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Old Nov 3, 2006 | 03:18 PM
  #16  
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From: Secret Volcano Island
Originally Posted by Ralph
That run wasn't that good, 99%~. Tis why my fuel pump is being installed tomorrow
I'm running 11.3-11.6 from end to end and I'm seeing IDCs hit 100% at 5600rpm and climb all the way to 111% by 7000rpm. And I even have a Walbro.
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Old Nov 3, 2006 | 03:19 PM
  #17  
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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?
Looking at my narrowband o2 values compared to wideband values..

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.
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Old Nov 3, 2006 | 03:20 PM
  #18  
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From: Butthole, MA
Originally Posted by Ralph
I meant to type 7000/200 change to 14.

My bad, typo.
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?
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Old Nov 3, 2006 | 03:22 PM
  #19  
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From: Butthole, MA
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
Right right. I'm just trying to figure out roughly the effect on injector time.
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.
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Old Nov 3, 2006 | 03:30 PM
  #20  
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From: N/A
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?
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 .
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Old Nov 3, 2006 | 03:40 PM
  #21  
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From: Secret Volcano Island
Could you just do a simpler formula:

=RPM*IPW/1200 ?
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Old Nov 3, 2006 | 03:45 PM
  #22  
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From: N/A
That does not work ^
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Old Nov 3, 2006 | 04:08 PM
  #23  
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From: Columubs, OH
Ralph - Thanks for posting that formula. I've been meaning to look it up for a while.
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Old Nov 3, 2006 | 04:09 PM
  #24  
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No prob.
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Old Nov 3, 2006 | 05:21 PM
  #25  
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From: Butthole, MA
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 .
Thanks. I thought there was an easier way of calculating it but I guess not.
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Old Nov 3, 2006 | 05:27 PM
  #26  
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From: N/A
Originally Posted by SophieSleeps
Thanks. I thought there was an easier way of calculating it but I guess not.
Its really simple, but I like to explain things in detail :
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Old Nov 3, 2006 | 05:36 PM
  #27  
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From: CA
Originally Posted by -=SPECTRE=-
Could you just do a simpler formula:

=RPM*IPW/1200 ?

Yes, you are absolutely correct.

IDC (%) = IPW*RPM/1200

IPW is in ms.


I posted this a while back.

Eric

Last edited by l2r99gst; Nov 3, 2006 at 05:41 PM.
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Old Nov 3, 2006 | 05:43 PM
  #28  
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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
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Old Nov 3, 2006 | 05:44 PM
  #29  
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Originally Posted by Ralph
That does not work ^
It works fine.
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Old Nov 3, 2006 | 05:44 PM
  #30  
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From: N/A
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

Divided by 12000 Works.. comes out to exactly what I had..
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