
>         Math.sin(dLon/2) * Math.sin(dLon/2); 
> var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
> var d = R * c;
> 
> Also, if you are using Java, you will want to preform all the operations 
> from previous equations sent to you from me, using floats or doubles.
> 
> newbie wrote:
>> Dear Chris, 
>>               My data is in Latitude/Longitude. Ok I have used the java
>> sample code giving in the exercise. Only thing now I need to confirm is
>> that
>> what is distance output in wat format ? Thank you.
>>
>> Chris Adams wrote:
>>   
>>> 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 ? >
>>>>>>     _______________________________________________ Users mailing
>>>>>> list
>>>>>>     Users at openlayers.org http://openlayers.org/mailman/listinfo/users
>>>>>>
>>>>>>
>>>>>> ------------------------------------------------------------------------
>>>>>> View this message in context: Re: Point in parrallel with line
>>>>>> formula 
>>>>>> <http://n2.nabble.com/Point-in-parrallel-with-line-formula-tp2742247p2746662.html>
>>>>>> Sent from the OpenLayers Users mailing list archive 
>>>>>> <http://n2.nabble.com/OpenLayers-Users-f1822463.html> at Nabble.com.
>>>>>> ------------------------------------------------------------------------
>>>>>>
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>>>>   
>>>>       
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