Problem in Tension/compression

[Darrel Taylor]

If LastW0.15 = 1 then the result is negative.

I'm not familiar with your display, so I don't know how to show the minus sign.
<br>

[precision]

Thanks a million Sir Darrel Taylor. My project is over.

i add this line,

If LastW0.15 = 1 then 
   E6 = $BF
    ELSE
   E6 = $FF
 ENDIF
  
        HC_Latch=0
        shiftout HC_data, HC_Clk, MSBFIRST,[E1,E2,E3,E4,E5,E6]        
        pauseus 1
         HC_Latch=1
                
        '---------------------------------------
       ' endif   
         GOTO LOOP

Now my next step is "Calibration Factor" with Pitch of machine's screw.
I will start it Using Floating Point
Thanks again.

[Darrel Taylor]

I doubt that you need to use Floating Point.
It's a real pain.

Do you know the numbers yet?
pulses/mm or something like that?
<br>

[precision]

Sir my plan is,

When motor's pully moves 68 turns then Screw( worm gear ) turn 1 time.
Screw = 6 TPI ( then plateform moves 1 inch = 25.40 mm )
so 1 turn of screw = 25.40/6 = 4.233333 mm.
so 68 turns of pully = 4.233 mm
so 1 turn of pully = 4.233/68 = 0.062205882 mm

I will attach encoder of 6 PPR direct to motor pully.

code
pitch var word
pitch = 10368 ; (62208 / 6 = 10368)
dummy var word

loop: dummy = (pitch * ABS(lastw0))
w1 = div32 10000

so w1 = .01 mm least count
motor's 68 turns = encoder's 408 turns (10368 * 408 )/10000

so plateform moves 4.230 mm

This is my calculation

.

[Darrel Taylor]

Oh, OK.

I thought you meant the microchip Floating Point routines.

Sounds like you got it covered.
<br>