Roomba Hack - Uart Questions

 
#1

Hi,

I have finally found a Roomba 620 I wish to use as a robot base. Everything works great, thanks to @RichardR examples and some very interesting threads.
Now I try to fully understand UART functions and the Roomba Open Interface, and sometimes it is a bit confusing... So, at this point I need some expert enlightenments.

Here is an example I don't understand:

Richard R made a script for object avoidance and it works great :

Quote:


:top
$drive=1
uartinit(0, 2, 57600) #initiate UART port 2 (pins 18 Tx and Pin 19 Rx)
sleep(20)
uartWrite(0, 2,128,132) #Init iRobot create and place in full mode
sleep(20)
uartWrite(0,2,137,0,125,128,0) #forward slow

$x=0
REPEATUNTIL($x>0)

uartWrite(0,2,142,7) #Roomba Bump and wheel drop sensor

IF (UARTAvailable(0,2)>0)
$x=UARTRead(0,2,1)
$x=GetByte($x)

ENDIF
print($x)

IF ($x=2) # left bumper hit
uartWrite(0,2,137,255,155,128,0) # back up
sleep(1000)
uartWrite(0,2,137,0,100,255,255) #turn right
sleep(2000)
ELSEIF ($x=1) # right bumper hit
uartWrite(0,2,137,255,155,128,0) # back up
sleep(1000)
uartWrite(0,2,137,0,100,0,1) #turn left
sleep(2000)
ELSEIF ($x>2) # centre bump or wheel drop
uartWrite(0,2,137,255,155,128,0) # back up
sleep(1000)
uartWrite(0,2,137,0,100,0,1,157,0,180,137,0,0,0,0) #turn around 180deg
sleep(2000)
ENDIF

sleep(20)
ENDREPEATUNTIL
goto(top)



But I don't understand the GetByte values: $x=2 correspond to "left bumper hit", $x=1 to "right bumper hit" and $x>2 to "centre bump or wheel drop"... but when I look at the OI documentation :

User-inserted image

I don't see the fit with the bit values and I don't see how the Range could be 0-15 with only one byte...

Thanks for the help and sorry if it is a stupid question (I am afraid it will not be the last Grin)

#2

The bit positions all have a decimal weight.

Bit position 0 set to binary 1 has a weight of 1

Bit position 1 set to binary 1 has a weight of 2

Bit position 2 set to binary 1 has a weight of 4

Bit position 3 set to binary 1 has a weight of 8


If all 4 bit positions are set then the decimal value would be 15. 1+2+4+8=15

#3

Thanks @Robot Doc.

All these "bit", "bytes", "16-bit values", "high bytes", "two’s complement" notions are completely new to me. But I think I get it...

So, to take the same example, if there is both a "Bump Left" and a "Wheel Drop Left", it will correspond to a bit position 1 (weight of 2) + a bit position 3 (weight of 8) and so, in the script, the $x value will be 10. Right?

#4

Use GetBit()

Code:



$x = 2

if (getbit($x, 0))

print("Bump Right")

endif

if (GetBit($x, 1))

print ("Bump Left")

endif

if (GetBit($x, 2))

print ("Wheel Drop Right")

endif

if (GetBit($x, 3))

print ("Wheel Drop Left")

endif

#5

I am at work now.... i will check this thread later when i get home.... i think this script i wrote for the 400 series roomba.... i think the baud rate is wrong for the 600 series. .. @Doc can help you too after all he is the roomba doctor Smile

#6

That script is a little rough... and the baud rate is wrong. It should be 115200 not 57600... The 600 series uses 115200 baud.... also you only need to initialize the UART once.....

#7

@Richard' you are right. I already have modify baud rate and, as I said, the script works just fine (that's why it is not "required assistance thread" Smile ).

But I want to better understand the way EZ Builder and Roomba OI works together to go further in my future scripts, and your script was a good example to illustrate what I understand and what I don't...

@DJ, thanks for the tip and the script example

#8

@fredebec... Ok no worries..... I have another script that can read 2 byte data like Roomba battery voltage.... I'll post it for you tomorrow (my time) afternoon when I get home from work...

#9

fredebec

Quote:

So, to take the same example, if there is both a "Bump Left" and a "Wheel Drop Left", it will correspond to a bit position 1 (weight of 2) + a bit position 3 (weight of 8) and so, in the script, the $x value will be 10. Right?



Yes, you got it !

#10

@DJ,
Ok, I see why it is interesting to use GetBit(). But now, I don't understand where the $x=2 come from....

The function is GetBit(Value, Bit). Why the value is $x=2 in that case ?

Thank you all for your help.

@Richard, thank you for your scripts, it helps me a lot to have concrete examples.