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Old 01-13-2004, 08:14 PM   #1 (permalink)
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I thought I would share this info/facts on "Calculating Injector Pulse Width". This was taken from another webiste...I know this may be grounds for a lock. But mods/admins please consider the information contained below ! This should/needs to be a sticky anyway..
Or added to the articles section. Priceless info !

The thread was written by Ben Strader from EFI University. You would probally have to pay a few $$$$ to get this info normally.

So, here goes...

If we know the size of a particular injector and we also know the displacement of an engine, we can readily calculate the exact amount of time {or Pulse Width} to open an injector for any amount of volumetric efficiency and engine speed.

To do this we must first determine the Volumetric Efficiency of the engine at the time we want to calculate how much fuel to use.

Essentially this is the assumption we make when we change the numbers in a base fuel table. We are assuming that a given amount of fuel added to the mixture will create the proper air/fuel ratio based on the theory that we can work backwards from fuel flow and measured A/F ratio to get volumetric efficiency.

It goes something like this:

First, lets convert our engine displacement that is given in Cubic Inches to Cubic Feet. We do this so that we can calculate the engine airflow and display it in Cubic Feet/ Minute, or CFM.

To do this, there are 1728 cubic inches in one cubic foot, so we divide the number or cubic inches the engine displaces by 1728 to get cubic feet of engine displacement.

Example: an engine that displaces 350 cu in would displace .2025 cubic feet.


Now if we multiply the engine displacement by the number of engine “cycles” we will obtain airflow numbers in Cubic Ft/min, or CFM.

To do this we divide the engine RPM by 2 because it takes two revolutions to make one engine cycle. So if our engine is operating at 6000 RPM, it will be completing 3000 cycles per minute.

on each cycle the engine displaces .2025 cubic feet of air, so 3000 times .2025 equals…

607.5 cubic feet per minute of airflow or 607.5 CFM.

*Note: it is important to remember that we are assuming that the engine is also operating at 100% VE. If our engine did not completely fill its cylinders on each cycle, it would not displace .2025 cubic feet per hour.

To calculate the air-flow for any other VE percentage, you must multiply your cubic feet of airflow number by the VE percentage you want to represent.

Example:

.2025 Cubic feet/hour x 80% VE = .162 cubic feet/hour

For Now, lets assume 100% VE though.


Okay moving on, we now know that our engine will displace 607.5 CFM at 6000 RPM.

We know that one cubic foot of air weighs .076 lbs at standard atmospheric pressure and temperature.

So, if we multiply the CFM by .076, we will get our airflow numbers in Lbs/min.

Example:

607.5 x .076 = 46.17 lbs/hr of airflow


Remember that we are looking at the total engine displacement, so we need to divide this number by the number of cylinders to find the airflow for one cylinder so that we know how long to turn on each injector.

So 46.17 lbs/min divided by 6 cylinders would be 7.695 lbs/min for one cylinder.

Now, if we know a desired A/F ratio that we want to obtain we can just divide the lbs/hr of airflow by that number to get the amount of fuel required in lbs/min.

Example:

If we wanted a 13:1 A/F ratio, we would divide 7.695 by 13

7.695 / 13 = .592 lbs/min of fuel flow to obtain a 13:1 A/F ratio


Now let’s take a look at our injector:

We have injectors that are rated in lbs/hour of fuel flow, so we need to convert them to lbs/min. To do this we simply divide by 60 because there are 60 seconds in one minute.

Example:

An injector that flows 60 lbs/hr divided by 60 seconds would flow
1 lb/min of fuel.

Now that we know the maximum amount of fuel we can get from our injector in one minute, we can move forward.

Next, lets see what the maximum amount of time is to actually hold our injector open during the engine cycle.

We know that RPM / 2 = cycles per minute

If we divide this number again by 60 seconds, we will get cycles per second.

Example:

6000 / 2 = 3000

3000 / 60 = 50 cycles per second.


Now if we divide one second by the number of cycles per second, we will find out how much time it takes for each cycle.

Example:

1 / 50 = .02 seconds per cycle

This means that the maximum amount of time we have for one engine cycle is .02 seconds, or 20 milliseconds {20ms}

So if we held our injector open for 20ms at 6000 RPM, we would have 100% duty cycle, which is the maximum amount of fuel we could pass through the injector.

Now we can divide the required amount of fuel in lbs/min by the total lbs/min of fuel available.

Example:

We discovered earlier that we would need .592 lbs/min of fuel to obtain our 13:1 A/F ratio

And then we determined that our injector was capable of flowing a total of 1 lb/min.

So we divide:

.592 / 1 = .592

This means that we need to turn on our injector for 59.2% of the total time available or a 59.2% duty cycle.

Now we can see how much actual time this represents by multiplying the actual amount of time in 100% of the cycle by the duty cycle to get an actual injector pulse width.

Example:

20 ms x .592 = .0118 OR 11.8 ms of injector pulse width!

There we did it! We now know that a 350 cubic inch engine operating at 100% VE at 6000 RPM would require 11.8 ms from a 60 lb/hr injector to obtain a 13:1 A/F ratio!

Now just for a second, lets say we calculated all this out and gave our injector a command of 11.8 ms, but in actual engine operation our A/F ratio ended up being 12:1 instead of 13:1.

This would indicate that the engine was not actually operating at 100% VE as we assumed.

Lets see how far off we were:

Divide 13 / 12 = 1.083

That means we had 8.3% too much fuel. All things being equal, we could take 100% and subtract 8.3% to realize that our actual VE was closer to 91.7%

So we could take our 11.8ms pulse width and multiply it by .917 to obtain the correct amount of fuel for 91.7% VE

Example:

11.8 x .917 = 10.82 ms of fuel required to get us back to 13:1 A/F ratio!


Enjoy! I hope this helps some of you out there!

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Old 01-13-2004, 11:13 PM   #2 (permalink)
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http://www.hondata.com/techwidebandtuning.html


http://www.hondata.com/techpartthrottle.html

hondata has their fuel values.

accel dfi calmap and motec m4 (% IJPU) has their own calibration that relates air fuel to % pulse width fuel table calibration matrices.



I'm sure Tyler can briefly fill us in on the aem ems' units.



most programs today use interpolation software and so the calculated % pulse widths may not be the same as yours after you drive around initially because it uses info from surrounding fuel table neighbouring cells in the matrix to determine it's value as well.

each brand standalone has their own way.

thanks for posting this. good introduction basics.
perhaps we can apply it to our situation with 109 cu in. We like to shoot for 97% VE minimum...
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Old 01-14-2004, 01:03 AM   #3 (permalink)
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aem ems can display fuel data in 3 units... "raw", pulse width (in ms), and duty cycle.

i find it very useful, as the different styles can come in handy depending on what you're doing.

a rough-cut fuel map can be done in pulse width, because if you have one rpm/load cell's AFR, you can copy it across that whole load row, and still be close enough AFR across the rpm board. this works under the assumption that the VE of the motor won't vary *that* much, for any given load level. works for rough-cut.

you can then switch to duty cycle to see how smooth the fuel table is, and see if there's any funny spikes, slopes, or holes that may not be easily seen with pulse width. this is because there is a relatively consistent postive slope with increasing rpm, at a given load.

only problem i see with the math above is that it assume the fuel injector is either "open" or "closed"... which isn't necessarily the case, especially with larger injectors. they take a while to open, and a while to close--in short, 20ms doesn't necessarily deliver twice as much fuel as 10ms.

hondata has a nice graph of that on their site.
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Old 01-14-2004, 04:34 AM   #4 (permalink)
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That's a really good way to figure out the VE of a motor. I like that information more than calculating injector pulse width.(besides that's why I pay $180 an hour for my tuner to figure out)
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