A-PAC Shift Function -

[Tiger-Heli]

I have a question regarding the A-PAC shift function - The graphic here: http://www.ultimarc.com/a-pac.html (also shown below), implies that every input can be shifted, giving 28 inputs and the 4 directionals per joystick.  The text also says that the C input is used as the shift function.  Normally the C button is read as Button 11.  How do I get the "Shifted C button" to register as button 27.  If it's a double press, then the A-PAC can't differentiate between sending a button 27 and a double button 11.   If it's a long key press, then wouldn't the A-PAC send 11 and then 27, rather than just 27.  Perhaps the Player 2 C button shifts the inputs for Player 1 C, but I assumed from everything else I have read that the two sides of the board were independent.

On the I-PAC the shift button cannot send a shifted input.  I would think this was true on the A-PAC as well, but the graphic is supposed to be an actual screen capture and shows 28 buttons and an analog joystick, so something is not adding up to me, or maybe I'm missing something really obvious due to caffienne (sp?) deprivation?

[AndyWarne]

At the top of the page, it is mentioned that the total number of "logical buttons" (ie actual plus shift) is 62. This is correct, because as you have mentioned, the "shifted shift" button is of no use and so is not counted. Yes it does appear on the screen capture as a button though. Maybe a clarification of this is need in the FAQ, I'll add this.
Andy

[Tiger-Heli]

Thanks Andy.

Maybe what would be better would be to change the text above the screen capture to say something like "Buttton 27 is recognized by Windows, but corresponds to the shifted shift button, and therefore cannot be generated or used".

You also might want to revise the graphic as I have shown below to de-emphasize the Button 27 input (but then it isn't a true screen capture, so maybe not).

Use your best judgement, but thanks as always for addressing this issue so quickly.