rec.autos.simulators

Good move, and also, in that other thread, I noticed a glimpse of the
old Byron.....:-)

Beers and cheers
(uncle) Goy

http://www.theuspits.com
http://www.teammirage.com

"A woman is an occasional pleasure but a cigar is always a smoke"
--Groucho Marx--

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  You're right about that na_biker.  However, it appears that Ruud and you are
speaking of two different definitions of engine braking.  Ruud is speaking of
engine braking as the torque itself, while you're speaking of the combined
effect of torque and inertia, i.e., the overall acceleration..  If you'd want
to calculate the acceleration and make comparisons betwen different cars, you'd
need the engine braking *torque* first, then you can calculate the acceleration
part you're talking about here...  

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  Understandable, and quite the reason arguments about it will happen.  I think
that anyone who is not trying to code a vehicle model or make the calculations
would think of it exactly the way you are.  A pro race car driver probably
would too, I'd imagine.  After all, when driving one car you might let off the
gas pedal and the car slows down at .2g's in top gear, while the next day after
installing a lighter flywheel, it slows down at .22g's.  You could certainly
say that the car has more engine braking now!  As far as the driver is
concerned, the effect of lightening the flywheel or making a camshaft/whatever
change that actually lowered the zero throttle torque itself is pretty much
identical.

 But, the hard headed folks like me would argue this definition a bit, and say
that since the engine's negative torque with the throttles closed didn't change
when you switched to the lighter flywheel, neither did the engine braking,
because I would define the engine braking as a torque, not an overall
acceleration.  Some people like to seperate the two effects, myself included
:0)

http://performancesimulations.com/partthrottle.htm

Here's a plot from an F1-2000 engine (don't know which one) at multiple
throttle positions.  At zero throttle, the bottom torque curve goes below zero.
 This is what I'd define as engine braking personally, but I don't know if
there's an official SAE definition, so yours is as good as mine :0)  However,
if we seperate the two this way, we can analyze the effects of your definition
of engine braking by looking at BOTH torque and engine/flywheel inertia.  And
the current driving gear, I suppose.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  I don't think you need a "for dummies" version, Gunnar! :0)  Let's define two
versions of "engine braking", since I think this is where the confusion in this
thread is coming from:

1.  The race car driver's seat-of-the-pants version:  Engine braking is how
quickly you slow down when you let off the throttle, regardless of what caused
it (minus aerodynamic drag effects).  

2.  The engineer's version:  Engine braking is the "torque" the engine makes
when you let off the throttle, and is seperate from the acceleration.  In this
definition, inertia is not part of the equation at all, so it's irrelevant.

 According to "the engineer's version", it would have no effect on engine
braking, however, you think in terms of the "race car driver's version" of
engine braking, as most people probaby do.  There's nothing wrong with that, of
course, it's intuitively quite right once you're behind the wheel :0)  

 So...  According to the race car driver's version, a heavier flywheel would do
just the opposite of what you said; it would *decrease* engine braking.  Why?
Suppose at 5000 rpm (or whatever speed) we shut the throttles.  The engine
makes (wild guess) negative 50 lbs-feet of torque (trying to slow the
car/engine down, the opposite of making power).  The engine's inertia with the
normal flywheel is 5 (another wild guess).  

 The engine's acceleration when you shut the throttle would be:
Acceleration = Torque/inertia
                   = -50lbs*feet / 5
                   = -10 radians/sec/sec
                   = 95 Rpm's/sec  

If we put a flywheel that had twice as much inertia (either it was heavier, or
bigger, or a combination of the two), the inertia is 10.  The acceleration
would be:
Acceleration = Torque/inertia
                   = -50lbs*feet / 10
                   = -5 radians/sec/sec
                   = 47 Rpm's/sec

Exactly 1/2 of what you had before.  So you'd wind up with less engine braking
(according to "the race car driver's definition").  

This is somewhat misleading though, because it appears that we've got 1/2 the
engine braking with the heavier flywheel, so if you let off the throttle, you'd
expect the car to slow down about 1/2 as quickly.  This doesn't happen though;
this "inertia" effect is rather small compared to what's happening at the
wheels.  I'll spare you the math on that part though.

The "engineer's" definition (I know I'm repeating myself here in different
posts and you got it by now) says that the -50lbs*feet _IS_ the engine braking.
 This didn't change from one flywheel to the next, so neither did the engine
braking.

Again, neither definition is really wrong imo, it's just a matter of how you
want to define engine braking.  Ruud, Jonny, and Collin prefer the "engineer's"
definition though.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

I wasn't speaking of any definitions per se, I was talking about how
more inertia affected the engine braking(whilst driving).

I mean, I know what you're saying, it's pretty obvious, but I think
that in general speaking 'engine braking' is meant as the immediate
torque which affects the _driving wheels_ in roadgoing conditions.
I.e. when actually driving. And I had a feeling that's what Gunnar
meant also. If that's not what you mean when you say that, it means
that you're a frickin' bookworm and should go out and smell some
petrol fumes. ;-)

I mean, in Real Life(tm). especially on a racetrack, steady state
engine braking isn't likely to be 'felt' other than when turning an
engine over in the workshop. :) If *I* would out my physics hat on, I
would be very careful to use the term 'engine braking' to express the
negative torque generated by friction/"pumping"/et.c. I'd prefer
something else.

      _
Mats Lofkvist

--
Gunnar
    #31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo: 2
                             gitaren er en sjingke

(Drag) force (proportional to) speed^2

Power = force x speed

Power (proportional to) speed^3

Jonny

When did I say "dramatically"?  It will increase acceleration
(in either direction) but whether it's measureable or useful is
another matter - that depends on the exact case in question.

Right.  Let's try again, 'cos I'm not disagreeing with your
observations - only your explanations for them.

Engine braking is caused by drag torque ("motoring torque", as
Doug called it).  That's a combination of engine friction,
compression, load from ancilliaries and so on.

What that torque does, if there's insufficient throttle applied
to cancel it out, is to slow down all the rotatings parts of the
drivetrain.  That includes everything from the crankshaft to the
brake discs and wheel rims.  The less inertia the driveline has,
the more effective the braking torque is (hence a lighter flywheel
makes TTO more likely, just as you say, since the driveline slows
down more quickly which means that the tyres develop a greater slip
ratio, and come closer to the limit of traction).

A lighter driveline does not alter the *amount* of braking torque
produced by the engine, but it does increase the deceleration
*caused* by that torque.

Now, where do you disagree in that lot?

I also added an aside, aimed mostly at the modelling guys, noting
that the gear ratios between various parts of the driveline change
the effective inertia of the upstream parts compared to road speed.
A car in first gear has a greater total inertia (rotating plus
linear) than a gear in fourth gear, since the engine has to
accelerate more rapidly for a given vehicle acceleration.

Do you disagree with that?

<shrug>  Well, for the moment I'm happy to enjoy the Auto Eng job
which has followed my Auto Eng degree...

Can we be friends now?

Jonny

  I bet it does..  Someday maybe I'll race something besides a rent-a-kart :-)

 If an F1 car made 10% less negative torque than a Cadillac when the throttles
were closed, but it had 1/3rd the rotating inertia, which one has more engine
braking?  Maybe my example is bad, maybe the F1 car has more negative torque
than the Caddy?  I don't know...?

  Sounds like a real handful.  I'm jealous :-)  Why in the world did you sell
it ?

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  Are you talking about what I said here?  If so, I don't understand...  I'm
agreeing with you... ?  

  In case you were... :0)  I know that a lightened driveline does everything
you've just said it does, and of course has an effect on trailing throttle
oversteer behavior.  That's the acceleration part of the equation.  What I said
was that the negative *torque* (at the flywheel) produced is not related to the
engine's mass in a steady state condition (meaning the engine is not changing
speed).  The overall acceleration is though of course, because it's torque
divided by inertia.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  There is more engine braking "acceleration", you're right, which is what a
driver (apparantly) is referring to when talking about engine braking.  But I
think the others' point is is that the engine braking "torque", as measured at
the flywheel, is the same.  If somebody's modelling this stuff, they have to be
kept track of seperately.  If you're driving the car, you don't need to know
there's a difference.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm

  Youch..  Good choice though...  I guess ;-)  

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm