Tim
What sort of accuracy do you need? You talk about less than 100km distances,
but is accuracy to +/- 20m or to +/- 1m what you can tolerate?

Ian Thomas
GeoSciSoft - Perth, Australia


> -----Original Message-----
> From: Tim Smith [mailto:[EMAIL PROTECTED]
> Sent: Friday, 2 July 2004 4:57 PM
> To: Nielsen, Erik R
> Cc: [EMAIL PROTECTED]
> Subject: RE: MI-L New lat/long from distance/bearing
> 
> Erik,
> 
> Thanks for your reply.
> 
> I am also working over small spaces (< 100km area), so I could use
> normal trig as you mention. But I must first convert the lat/long to a
> meter grid because my distance is in meters. This is because X2= x1 + L
> * sin(ang) will mean nothing if x1 is in decimal degrees and L is in
> meters.
> 
> Kind regards
> 
> Tim
> 
> -----Original Message-----
> From: Nielsen, Erik R [mailto:[EMAIL PROTECTED]
> Sent: 02 July 2004 09:43
> To: Tim Smith
> Subject: RE: MI-L New lat/long from distance/bearing
> 
> 
> Tim,
> I've been following this with interest. I don't know MapX that well and
> I'm normally working with smaller areas, that can be assumed euklidean
> spaces (even if in lat-long). I liked Uffe's link, and those
> calculations should work fine. Sorry not to be of more help.
> Cheers
> Erik
> 
> 
> -----Original Message-----
> From: Tim Smith [mailto:[EMAIL PROTECTED]
> Sent: 02 July 2004 09:36
> To: [EMAIL PROTECTED]
> Subject: RE: MI-L New lat/long from distance/bearing
> 
> 
> Hi Erik,
> 
> This is fine. But I need to know how to convert from WGS84 lat-long into
> a meter grid so I can do the trig.
> 
> Tim
> 
> 
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> 
> X2= x1 + L * sin(ang)
> Y2 = Y1+ L* cos(ang)
> 
> Or switch sin or Cos if it doesn't look right, don't test with 45 degree
> angle, test with zero or 90. Best regards Erik
> 
> -----Original Message-----
> From: Tim Smith [mailto:[EMAIL PROTECTED]
> Sent: 29 June 2004 15:56
> To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: RE: MI-L New lat/long from distance/bearing
> 
> 
> Hi,
> Thanks for people who have helped so far with this,
> 
> I want to draw a line in MapX between two points. The function in MapX
> to do this takes two WGS-84 lat/long points. I don't have two lat/long
> points. I have one lat/long, an angle in degrees and a length in meters.
> I need to calculate the second lat/long (unless there is a MapX function
> that can draw a line - given a position, angle and distance).
> 
> A C++ algorithm to do it would be fine. I can't believe MapX doesn't
> support such a calculation.
> 
> Tim
> 
> 
> 
> -----Original Message-----
> From: Terry McDonnell [mailto:[EMAIL PROTECTED]
> Sent: 29 June 2004 15:42
> To: Tim Smith
> Subject: RE: MI-L New lat/long from distance/bearing
> 
> 
> Is your problem what algorithm to create to do this or how to do it Mapx
> or code it in C++?  IOW do you not know how to go about it?
> 
> Terry
> 
> -----Original Message-----
> From: Tim Smith [mailto:[EMAIL PROTECTED]
> Sent: 29 June 2004 12:42
> To: [EMAIL PROTECTED]
> Subject: MI-L New lat/long from distance/bearing
> 
> 
> Hi List,
> 
> I have a latitude and longitude in WGS-84.
> I want to calculate a new point based on a bearing and distance from
> this position.
> 
> Difficult bit -
> I want to calculate this using MapX, or alternatively I could use some
> C/C++ code.
> 
> Thanks for any help with this
> 
> Kind regards
> 
> Tim Smith
> 
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