Re: 16-bit value in 3 bytes?

"Andre-John Mas" <andrejohn.mas@xxxxxxxxx> wrote in message news:3d4f6171-00fb-4047-a8ce-e756ce4917d3@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Hi,

I recently got myself a 'Spaceball 5000' and since there is no driver
for my platform, I am trying to reverse engineer the protocol. The
place where I am stuck is that a given value for a rotation is
represented as a 16-bit signed integer, using four bytes. From what I
can tell the first byte represents the sign and the remaining 3 bytes
are used to encode the value using the characters ".023569:<?
ABCDEGHKLMNPSenors", so no more than 7 bits are used in each 8 bit
value. Additionally the first of the four bytes either has a value H
or G, which I believe might represent the sign.

Can anyone help me working out how the 16-bit value is being encoded?
Is there a better place to ask this question?

Andre

Sounds like BCD

.

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