50 replies to this topic

[Lukin]

For years everyone I spoke to said that the reason a car decelerates when you change from, say, 5th to 2nd at a high speed is due the compression of the engine slowing the car down, or to put it another way, the car driving the engine until the RPM drops to a 'suitable' level (whatever that level is).

I was reading a Carrol Smith book the other day and he says this is due to engine friction, not compression. Which kinda makes sense.

The way I see it from my relatively limited experience of driving (just road cars) and crashing (just road cars), when you downshift a few gears, there will be a deceleration as the axle and engine speed are different (and each 'accelerates' until they are the same speed), but this is only brief once drive is engaged. This explains the 'kickback' you get when you don't touch the throttle on downshifts.

Another observation. Driving at 85 km/h in second gear and coming off the throttle gives a much greater deceleration than coming off the throttle at 85 km/h in 4th gear.

So when Smith says it's due to engine friction, I assume he means the friction in the engine at higher revs is much more than in low revs, irrespective of your gearshifting habits or rituals? So a quadratic or higher relationship? But if that is the case, won't the losses in an engine doing 20000 RPM be massive?

Once again Im proving I know bugger all about engines. Cheers guys.

[prty]

Originally posted by Lukin
But if that is the case, won't the losses in an engine doing 20000 RPM be massive?

I think it was de la Rosa who said that if they get off the throttle in a formula one, just the downforce at those speeds and the "engine brake" produce equivalent decelerations of an emergency brake in a road car. No wonder why he said that to take turn 8 in Turkey he just releases the throttle, and braking would kill all the speed.

[zac510]

Perhaps the vacuum in the cylinders causes more friction in rings and bearings as it acts against the motion of the piston, thus the deceleration is calculated as friction based.

[shaun979]

I would say it is a combination of vacuum and friction but more so vacuum. In motoring tests on 5-7 litre engines, if the cylinder is left open to pump air, some lower power electric motors (<100hp) can barely overcome the pumping losses even at low RPM.

The lower a gear you're in the more violent the deceleration feels because of larger negative torque multiplication due to gear ratios.

F1 deceleration at high RPM ignoring aerodynamics, must be pretty strong too considering their throughput is large and friction high. I'm not sure about throtttle rules but perhaps they don't close the throttle completely or they program variable rates of closing for a given pedal let-off.

[Goran Malmberg]

Originally posted by Lukin
But if that is the case, won't the losses in an engine doing 20000 RPM be massive?

Once again Im proving I know bugger all about engines. Cheers guys.

With the stroke I am using, piston speed equal that of a F 1 car at 7000 rpm, and true, the deceleration is great when going off the trottle.
Goran Malmberg

[rhm]

"engine braking" is due to a combination of losses due to internal friction, having to drive anciliaries (oil, water pumps etc.) and pumping losses.

It's the pumping losses that are the most interesting as the others are always there even when you're accelerating. The pumping loss is due to the engine having to suck D (for displacement) litres of air past a partially open throttle, through the inlet port and then pumping it out the exhaust port and through the exhaust system. That uses up a considerable amount of energy if the throttle is in the idle position.

In between the intake and exhaust cycles the air in the cylinder is compressed and then expanded again. This uses very little energy because the air acts like a spring - the energy used in the compression cycle is returned to the crank in the combustion cycle.

In road car engines that 'disable' cylinders to save fuel, they do so by keeping the valves closed through all cycles for the disabled cylinders, thus in partial throttle conditions the pumping loss for those cylinders is saved. And that's were the main saving comes from because the frictional losses stay the same.

[Greg Locock]

But if the throttle is closed then where does this pumping loss come from? Imagine if the intake valves were disabled. There's be a bit of air in the cylinder, presumably less as time goes on, with less and less work being done on it as the pressure drops.

Now allow the intake valves to work, but assume no plenum, then we have a bit of airr in the runner and the bit in the cylinder. Now add in the plenum, yet that would be emptied in a few revolutions.

Is the leakage rate past the closed butterfly really so great as to provide enough airflow to absorb significant amounts of power?

Good question.

[J. Edlund]

Originally posted by Greg Locock
But if the throttle is closed then where does this pumping loss come from? Imagine if the intake valves were disabled. There's be a bit of air in the cylinder, presumably less as time goes on, with less and less work being done on it as the pressure drops.

Now allow the intake valves to work, but assume no plenum, then we have a bit of airr in the runner and the bit in the cylinder. Now add in the plenum, yet that would be emptied in a few revolutions.

Is the leakage rate past the closed butterfly really so great as to provide enough airflow to absorb significant amounts of power?

Good question.

If you close the throttle at high speeds you will reach a pressure after the throttle of perhaps .3 bar. In order to keep the engine turning, this low pressure will require additional work input. Since there is some pressure we must also have some sort of airflow (but a very low flow), if the plenum were emptied the pressure should be close to zero but a piston isn't capable of reaching such low pressures (only some special pumps can reach pressures very close to a complete vacuum).

On diesel engines we do not have any throttle, so it will not offer the braking performance of a SI engine, hence the use of exhaust brakes. The principle is the same but instead of a low pressure before the engine we have a high pressure after.

[NTSOS]

Engine braking comes primarily from the power needed to maintain a pressure drop in the plenum when the throttle is before the plenum and the throttle is closed.

Internal engine friction and accessory frictional losses contribute to engine braking but not nearly as much as the power drag it takes to keep a pressure drop within the manifold when the thottle is closed. You can eliminate compression, because if the throttle is closed, there is very little to compress and even if there was, the trapped air/spring effect would pretty much equalize the effort......like a diesel!

So......think diesel engine.....no throttle, not much engine braking.....a "Jake Brake" is somewhat helpfull though!

John

[Greg Locock]

MAP with a closed throttle is at most -80 kPA, according to my gauge. So there is no air in the throttle to do any work on.

[NTSOS]

Greg, first, you have a small inlet oriface that is the throttle plate at idle. Then you have valve overlap...a direct link to outside air pressure via the exhaust valve. Each time an inlet valve opens, air trys to re-pressurize the inlet/plenum chamber especially when the throttle is closed. To achieve and maintain a pressure drop takes effort/work.....otherwise according to your reasoning, a vacuum pump would not need a motor to achieve or maintain a pressure drop.

John

[Greg Locock]

I agree about the overlap as a source of extra air, but, by direct measurement, not /much/ is getting back into the plenum. Wish I had a dyno!.

[NTSOS]

Wish I had a dyno!.

If you had an SI vehicle with a non-lockable steering column.........turn off the ignition at 50mph and keep you foot on the throttle......record 50 - 0 time.

Same thing with foot off the throttle.

A CI vehicle doesn't have a throttle......it also does not stop as well.

John

[J. Edlund]

Originally posted by Greg Locock
MAP with a closed throttle is at most -80 kPA, according to my gauge. So there is no air in the throttle to do any work on.

Sure there is air, you even claim that yourself when you say that you reach -80 kPa at most. There is no such thing as a negative pressure, so what you actually mean is that the pressure in the manifold is about 20 kPa since those gauges are usually graded from zero at atmospheric pressure which is about 100 kPa. So, in order to reach a pressure of 20 kPa there must be some gas there as the pressure would be 0 kPa otherwise.

Originally posted by NTSOS
If you had an SI vehicle with a non-lockable steering column.........turn off the ignition at 50mph and keep you foot on the throttle......record 50 - 0 time.

Same thing with foot off the throttle.

Many newer cars have electronic throttles, so they probably close the throttle when the engine is turned off independent of throttle position. So, this should be done with an older car.

[Greg Locock]

Did it, in a less scientific way. Subjectively I could not tell the difference in retardation throttle open and throttle closed.


Which may tell you more about the condition of the engine than any proof either way!

[McGuire]

The common term "compression braking" is a bit of a misnomer. When the throttle slams shut on decel, the engine becomes in effect a vacuum pump. That's engine braking. This high vacuum @low load also suctions the oil off the rings, cylinder walls and valve guides and blows it out the exhaust when the engine accelerates again. If an engine has weak valve guides or seals, that's when it will smoke: on reapplication of throttle after a long coastdown. (Some engines all visibly smoke here, like Ferrari V12's.)

If you you shut off the ignition and hold the thottle open on decel, the pressure in the intake manifold will be near atmospheric and there will be little/no engine braking, just as with a diesel, which has no throttle. On diesels engine braking can be accomplished with the addition of an exhaust brake, or via a Jacobs brake. The Jake brake opens the exhaust valve as the piston nears TDC, venting the cylinder to the atmosphere. So there is pumping work on the compression stroke, but only atmospheric pressure on the power stroke...voila, engine braking. The sound of a Jake brake on a big diesel rig is loud and distinctive (like a demonic air tool) but due to local noise ordinances you seldom hear them anymore. Two-stroke SI engines offer poor engine braking as well.

Engine braking is part of the driveability picture... throttle closing rates are manipulated to handle driveline snatch and fishbite etc so every year cars seem to have more "sail-on" dialed into the calibration...to older folks not acclimated to this, it seems like the throttle is hanging up. On vehicles with electric regenerative braking systems (hybrids) engine braking is wasteful. So Honda and others disable the valves on three of the four cylinders on decel to kill off engine braking and maximize regenerative braking.

It might be difficult to monitor MAP via the data stream value sufficiently to study engine braking, due to the data rate and fenceposting, but if you put a scope (or a Fluke meter set on snapshot) on the MAP signal line it should show up nicely. Also, depending how fast and sensitive is the MAP sensor, you can throw a cylinder parade up on the scope, similar to a conventional primary/secondary ignition scope trace but showing manifold pressure, cool beans. Depending on the type of pressure sensor, it may work better using high reference (battery positive) instead of ground. This MAP cylinder display can be very useful for isolating valve problems, weird vacuum leaks etc.

[McGuire]

Originally posted by NTSOS
To achieve and maintain a pressure drop takes effort/work.....otherwise according to your reasoning, a vacuum pump would not need a motor to achieve or maintain a pressure drop.
John

It sure does. With a vacuum pump we are trying to pump low-pressure air into a high pressure zone and there is considerable work involved, even if the high pressure zone is "only" atmospheric.

14.7 PSI or 100 kpa may not sound like much, but if we suck the air out of a soft drink can, atmospheric pressure will crush it. When vehicle bodywork distorts at high speed, it's not so much the high pressure on the outside but the low pressure on the inside.

When we operate say a small nylon medical syringe, notable hand effort is required on the suction stroke to overcome atmospheric pressure on the back side of the syringe. The air can't rush into the cylinder fast enough through the tiny entry/exit orifice to equalize the pressure on both sides of the piston. And that's what happens with a piston engine when we slam the throttle shut, just as you say.

A constant displacement pump can evacuate a large volume of air, but it cannot produce a very impressive vacuum. To see some finely made machines (judging by the stuff you build, you obviously appreciate fine machinery) check out turbomolecular vacuum pumps. Beautiful things, miniature multi-stage axial turbines.

[NTSOS]

It sure does. With a vacuum pump we are trying to pump low-pressure air into a high pressure zone and there is considerable work involved, even if the high pressure zone is "only" atmospheric.

Oh, I agree....it was a bad example on my part, but the key word was "YOUR", as opposed to not me, not me!

check out turbomolecular vacuum pumps. Beautiful things, miniature multi-stage axial turbines.

I shall.....thanks!

John

[flannel]

Diesels do have some engine braking due to the high inertia of the moving components.

[hydra]

and by inertia you mean friction...

[kNt]

Is this different for super charged diesels?

[GSX-R]

Hello,

Higher inertia will be, one day or another, converted to friction at engine braking (except if you re-charge engine just after the revs up of the engine braking. This way is not a real engine braking).

I don't see why the turbocharged diesel would act as a different way. To my knowledge, the turbo is never used as an engine braker.

[Halfwitt]

Originally posted by hydra
and by inertia you mean friction...

..... perhaps he meant inertia....

Imagine a huge flywheel on the end of the crank, or even better geared from the crank to give a speed increase. If you change down a gear so that the engine speeds up, the flywheel will require energy to give it an the higher angular velocity. Therefore the car slows down more quickly than without the big flywheel. Hence, an engine or drivetrain with higher inertia will lead to increased engine brake effect.

[hydra]

Originally posted by Halfwitt


..... perhaps he meant inertia....

If you change down a gear so that the engine speeds up, the flywheel will require energy to give it an the higher angular velocity. Therefore the car slows down more quickly than without the big flywheel. Hence, an engine or drivetrain with higher inertia will lead to increased engine brake effect.

Umm... You just contradicted yourself there, increased inertia will DECREASE engine braking not increase it. Frictional losses, in the form of frictional horsepower are effectively equivalent to applying the brakes on the driveline or, in other words, engine braking...

[Greg Locock]

no, as you change down, a bigger flywheel will slow the car more as it comes up to speed. blleding obvious.

[desmo]

Increased engine inertia/flywheel mass will decrease engine braking effect, until one downshifts and requires the engine speed to accelerate, at which point as the clutch is reengaged one will see an increased braking effect as a portion the car's forward momentum will have to be used to reaccelerate the increased inertia/flywheel mass. Right?

Put another way, increased inertia only increases "engine braking" if a downshift is implemented. But is this effect properly called "engine braking" if it only is seen on downshifts? And if not, what shall we call it? I had thought of engine braking as the relative decelerative effect resulting from closing the throttles- SI here obviously- independant of downshifting. Gota love semantics

[GSX-R]

Desmo, with your "giant" flywheel you're tryng to make a real scale "friction car" !

To speak true, I would tell you that even if you don't down shift, you will keep exactly the same engine braking effort. Be careful : engine braking i said, not car braking.

But, without downshift your car will brake slower because the engine friction is opposed to [ energy stored in the flywheel (and inside the engine sparts) + car inertia ] whereas with downshift the friction is only opposed to the car inertia and, plus, the engine will acquire inertia from car inertia so the global car inertia is more reduced. So, and if you're an excellent driver, you use the car brakes, the engine friction, and at stop your engine is still spinning at high rotation speed !

So, without downshift it's like if you would have one "friction engine" with one friction brake together....

The real braker is the friction. If we want to exactly know, we'll have to figure exactly the energy that can contain an engine of 2 liters and 4.000 rpm for instance, to determinate, regarding the car weight, if the global "flywheel" effect of the engine and trasnmission regarding the car inertia.

That is not really the pain i guess.

A good test to make your mind is : Let's try to accelerate without any gear set until 3.000 rpm and release your pedal and wait to the idle rpm. You'll see that the friction effect is a lot more than the inertia of the engine... ! The engine will reach quite fast the idle time.

So you will conclude that the inertia of the engine is quite nothing compared to the car inertia.

Regards.

P.S :Hope the friction car means the same than in French. A toy that contains a flywheel as an engine.

[GSX-R]

For those that ares interested into,

Those samples are extracted from the excellent free software lotus engine simulator :



I've provided an example from a mono cylinder with quite squares bore & stroke.

Ther're several friction models developped by engine builders. Results are not always similar. Notice especially IDI DI comparison models values for diesel engines, higher values for diesel.

Enjoy : Bars are FMEP bars. (friction mean effective pressure that acts against IMEP)

================================================== ================================================
LOTUS ENGINE FRICTION PROGRAM - FRIC
================================================== ================================================
INPUT DATA
~~~~~~~~~~

Bore . . . . . . . . . 87.00 mm Stroke . . . . . . . . 84.00 mm
Compression Ratio . . 11.00
No. of Cylinders . . . 1 No. of Main Bearings . 2

Main Bearing Dimensions Default for Inline Engine
Bearing Diameter . . . 52.20 mm Bearing Width. . . . . 19.31 mm

Big End Bearing Dimensions Default for Inline Engine
Bearing Diameter . . . 49.59 mm Bearing Width. . . . . 20.33 mm

Valve Gear DOHC with Direct Acting Follower
Flat Follower
Number of Valves/Cyl . 4 Maximum Valve Lift . . 10.00 mm

Cam Bearing Dimensions Default Dimensions
Bearing Diameter . . . 29.58 mm Bearing Width. . . . . 19.52 mm

Load Fraction . . . . 1.00

Results
~~~~~~~

Patton and Heywood Component Based
Engine Speed Rotating Recip. Valve Train Auxilary Total
(RPM) (Bar) (Bar) (Bar) (Bar) (Bar)
1000. 0.1714 0.5474 0.2822 0.3294 1.330
1500. 0.2125 0.5666 0.2430 0.3431 1.365
2000. 0.2574 0.5963 0.2262 0.3564 1.436
2500. 0.3059 0.6310 0.2183 0.3691 1.524
3000. 0.3580 0.6688 0.2149 0.3814 1.623
3500. 0.4139 0.7088 0.2139 0.3931 1.730
4000. 0.4734 0.7505 0.2146 0.4044 1.843
4500. 0.5366 0.7937 0.2163 0.4151 1.962
5000. 0.6035 0.8382 0.2187 0.4254 2.086
5500. 0.6741 0.8837 0.2217 0.4351 2.215
6000. 0.7483 0.9302 0.2250 0.4444 2.348
6500. 0.8263 0.9776 0.2287 0.4531 2.486
7000. 0.9079 1.026 0.2326 0.4614 2.628

Sandoval and Heywood Component Based
Engine Speed Rotating Recip. Valve Train Auxilary Total
(RPM) (Bar) (Bar) (Bar) (Bar) (Bar)
1000. 0.1493 0.2965 0.2003 0.5462E-01 0.7006
1500. 0.1793 0.2980 0.1944 0.6393E-01 0.7357
2000. 0.2131 0.3136 0.1980 0.7510E-01 0.7998
2500. 0.2505 0.3421 0.2052 0.8813E-01 0.8860
3000. 0.2916 0.3828 0.2143 0.1030 0.9917
3500. 0.3364 0.4349 0.2245 0.1198 1.116
4000. 0.3849 0.4981 0.2352 0.1384 1.257
4500. 0.4370 0.5719 0.2464 0.1589 1.414
5000. 0.4928 0.6559 0.2579 0.1812 1.588
5500. 0.5523 0.7500 0.2695 0.2054 1.777
6000. 0.6155 0.8538 0.2814 0.2315 1.982
6500. 0.6823 0.9672 0.2933 0.2594 2.202
7000. 0.7529 1.090 0.3053 0.2892 2.438
================================================== ================================================

Comparision of Friction Models

Engine Speed H.B.MOSS M & H M & H Pat.&Hey Chen&Flynn Honda Mod.Honda Sand&Hey Mean
(RPM) (Bar)
1000. 0.8007 0.9655 1.034 1.330 .9444 1.193 .9246 .7006 .9868
1500. 0.9011 1.207 1.276 1.365 1.173 1.208 .9859 .7357 1.106
2000. 1.001 1.448 1.517 1.436 1.401 1.230 1.057 .7998 1.236
2500. 1.102 1.690 1.759 1.524 1.629 1.257 1.137 .8860 1.373
3000. 1.202 1.931 2.000 1.623 1.857 1.290 1.226 .9917 1.515
3500. 1.302 2.172 2.241 1.730 2.085 1.329 1.326 1.116 1.663
4000. 1.403 2.414 2.483 1.843 2.314 1.375 1.434 1.257 1.815
4500. 1.503 2.655 2.724 1.962 2.542 1.426 1.552 1.414 1.972
5000. 1.604 2.897 2.965 2.086 2.770 1.484 1.679 1.588 2.134
5500. 1.704 3.138 3.207 2.215 2.998 1.547 1.816 1.777 2.300
6000. 1.804 3.379 3.448 2.348 3.226 1.617 1.962 1.982 2.471
6500. 1.905 3.621 3.690 2.486 3.455 1.693 2.118 2.202 2.646
7000. 2.005 3.862 3.931 2.628 3.683 1.774 2.283 2.438 2.825
================================================== ================================================

2 first M&H stand for DI and IDI diesel engines

[GeorgeTheCar]

This discussion reminds me of what Ford learned at LeMans with the MKll cars, that downshifting did not increase deceleration but hampered it.

They had to teach the drivers to use only the brakes for slowing and downshift just to be in the correct gear to accelerate what that was appropriate.

I think it was Carrol Smith who said brake pads and even rotors are much cheaper than clutches and gearboxes!

I beleive that the Penske team even used "suck back" calipers in Trans Am to allow the brakes to be used to the full extent! I know that Ford had them in endurance racing.

[phantom II]

Only if you don't know how to heel & Toe.

Originally posted by Greg Locock
no, as you change down, a bigger flywheel will slow the car more as it comes up to speed. blleding obvious.

[shaun979]

Originally posted by GeorgeTheCar
This discussion reminds me of what Ford learned at LeMans with the MKll cars, that downshifting did not increase deceleration but hampered it.

Can you share more detail on this? The only way I see this happening is by upsetting brake balance, or by the downshifting affecting the driver's ability to apply consistent brake force at lockup threshold. FWIW when I track drive, I use the brakes and not the engine for braking, but as a purely academic exercise I'm just wondering why downshifting would not aid deceleration, assuming above 2 factors are catered for or do not occur. Thanks

Shaun

[Greg Locock]

Phantom, I was being a bit cryptic/sarcastic. During the downshift, having a large flywheel does slow the car more. However, over a complete deceleration event, on average, it will have a negative or zero effect, depending on how you define your endpoints and relativities.

[Joe Bosworth]

I can share some old data that I once developed. The data was for a six cylinder engine using 4 ring pistons. I don't have any way of exrapolating the data so that it is applicable to modern race engine design, however the information is of general use to the present discussion.

The data was developed by running the engine on a dyno to determine BHP output. I then hooked up a variable speed DC motor to measure friction HP and pumping losses. The DC motor meant I was limited to spinning the engine to about 2400 rpm.

Friction and pumping losses were measured through the manifolds and carb in wide open throttle position. Friction losses alone were measured by removing the head, leaving only piston, bearings and oil circulation loads.

Total friction and pumping losses increased from nil to about 17% of BHP output at 2400 rpm. The increase was a little greater than linear. I estimate that friction and pumping losses would have been close to 50% of BHP at 6000 rpm.

The losses at all points measured about 15% pumping loss and 85% friction losses.

The friction losses would have been a bit higher percentage of the total losses if the valves had been actuating, which they weren't due to head removal.

Make of this data what you wish.

Regards

[hydra]

Let me throw in my $0.02,
State of the art hi-perf and racing engines typically have frictional losses (not counting PMEP) of around 12-16% at peak bhp rpm , and pumping losses of around 6-9%...

[phantom II]

Iskendarian spun a NASCAR engine at 8000 rpm without pistons and rods, pumps, etc. It took 100hp to run the valve train.


The data was developed by running the engine on a dyno to determine BHP output. I then hooked up a variable speed DC motor to measure friction HP and pumping losses. The DC motor meant I was limited to spinning the engine to about 2400 rpm.

Total friction and pumping losses increased from nil to about 17% of BHP output at 2400 rpm. The increase was a little greater than linear. I estimate that friction and pumping losses would have been close to 50% of BHP at 6000 rpm.

The friction losses would have been a bit higher percentage of the total losses if the valves had been actuating, which they weren't due to head removal.

Make of this data what you wish.

Regards [/B][/QUOTE]

[desmo]

Desmo,

Sorry it's taken so long to get back but I've been tapping a source of info on valve gear. His estimations are that power consumption on a F1 engine at 17,000rpm with 4 valves per cylinder and 10 cylinders is that power absorption will be i.t.r.o 120Bhp (90Kw) - he bases this on knowing valve sizes, weights, air spring pressure, and a load of other bits of info which is not normally in the public domain.

The interesting point here is that recovered energy is i.t.r.o 100Bhp (75Kw) - this he said is proven by reducing air spring pressure on an engine (or more correctly tested on a head rig) to get to the point of valve float, at this point recovered energy is lost along with about the 100Bhp off the dyno (or similar increase in power required to drive a rig) - it's one way of telling if you have a valve float problem, when a sudden massive loss of power is detected.

[Joe Bosworth]

Desmo

Some days I am thick.

Are you saying that the energy to open the valves is 220 bhp of which 100 bhp is recovered by the valve closing process so that the net energy used is 120 bhp?

Or what? Very interesting!

Regards

[shaun979]

Originally posted by Joe Bosworth
Desmo

Some days I am thick.

Are you saying that the energy to open the valves is 220 bhp of which 100 bhp is recovered by the valve closing process so that the net energy used is 120 bhp?

Or what? Very interesting!

Regards

I think he is saying effectively on 20hp to drive it. 120 - 100 = 20

Phantom's previous number of 100 hp required to drive Cup valvetrain at 8000 RPM sounds a bit high.. that's a lot of heat to dissipate!

[J. Edlund]

Originally posted by shaun979


I think he is saying effectively on 20hp to drive it. 120 - 100 = 20

Phantom's previous number of 100 hp required to drive Cup engine at 8000 RPM sounds a bit high.. that's a lot of heat to dissipate!

20 hp seems to be in the ballpark.

One can estimate the friction losses in an road car engines with the following formula:

FMEP = 0.3 + 0.04*(vP)^1.3

vP is piston velocity in meters per second and FMEP is given in bar.

Given a piston velocity of 18 metres per second that equals to a FMEP of 2 bar; or about 30 Nm for 2 litre engine. Given a torque output in the region of 200 Nm at peak power speed this indicate that about 13% of the total output, or 26 hp is lost due to friction.
On a racing engine I would guess that the friction losses are somewhat smaller.

[shaun979]

J.Edlund, I just edited my post to correct a typographical error "engine" to "valvetrain".
=============

Thanks for the equation.

The FMEP you refer to is totally separate from PMEP correct?

How do you see racing engines as having roughly the same FMEP percentage wise when they generally run much higher engine speeds that increase FMEP? Trying to understand.

[GSX-R]

FMEP is opposed to IMEP. Total result including pumping produces BMEP.

Race engines use best materials are better machined and adjusted and designed, thin piston rings. Logically FMEP should be less, at least at same rpm i guess

[GSX-R]

90kW on a F1 engine at 17.000 rpm would be approximately be 2,11 bars FMEP, just for valve gear...


Total FMEP values issued from all lotus simulator models are beetween 2,4 and... 4,2 bars

[McGuire]

Friction is one area where models are not usefully accurate.

[GSX-R]

Hello Mr Mc guire

Better than nothing..

[desmo]

Originally posted by McGuire
Friction is one area where models are not usefully accurate.

One of the SAE papers done by Ferrari guy Martinelli makes this same point. I wonder why?

[GSX-R]

Because friction is one of the most complex phenomenon to figure ... so many parts, so many parameters.

[GSX-R]

Originally posted by desmo


One of the SAE papers done by Ferrari guy Martinelli makes this same point. I wonder why?

Do you have any extract of that paper ?

[desmo]

"Friction values are obtained by using a certainly less reliable empirical
method. Such technique represents typical rapid methods
that are often used to obtain a feel for a complex subject
based on a limited amount of experimental data.
Consequently, results in term of friction and brake values
have to be considered very carefully."

Comparison of V10 and V12 F1 Engines
Alberto A. Boretti
Fiat Research Center, Engines Dept.
Giuseppe Cantore
University of Modena

[desmo]

This, from the Tech Papers thread, is worth a read:

http://me.engin.umic...be/P2000_08.PDF

[GSX-R]

Thank you.