How to convert my binary (hex) data to latitude an

问题:

I have some binary data stream which passes geo location coordinates - latitude and longitude. I need to find the method they are encoded.

4adac812 = 74°26.2851' = 74.438085
2b6059f9 = 43°0.2763'  = 43.004605

4adaee12 = 74°26.3003' = 74.438338
2a3c8df9 = 42°56.3177' = 42.938628

4ae86d11 = 74°40.1463' = 74.669105
2afd0efb = 42°59.6263' = 42.993772

1st value is hex value. 2nd & 3rd are values that I get in output (not sure which one is used in conversion).

I've found that first byte represents integer part of value (0x4a = 74). But I cannot find how decimal part is encoded.

I would really appreciate any help!

Thanks.

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Upd: This stream comes from some "chinese" gps server software through tcp protocol. I have no sources or documentation for clent software. I suppose it was written in VC++6 and uses some standard implementations.

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Upd: Here is packets I get:

Hex data:
41 00 00 00  13 bd b2 2c
4a e8 6d 11  2a 3c 8d f9
f6 0c ee 13

Log data in client soft:
[Lng] 74°40.1463', direction:1
[Lat] 42°56.3177', direction:1
[Head] direction:1006, speed:3318, AVA:1
[Time] 2011-02-25 19:52:19

Result data in client (UI):
74.669105
42.938628
Head 100 // floor(1006/10)
Speed 61.1 // floor(3318/54.3)


41 00 00 00  b1 bc b2 2c
4a da ee 12  2b 60 59 f9
00 00 bc 11
[Lng] 74°26.3003', direction:1
[Lat] 43°0.2763', direction:1
[Head] direction:444, speed:0, AVA:1
[Time] 2011-02-25 19:50:49
74.438338
43.004605


00 00 00 00  21 bd b2 2c
4a da c8 12  aa fd 0e fb
0d 0b e1 1d
[Lng] 74°26.2851', direction:1
[Lat] 42°59.6263', direction:1
[Head] direction:3553, speed:2829, AVA:1
[Time] 2011-02-25 19:52:33
74.438085
42.993772

I don't know what first 4 bytes mean.

I found the lower 7 bits of 5th byte represent the number of sec. (maybe 5-8 bits are time?) Byte 9 represent integer of Lat.

Byte 13 is integer of Lng.

Bytes 17-18 reversed (word byte) is speed.

Bytes 19-20 reversed is ava(?) & direction (4 + 12 bits). (btw, somebody knows what ava is?)

And one note. In 3rd packet 13th byte you can see only lower 7 bits are used. I guess 1st bit doesnt mean smth (I removed it in the beginning, sorry if I'm wrong).

回答1:

I have reordered your data so that we first have 3 longitures and then 3 latitudes:

74.438085, 74.438338, 74.669105, 43.004605, 42.938628, 42.993772

This is the best fit of the hexadecimals i can come up with is:

74.437368, 74.439881, 74.668392, 42.993224, 42.961388, 42.982391

The differences are: -0.000717, 0.001543, -0.000713, -0.011381, 0.022760, -0.011381

The program that generates these values from the complete Hex'es (4 not 3 bytes) is:

int main(int argc, char** argv) {
    int a[] = { 0x4adac812, 0x4adaee12, 0x4ae86d11, 0x2b6059f9, 0x2a3c8df9, 0x2afd0efb };
    int i = 0;
    while(i<3) {
        double b = (double)a[i] / (2<<(3*8)) * 8.668993 -250.0197;
        printf("%f\n",b);
        i++;
    }
    while(i<6) {
        double b = (double)a[i] / (2<<(3*8)) *  0.05586007 +41.78172;
        printf("%f\n",b);
    i++;
    }
    printf("press key");
    getch();
}

回答2:

Brainstorming here.

If we look at the lower 6 bits of the second byte (data[1]&0x3f) we get the "minutes" value for most of the examples.

0xda & 0x3f = 0x1a = 26; // ok
0x60 & 0x3f = 0; // ok
0xe8 & 0x3f = 0x28 = 40; // ok
0x3c & 0x3f = 0x3c = 60; // should be 56
0xfd & 0x3f = 0x3d = 61; // should be 59

Perhaps this is the right direction?

回答3:

I have tried your new data packets:

74+40.1463/60 74+26.3003/60 74+26.2851/60 42+56.3177/60 43+0.2763/60 42+59.6263/60

74.66910, 74.43834, 74.43809, 42.93863, 43.00460, 42.99377

My program gives:

74.668392, 74.439881, 74.437368, 42.961388, 42.993224, 39.407346

The differences are:

-0.000708,  0.001541,  -0.000722,  0.022758, -0.011376, -3.586424

I re-used the 4 constants i derived from your first packet as those are probably stored in your client somewhere. The slight differences might be the result of some randomization the client does to prevent you from getting the exact value or reverse-engineering their protocol.