Hi all,
Just getting started with TP and tuning.. very impressed so far.
I've an issue with getting the GM 1bar MAP sensor setup as a source.. I want to represent InHg as presented in the Aussie Ford EEC-IV/-EEC-V defs, which is below barometric eg 0 Ih Hg = WOT, 30 In Hg=max idle vac.
I have to use a GM MAP as the Ford MAP uses a frequency-based signal which I can't do anything with. I'm logging via my Tech Edge WBO2, as there's no real-time available (lack of defs) for these bins.
I've searched far and wide without any real success..
Anyone able to help me out with an equation to convert the MAP to InHg?
Thanks!
Ben
GM 1 Bar MAP Conversion Equation (Volts to In Hg)?
Moderators: Mangus, robertisaar, dex
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- Author of Defs
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- Joined: Sat Feb 21, 2009 3:18 pm
- Location: Camden, MI
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- Author of Defs
- Posts: 962
- Joined: Sat Feb 21, 2009 3:18 pm
- Location: Camden, MI
if it is 10 bit, then:
X * 0.0048828125 = MAP Volts
30 - (X * 0.029296875) = in-hg
examples:
X = 1024 (5 volts, 0 vacuum)
30 - (1024 * 0.029296875 ) = 0 in-hg
and for high vacuum... X = 0
30 - (0 * 0.029296875) = 30 in-hg
that is, assuming 0 volts = 30 in-hg... i know with the GM 1BAR sensors i'm used to, 0 volts = 10.354kPa.
X * 0.0048828125 = MAP Volts
30 - (X * 0.029296875) = in-hg
examples:
X = 1024 (5 volts, 0 vacuum)
30 - (1024 * 0.029296875 ) = 0 in-hg
and for high vacuum... X = 0
30 - (0 * 0.029296875) = 30 in-hg
that is, assuming 0 volts = 30 in-hg... i know with the GM 1BAR sensors i'm used to, 0 volts = 10.354kPa.
Probably best to search for a datasheet for the part number you are using. For example the 12247571 two bar map sensor gives this for its transfer:
At 25 deg C and 5.1 v supply the output is:
20 kPa = 0.199 v to 0.413 v
40 kPa = 0.755 v to 0.877 v
160 kPa = 3.815 v to 3.937 v
200 kPa = 4.814 v to 4.978 v
As you can see, it's not quite a linear output so you generally take the best fit line for the equation. Also, quite a few A-D convertors give the steps as ranging from 0 v to 5.12 v as these numbers are easily scaled (I suspect the 5.1 v quoted above is rounded to 1 decimal place).
At 25 deg C and 5.1 v supply the output is:
20 kPa = 0.199 v to 0.413 v
40 kPa = 0.755 v to 0.877 v
160 kPa = 3.815 v to 3.937 v
200 kPa = 4.814 v to 4.978 v
As you can see, it's not quite a linear output so you generally take the best fit line for the equation. Also, quite a few A-D convertors give the steps as ranging from 0 v to 5.12 v as these numbers are easily scaled (I suspect the 5.1 v quoted above is rounded to 1 decimal place).
Thanks for the reply Rob.
Using your equation, I get -196 (atmosphere) to +30 (full vac).
My WBO2 uses the following in the ADX (kindly supplied by TP) for its User1Raw to User1Volts conversion: x*5/8184
Don't know why it's not the 10-bit 1024, something about legacy hardware...
While I'm good with cars and tech, sadly my maths always sucked, just not wired for it!
Given the 8184 scale, are you able to suggest a formula that will work?
Also, the InHg value is the other way around - it's +30InHg for full vacuum and 0 for WOT. No idea why it's unsigned.. but the graphs look funny with a negative value. I can't come up with a way to un-sign it from your formula.. I know, I'm dim
Using your equation, I get -196 (atmosphere) to +30 (full vac).
My WBO2 uses the following in the ADX (kindly supplied by TP) for its User1Raw to User1Volts conversion: x*5/8184
Don't know why it's not the 10-bit 1024, something about legacy hardware...
While I'm good with cars and tech, sadly my maths always sucked, just not wired for it!
Given the 8184 scale, are you able to suggest a formula that will work?
Also, the InHg value is the other way around - it's +30InHg for full vacuum and 0 for WOT. No idea why it's unsigned.. but the graphs look funny with a negative value. I can't come up with a way to un-sign it from your formula.. I know, I'm dim
Thanks Dex, yes I've had a look at the spec and I saw it wasn't quite linear. Didn't know what to do about that, but if best-fit works okay I'll use something that approximates.dex wrote:Probably best to search for a datasheet for the part number you are using. -snip-
As you can see, it's not quite a linear output so you generally take the best fit line for the equation. Also, quite a few A-D convertors give the steps as ranging from 0 v to 5.12 v as these numbers are easily scaled (I suspect the 5.1 v quoted above is rounded to 1 decimal place).
Unfortunately I'm not smart enough with numbers to make the leap from the kpa conversion to InHg.. that's why I've come to the experts for help!
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- Author of Defs
- Posts: 962
- Joined: Sat Feb 21, 2009 3:18 pm
- Location: Camden, MI
8184?
kind of odd, but that would imply VERY close to 14-bit resolution... 14 bit being 8192.
are you following the order of operations with that equation? i can't see a way to get negative numbers out of that equation without switching steps around.
30 - (X * 0.029296875) = in-hg
(X * 0.029296875) is done first
then take 30 and subtract that result from it.
but since we're dealing with what appears to be 14 bit resolution, then the equation changes a bit...
30 - (X * 0.003665689) = in-hg
that's using the 8184. if it is a true 14 bits, then
30 - (X * 0.003662109375) = in-hg
though at that point, you're probably well within sensor error range.
kind of odd, but that would imply VERY close to 14-bit resolution... 14 bit being 8192.
are you following the order of operations with that equation? i can't see a way to get negative numbers out of that equation without switching steps around.
30 - (X * 0.029296875) = in-hg
(X * 0.029296875) is done first
then take 30 and subtract that result from it.
but since we're dealing with what appears to be 14 bit resolution, then the equation changes a bit...
30 - (X * 0.003665689) = in-hg
that's using the 8184. if it is a true 14 bits, then
30 - (X * 0.003662109375) = in-hg
though at that point, you're probably well within sensor error range.