[OpenLayers-Users] Get point on line?

[shane_china]

Bill Thoen, Thank you! I'll try to complete the algorithm in javascript by
myself reference the link you given me.

And also thank paweluz. Your method is my second choise. If I can't finish
the algorithm in javascript, I'll do it with the help of postgis db but also
with another ajax request to server that's the reason why it is the second
choise.

I regret why I haven't learned Math so well.


Bill Thoen wrote:
> 
> I put an article about great circle algorithms in the wiki some time ago 
> that might help you.  The links to the demos are down right now (that 
> server needs a power supply I think) but the algorithms and formulas are 
> shown on the wiki page at 
> http://trac.openlayers.org/wiki/GreatCircleAlgorithms. Hopefully that 
> will give you enough help to complete your task.
> 
> Unfortunately, I'm swamped with my own OpenLayers problems right now, so 
> I don't have any free time to help you much more than that. JavaScript 
> and OpenLayers is so much fun that it's all too easy to program yourself 
> into places you have no business being. 
> 
> But if you got the answers to all your problems immediately, your heart 
> wouldn't become purified by the long quest.
> 
> Good luck!
> 
> - Bill Thoen
> 
> 
> shane_china wrote:
>> Hi, Bill Thoen, I go to the website and find the formulary I want. As
>> following:
>> Point(s) known distance from a great circle
>> Let points A and B define a great circle route and D be a third point.
>> Find
>> the points on the great circle through A and B that lie a distance d from
>> D,
>> if they exist. 
>>    A = crs_AD - crs_AB
>>
>> ( crs_AB and crs_AD are the initial GC bearings from A to B and D,
>> respectively. Compute using Course between points) 
>>
>>    b = dist_AD
>>
>> (dist_AD is the distance from A to D. Compute using Distance between
>> points) 
>>
>>    r=(cos(b)^2+sin(b)^2*cos(A)^2)^(1/2)
>>
>> (acos(r) is the XTD) 
>>
>>    p=atan2(sin(b)*cos(A),cos(b))
>>
>> (p is the ATD) 
>>
>>    IF (cos(d)^2 > r^2) THEN
>>       No points exist
>>    ELSE
>>       Two points exist
>>      dp = p +- acos(cos(d)/r)
>>    ENDIF
>>
>> dp are the distances of the desired points from A along AB. Their
>> lat/lons
>> can be computed using Lat/lon given radial and distance 
>>
>> But I want to know, Is there any such fomulary in OpenLayers to help me
>> simplify the algorithm? For example, I know
>> "OpenLayers.Util.distVencenty",
>> this method can calculate the distance between two points. Does
>> openlayers
>> also have such functions can help me? Thank you.
>>
>> My math is not so good. Write the algorithm in js all by myself is so
>> difficult for me.
>>
>>
>>
>>
>> Bill Thoen wrote:
>>   
>>> Yes, it's a little different for the spherical (Lon/Lat) case. For those 
>>> formulas, check out Ed Williams' Aviation Formulary website at 
>>> http://williams.best.vwh.net/avform.htm. What you're looking for is 
>>> determining waypoint locations, i think.
>>>
>>> - Bill Thoen
>>>
>>> shane_china wrote:
>>>     
>>>> Bill Thoen wrote:
>>>>   
>>>>       
>>>>> shane_china wrote:
>>>>>     
>>>>>         
>>>>>> In openlayers,
>>>>>> I have a straight line with two endpoints. How could I get a point on
>>>>>> the
>>>>>> line with given distance to one endpoint? Thank you. 
>>>>>>
>>>>>>       
>>>>>>           
>>>>> You can find a point anywhere on a line by using a ratio calculation.  
>>>>> Start with the distance to your point divided by the total distance of 
>>>>> your line, then your new X coordinate = the difference between the X 
>>>>> coordinates of the endpoints times the ratio (plus the X coordinate of 
>>>>> the starting line endpoint. For example, if your endpoints are at 
>>>>> coordinates (X0, Y0) and (X1,Y1) then the Cartesian distance is
>>>>>
>>>>> D = sqrt((X1-X0)*(X1-X0) + (Y1-Y0)*(Y1-Y0))
>>>>>
>>>>> Then if the ratio between this and the distance to your point is d,
>>>>> your 
>>>>> new point's coordinates will be:
>>>>>
>>>>> X= X0+(X1-X0)*d/D
>>>>> Y=Y0+(Y1-Y0)*d/D
>>>>>     
>>>>>         
>>>> So thank you for your answer. Your advise is in X-Y coodinates. Does
>>>> this
>>>> method also apply to lon lat coodinates? Our earth is a ball, so is it
>>>> different?
>>>> I haven't demonstrate it yet.
>>>>   
>>>>       
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>>>
>>>
>>>     
>>
>>   
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