Im dumb, Help me with my wideband question and I'll give you a discount!
12-13-2005, 08:04 PM
#1
1.5 BAR
Thread Starter
Join Date: Sep 2003
Posts: 1,489
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.
12-13-2005, 08:16 PM
#2
1.5 BAR
Join Date: Jan 2005
Posts: 1,440
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
12-13-2005, 08:31 PM
#3
3.0 BAR
Join Date: Jan 2003
Posts: 4,438
Re: Im dumb, Help me with my wideband question and I'll give you a discount!
12-13-2005, 10:41 PM
#4
0.0 BAR
Join Date: Oct 2005
Posts: 0
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
+7.35Example: R=2.45
X=(2.45*3.00
+7.35X=14.72
This should provide you with the values you are seeking. Good luck!
And where's my discount
12-14-2005, 08:51 PM
#6
3.0 BAR
Join Date: Jul 2003
Posts: 6,449
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
12-14-2005, 11:27 PM
#7
1.5 BAR
Thread Starter
Join Date: Sep 2003
Posts: 1,489
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
12-15-2005, 12:37 AM
#8
1.0 BAR
Join Date: Aug 2005
Posts: 327
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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