PbPro Maths Help

[retepsnikrep]

In the first example it may help if i explain the range of values my forumla will encounter, for BD[4] it is $00-$18 and for BD[5] it is $00-$7F. The application is a current sensor data stream which measures from -100 to +50A

This i think will allow me multiply the result by 10 and still fit in a WORD before performing the divide by 20.48 increasing integer accuracy as I can divide by 205

A problem however is that the result of a multiplication can be negative

If the result of a multiplication could possibly be negative, it should be
stored to a long-sized variable type to preserve the sign. If a negative
result is placed in a variable type other than long, subsequent
calculations using this value will interpret it as a positive number.

I don't know how to manage that PBL?

For the second example the range of values for BD[2] is $21 -$34 and for BD[3] it's again $00-$7F This isnt a current sensor but relates to a SOC reading probably varying from 0-100%

[HenrikOlsson]

Hi,
I don't get it.... $18 and $7F is 24 and 127 respectively. Putting these in the formula should result in "real life values" ranging from -100 to 50 representing the current in amperes, correct?

((0*128)+0-2048) / 20.48 = -100 OK
((24*128)+127-2048) / 20.48 = 56 ??

Next question is what you're going to do with the values? If you're displaying them to a human then it makes sense to scale them properly but of you're continuing to do massage the numbers it doesn't matter what "units" they are in.

Anyway, how about:

Result var WORD
Sign VAR BIT
 
Result = (BD[4] * 128) + BD[5] - 2048
Sign = Result.15
Result = ABS Result
Result = Result * 2500
Result = DIV32 512
If Sign = 1 Then
LCDOUT $FE, 1, "-", #RESULT/100, ".", DEC2 RESULT//100
ELSE
LCDOUT $FE, 1, #RESULT/100, ".", DEC2 RESULT//100 
ENDIF

In the above, if BD[4]=3 and BD[5]=90 the correct result is -76.855 and the code displays -76.85. If BD[4]=19 and BD[5]=89 the correct result is 23.096 and the code displays 23.09. Is that close enough?

/Henrik.

[retepsnikrep]

Hi,
I don't get it.... $18 and $7F is 24 and 127 respectively. Putting these in the formula should result in "real life values" ranging from -100 to 50 representing the current in amperes, correct?

((0*128)+0-2048) / 20.48 = -100 OK
((24*128)+127-2048) / 20.48 = 56 ??
/Henrik.

Henrik and others thanks for your help.

Sorry yes the maximum value for $18 is actually $17 so thats puts the range covered within -100 to + 50.

That code looks really helpful, and is easily accurate enough, yes the answer is being displayed via a serial lcd device that's not a problem.

The second issue with the seperate lower formula is

Also for a second formula involving BD [2] & BD [3] I need to

(Multiply the right hand digit of 1st hex number by 128, add result to 2nd number)

I'm stuck on extracting a value for the right hand digit of the hex number?

As an aside it may be my formula for BD[2] & BD [3] is incorrect I suspect it should resolve to a number between 0-100% at it represents battery state of charge.
My captured range of values for BD[2] is $21 -$34 and for BD[3] it's again $00-$7F

Ignoring the first digit of BD [2] gives a range of values that seems to work

[HenrikOlsson]

Hi,
I'm confused about that second calculation....

My captured range of values for BD[2] is $21 -$34 and for BD[3] it's again $00-$7F
Ignoring the first digit of BD [2] gives a range of values that seems to work.

If you just look at the "right hand" digit of BD[2] your formula will return the same result when BD[2]=$21 and $31 and so on. Let's say BD[2] is $22 and BD[3] is $10 then your formula would be 2*128+16=272. Now lets say BD[2] is $32 and BD[3] is $10, your formula returns 2*128+16= 272 - is that really correct?

/Henrik.

[retepsnikrep]

OK thanks to all i believe I have the answers now.

For the last issue masking the four bits works.

BD[2] = bd[2] AND $0F

[retepsnikrep]

Forum members kindly helped me with this code a while back.

Amps = ((BCM87[4] * 128) + BCM87[5]) - 2048			'Calculate Battery Current
	if Amps.15 = 1 then						'Calculate AmpSign
	AmpSign = 45							'Set AmpSign to 45 (-)
	else
	AmpSign = 43							'Set AmpSign to 43 (+)
	endif
	Amps = ABS Amps
	Amps = Amps * 2500
	Amps = DIV32 512

It evaluates a 16 bit hex number from an array and produces a current reading to two decimal places with a + or - sign to denote charging or discharging. It work fine.

Now I need to manipulate the current, so say the current is +10.00A i need to be able to reduce it by say 8 or 16% and then generate two new hex bytes in the correct format to overwrite those it came from.

Imagine the device sits passing through the traffic current data. it now needs to get the data evaluate, modify it by X% and squirt it out the other side.

I would be grateful for ideas as maths is not my strong point. Thanks Peter

[aratti]

The following code should work.

Cheers

Al.

Percent  var byte
R_Amp var Word
I_Amp var Byte
D_Amp var Byte

main:
Percent = 8 ' set variable with your percent
gosub Calc
LDCOUT I_Amp, ".", D_Amp ' display values
goto main

Calc:
R_Amp = (100 - Percent) * Amps ' here Amps is your variable with the primitive value
I_Amp = R_Amp/100 ' integerpart
D_Amp = R_Amp//100 ' decimal part
Return

[retepsnikrep]

Thanks i think that helps with the percentage, but it's reconstructing the two hex bytes with the new data to forward on which is really tricky? If you get me?