Im dumb, Help me with my wideband question and I'll give you a discount!
#1
Im dumb, Help me with my wideband question and I'll give you a discount!
Ok so I finally got my 3rd wideband (2 bad widebands in a row from Innovate). Well it works and I am datalogging with ecucontrol but I have to log the raw wideband voltage. I need a formula to convert the voltages into a/f ratio's. I havent taken math for years and can't remember.
Example:
0v = 7.35afr
5v = 22.39afr
So I need to figure out what the incriments of voltage would correspond to as afr's.....like 3.78v would be ?? I would think there is a formula to work it out in excel, any input would be helpful, I'll throw a discount on any of my products to whoever can do this for me.
Example:
0v = 7.35afr
5v = 22.39afr
So I need to figure out what the incriments of voltage would correspond to as afr's.....like 3.78v would be ?? I would think there is a formula to work it out in excel, any input would be helpful, I'll throw a discount on any of my products to whoever can do this for me.
#2
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
make a X | Y table
EX:
X | Y
0 | 0
1 | 4
2 | 8
x=4y, or 1v = 4a (in this example)
dunno if that example was real or not. i got the table so i get points right? i gotta get back to work so i'll check back afterwords and get shopping
EX:
X | Y
0 | 0
1 | 4
2 | 8
x=4y, or 1v = 4a (in this example)
dunno if that example was real or not. i got the table so i get points right? i gotta get back to work so i'll check back afterwords and get shopping
#3
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
#4
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
Use the proportion formula.
X1=Y1*X2/Y1
example: X=3.78*22.39/5
X=16.93
The following is true only when using Innovate's factory settings.
Where R=Reading in Volts and X= measured A/F
X=(R*3.00+7.35
Example: R=2.45
X=(2.45*3.00+7.35
X=14.72
This should provide you with the values you are seeking. Good luck! And where's my discount
X1=Y1*X2/Y1
example: X=3.78*22.39/5
X=16.93
The following is true only when using Innovate's factory settings.
Where R=Reading in Volts and X= measured A/F
X=(R*3.00+7.35
Example: R=2.45
X=(2.45*3.00+7.35
X=14.72
This should provide you with the values you are seeking. Good luck! And where's my discount
#6
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
Yes you can program its output to any 0-5v voltage range...linear.
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
#7
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
Originally Posted by xenocron
Yes you can program its output to any 0-5v voltage range...linear.
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
#8
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
Originally Posted by xenocron
Yes you can program its output to any 0-5v voltage range...linear.
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
So you can use the slope/line formulas.
Y =mX + B
where M is the slope and B is the Y intercept
m = (Y2 - Y1)/(X2 - X1)
Remember 10th grade
So if you make .5v = .5 lambda and 4.5v = 1.5 lambda...what voltage is 1 lambda (14.7 AFRs)?
m = (4.5 - .5) / (1.5 - .5)
slope = 4
b = Y - mX
b = 4.5 - 4(1.5)
b = -1.5
Y = 4(1) - 1.5
Y = 2.5
So at approx 2.5v you will have 1 lambda or 14.7
I think I'm thinking straight today :P
damn u guys are smart!
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