in my LM-1 software i input the "xx psi = xx volts" parameters so, being an absolute sensor the 0v needs to have atmosphereic pressure (14.7psi) subtracted from it(ie: -14.7psi = 0v), so my question is this: following that logic, since the sensor charactoristics are 2.9psi @ 0v, do i need to take an additional 2.9 psi from that giving me -17.6psi to = 0v, or the other way around getting me -11.8psi to = 0v? whatever that answer is, i would take that amount off the top values as well, right? so 36.6 @ 5v would be 36.3 minus either 17.6 or 11.8 to give me "corrected" max pressure at 5v.
thanx...!
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#76156, "RE: LAST ONE...i promise...can ya help me with a Basic math conversion...? (psi vs. volts)" In response to Reply # 0 Sep-07-04 07:03 AM by Avenger
Just substract 14.7 from both numbers ... 2.9-14.7 and 36.3-14.7
#76157, "RE: LAST ONE...i promise...can ya help me with a Basic math conversion...? (psi vs. volts)" In response to Reply # 1
DOH!!!!!!!!!
(sheepishly... thanx)
USED TO HAVE...Everything you wish you had, and then some...
You are somebody special Tim, you matter! - from the admins (but nobody likes you, because your username is just like that other guy's, and you whine a lot)
The (250 - 20) is the pressure range and you have to -20 from your manifold pressure to account for the fact that the sensor doesn't see anything till 20kpa. The (4.9 - .2) is the voltage range. And the +.2 is the voltage start.
0 pounds of boost is gonna be ~1.83v max boost is gonna be ~22 pounds of boost - and it should hit 4.9v at that point
#76172, "RE: LAST ONE...i promise...can ya help me with a Basic math conversion...? (psi vs. volts)" In response to Reply # 3
^^^^^^ is THE MAN!!!!!!!!!!!!
thanx jim
USED TO HAVE...Everything you wish you had, and then some...
You are somebody special Tim, you matter! - from the admins (but nobody likes you, because your username is just like that other guy's, and you whine a lot)