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  • 3 weeks later...

I was talking to John Boston the other day and he told me that there is a small possibility that this event might not even go ahead. !!!! :happy:

Please tell me that this isn't true, I'm pretty convinced that we have a good chance of winning one of those wrx's.

Gee I hope that's not true! I've booken my flights and accomodation............... and I've also built a new WRX to the same specification to practice in, may sound silly but there half a mill on the line here!

yeah, 300 of them, they all paid up $13k in the first week out of their poket money, Jason can't be contacted as he is busy building 40 wrx's for us to thrash, and not only that, he has a lot of emials to answer, 287 of them, as to why the last 13 people got in for just $1300, lol.

Anyone bought my skyline yet? I have to pay $13k cause I'm not special!

Oh really! How awesome!

I've also heard that in keeping with the "Billy Big Time-edness" of this event, you know. AMG, TAG watches etc, no expence has been spared and the WRX will be delivery by;

ufo.jpg

Skillfully piloted by Jason himself; now you see him, now you don't.

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    • Yeah, that's fine**. But the numbers you came up with are just wrong. Try it for yourself. Put in any voltage from the possible range and see what result you get. You get nonsense. ** When I say "fine", I mean, it's still shit. The very simple linear formula (slope & intercept) is shit for a sensor with a non-linear response. This is the curve, from your data above. Look at the CURVE! It's only really linear between about 30 and 90 °C. And if you used only that range to define a curve, it would be great. But you would go more and more wrong as you went to higher temps. And that is why the slope & intercept found when you use 50 and 150 as the end points is so bad halfway between those points. The real curve is a long way below the linear curve which just zips straight between the end points, like this one. You could probably use the same slope and a lower intercept, to move that straight line down, and spread the error out. But you would 5-10°C off in a lot of places. You'd need to say what temperature range you really wanted to be most right - say, 100 to 130, and plop the line closest to teh real curve in that region, which would make it quite wrong down at the lower temperatures. Let me just say that HPTuners are not being realistic in only allowing for a simple linear curve. 
    • I feel I should re-iterate. The above picture is the only option available in the software and the blurb from HP Tuners I quoted earlier is the only way to add data to it and that's the description they offer as to how to figure it out. The only fields available is the blank box after (Input/ ) and the box right before = Output. Those are the only numbers that can be entered.
    • No, your formula is arse backwards. Mine is totally different to yours, and is the one I said was bang on at 50 and 150. I'll put your data into Excel (actually it already is, chart it and fit a linear fit to it, aiming to make it evenly wrong across the whole span. But not now. Other things to do first.
    • God damnit. The only option I actually have in the software is the one that is screenshotted. I am glad that I at least got it right... for those two points. Would it actually change anything if I chose/used 80C and 120C as the two points instead? My brain wants to imagine the formula put into HPtuners would be the same equation, otherwise none of this makes sense to me, unless: 1) The formula you put into VCM Scanner/HPTuners is always linear 2) The two points/input pairs are only arbitrary to choose (as the documentation implies) IF the actual scaling of the sensor is linear. then 3) If the scaling is not linear, the two points you choose matter a great deal, because the formula will draw a line between those two points only.
    • Nah, that is hella wrong. If I do a simple linear between 150°C (0.407v) and 50°C (2.98v) I get the formula Temperature = -38.8651*voltage + 165.8181 It is perfectly correct at 50 and 150, but it is as much as 20° out in the region of 110°C, because the actual data is significantly non-linear there. It is no more than 4° out down at the lowest temperatures, but is is seriously shit almost everywhere. I cannot believe that the instruction is to do a 2 point linear fit. I would say the method I used previously would have to be better.
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