Hi Erik,

I understand what you're saying. I still need to be able to convert from
my lat/long to meters before I can use the trig.

E.g.
My latitude is 50.9051 degrees and longitude is -1.0649 degrees. My
angle is 45 degrees and my distance is 500m.

If I use the trig as it stands I would get a target of.
Latitude X2 = 50.9051 + 500 * sin(45) = 404.46

This is incorrect.

What I need to do is convert my lat/long into a meter grid - lets say
UTM.
I now have 636056 easting and 5641055 northing

Now the trig works

X2 = 5641055 + 500 * sin(45) = 5641409

Now convert back to degrees.

50.9083

What I need to be able to do is convert from my position in degrees to a
meter system.


Cheers

Tim




-----Original Message-----
From: Nielsen, Erik R [mailto:[EMAIL PROTECTED] 
Sent: 02 July 2004 11:21
To: Tim Smith
Subject: RE: MI-L New lat/long from distance/bearing


Tim,
Here's an idea.
Can you just create a line of any odd length but pointing in the right
direction 
( that's my X2= x1 + L * sin(ang) which as you point out isn't making a
lot of sense (but use Deg_2_RAD conversion and it will)) Now get the
length MeterL of this object in meters (that should be possible even
with MapX) Now you wil have a L to meter conversion in the part of the
globe where you are (L/MeterL) You can now create a new line with the
right L and the right Direction.)

Is this an idea you can use?
Best regards
Erik

-----Original Message-----
From: Tim Smith [mailto:[EMAIL PROTECTED] 
Sent: 02 July 2004 10:48
To: SCISOFT; Nielsen, Erik R
Cc: [EMAIL PROTECTED]
Subject: RE: MI-L New lat/long from distance/bearing


Ian,

The further the distance, the less accurate it needs to be.
So +/- 1 degree is probably the best way to represent the accuracy I
need. e.g. roughly +/- 1.7km @ 100km and 1.7m @ 100m etc.

Cheers

Tim

-----Original Message-----
From: SCISOFT [mailto:[EMAIL PROTECTED] 
Sent: 02 July 2004 10:15
To: Tim Smith; 'Nielsen, Erik R'
Cc: [EMAIL PROTECTED]
Subject: RE: MI-L New lat/long from distance/bearing


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
> 
> 
> This email and any attached files are confidential and copyright 
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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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