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auto80
12th March, 2024, 08:07 PM
Hi Clusters

Can you tell me what is the calculation to get 26000 km, hex ( A8D4 1182 ), on this PSA Boxer file ? THANKS


Xeers

Andresolivares
12th March, 2024, 08:35 PM
A8 D4 11 82 57 2B EE 7D

jomberykaso
12th March, 2024, 09:14 PM
26000 *100 = 2600000 to hex : 00 27 AC 40 xor FF FF FF FF : FF D8 53 BF

Then you must encrypt data using this table and invert all bytes

00 = 82
01 = 80
02 = e0
03 = 03
04 = 20
05 = 00
06 = 81
07 = 40
08 = a1
09 = 06
0a = 87
0b = c1
0c = 84
0d = 01
0e = 61
0f = 05
10 = a2
11 = 0a
12 = 8b
13 = c2
14 = 88
15 = 02
16 = 62
17 = 09
18 = 8e
19 = 83
1a = e3
1b = 0f
1c = 23
1d = 0c
1e = 8d
1f = 43
20 = a4
21 = 12
22 = 93
23 = c4
24 = 90
25 = 04
26 = 64
27 = 11
28 = 96
29 = 85
2a = e5
2b = 17
2c = 25
2d = 14
2e = 95
2f = 45
30 = 9a
31 = 86
32 = e6
33 = 1b
34 = 26
35 = 18
36 = 99
37 = 46
38 = a7
39 = 1e
3a = 9f
3b = c7
3c = 9c
3d = 07
3e = 67
3f = 1d
40 = a8
41 = 22
42 = a3
43 = c8
44 = a0
45 = 08
46 = 68
47 = 21
48 = a6
49 = 89
4a = e9
4b = 27
4c = 29
4d = 24
4e = a5
4f = 49
50 = aa
51 = 8a
52 = ea
53 = 2b
54 = 2a
55 = 28
56 = a9
57 = 4a
58 = ab
59 = 2e
5a = af
5b = cb
5c = ac
5d = 0b
5e = 6b
5f = 2d
60 = b2
61 = 8c
62 = ec
63 = 33
64 = 2c
65 = 30
66 = b1
67 = 4c
68 = ad
69 = 36
6a = b7
6b = cd
6c = b4
6d = 0d
6e = 6d
6f = 35
70 = ae
71 = fc
72 = bb
73 = ce
74 = b8
75 = 0e
76 = 6e
77 = 39
78 = be
79 = 8f
7a = ef
7b = 3f
7c = 2f
7d = 3c
7e = bd
7f = 4f
80 = b0
81 = 42
82 = c3
83 = d0
84 = c0
85 = 10
86 = 70
87 = 41
88 = c6
89 = 91
8a = f1
8b = 47
8c = 31
8d = 44
8e = c5
8f = 51
90 = ca
91 = 92
92 = f2
93 = 4b
94 = 32
95 = 48
96 = c9
97 = 52
98 = b3
99 = 4e
9a = cf
9b = d3
9c = cc
9d = 13
9e = 73
9f = 4d
a0 = d2
a1 = 94
a2 = f4
a3 = 53
a4 = 34
a5 = 50
a6 = d1
a7 = 54
a8 = b5
a9 = 56
aa = d7
ab = d5
ac = d4
ad = 15
ae = 75
af = 55
b0 = b6
b1 = 5a
b2 = db
b3 = d6
b4 = d8
b5 = 16
b6 = 76
b7 = 59
b8 = de
b9 = 97
ba = f7
bb = 5f
bc = 37
bd = 5c
be = dd
bf = 57
c0 = e2
c1 = 98
c2 = f8
c3 = 63
c4 = 38
c5 = 3a
c6 = e1
c7 = 58
c8 = b9
c9 = 66
ca = e7
cb = d9
cc = e4
cd = 19
ce = 79
cf = 65
d0 = ba
d1 = 6a
d2 = eb
d3 = da
d4 = e8
d5 = 1a
d6 = 7a
d7 = 69
d8 = ee
d9 = 9b
da = fb
db = 6f
dc = 3b
dd = 6c
de = ed
df = 5b
e0 = bc
e1 = 72
e2 = f3
e3 = dc
e4 = f0
e5 = 1c
e6 = 7c
e7 = 71
e8 = f6
e9 = 9d
ea = fd
eb = 77
ec = 3d
ed = 74
ee = f5
ef = 5d
f0 = fa
f1 = 9e
f2 = fe
f3 = 7b
f4 = 3e
f5 = 78
f6 = f9
f7 = 5e
f8 = bf
f9 = 7e
fa = ff
fb = df
fc = 60
fd = 1f
fe = 7f
ff = 7d

ex: 00 27 AC 40 XOR: FF D8 53 BF
Encrypted: 82 11 D4 A8 7D EE 2B 57

Then bytes inverted :A8 D4 11 82 57 2B EE 7D


GOOD LUCK WITH YOUR PROJECT

auto80
12th March, 2024, 09:25 PM
Many Thanks jomberykaso (https://www.digital-kaos.co.uk/forums/member.php/390353-jomberykaso) for the great explanation :friendly_wink:

Xeers