a/f ratio concerns
- Thread starter norbs
- Start date
Waitasecond.....
Most cars do accelerate slower. Its called a torque CURVE. If you made the SAME torque from idle to 6000 RPM, your car SHOULD accelerate from 0 to whatever the mechanical speed limit is at a constant rate. But since not many cars curves are that flat, there will be a point, where the power made wont as easily overcome, wind/rolling etc resistance, the acceleration rate wont be constant all the way thru the RPM range. Heck, if it was, we wouldnt NEED transmissions. Just slap a clutch or converter on it and go.
Anyway, back to the point of the topic, it must take more fuel on the track, or, less precisely controlled conditions, or SOMETHING, that required more fuel. If the cars making 600hp on the dyno, but needs more fuel to do the same at the track, SOMETHING changed.
Valid enough, but here are a few guesses:
At the track, you have air constantly cooling a lot of stuff- intercooler, underhood air, etc. On a dyno, the air around the engine is static, and you don't get cooling like you would at 50-100 MPH of things like the intake manifold, throttle body, up-pipe, etc. (Also perhaps a bit of a ram-air effect on some cars, although in theory the A/F ratio shouldn't change because of that, pressure is pressure).
Under longer-term conditions (12 seconds vs. 4), the engine will build up more heat, which requires you to run a bit richer to suppress detonation and keep the exhaust valves cool by the end of the track.
Exhaust is getting sucked out the tailpipes at high speeds (Bernoulli was the guy on this one, I think), causing the cylinders to be empty a bit better- same effect as when you reduce exhaust backpressure on a NA car and have to jet a little bit richer. (In this case, you *are* making more power)
I'm not trying to fight with anyone here- I am just trying to learn more, so I am expressing my understanding as best I can, with the hope of someone pointing out a flaw in my thinking so I can go tune my car to be faster! (I've already nudged my target A/F a bit because of this discussion). Thanks to everyone for helping me to think a bit more. I generally don't accept old wives tales without understanding them before applying them, but I'm trying this anyhow...
-Bob Cunningham
bobc@gnttype.org
At the track, you have air constantly cooling a lot of stuff- intercooler, underhood air, etc. On a dyno, the air around the engine is static, and you don't get cooling like you would at 50-100 MPH of things like the intake manifold, throttle body, up-pipe, etc. (Also perhaps a bit of a ram-air effect on some cars, although in theory the A/F ratio shouldn't change because of that, pressure is pressure).
Under longer-term conditions (12 seconds vs. 4), the engine will build up more heat, which requires you to run a bit richer to suppress detonation and keep the exhaust valves cool by the end of the track.
Exhaust is getting sucked out the tailpipes at high speeds (Bernoulli was the guy on this one, I think), causing the cylinders to be empty a bit better- same effect as when you reduce exhaust backpressure on a NA car and have to jet a little bit richer. (In this case, you *are* making more power)
I'm not trying to fight with anyone here- I am just trying to learn more, so I am expressing my understanding as best I can, with the hope of someone pointing out a flaw in my thinking so I can go tune my car to be faster! (I've already nudged my target A/F a bit because of this discussion). Thanks to everyone for helping me to think a bit more. I generally don't accept old wives tales without understanding them before applying them, but I'm trying this anyhow...
-Bob Cunningham
bobc@gnttype.org
Just saw your post, apparently we were typing at the same time...
If you had a perfectly flat torque curve, you'd still need a tranny. A tranny *multiplies* torque at low speeds, so if your engine produced 500 ft-lbs all day long, a tranny would enable you to get 1500 ft-lbs at low speeds, causing more acceleration (at low speeds) and stay in a reasonable RPM range at high vehicle speeds (on the highway).
I think you're getting confused with the torque output of the engine vs. the torque output of the transmission and so on.
FWIW, there is a division at the local roundy-rounds that uses no transmission, just a clutch- they are only concerned with one speed range, so as long as they don't stall coming out of the pits they will be fine for the whole race. Their rearend gear is changed according to the speeds they'll see on that track. Very efficient power coupling, but wouldn't be very good in every day driving situations or even at the dragstrip.
-Bob Cunningham
bobc@gnttype.org
If you had a perfectly flat torque curve, you'd still need a tranny. A tranny *multiplies* torque at low speeds, so if your engine produced 500 ft-lbs all day long, a tranny would enable you to get 1500 ft-lbs at low speeds, causing more acceleration (at low speeds) and stay in a reasonable RPM range at high vehicle speeds (on the highway).
I think you're getting confused with the torque output of the engine vs. the torque output of the transmission and so on.
FWIW, there is a division at the local roundy-rounds that uses no transmission, just a clutch- they are only concerned with one speed range, so as long as they don't stall coming out of the pits they will be fine for the whole race. Their rearend gear is changed according to the speeds they'll see on that track. Very efficient power coupling, but wouldn't be very good in every day driving situations or even at the dragstrip.
-Bob Cunningham
bobc@gnttype.org
Load,
ahhhhh,
Load does vary.
(in brief)
It's a function of the mechanical advantge
Lets start from a standstill.
As the car first moves, you have the the torque converter multiplication, of about 2 to 1. First gear is 2.5 to one, the rear axle gear of 3.42. and just to keep the math easy we will say 24 in tires. So the engine has a mechanical advantage of 17 to one.
Now in 3rd gear all you have is the final gearing of 3.4.
Plus as you get to 80 MPH the aero loading is going way up.
Now your still asking the engine to acclerate the car. Which do you think allows the engine to work the least?. At 17 to one or 3.4 to one?.
And in the lower gears while you might not be melting the tires, with massive spin, there can be a fair amount of slippage which limits the load the engine can actually develope.
Bruce-
How does that differ from torque?
Sure torque can be multiplied at the expense of RPM, but from the engine's perspective it is under the same operating conditions in 2nd gear or OD when you're at WOT.
-Bob Cunningham
bobc@gnttype.org
How does that differ from torque?
Sure torque can be multiplied at the expense of RPM, but from the engine's perspective it is under the same operating conditions in 2nd gear or OD when you're at WOT.
-Bob Cunningham
bobc@gnttype.org
Let's start in a linear world, where Newton says F=ma. That's force = mass times acceleration. If you have a fixed force and you measure some acceleration rate for a mass, and then the observed acceleration rate increases and you are sure that the force hasn't changed, the only possibility is that the mass has decreased. Or, to pound it into our vernacular, the load (mass) has decreased. Now, torque is force applied around a circle (such as the flywheel). The acceleration is now the rate of change of angular velocity, or how fast the rpm is changing. The load is all the drag trying to make the flywheel stop rotating. That means that (roughly) torque = load * rate of increse of rpm. Now insert my previous paragraph on how since the engine output (torque) is the same on a dyno and a track, and since the rpms are increasing faster on the inertial dyno the load must be less.
Yes, the engine torque is multiplied by the gear ratio, but it works in reverse too. The load at the flywheel is the load at the rear tires divided by the gear ratio.
Yes it's true that the engine will be hotter at the track since it's at wot longer, and you might have to run richer to compensate for that. However, that doesn't explain the observation of higher O2 volts (or richer on a wideband O2 sensor) with the same fueling conditions (chip, boost, fuel pressure, air temp), on a chassis inertial dyno versus on a track. Like you said, Bob, just a good discussion of something none of us are crystal clear on.
Yes, the engine torque is multiplied by the gear ratio, but it works in reverse too. The load at the flywheel is the load at the rear tires divided by the gear ratio.
Yes it's true that the engine will be hotter at the track since it's at wot longer, and you might have to run richer to compensate for that. However, that doesn't explain the observation of higher O2 volts (or richer on a wideband O2 sensor) with the same fueling conditions (chip, boost, fuel pressure, air temp), on a chassis inertial dyno versus on a track. Like you said, Bob, just a good discussion of something none of us are crystal clear on.
Holy molly Carl I don't know if you got me or not on that one, or if we agree. Try this one on for size.
Do a data log on the dyno and at the strip. Or just do a run down a country road or something through the gears. If we agree that the load on the motor, which is what I am talking about is the variable than when you overlay the data log on your VE screen the trace line should move up on the Kpa line for each gear. If I remember mine seems to lay on top of one another but I have none to look at right now. Now irregardless of what this shows lay the datalog of the dyno run over the VE screen. That should answer the question as the VE screen tells you how the motor is loaded.
What I am trying to concentrate on is what does the motor see, more load in first,second or third on the dyno or strip. I say it is basically equal (trying to hedge my bets here!!!), street or strip. It only has so much it can put out. The only difference on the strip or street with load as far as the motor is concerned is how long it is at that load and does that load mimic itself on the dyno and on the strip. In the f=ma the f is fixed, as you pointed out. Mass is increasing as gearing is changing, wind resistance and so forth. But the load (torque output) on the motor is fixed, again it can only put out so much. That load is the same on the dyno as on the road. Only dif is you are loading it longer on the road/strip than on the dyno.
Now maybe on the chassis dyno the mass of the roller is less than what you would have on the street. Therefore the calculation of peak power/torque is a calculation with variables above and beyond what is recorded. I don't know. That would mean you are dialing in cells on the dyno that are different than what are being referenced on the strip. Because the dyno is just not mechanically loading the motor to its max. The question I would have is if it takes 4 seconds to make a run on the dyno in second gear how long does your car take to blow through second gear at the strip or on the street? I would think that on the dyno you would take longer than at the strip. So from that standpoint using the f=ma their is more load on the dyno than on the strip.
On a water brake dyno the duration of the run is short by design. It is still loaded to the max the motor can take without pulling it down but just programmed in short steps. On a Stutzka you do pulls from max RPM down to a set point. Those can be as slow and brutal as the customer wants. Open up the throttle, load the water brake and keep loading and adding throttle until your low rpm cutoff point. Some claim a more accurate way to dyno a motor.
Just a bunch of blubering thoughts, while I really should be working.
Do a data log on the dyno and at the strip. Or just do a run down a country road or something through the gears. If we agree that the load on the motor, which is what I am talking about is the variable than when you overlay the data log on your VE screen the trace line should move up on the Kpa line for each gear. If I remember mine seems to lay on top of one another but I have none to look at right now. Now irregardless of what this shows lay the datalog of the dyno run over the VE screen. That should answer the question as the VE screen tells you how the motor is loaded.
What I am trying to concentrate on is what does the motor see, more load in first,second or third on the dyno or strip. I say it is basically equal (trying to hedge my bets here!!!), street or strip. It only has so much it can put out. The only difference on the strip or street with load as far as the motor is concerned is how long it is at that load and does that load mimic itself on the dyno and on the strip. In the f=ma the f is fixed, as you pointed out. Mass is increasing as gearing is changing, wind resistance and so forth. But the load (torque output) on the motor is fixed, again it can only put out so much. That load is the same on the dyno as on the road. Only dif is you are loading it longer on the road/strip than on the dyno.
Now maybe on the chassis dyno the mass of the roller is less than what you would have on the street. Therefore the calculation of peak power/torque is a calculation with variables above and beyond what is recorded. I don't know. That would mean you are dialing in cells on the dyno that are different than what are being referenced on the strip. Because the dyno is just not mechanically loading the motor to its max. The question I would have is if it takes 4 seconds to make a run on the dyno in second gear how long does your car take to blow through second gear at the strip or on the street? I would think that on the dyno you would take longer than at the strip. So from that standpoint using the f=ma their is more load on the dyno than on the strip.
On a water brake dyno the duration of the run is short by design. It is still loaded to the max the motor can take without pulling it down but just programmed in short steps. On a Stutzka you do pulls from max RPM down to a set point. Those can be as slow and brutal as the customer wants. Open up the throttle, load the water brake and keep loading and adding throttle until your low rpm cutoff point. Some claim a more accurate way to dyno a motor.
Just a bunch of blubering thoughts, while I really should be working.
Craig Smith
That was easy!
- Joined
- May 25, 2001
Here's more food for thought on the loading theory.
One concept being kicked around here is that WOT = max load, period. I disagree. If I go set my engine dyno to hold the engine at 5000 RPM, open the throttle all the way, and then start turning the brake RPM down, am I not increasing the load on the motor, even though it is still at WOT? Engine load is really the variable responsible for the topic of this discussion IMO.
I don't know that you would necessarily see the difference in the MAP reading at WOT with more or less load. At WOT, unless the throttle body is too small, there should be no pressure difference between the air over the TB and the air in the plenum... or at least that's what I think. Any thoughts?
One concept being kicked around here is that WOT = max load, period. I disagree. If I go set my engine dyno to hold the engine at 5000 RPM, open the throttle all the way, and then start turning the brake RPM down, am I not increasing the load on the motor, even though it is still at WOT? Engine load is really the variable responsible for the topic of this discussion IMO.
I don't know that you would necessarily see the difference in the MAP reading at WOT with more or less load. At WOT, unless the throttle body is too small, there should be no pressure difference between the air over the TB and the air in the plenum... or at least that's what I think. Any thoughts?
First, to Craig: we were talking about inertial chassis dynos, not waterbrake dynos where you are correct, you can get to a steady state. My theory is that that steady state should be closer to the track tune than an inertial chasis dyno tune.
.
. If my theory about the reason chassis dynos give a leaner tune than the track is right, the waterbrake dyno should be more accurate. Also, if my theory is right the engine will make a little more power on the track than on the dyno, so maybe the bsfc isn't changing?
That's a very good point about looking at the data traces. If I had a FAST and a dyno run I'd do it
.Here's where we disagree. I say the load is less on the dyno, because in the same gear at the same rpm the engine will rev faster (accelerate faster) on the dyno than on the track. The motor is making the same torque so if the acceleration changes then the load HAS to have changed, right?
I've only seen one dyno session but the car definitely went through the gears faster on the dyno than on the track. The dyno is calibrated for the weight of the roller and you tell it what the gear ratios and tire height are so it can correct for all that to get rwhp assuming 1:1 gearing.
Me too
. If my theory about the reason chassis dynos give a leaner tune than the track is right, the waterbrake dyno should be more accurate. Also, if my theory is right the engine will make a little more power on the track than on the dyno, so maybe the bsfc isn't changing?OK, Just for ****'s and giggles, as I don't know what it really means actually I went out tonight and did a data log in second third and fourth at WFO. Kpa's were constant from the beginning of the run to the end of the run in each gear. I short shifted at 5800 in each gear. Just bounced up and down one number. What does that tell me? I dunno. What I think though is durgs are a wonderful thing.....no strike that.... I think that the loading on a chassis dyno sounds like it must be different than when running on the street or strip. That seems to make sense to me now that you explained what you have seen and thinking about the fixed mass the roller represents. I think this is a question for Dyno Jet.
Hey Craig, don't know if I want to try it with my engine but I wonder if you put no load on a motor and give if full throttle will it have zero pressure differential? Never done it so I don't know. I know what you are thinking though as far as load is not load. On the water brake it is easy to see why that is true, BAAAAAHHHHHHHRRRRR pop. So given that is true then and load is not load even at WFO and zero pressure differential how would we account for that when dialing a fuel curve on the VE table, on the dyno? Or is there any anecdotal information on combos that have been dialed in on a water brake (or eddy current) dyno and then bolted into the car and run? On the water brake we basically are loading that bad boy to the max. On the street possibly due to factors that are right in front of our eyes the motor is not loaded as much, or is just loaded different in each cell. Example, based on a preliminary consensus, there will be more load on the motor at WFO going up a steep hill than if it was going down a steep hill. That is the logic that seems to be playing out here. I still don't know though if it is apples to apples as it relates to the tune up or not though.
Hey Craig, don't know if I want to try it with my engine but I wonder if you put no load on a motor and give if full throttle will it have zero pressure differential? Never done it so I don't know. I know what you are thinking though as far as load is not load. On the water brake it is easy to see why that is true, BAAAAAHHHHHHHRRRRR pop. So given that is true then and load is not load even at WFO and zero pressure differential how would we account for that when dialing a fuel curve on the VE table, on the dyno? Or is there any anecdotal information on combos that have been dialed in on a water brake (or eddy current) dyno and then bolted into the car and run? On the water brake we basically are loading that bad boy to the max. On the street possibly due to factors that are right in front of our eyes the motor is not loaded as much, or is just loaded different in each cell. Example, based on a preliminary consensus, there will be more load on the motor at WFO going up a steep hill than if it was going down a steep hill. That is the logic that seems to be playing out here. I still don't know though if it is apples to apples as it relates to the tune up or not though.
Craig Smith
That was easy!
- Joined
- May 25, 2001
For what this is worth, Lance's V6 made peak horsepower on the water brake engine dyno at 12.5:1 but his Buick ran fastest on the 1/4 mile at 11.5:1 with that motor.
- Joined
- May 28, 2001
FWIW Dept!!
My take.. from an old buzzard's perspective.
Take a look at the T/F and F/C folks... Engine loading is a MAJOR tool in tuning those cars.
2 items come into play there.[Remember, no trans involved]
1. The lower, [numerically] the rear gear is,[3.23, 3.08, etc] the faster the car should run.[given they have an excess of useable HP] That's 1 reason NHRA put a limit on the gear ratio they can run.[ Currently 3.20, I think]
2. The teams were utilizing these real low number ratios to LOAD the motor. More load, more fuel burned, higher cyl pressures, all equate to higher HP...Item 1 was put into force to limit top speeds. Now that the HP numbers have been raised, NHRA put tire size limits on the cars.
More fuel required is confirmed as they all utilize multiple pumps... Now NHRA has stepped in there too!!
Now to the "real world"...
Dyno vs track tuneups.
There are SIGNIFICANT differences. As has already been mentioned, the motion of the car is involved as are alot of other factors. Some of them are:
Dynamic pressure: @20mph this press. is almost zero..At 120mph, the press is 38#/sq ft.
Rolling resistence: this factor is expressed in #'s per 1000# of vehicle wt. At 20mph this is 18. Therefore a 3700# brick has a resistence of 66.6#. BUT at 120mph this value goes to 27# for a total of 99.9#
You add these 2 to together and find some significant drag to be overcome.. The method of doing this is to burn more fuel, to make more power.
Now we add in the drag vs speed and it becomes REAL interesting.
Aero drag at 20mph is negligible. At 120 mph it becomes 300# assuming a Cd of .4.. I doubt our bricks are .4!!
Now the car weighs 4000# as far as the engine is concerned..
Want to see the HP required to propel YOUR car at a given speed?? Try the coastdown test.
Bottom line is: dyno tests cannot take these added HP requirements into account. Therefore the BSFC on the road at max engine load will be higher than on the dyno...As someone said.. correct A/F on the dyno lean on the track...
My story and I'm stickin w/ it!!
My take.. from an old buzzard's perspective.
Take a look at the T/F and F/C folks... Engine loading is a MAJOR tool in tuning those cars.
2 items come into play there.[Remember, no trans involved]
1. The lower, [numerically] the rear gear is,[3.23, 3.08, etc] the faster the car should run.[given they have an excess of useable HP] That's 1 reason NHRA put a limit on the gear ratio they can run.[ Currently 3.20, I think]
2. The teams were utilizing these real low number ratios to LOAD the motor. More load, more fuel burned, higher cyl pressures, all equate to higher HP...Item 1 was put into force to limit top speeds. Now that the HP numbers have been raised, NHRA put tire size limits on the cars.
More fuel required is confirmed as they all utilize multiple pumps... Now NHRA has stepped in there too!!
Now to the "real world"...
Dyno vs track tuneups.
There are SIGNIFICANT differences. As has already been mentioned, the motion of the car is involved as are alot of other factors. Some of them are:
Dynamic pressure: @20mph this press. is almost zero..At 120mph, the press is 38#/sq ft.
Rolling resistence: this factor is expressed in #'s per 1000# of vehicle wt. At 20mph this is 18. Therefore a 3700# brick has a resistence of 66.6#. BUT at 120mph this value goes to 27# for a total of 99.9#
You add these 2 to together and find some significant drag to be overcome.. The method of doing this is to burn more fuel, to make more power.
Now we add in the drag vs speed and it becomes REAL interesting.
Aero drag at 20mph is negligible. At 120 mph it becomes 300# assuming a Cd of .4.. I doubt our bricks are .4!!
Now the car weighs 4000# as far as the engine is concerned..
Want to see the HP required to propel YOUR car at a given speed?? Try the coastdown test.
Bottom line is: dyno tests cannot take these added HP requirements into account. Therefore the BSFC on the road at max engine load will be higher than on the dyno...As someone said.. correct A/F on the dyno lean on the track...
My story and I'm stickin w/ it!!
Only in the most general of terms is the engine in the same operating conditions. Underhood airflow is much diffferent at 2 MPH then at 100MPH. Oil temps, and piston dome temps have risen. Combustion chambers are hotter. Ignition coils are hotter, and the spark energy starts to decay. The localized hot spots in the head are getting larger, hardly the same operting conditions going thru the traps as when doing the 60 ft..
Load is about leverage.
Once you get to where there is no leverage, the load approaches infinity, and in the case of an engine it stalls. Take only engine strap it to a dyno, lock up the brake, and you will stall the engine, in that case the LOAD was enough to drag the engine down to where it couldn't operate. Loading an engine on a dyno to get a HP reading is just taking the engine to where it won't acclerate, and goes about steady state, RPM wise.
try diging with a 3' shovel, and then with a 6' shovel.
The world revolves around making loads manageable.
Carl,
When we look at F=MA, we can also look at torque = force (F) * Length. In this case length is the effective gear ratio, a combination of tire diameter, rearend ratio, transmission gear ratio, and torque convertor slippage / multiplication.
So in english, we can change F=MA to torque*gear ratio= MA.
(Aside: If you want, you can plug your torque, gear ratios, tire diameter, shift points, redline, and weight (mass) into a big formula and come up with a calculation for your 1/4 ET and MPH. I have done this with a spreadsheet, it is amazingly accurate. (TurboJim- Because torque comes in a curve, I put in peak torque and multiply by 0.85 or something to account for the lower spots on the torque curve, then I multiply again by 0.85 for drivetrain frictional losses). This is the same way the Desktop Dyno's work, except they get a bit fancier when they consider your torque convertor, etc.)
On they dyno, since we have less mass (M) to accelerate (I assume - I've never built a dyno), the RPM's will increase more quickly in a given gear than on a road/track.
So far, agreed?
And if you want to define "load" as torque / rotational acceleration, I guess I can live with that. But please explain (guesses are fine!) to me how that would would affect anything inside the engine - amount of fuel required, etc. Also, we might want to check our formula a bit, because when the tranny shifts, we have a negative load (torque production is still positive, but now we have a negative rotational acceleration).
RickWI,
A very interesting experiment. Because we had the same operating conditions (timing, fuel) regardless of gear, I would have to assume that the engine didn't care about the "load", it is irrelevant to engine performance. And kudos for running all the way through 4th gear up to 5800 RPM! We don't want any wussy experiments here...
And just to clarify, your mass is not increasing when you have more aerodynamic resistance, your F is decreasing - you have to look at it as the "F" = F(due to engine torque) minus F (due to wind resitance).
If you have *no* load on the motor (by any proposed definition so far), it won't be running, will it? And with *no* pressure differential, there will be *no* flow.
And if you are going uphill, should you tune the motor differently than going downhill?
Craig,
Your definition of "load" is different than Carls (but also perfectly acceptable). You apparently define load as the resistance to the engine turning. When installed in a car, this resistance is provided by the weight of the car, aerodynamics, etc., as modified by the effective gear ratio.
On the stand, and engine is producing torque and the dyno is absorbing torque (creating a negative torque). If the dyno starts to absorb more torque than the engine can produce, you will get a negative rotational acceleration. By your definition, this is fine, but again I don't think that your "load" should affect engine operating parameters (fuel, timing, etc.). Will it?
And FWIW, if the pressure were the same on both sides of the TB, no air would flow- there must be a pressure gradient of some sort to get air to flow. It's probably not a very large pressure drop, but there has to be one.
I think that everyone here has their own definition of load, which is creating some confusion. I still want to know why my engine will used to ping during a tranny shift (before I got SpeedPro) when it's fine during the rest of the run!!! It could be related to this load thing, but I am still confused...
-Bob Cunningham
bobc@gnttype.org
When we look at F=MA, we can also look at torque = force (F) * Length. In this case length is the effective gear ratio, a combination of tire diameter, rearend ratio, transmission gear ratio, and torque convertor slippage / multiplication.
So in english, we can change F=MA to torque*gear ratio= MA.
(Aside: If you want, you can plug your torque, gear ratios, tire diameter, shift points, redline, and weight (mass) into a big formula and come up with a calculation for your 1/4 ET and MPH. I have done this with a spreadsheet, it is amazingly accurate. (TurboJim- Because torque comes in a curve, I put in peak torque and multiply by 0.85 or something to account for the lower spots on the torque curve, then I multiply again by 0.85 for drivetrain frictional losses). This is the same way the Desktop Dyno's work, except they get a bit fancier when they consider your torque convertor, etc.)
On they dyno, since we have less mass (M) to accelerate (I assume - I've never built a dyno), the RPM's will increase more quickly in a given gear than on a road/track.
So far, agreed?
And if you want to define "load" as torque / rotational acceleration, I guess I can live with that. But please explain (guesses are fine!) to me how that would would affect anything inside the engine - amount of fuel required, etc. Also, we might want to check our formula a bit, because when the tranny shifts, we have a negative load (torque production is still positive, but now we have a negative rotational acceleration).
RickWI,
A very interesting experiment. Because we had the same operating conditions (timing, fuel) regardless of gear, I would have to assume that the engine didn't care about the "load", it is irrelevant to engine performance. And kudos for running all the way through 4th gear up to 5800 RPM! We don't want any wussy experiments here...
And just to clarify, your mass is not increasing when you have more aerodynamic resistance, your F is decreasing - you have to look at it as the "F" = F(due to engine torque) minus F (due to wind resitance).
If you have *no* load on the motor (by any proposed definition so far), it won't be running, will it? And with *no* pressure differential, there will be *no* flow.
And if you are going uphill, should you tune the motor differently than going downhill?
Craig,
Your definition of "load" is different than Carls (but also perfectly acceptable). You apparently define load as the resistance to the engine turning. When installed in a car, this resistance is provided by the weight of the car, aerodynamics, etc., as modified by the effective gear ratio.
On the stand, and engine is producing torque and the dyno is absorbing torque (creating a negative torque). If the dyno starts to absorb more torque than the engine can produce, you will get a negative rotational acceleration. By your definition, this is fine, but again I don't think that your "load" should affect engine operating parameters (fuel, timing, etc.). Will it?
And FWIW, if the pressure were the same on both sides of the TB, no air would flow- there must be a pressure gradient of some sort to get air to flow. It's probably not a very large pressure drop, but there has to be one.
I think that everyone here has their own definition of load, which is creating some confusion. I still want to know why my engine will used to ping during a tranny shift (before I got SpeedPro) when it's fine during the rest of the run!!! It could be related to this load thing, but I am still confused...
-Bob Cunningham
bobc@gnttype.org
Sheesh! Three other replies while I was typing mine. I gotta learn to type faster or something...
Bruce,
You seem to be saying that load (force) = torque / moment arm. If so, I agree completely. But you are defining load as force, which is again different than everyone elses (not that I'm trying to pick favorites here).
You also seem to define power as the amount of work an engine can do at a constant RPM. Also agreed.
And we agree that heat builds up over the course of the 1/4 mile differently than on a dyno. 'Scool by me. I proposed earlier that heat buildup requires some extra fuel to suppress detonation and keep exhaust valves cool, and you seem to concur.
Chuck,
Seems like you're saying that the Top Fuel cars have to add more fuel to make more HP/Torque to overcome a loss in mechanical advantage from the rearend ratio (so they can obtain a higher top end speed). Yes? If so, we could say that if the engine made less torque but you could use a shorter rearend gear, you could get the same acceleration (assuming you don't run out of RPM by the 1/4 mile)? Then, could we say that from the engine's perspective, load = torque?
-Bob Cunningham
bobc@gnttype.org
Bruce,
You seem to be saying that load (force) = torque / moment arm. If so, I agree completely. But you are defining load as force, which is again different than everyone elses (not that I'm trying to pick favorites here).
You also seem to define power as the amount of work an engine can do at a constant RPM. Also agreed.
And we agree that heat builds up over the course of the 1/4 mile differently than on a dyno. 'Scool by me. I proposed earlier that heat buildup requires some extra fuel to suppress detonation and keep exhaust valves cool, and you seem to concur.
Chuck,
Seems like you're saying that the Top Fuel cars have to add more fuel to make more HP/Torque to overcome a loss in mechanical advantage from the rearend ratio (so they can obtain a higher top end speed). Yes? If so, we could say that if the engine made less torque but you could use a shorter rearend gear, you could get the same acceleration (assuming you don't run out of RPM by the 1/4 mile)? Then, could we say that from the engine's perspective, load = torque?
-Bob Cunningham
bobc@gnttype.org
Here is the response I got from Shawn Arnold at Dynojet:
It is company policy that we not respond to web rings or chats but if you want some good explanations you can email Paul from Paul's High Performance. He is a chassis dyno owner who has a lot of experience in this field and would be happy to answer some questions. Thank you for the interest.
paul@paulshp.com
I emailed Paul just a minute ago
Here is what I said:
Good Day, On the Turbo Buick web site we have been having a "heated" debate as to why when we dyno and tune our electronic fuel injected motors on the chassis dyno for best power and then go to the track the tune-ups change Typically what is found is a need to tune richer. The thread this discussion is taking place on is here http://www.turbobuick.com/forums/showthread.php?s=&threadid=49628&perpage=25&pagenumber=1 Since most of us are doing our tuning on a chassis dyno, as it lends itself well to our stand alone fuel injection management systems we run, I would really appreciate you helping us out in as to why this is. If you go into the thread, which is two pages long right now, you can see the circles we are going in. It is getting fairly technical in many areas we are not to sure of ourselves in. I figured you folks are the experts and could quickly settle this for us. Feel free to sign in and respond on the thread or just email back after you review the thread and I will post your response.
Rick Pettibone
Ph: 608-837-5726
It is company policy that we not respond to web rings or chats but if you want some good explanations you can email Paul from Paul's High Performance. He is a chassis dyno owner who has a lot of experience in this field and would be happy to answer some questions. Thank you for the interest.
paul@paulshp.com
I emailed Paul just a minute ago
Here is what I said:
Good Day, On the Turbo Buick web site we have been having a "heated" debate as to why when we dyno and tune our electronic fuel injected motors on the chassis dyno for best power and then go to the track the tune-ups change Typically what is found is a need to tune richer. The thread this discussion is taking place on is here http://www.turbobuick.com/forums/showthread.php?s=&threadid=49628&perpage=25&pagenumber=1 Since most of us are doing our tuning on a chassis dyno, as it lends itself well to our stand alone fuel injection management systems we run, I would really appreciate you helping us out in as to why this is. If you go into the thread, which is two pages long right now, you can see the circles we are going in. It is getting fairly technical in many areas we are not to sure of ourselves in. I figured you folks are the experts and could quickly settle this for us. Feel free to sign in and respond on the thread or just email back after you review the thread and I will post your response.
Rick Pettibone
Ph: 608-837-5726
http://www.dynojet.com/aupgrades.shtml
DYNOTRAC LOAD CONTROL OPTION (model 248 only)
This option allows steady state load testing at a preset RPM, speed or percentage of brake pressure.
Does the above look like what might be needed to "load" the car properly on a chassis dyno?
Basically I wonder about a coupe of things. When you do a dyno pull on the chassis dyno are you programming the same cells that are referenced on the dyno as what are used going down the strip. Let me explain what may be happening. Let us assume you make a pull on the chassis dyno and the map reading from 3000 to 6000 is steady at 95. So you dial in that slection of cells. Now lets say you make a run at the strip and the motor pulls 98 Kpa. Could that be true? It still does not explain why in the case of Lance's situation it needed a richer A/F ratio. I think it boils down to the amount of dwell time the motor is loaded on the dyno versus the strip that causes the difference. The above tool seems to correct this situation.
DYNOTRAC LOAD CONTROL OPTION (model 248 only)
This option allows steady state load testing at a preset RPM, speed or percentage of brake pressure.
Does the above look like what might be needed to "load" the car properly on a chassis dyno?
Basically I wonder about a coupe of things. When you do a dyno pull on the chassis dyno are you programming the same cells that are referenced on the dyno as what are used going down the strip. Let me explain what may be happening. Let us assume you make a pull on the chassis dyno and the map reading from 3000 to 6000 is steady at 95. So you dial in that slection of cells. Now lets say you make a run at the strip and the motor pulls 98 Kpa. Could that be true? It still does not explain why in the case of Lance's situation it needed a richer A/F ratio. I think it boils down to the amount of dwell time the motor is loaded on the dyno versus the strip that causes the difference. The above tool seems to correct this situation.
Similar threads
