Here's a little something for all the "tuners" out there
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Here's a little something for all the "tuners" out there
Some people have asked if EFI 101 is a pre-requisite course for taking EFI advanced.
Simply, yes.
Some tuners have already installed and tuned several systems and feels that they are beyond the level of instruction provided by the introductory class.
While the EFI 101 class is entry level, we feel that many of the topics covered typically dispell many rumors and open up many truths about the how- and why's of engine's and fuel injection systems.
It is our opinion that even those with lots of hands-on practical experience in tuning may not have ever learned the fundamental scientific properties behind WHY they do things.......the point is that we feel that while there may be a lot of tuners putting the right numbers into the maps, there are few who actually can explain WHY those are the right numbers.
We know that not everything needed to be a profesional tuner is taught in our EFI 101 class, especially the number one ingredient.......many hours of practice.
However, by taking the EFI 101 course, we find that many learn much about what is actually taking place inside the engine and this leads to totally new tuning techniques and a more complete understanding of engine tuning in general.
Still, there are those who feel they have progressed beyond the need for rudimentary training and think that their time and money would be wasted on such a course. For this reason, we will be offering a "test-out" option where folks can pay a fee to take a test online. If they pass the test, then they can qualify for automatic entrance into the advanced class.
This test will NOT be easy and is not reccomended for most beginner and intermediate tuners. The test will only be allowed to be taken once to prevent someone from looking up the answers afterwards and trying again. The test is designed to test the very limits of one's tuning knowledge and some topics in the test will not have been covered in the EFI 101 class, so they can only have been learned thorugh one's own study of science, physics, and math in addition to large amounts of actual tuning experience.
Here are some examples of questions that may be on the test:
1} Explain in detail how an engine's BSFC affects the injector selection process?
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
3} If an engine had a wedge shaped combustion chamber with the sparkplug located near the exhaust valve, would it require more or less advance than a combustion chamber with a pent-roof design with the sparkplug in the center of the chamber, and why?
4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?
5} At the same operating point, what would be the lowest acheiveable air fuel ratio with the given parameters?
Well, these and about 20 other questions of varying difficulty will be on the test, so if you think you're up to it, then keep watching for the upcomming link to take the challenge!
A candidate must correctly answer at least 80% of the test questions in order to bypass the EFI 101 class.
By the way..............the test WON"T be open book!
Good Luck!
Simply, yes.
Some tuners have already installed and tuned several systems and feels that they are beyond the level of instruction provided by the introductory class.
While the EFI 101 class is entry level, we feel that many of the topics covered typically dispell many rumors and open up many truths about the how- and why's of engine's and fuel injection systems.
It is our opinion that even those with lots of hands-on practical experience in tuning may not have ever learned the fundamental scientific properties behind WHY they do things.......the point is that we feel that while there may be a lot of tuners putting the right numbers into the maps, there are few who actually can explain WHY those are the right numbers.
We know that not everything needed to be a profesional tuner is taught in our EFI 101 class, especially the number one ingredient.......many hours of practice.
However, by taking the EFI 101 course, we find that many learn much about what is actually taking place inside the engine and this leads to totally new tuning techniques and a more complete understanding of engine tuning in general.
Still, there are those who feel they have progressed beyond the need for rudimentary training and think that their time and money would be wasted on such a course. For this reason, we will be offering a "test-out" option where folks can pay a fee to take a test online. If they pass the test, then they can qualify for automatic entrance into the advanced class.
This test will NOT be easy and is not reccomended for most beginner and intermediate tuners. The test will only be allowed to be taken once to prevent someone from looking up the answers afterwards and trying again. The test is designed to test the very limits of one's tuning knowledge and some topics in the test will not have been covered in the EFI 101 class, so they can only have been learned thorugh one's own study of science, physics, and math in addition to large amounts of actual tuning experience.
Here are some examples of questions that may be on the test:
1} Explain in detail how an engine's BSFC affects the injector selection process?
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
3} If an engine had a wedge shaped combustion chamber with the sparkplug located near the exhaust valve, would it require more or less advance than a combustion chamber with a pent-roof design with the sparkplug in the center of the chamber, and why?
4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?
5} At the same operating point, what would be the lowest acheiveable air fuel ratio with the given parameters?
Well, these and about 20 other questions of varying difficulty will be on the test, so if you think you're up to it, then keep watching for the upcomming link to take the challenge!
A candidate must correctly answer at least 80% of the test questions in order to bypass the EFI 101 class.
By the way..............the test WON"T be open book!
Good Luck!
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (EFIGUY)
First 1 done, someone tell me if Im stupid.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">
1} Explain in detail how an engine's BSFC affects the injector selection process?
BSFC affects the injector selection process because if an engine has a BSFC of .40, that engine would consume approximately .4lbs of fuel per hour per horsepower. Using this base number, you can calculate the maxiumu injector needs of the engine at peak torque.
</TD></TR></TABLE>
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">
1} Explain in detail how an engine's BSFC affects the injector selection process?
BSFC affects the injector selection process because if an engine has a BSFC of .40, that engine would consume approximately .4lbs of fuel per hour per horsepower. Using this base number, you can calculate the maxiumu injector needs of the engine at peak torque.
</TD></TR></TABLE>
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (Suprdave)
Numba 2
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
The rule of thumb is that an engine will make ~1 bhp per 1.4SCFM of air, right?
So the engine should make ~1070bhp....
I was alittle stumped on lb\hr and cfm conversions...i was confusing myself.
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
The rule of thumb is that an engine will make ~1 bhp per 1.4SCFM of air, right?
So the engine should make ~1070bhp....
I was alittle stumped on lb\hr and cfm conversions...i was confusing myself.
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (EFIGUY)
3} If an engine had a wedge shaped combustion chamber with the sparkplug located near the exhaust valve, would it require more or less advance than a combustion chamber with a pent-roof design with the sparkplug in the center of the chamber, and why?
[b] The Wedge-Shaped Chamber will require less timing, because on the compression stroke the mixture is pushed toward the exhaust valve...Ina wedge design with the sparkplug being closest to the exhaust valve, it will require less time to complete the burn. Ona side note, the spark plug in the middle of a pentroof chamber will be more efficient because it will allow a more complete burn. If the mixture is ignited close to the exhaust port, you run the risk of some of the mixture being blown out the exhaust valve without full combustion.[b]
[b] The Wedge-Shaped Chamber will require less timing, because on the compression stroke the mixture is pushed toward the exhaust valve...Ina wedge design with the sparkplug being closest to the exhaust valve, it will require less time to complete the burn. Ona side note, the spark plug in the middle of a pentroof chamber will be more efficient because it will allow a more complete burn. If the mixture is ignited close to the exhaust port, you run the risk of some of the mixture being blown out the exhaust valve without full combustion.[b]
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (Suprdave)
4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?
CFM of Engine = 120 x 6500\3456 = 225.7 cfm, assume 1.4cfm per hp...so 161bhp? Assume a BSFC of .5, so .5lb of fuel per hp per hour...The engine requires
80.5lb of fuel per hour of running. 80.5\96 = 84% duty cycle on the injector. I believe 84% would only achieve stoich, 14.7:1. So ****...im stuck there
Am I way off on this?
CFM of Engine = 120 x 6500\3456 = 225.7 cfm, assume 1.4cfm per hp...so 161bhp? Assume a BSFC of .5, so .5lb of fuel per hp per hour...The engine requires
80.5lb of fuel per hour of running. 80.5\96 = 84% duty cycle on the injector. I believe 84% would only achieve stoich, 14.7:1. So ****...im stuck there
Am I way off on this?
Trending Topics
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (Suprdave)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Suprdave »</TD></TR><TR><TD CLASS="quote">Numba 2
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
The rule of thumb is that an engine will make ~1 bhp per 1.4SCFM of air, right?
So the engine should make ~1070bhp....
I was alittle stumped on lb\hr and cfm conversions...i was confusing myself. </TD></TR></TABLE>
Lets look at it this way Dave,
one cubic foot of air weighs .076 lbs, so 1500 cubic feet/min would weigh 114 lbs/min, so if our BSAC allowed us to make 7.5 Hp for every LB of air/min, that would be .....855 Hp
Good try though...this is a shining example of why those "rule of thumb" calcs are always so far off.
-ben
2} If an engine had a BSAC of 7.5, and consumed 1,500 cfm of air.......how much power could it be expected to make at standard atmospheric conditions?
The rule of thumb is that an engine will make ~1 bhp per 1.4SCFM of air, right?
So the engine should make ~1070bhp....
I was alittle stumped on lb\hr and cfm conversions...i was confusing myself. </TD></TR></TABLE>
Lets look at it this way Dave,
one cubic foot of air weighs .076 lbs, so 1500 cubic feet/min would weigh 114 lbs/min, so if our BSAC allowed us to make 7.5 Hp for every LB of air/min, that would be .....855 Hp
Good try though...this is a shining example of why those "rule of thumb" calcs are always so far off.
-ben
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (Suprdave)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Suprdave »</TD></TR><TR><TD CLASS="quote">3} If an engine had a wedge shaped combustion chamber with the sparkplug located near the exhaust valve, would it require more or less advance than a combustion chamber with a pent-roof design with the sparkplug in the center of the chamber, and why?
[b] The Wedge-Shaped Chamber will require less timing, because on the compression stroke the mixture is pushed toward the exhaust valve...Ina wedge design with the sparkplug being closest to the exhaust valve, it will require less time to complete the burn. Ona side note, the spark plug in the middle of a pentroof chamber will be more efficient because it will allow a more complete burn. If the mixture is ignited close to the exhaust port, you run the risk of some of the mixture being blown out the exhaust valve without full combustion.[b]</TD></TR></TABLE>
Dave,
I'm glad to see that this one got the gears turning in that overly enormous head of yours..{just F'ing with ya}, but typically a wedge shaped combustion chamber does not promote very efficient burn and because of this has a slower flame propagation speed. This results in the need for earlier ignition to give enough time to build cylinder pressure adequately before the piston gets away.
again.....good try
-B
[b] The Wedge-Shaped Chamber will require less timing, because on the compression stroke the mixture is pushed toward the exhaust valve...Ina wedge design with the sparkplug being closest to the exhaust valve, it will require less time to complete the burn. Ona side note, the spark plug in the middle of a pentroof chamber will be more efficient because it will allow a more complete burn. If the mixture is ignited close to the exhaust port, you run the risk of some of the mixture being blown out the exhaust valve without full combustion.[b]</TD></TR></TABLE>
Dave,
I'm glad to see that this one got the gears turning in that overly enormous head of yours..{just F'ing with ya}, but typically a wedge shaped combustion chamber does not promote very efficient burn and because of this has a slower flame propagation speed. This results in the need for earlier ignition to give enough time to build cylinder pressure adequately before the piston gets away.
again.....good try
-B
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (Suprdave)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Suprdave »</TD></TR><TR><TD CLASS="quote">4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?
CFM of Engine = 120 x 6500\3456 = 225.7 cfm, assume 1.4cfm per hp...so 161bhp? Assume a BSFC of .5, so .5lb of fuel per hp per hour...The engine requires
80.5lb of fuel per hour of running. 80.5\96 = 84% duty cycle on the injector. I believe 84% would only achieve stoich, 14.7:1. So ****...im stuck there
Am I way off on this?
</TD></TR></TABLE>
Okay, this one takes some math......look at it agin, and tomorrow if you haven't got it, I'll show you the way....right now its almost 2 am, and my white bread *** is hittin the sack.
I'll give you a hint:
forget about BSFC.
Look at the "TIME" factor and make all the units work together....we have rev's per "MINUTE" and Lbs per "Hour"
Think about how long it takes to complete one engine cycle....this would be the same amount of time as 100% duty cylce......you can't turn on the injector longer than that....soooooo how much fuel could that injector put out in that amount of time? What percentage would that be compared to the airflow....remeber, we want a ratio of air to fuel of 13:1
That should help you a bit...
-Ben
-B
Modified by EFIGUY at 1:30 AM 1/8/2004
CFM of Engine = 120 x 6500\3456 = 225.7 cfm, assume 1.4cfm per hp...so 161bhp? Assume a BSFC of .5, so .5lb of fuel per hp per hour...The engine requires
80.5lb of fuel per hour of running. 80.5\96 = 84% duty cycle on the injector. I believe 84% would only achieve stoich, 14.7:1. So ****...im stuck there
Am I way off on this?
</TD></TR></TABLE>
Okay, this one takes some math......look at it agin, and tomorrow if you haven't got it, I'll show you the way....right now its almost 2 am, and my white bread *** is hittin the sack.
I'll give you a hint:
forget about BSFC.
Look at the "TIME" factor and make all the units work together....we have rev's per "MINUTE" and Lbs per "Hour"
Think about how long it takes to complete one engine cycle....this would be the same amount of time as 100% duty cylce......you can't turn on the injector longer than that....soooooo how much fuel could that injector put out in that amount of time? What percentage would that be compared to the airflow....remeber, we want a ratio of air to fuel of 13:1
That should help you a bit...
-Ben
-B
Modified by EFIGUY at 1:30 AM 1/8/2004
Member
Joined: May 2003
Posts: 1,609
Likes: 1
Re: Here's a little something for all the "tuners" out there (EFIGUY)
AW man it's 4:30 am and I was hoping to find out,now I have to wait to tomorrow now DAMN.Ben is the man if you want to learn all(well what he's willing to tell anyway)about EFI he's who you go to.
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (EFIGUY)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">
Lone cubic foot of air weighs .076 lbs, so 1500 cubic feet/min would weigh 114 lbs/min,
</TD></TR></TABLE>
This was the number I was looking for....I couldn't find a weight to volume of air conversion anywhere.
Lone cubic foot of air weighs .076 lbs, so 1500 cubic feet/min would weigh 114 lbs/min,
</TD></TR></TABLE>
This was the number I was looking for....I couldn't find a weight to volume of air conversion anywhere.
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (EFIGUY)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">
Dave,
I'm glad to see that this one got the gears turning in that overly enormous head of yours..{just F'ing with ya}, but typically a wedge shaped combustion chamber does not promote very efficient burn and because of this has a slower flame propagation speed. This results in the need for earlier ignition to give enough time to build cylinder pressure adequately before the piston gets away.
again.....good try
-B</TD></TR></TABLE>
Yea, I kinda thought I was wrong on this one...it's just that having the spark plug by the exhaust port seems *** backwards, because the flame wall is fighting the nature direction the mixture is pushing. My 2am no sleep thinking was that combustion was somehow happening in half of the cylinder (Which I would think could maybe be possible depending on piston design) and therefore the wedge would burn faster, but would just burn really shitty and make less power. *Shrugs*
Dave,
I'm glad to see that this one got the gears turning in that overly enormous head of yours..{just F'ing with ya}, but typically a wedge shaped combustion chamber does not promote very efficient burn and because of this has a slower flame propagation speed. This results in the need for earlier ignition to give enough time to build cylinder pressure adequately before the piston gets away.
again.....good try
-B</TD></TR></TABLE>
Yea, I kinda thought I was wrong on this one...it's just that having the spark plug by the exhaust port seems *** backwards, because the flame wall is fighting the nature direction the mixture is pushing. My 2am no sleep thinking was that combustion was somehow happening in half of the cylinder (Which I would think could maybe be possible depending on piston design) and therefore the wedge would burn faster, but would just burn really shitty and make less power. *Shrugs*
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (EFIGUY)
I think I have it now:
120cu in @ 6500 would be 226cfm, 96lb\h injectors would put out 1.6lb\m...
Theres .076lb per cu ft of air, so 17.2lb\m of air. So we have 17.2lb\m of air and 1.6 lb\m of fuel. Having a Ratio of 17.2:1.6 which is 10.3:1 A\F Ratio. Air Flow is static, so we need less fuel to get 13:1. a 17.2:.755 Ratio...which .755 would be 47% of 1.6, so there ya go... .47
I hope I didnt fuckup that final math, but I understand now.
120cu in @ 6500 would be 226cfm, 96lb\h injectors would put out 1.6lb\m...
Theres .076lb per cu ft of air, so 17.2lb\m of air. So we have 17.2lb\m of air and 1.6 lb\m of fuel. Having a Ratio of 17.2:1.6 which is 10.3:1 A\F Ratio. Air Flow is static, so we need less fuel to get 13:1. a 17.2:.755 Ratio...which .755 would be 47% of 1.6, so there ya go... .47
I hope I didnt fuckup that final math, but I understand now.
Honda-Tech Member
Joined: Mar 2001
Posts: 6,183
Likes: 1
From: South Beach and Chicago, FL, USA
Re: Here's a little something for all the "tuners" out there (Suprdave)
5} At the same operating point, what would be the lowest acheiveable air fuel ratio with the given parameters?
10.3:1 with the Engine at 100% VE and the Injector @ 100%
10.3:1 with the Engine at 100% VE and the Injector @ 100%
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (Suprdave)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Suprdave »</TD></TR><TR><TD CLASS="quote">I think I have it now:
120cu in @ 6500 would be 226cfm, 96lb\h injectors would put out 1.6lb\m...
Theres .076lb per cu ft of air, so 17.2lb\m of air. So we have 17.2lb\m of air and 1.6 lb\m of fuel. Having a Ratio of 17.2:1.6 which is 10.3:1 A\F Ratio. Air Flow is static, so we need less fuel to get 13:1. a 17.2:.755 Ratio...which .755 would be 47% of 1.6, so there ya go... .47
I hope I didnt fuckup that final math, but I understand now. </TD></TR></TABLE>
Alright Dave! You are real close here, but go back and read the original question.......you gotta tell me what the ACTUAL pulse width is............. to do that you'll have to find out the "time per cycle" and figure out what percentage of that time your injector is open....that will give you an actual Pulse Width.
-Ben
120cu in @ 6500 would be 226cfm, 96lb\h injectors would put out 1.6lb\m...
Theres .076lb per cu ft of air, so 17.2lb\m of air. So we have 17.2lb\m of air and 1.6 lb\m of fuel. Having a Ratio of 17.2:1.6 which is 10.3:1 A\F Ratio. Air Flow is static, so we need less fuel to get 13:1. a 17.2:.755 Ratio...which .755 would be 47% of 1.6, so there ya go... .47
I hope I didnt fuckup that final math, but I understand now. </TD></TR></TABLE>
Alright Dave! You are real close here, but go back and read the original question.......you gotta tell me what the ACTUAL pulse width is............. to do that you'll have to find out the "time per cycle" and figure out what percentage of that time your injector is open....that will give you an actual Pulse Width.
-Ben
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (EFIGUY)
anybody else wanna take a crack at it before I spill the beans?
Come on guys................didn't we just have a several-page long thread about who the best tuners were? Where's all the tuning gods?
Alright, now I'm just being plain arrogant.....
Come on guys................didn't we just have a several-page long thread about who the best tuners were? Where's all the tuning gods?
Alright, now I'm just being plain arrogant.....
Honda-Tech Member
Joined: Oct 2003
Posts: 468
Likes: 0
Re: Here's a little something for all the "tuners" out there (EFIGUY)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?</TD></TR></TABLE>
6500rpm/60 = 108.3333rps, and since time is the reciprocal of frequency, 1/108.3333 = 9.2308ms for 100% pulse width at 6500rpms.
96lbs/hr = 1.6lbs/min = 12.0958grams of fuel/sec
those are facts, but here is where it's late, and rampant speculation takes over, superdave gimme a hand here :-0
12.0958grams fuel/13:1 = 93%, * .0092308sec = 8.59ms?
6500rpm/60 = 108.3333rps, and since time is the reciprocal of frequency, 1/108.3333 = 9.2308ms for 100% pulse width at 6500rpms.
96lbs/hr = 1.6lbs/min = 12.0958grams of fuel/sec
those are facts, but here is where it's late, and rampant speculation takes over, superdave gimme a hand here :-0
12.0958grams fuel/13:1 = 93%, * .0092308sec = 8.59ms?
Thread Starter
Honda-Tech Member
Joined: Jan 2003
Posts: 740
Likes: 0
From: NotCal
Re: Here's a little something for all the "tuners" out there (danimal)
Hey Guys,
good job.......I've SUPER busy lately, and haven't had time to double check your math, but you've got the right idea now....
maybe on monday when I have a few minuets I'll show you the whole way to do it stright through and make it a little easier....but for now.....good work!
good job.......I've SUPER busy lately, and haven't had time to double check your math, but you've got the right idea now....
maybe on monday when I have a few minuets I'll show you the whole way to do it stright through and make it a little easier....but for now.....good work!
Junior Member
Joined: Oct 2002
Posts: 453
Likes: 0
From: Ventura, California, USA
Re: (SCD16A6)
Damn I have no idea in what you guys are talking about 
. I wish I did, maybe ill take that class. Still probly wouldnt understand 3/4 of the stuff you were teaching tho lol.
. I wish I did, maybe ill take that class. Still probly wouldnt understand 3/4 of the stuff you were teaching tho lol.
Honda-Tech Member
Joined: Oct 2003
Posts: 468
Likes: 0
Re: Here's a little something for all the "tuners" out there (EFIGUY)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">4} If a 120 cubic inch, single cylinder engine had a VE of 100%, and an injector that was capable of flowing 96lbs of fuel per hour at 100% duty cycle, then what pulse width would be required of it at 6,500 RPM under normal conditions of atmosphere and fuel pressure to obtain an air/fuel ratio of 13:1?</TD></TR></TABLE>
i'm getting closer, lol! we needed to solve for the air, not the fuel.
13:1 a/f ratio is a weight measurement, first and foremost.
so .076lb per cu ft of air * 226cfm = 17.176lbs/min of air as the actual weight in the cylinder.
17.176/13:1 = 1.321lb/m(or 9.987grams/s) of fuel required to get the specified 13:1 a/f ratio.
6500rpm/60 = 108.3333rps, and since time is the reciprocal of frequency, 1/108.3333 = 9.2308ms for 100% pulse width at 6500rpms(one 360 degree crank rotation).
96lbs/hr = 1.6lbs/min = 12.0958grams of fuel/sec is the max injector flow.
9.987gm/sec target ratio / 12.0958gm/sec max injector flow = 82.6%, * 9.2308ms = 7.62ms pulse width to get 13:1 a/f ratio.
and oh yeah, an entire 4-stroke engine cycle is 720 degrees(two crankshaft rotations), which allows for twice as long of an injection period because you can inject fuel thruout that duration... but i only calculated for one max injection pulse(a 360 degree rotation)... the rest of it depends on the efi system design.
why 720 degrees? camshaft only turns half as much as the crank does, see the diagram at the middle of this page: http://users.spec.net/home/emx....html
Modified by danimal at 11:57 PM 1/11/2004
i'm getting closer, lol! we needed to solve for the air, not the fuel.
13:1 a/f ratio is a weight measurement, first and foremost.
so .076lb per cu ft of air * 226cfm = 17.176lbs/min of air as the actual weight in the cylinder.
17.176/13:1 = 1.321lb/m(or 9.987grams/s) of fuel required to get the specified 13:1 a/f ratio.
6500rpm/60 = 108.3333rps, and since time is the reciprocal of frequency, 1/108.3333 = 9.2308ms for 100% pulse width at 6500rpms(one 360 degree crank rotation).
96lbs/hr = 1.6lbs/min = 12.0958grams of fuel/sec is the max injector flow.
9.987gm/sec target ratio / 12.0958gm/sec max injector flow = 82.6%, * 9.2308ms = 7.62ms pulse width to get 13:1 a/f ratio.
and oh yeah, an entire 4-stroke engine cycle is 720 degrees(two crankshaft rotations), which allows for twice as long of an injection period because you can inject fuel thruout that duration... but i only calculated for one max injection pulse(a 360 degree rotation)... the rest of it depends on the efi system design.
why 720 degrees? camshaft only turns half as much as the crank does, see the diagram at the middle of this page: http://users.spec.net/home/emx....html
Modified by danimal at 11:57 PM 1/11/2004