Some advice with knock and adjustment [pics]
Nov 3, 2006, 03:18 PM
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Originally Posted by Ralph
That run wasn't that good, 99%~. Tis why my fuel pump is being installed tomorrow 
And I even have a Walbro.
Nov 3, 2006, 03:19 PM
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Originally Posted by SophieSleeps
Ahh. I see. You're trying to gauge the AF from the narrowband.
Yeah I suggested for him to get a wideband in there ASAP.
In the meantime, we may make richening adjustments asap.
I'm trying to finger out how much of a change will correspond to what sort of change in AF.
So if I go from a value of 9.0 (Af = 11.5) to 8.9, what sort of changes in AF will I be expecting?
Yeah I suggested for him to get a wideband in there ASAP.
In the meantime, we may make richening adjustments asap.
I'm trying to finger out how much of a change will correspond to what sort of change in AF.
So if I go from a value of 9.0 (Af = 11.5) to 8.9, what sort of changes in AF will I be expecting?
You could not really say .8x or .9x = xx.x AFR 100%..
One of my logs shows .975 = 11.0
Another 0.897 = 11.5
But comparing the same log, if the narrowband reading is decreasing, the wideband AFR's are increasing..
Food for thought
Last edited by Ralph; Nov 3, 2006 at 03:48 PM.
Nov 3, 2006, 03:20 PM
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Originally Posted by Ralph
I meant to type 7000/200 change to 14.
My bad, typo.
My bad, typo.
Could I go one step further and ask where you are picking up your injector duty cycle from? Is that a column that EvoScan logs?
Nov 3, 2006, 03:22 PM
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Originally Posted by Ralph
Looking at my narrowband o2 values compared to wideband values..
You could not really say 9.x = xx.x AFR 100%..
One of my logs shows .975 = 11.0
Another 0.897 = 11.5
But comparing the same log, if the narrowband reading is decreasing, the wideband AFR's are increasing..
Food for thought
You could not really say 9.x = xx.x AFR 100%..
One of my logs shows .975 = 11.0
Another 0.897 = 11.5
But comparing the same log, if the narrowband reading is decreasing, the wideband AFR's are increasing..
Food for thought
I don't want to make a change from 8.9 to 9.0 if it's gonna bump my AF ratio's up by a full point. I guess I'll just do some experimentation.
Nov 3, 2006, 03:30 PM
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Originally Posted by SophieSleeps
Very cool. Thanks for explaining.
Could I go one step further and ask where you are picking up your injector duty cycle from? Is that a column that EvoScan logs?
Could I go one step further and ask where you are picking up your injector duty cycle from? Is that a column that EvoScan logs?
InjectorPulseWidth to Injector Duty Cycle
((IPW*RPM/60)/2)/1000
In Excel, highlight the cell box you want to add the IDC% Value to.. then in the function box type =SUM(((IPW*RPM/60)/2)/1000)
IPW = Whatever column shows "Injector Pulse Width" (Example X2)
RPM = The RPM column.. (Example H2)
Then press enter, after this double click on the bottom right of the box you just entered this formula in. This should expand that formula in the entire column..
Hope it wasn't too confusing.
AFTER this, right click on the letter of your column .. then click Format Cells. Go to Percentage and it should default to 2 decimal places, perfect. Press OK
.
Nov 3, 2006, 05:21 PM
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Originally Posted by Ralph
I use this formula
InjectorPulseWidth to Injector Duty Cycle
((IPW*RPM/60)/2)/1000
In Excel, highlight the cell box you want to add the IDC% Value to.. then in the function box type =SUM(((IPW*RPM/60)/2)/1000)
IPW = Whatever column shows "Injector Pulse Width" (Example X2)
RPM = The RPM column.. (Example H2)
Then press enter, after this double click on the bottom right of the box you just entered this formula in. This should expand that formula in the entire column..
Hope it wasn't too confusing.
AFTER this, right click on the letter of your column .. then click Format Cells. Go to Percentage and it should default to 2 decimal places, perfect. Press OK.
InjectorPulseWidth to Injector Duty Cycle
((IPW*RPM/60)/2)/1000
In Excel, highlight the cell box you want to add the IDC% Value to.. then in the function box type =SUM(((IPW*RPM/60)/2)/1000)
IPW = Whatever column shows "Injector Pulse Width" (Example X2)
RPM = The RPM column.. (Example H2)
Then press enter, after this double click on the bottom right of the box you just entered this formula in. This should expand that formula in the entire column..
Hope it wasn't too confusing.
AFTER this, right click on the letter of your column .. then click Format Cells. Go to Percentage and it should default to 2 decimal places, perfect. Press OK
.
Nov 3, 2006, 05:27 PM
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Originally Posted by SophieSleeps
Thanks. I thought there was an easier way of calculating it but I guess not.
Nov 3, 2006, 05:43 PM
#28
From what I read the formula is IDC=IPW*RPM/1200
"The fuel injector duty cycle (IDC) is the percentage of time the injector is supplied with power. The time during which the injector is powered (or activated) is called the injector pulse width (IPW). During normal engine operation, the fuel injector fires once during the four strokes of the Otto cycle, which last for 2 revolutions of the engine. As an example, at 3000 rpm it takes 0.040 seconds or 40 milliseconds (ms) for the engine to complete 2 revolutions (3000 rpm divided by 60 equals 50 revs per second; invert to get 0.02 sec per rev or 0.04 second for 2 revs). At 6000 rpm it takes 20 ms for two revolutions. If a fuel injector is activated for 15 ms (the IPW) at 3000 rpm the duty cycle is 37.5% (15 ms/40 ms), or rpm times IPW divided by 1200 equals IDC in percent. If an injector is powered for 15 ms at 6000 rpm, then IDC is 75% (15 ms/20 ms). If you know the engine speed (rpm) and the IPW (dataloggers can provide this information), then it is easy to calculate the IDC."
http://www.stealth316.com/2-calc-idc.htm
"The fuel injector duty cycle (IDC) is the percentage of time the injector is supplied with power. The time during which the injector is powered (or activated) is called the injector pulse width (IPW). During normal engine operation, the fuel injector fires once during the four strokes of the Otto cycle, which last for 2 revolutions of the engine. As an example, at 3000 rpm it takes 0.040 seconds or 40 milliseconds (ms) for the engine to complete 2 revolutions (3000 rpm divided by 60 equals 50 revs per second; invert to get 0.02 sec per rev or 0.04 second for 2 revs). At 6000 rpm it takes 20 ms for two revolutions. If a fuel injector is activated for 15 ms (the IPW) at 3000 rpm the duty cycle is 37.5% (15 ms/40 ms), or rpm times IPW divided by 1200 equals IDC in percent. If an injector is powered for 15 ms at 6000 rpm, then IDC is 75% (15 ms/20 ms). If you know the engine speed (rpm) and the IPW (dataloggers can provide this information), then it is easy to calculate the IDC."
http://www.stealth316.com/2-calc-idc.htm
Nov 3, 2006, 05:44 PM
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Originally Posted by nj1266
From what I read the formula is IDC=IPW*RPM/1200
"The fuel injector duty cycle (IDC) is the percentage of time the injector is supplied with power. The time during which the injector is powered (or activated) is called the injector pulse width (IPW). During normal engine operation, the fuel injector fires once during the four strokes of the Otto cycle, which last for 2 revolutions of the engine. As an example, at 3000 rpm it takes 0.040 seconds or 40 milliseconds (ms) for the engine to complete 2 revolutions (3000 rpm divided by 60 equals 50 revs per second; invert to get 0.02 sec per rev or 0.04 second for 2 revs). At 6000 rpm it takes 20 ms for two revolutions. If a fuel injector is activated for 15 ms (the IPW) at 3000 rpm the duty cycle is 37.5% (15 ms/40 ms), or rpm times IPW divided by 1200 equals IDC in percent. If an injector is powered for 15 ms at 6000 rpm, then IDC is 75% (15 ms/20 ms). If you know the engine speed (rpm) and the IPW (dataloggers can provide this information), then it is easy to calculate the IDC."
http://www.stealth316.com/2-calc-idc.htm
"The fuel injector duty cycle (IDC) is the percentage of time the injector is supplied with power. The time during which the injector is powered (or activated) is called the injector pulse width (IPW). During normal engine operation, the fuel injector fires once during the four strokes of the Otto cycle, which last for 2 revolutions of the engine. As an example, at 3000 rpm it takes 0.040 seconds or 40 milliseconds (ms) for the engine to complete 2 revolutions (3000 rpm divided by 60 equals 50 revs per second; invert to get 0.02 sec per rev or 0.04 second for 2 revs). At 6000 rpm it takes 20 ms for two revolutions. If a fuel injector is activated for 15 ms (the IPW) at 3000 rpm the duty cycle is 37.5% (15 ms/40 ms), or rpm times IPW divided by 1200 equals IDC in percent. If an injector is powered for 15 ms at 6000 rpm, then IDC is 75% (15 ms/20 ms). If you know the engine speed (rpm) and the IPW (dataloggers can provide this information), then it is easy to calculate the IDC."
http://www.stealth316.com/2-calc-idc.htm
Divided by 12000 Works.. comes out to exactly what I had..
