0x3C 0x7B 0x30 0x3d
0.0430863

0x7B - 0x7F = -4
2^-4 = 1/16 = 0.0625

0x7B - 0x80 = -5
2^-5 = 1/32 = 0.03125

0.0430863 = 0x00B07B3C / 2^28

So, 3rd byte 0x30 = ms byte of mantissa, implied set MSB of
mantissa (so 0x30 means 0xB0), sign bit of mantissa stored
in MSB, 0 = positive.

Multiplying 0.0430863 by the power of 2 making the result
closest to, but less than, 1.0 gives 0.6893804 * 2^-4.

So, the 0x3d byte must contain the exponent, probably biased
somehow. 0x3d = 61 = 00111101. -4 = 0xFC, or 11111100.

If the biasing is simple, the simplest way to obtain the
correct result might be to build a long using the three
mantissa bytes as the three low bytes (with implied sign),
then multiply by 2**(exp. byte - 89).

