CODEWITHSUNDEEP
geometry2d
svanerik
svanerik

Reputation: 123

Perpendicular line in lat/long

I am trying to implement the equations presented here regarding finding a point on a given perpendicular distance from a line. Unfortunately the lines I receive are not straight. Is this due to me mixing Lat/long with regular x/y coordinates or have I done something else wrong?! 

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;  
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLat;  
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLong;

 double newLat2 = temp2.Latitude -  (distPoint / 2) * uLat;  
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLong;

Have changed the code now, and the variable names. Still receiving the fault :-(

double dist = 0.02;

double latDiff = secondTestPoint.Latitude - firstTestPoint.Latitude;
double longDiff = secondTestPoint.Longitude - firstTestPoint.Longitude;

double length = Math.Sqrt(latDiff * latDiff + longDiff * longDiff);
double uLat = latDiff / length;
double uLong = longDiff / length;

double newLat1 = secondTestPoint.Latitude + (dist / 2) * uLong;
double newLong1 = secondTestPoint.Longitude - (dist / 2) * uLat;

double newLat2 = secondTestPoint.Latitude - (dist / 2) * uLong;
double newLong2 = secondTestPoint.Longitude + (dist / 2) * uLat;

Here are the variable values:

latDiff = -0.0046187639236450195
longDiff = -0.0058203935623168945

length = 0.0074303405980227239
uLat = -0.62160864131505777
uLong = -0.78332796263279647

newLat1 = 58.39273776863341
newLong1 = 15.558675147558933

newLat2 = 58.408404327886061
newLong2 = 15.546242974732632

UPDATE: I have come to the conclusion that the fault is due to lat/long issues. It seems reasonable that it would generate faults to think that the lat/long are equivalent to squares when they in fact are not. Especially when working with northern europe.

Upvotes: 3

Views: 3012

Answers (3)

MyItchyChin
MyItchyChin

Reputation: 14041

Latitude and Longitude degrees are not 1:1.

Upvotes: 0

Gavin Miller
Gavin Miller

Reputation: 43875

In the formula you linked, the last 4 lines have a multiplication sequence as:

[dy, dx, dy, dx] == [uLong, uLat, uLong, uLat]

however you've used:

[dx, dy, dx, dy] == [uLat, uLong, uLat, uLong]

Your last 4 lines should be this:

double newLat1 = temp2.Latitude + (distPoint / 2) * uLong;   //dy appears first
double newLong1 = temp2.Longitude - (distPoint / 2) * uLat;  //then dx

double newLat2 = temp2.Latitude -  (distPoint / 2) * uLong;  
double newLong2 = temp2.Longitude + (distPoint / 2) * uLat;

Upvotes: 0

Daniel Martin
Daniel Martin

Reputation: 23578

At such a small scale, the difference between x/y and lat/long isn't relevant. You've done something else wrong; what you should have is:

 double distPoint = 0.02;

 double latDiff = temp2.Latitude - temp.Latitude;
 double longDiff = temp2.Longitude - temp.Longitude;
 double length = Math.Sqrt(latDiff * latDiff + > longDiff * longDiff); 
 double uLat = latDiff / length; 
 double uLong = longDiff / length;

 double newLat1 = temp2.Latitude + (distPoint / 2) * uLong;
 double newLong1 = temp2.Longitude - (distPoint / 2) * uLat;

 double newLat2 = temp2.Latitude - (distPoint / 2) * uLong;
 double newLong2 = temp2.Longitude + (distPoint / 2) * uLat;

That is, the vector (uLat, uLong) is a unit vector in the direction of your line, so a perpendicular unit vector is (uLong, -uLat) - note that the coordinates swapped position, in addition to one being negated.

Upvotes: 2

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