Android Question Issue with DistanceTo

[Keith Flanagan]

Hi All

I think there is a problem with the DistanceTo method in Location.

When I run the following code I get a return value of 7482515 for something that is only 52ish meters away.

Any ideas what may be wrong here? Thanks

    Dim BigBenLongitude As Double = 51.500770
    Dim BigBenLatitude As Double = -0.124622
       
    Dim TescoLat As Double = 51.501137
    Dim TescoLong As Double = -0.125092

   
    Dim L1, L2 As Location
    L1.Initialize2(TescoLat, TescoLong)
    L2.Initialize2(BigBenLatitude, BigBenLongitude)
   
    Dim DistanceInMeters As Long = L1.DistanceTo(L2)
   
    ToastMessageShow("Distance: " & CRLF & DistanceInMeters,True) '7482515

See Google maps showing this:

Version Details:

 

[Jorge M A]

There is no reason to update B4A to 9.8.
Now it is 100% Free!

B4A – The simple way to develop native Android apps

B4A includes all the features needed to quickly develop any type of Android app. B4A is used by tens of thousands of developers from all over the world, including companies such as NASA, HP, IBM and others.

www.b4x.com

Then, maybe you want see this: https://www.b4x.com/android/forum/t...tween-two-different-points.108179/post-676401

 

[Keith Flanagan]

Thanks Jorge

I just updated and rechecked the output but it is still giving 7482515?

 

[drgottjr]

i get 66.1 meters.

here's the deal:
1) you NEED to read the documentation relating to the Location object
2) prepare moist towel for some head banging and blood sopping
3) you need to feed location2() 2 strings, not 2 variables declared as strings

specifically:

   ' works:
    L1.Initialize2("-0.124622","51.500770" )     ' you can use 3 formats: degrees, minutes or seconds
    L2.Initialize2("-0.125092", "51.501137")    ' you can use 3 formats: degrees, minutes or seconds
    
    ' she no work:
    L1.Initialize2(BigBenLatitude,BigBenLongitude )     ' no, no no
    L2.Initialize2(TescoLat, TescoLong)                          ' no, no, no

why, you ask? good question.

 

[Keith Flanagan]

Thanks drgottjr

It turns out I had swapped the values for Log and Lat for Big Ben in my first code.

Below is the code with the variables the correct way and now I get 52, which is as close to google maps.

    Dim BigBenLatitude As Double = 51.500770
    Dim BigBenLongitude As Double = -0.124622
        
    Dim TescoLat As Double = 51.501137
    Dim TescoLong As Double = -0.125092
    
    Dim BigBenLoc, TescoLoc As Location
    TescoLoc.Initialize2(TescoLat, TescoLong)
    BigBenLoc.Initialize2(BigBenLatitude, BigBenLongitude)
'   
    Dim DistanceInMetersFromBigBen As Long = BigBenLoc.DistanceTo(TescoLoc)
    ToastMessageShow("Distance: " & CRLF & DistanceInMetersFromBigBen,True) '52

 

[Jorge M A]

Just for additional information:

you need to feed location2() 2 strings,

why, you ask?

That's because the formats that the object accepts:

Values can be formatted in any of the three formats:
Degrees: [+-]DDD.DDDDD
Minutes: [+-]DDD:MM.MMMMM (Minute = 1 / 60 of a degree)
Seconds: [+-]DDD:MM:SS.SSSSS (Second = 1 / 3600 of a degree)
Example:

Dim L1 As Location
L1.Initialize2("45:30:30", "45:20:15")

 

[barx]

Maybe a road was closed for works and had to take a diversion ?

 

[drgottjr]

Just for additional information:
That's because the formats that the object accepts:

sorry, you misunderstood the question. the format supplied was one of the accepted formats. the question was related to why passing a value as a string variable gave a different answer than when the same value as a literal string was passed.

 

[Jorge M A]

You are right!

 

[klaus]

Just to clarify.
The problem was not the variable !
But the inversion between latitude and longitude values !!!
As already reported by Keith Flanagan in post #5.
The code in post #1 is OK, with the correct lat / lng values!
I tested it!

Just as a reminder:

Dim BigBenLongitude As Double = 51.500770 ' longitude'
Dim BigBenLatitude As Double = -0.124622
      
Dim TescoLat As Double = 51.501137 '<<<< latitude
Dim TescoLong As Double = -0.125092

 

[netsistemas]

I use this code (i'm not sure that mi code shared are ok). i don't know this other funcion.

Sub  Distancia(lat1 As Double, lon1 As Double, lat2 As Double, lon2 As Double)  As Double
  Dim  p As Double = 0.017453292519943295  ' // Math.PI / 180
  Dim a  As Double = 0.5 - Cos((lat2 - lat1) * p) / 2 +     Cos(lat1  * p) * Cos(lat2 * p) *     (1 - Cos((lon2 - lon1) * p)) / 2

  Return 12742 * Sin(Sqrt(a))' // 2 * R; R = 6371 km
End Sub

 

[Keith Flanagan]

Thanks All

The issue was with my variables rather then the DistanceTo method.

In the original code in post 1 I have Lat and Long the wrong way around for Big Ben

My bad!

 

[klaus]

I use this code (i'm not sure that mi code shared are ok). i don't know this other funcion.

Your code is OK for small distances.
It will give less precise values than DistanceTo.

 

[klaus]

The issue was with my variables rather then the DistanceTo method.

That's what I confirmed in post #10 .