Hi all, I've just finished the new demo which can be downloaded from
the dynamics/downloads section of my homepage :-
http://www.cersoft.com
Due to request I have implemented fixed cameras, mouse control and
preliminary sound (only engine at the moment).
Please read the readme.txt as there are options that allow the use of
mouse control, are some rendering options.
There are still some problems :-
No controller support yet, I do have it working but I want to tweak it
more and make some configuration screens before releasing to the
public.
I have not yet implemented relaxation lengths for the tires and hence
the wheels rotate when at low speeds.
Still no keyboard configuration, sorry to anyone with non QWERTY
keyboards, you can always use the mouse :-)
Enjoy,
Colin
Notable threadColin Reed's car sim demo and an engine braking debate
Colin Reed released a new demo of his hobby driving simulator, which testers liked but found hard to catch once sliding. He traced this to a bug that gave the engine far too much engine braking, which sparked a long technical debate between modellers Ruud van Gaal, Jonny Hodgson and Todd Wasson and a former racer over engine braking torque, flywheel inertia and drag on F1 cars. Doug Milliken and Matt Jessick of Motorsims added notes on motoring torque and aerodynamics.
With Matt Jessick (Motorsims)
physics/realism debaterelease/news
Automated summary.
Looks very promising.. I actually managed to get myself caught up in one of
those "gotta try this just one more time" things, that sort of implies that
it's already fun to mess around on the track.
One thing I find a bit off the mark is fact that it's just about impossible
to save the car once it starts powersliding. Maybe it's because of the
keyboard input ?? It felt like there was a ton of bricks in the trunks and
when the rearend broke loose there was no way to countersteer and catch the
slide.
Am I wrong here ??
Mikkel Gram-Hansen
http://www.racesimcentral.com
Colin Reed <[email hidden]> wrote in message
news:[email hidden]...
Quoted text (25 lines)
> Hi all, I've just finished the new demo which can be downloaded from > the dynamics/downloads section of my homepage :- > > http://www.cersoft.com > > Due to request I have implemented fixed cameras, mouse control and > preliminary sound (only engine at the moment). > > Please read the readme.txt as there are options that allow the use of > mouse control, are some rendering options. > > There are still some problems :- > > No controller support yet, I do have it working but I want to tweak it > more and make some configuration screens before releasing to the > public. > > I have not yet implemented relaxation lengths for the tires and hence > the wheels rotate when at low speeds. > > Still no keyboard configuration, sorry to anyone with non QWERTY > keyboards, you can always use the mouse :-) > > Enjoy, > ColinReally nice demo, from both the physics and graphics point of views.
But beware: posts asking for a cart sim should not be long in coming :-)
Xavier.
"Xavier" <[email hidden]> skrev i en meddelelse
news:a3ee2e$597$[email hidden]...
> Really nice demo, from both the physics and graphics point of views.
> But beware: posts asking for a cart sim should not be long in coming :-)
>
KART, KART, KART!!!
CART SIm!!! MAKE A CART SIM!!!!
> Looks very promising.. I actually managed to get myself caught up in one of
> those "gotta try this just one more time" things, that sort of implies that
> it's already fun to mess around on the track.
:-) This is good to here.
Yes and no. No, I haven't got any code in there to simulate bricks in
the trunk ;-) but I haven't had time to setup the car properly yet and
so it is very tail heavy, even when not under power if you steer from
left to right and back repeatedly it will start to oversteer which is
something that I have to look at. Also, try the mouse controls, it is
much easier to catch a slide (although still hard!).
Colin Reed
Quoted text (7 lines)
> One thing I find a bit off the mark is fact that it's just about impossible > to save the car once it starts powersliding. Maybe it's because of the > keyboard input ?? It felt like there was a ton of bricks in the trunks and > when the rearend broke loose there was no way to countersteer and catch the > slide. > Am I wrong here ??> Really nice demo, from both the physics and graphics point of views.
Thanks, still much more to do, I really want to get the controller
support in there so that I can really see how it is really handling
and how much I have to work on.
> But beware: posts asking for a cart sim should not be long in coming :-)
Aaaarrrrrggghhhhh, you're gonna start them off again !!!!
Colin
But, we can make our own cars and tracks, right!?!? Get the controller stuff
done!! Get the controller stuff done!!Get the controller stuff done!! :)
Nice piece of work.
Charlie
"Colin Reed" <[email hidden]> wrote in message
news:[email hidden]...
Quoted text (12 lines)
> > Really nice demo, from both the physics and graphics point of views. > > Thanks, still much more to do, I really want to get the controller > support in there so that I can really see how it is really handling > and how much I have to work on. > > > But beware: posts asking for a cart sim should not be long in coming :-) > > Aaaarrrrrggghhhhh, you're gonna start them off again !!!! > > Colin >> > One thing I find a bit off the mark is fact that it's just aboutimpossible
> > to save the car once it starts powersliding. Maybe it's because of the
> > keyboard input ?? It felt like there was a ton of bricks in the trunksand
> > when the rearend broke loose there was no way to countersteer and catchthe
Ahh mousecontrol did the trick... can't fool an old GPL driver, yessir I
know how to slide a car ;)
It's still extremely difficult to control probably because I'm not used to
mousecontrol, but the physics feel right.. sorry I cant pinpoint it better
than that.
Now I want you to forget what you was doing, and fix me up a track with some
some jumps.... jumps NOW *lol*
I know it's a serious racing thing, but please make a sideproject for us
wannabe stuncar racers.
Mikkel Gram-Hansen
http://www.RaceSimCentral.com
Quoted text (10 lines)
> > slide. > > > Am I wrong here ?? > > Yes and no. No, I haven't got any code in there to simulate bricks in > the trunk ;-) but I haven't had time to setup the car properly yet and > so it is very tail heavy, even when not under power if you steer from > left to right and back repeatedly it will start to oversteer which is > something that I have to look at. Also, try the mouse controls, it is > much easier to catch a slide (although still hard!).
> Now I want you to forget what you was doing, and fix me up a track with some
> some jumps.... jumps NOW *lol*
> I know it's a serious racing thing, but please make a sideproject for us
> wannabe stuncar racers.
Anyone remember Whiplash? That was a fun racing game with great jumps in
it. Wonder if it'll run on todays hardware? If I remember correctly it
was a glide only game?
> It's still extremely difficult to control probably because I'm not used to
> mousecontrol, but the physics feel right.. sorry I cant pinpoint it better
> than that.
I've been thinking about this some more, it could be one of the
following things :-
1) The suspension currently moves in a perpendicular line to the body
(yes I know, very lame) and so because 'track camber' is modelled as
soon as you turn and the weight transfers to the outside of the car,
the side force afforded by the -ve static camber is going to reduce.
2) The inertia of the body is too high resulting in unstoppable spins.
3) the setup of the body (CG location, front/rear weight ratio, etc)
are wrong, if someone wants to have a fiddle with these, please feel
free, you can edit the file dataDir/carModel/carSetup.ini with a text
editor.....
> I know it's a serious racing thing, but please make a sideproject for us
> wannabe stuncar racers.
As of the moment, 'raceDemo' (really must think of a cool name for it)
is an open book, I do intend for MODs to be easily makeable with the
possibility of an SDK in the future ....
Colin
On 1 Feb 2002 01:35:56 -0800, [email hidden] (Colin Reed) wrote:
>Hi all, I've just finished the new demo which can be downloaded from
>the dynamics/downloads section of my homepage :-
>
>http://www.cersoft.com
Cool, gotta get to see this...
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
> >Hi all, I've just finished the new demo which can be downloaded from
> >the dynamics/downloads section of my homepage :-
This is kind of embarassing, I've found a HUGE handling affecting bug
that has now been fixed, please wait until the new version later today
;-)
Colin
[email hidden] (Colin Reed) wrote in message news:<[email hidden]>...
Great demo! What's the bug? I haven't noticed it.
/Jerry
Quoted text (8 lines)
> > >Hi all, I've just finished the new demo which can be downloaded from > > >the dynamics/downloads section of my homepage :- > > This is kind of embarassing, I've found a HUGE handling affecting bug > that has now been fixed, please wait until the new version later today > ;-) > > Colin> Great demo! What's the bug? I haven't noticed it.
Ok, in the 0.12 demo, drive along in 3rd gear at around 6000rpm, then
suddenly let of the gas totally and try to steer, you will get a
massive amount of oversteer and probably crash.
Ever since the first raceDemo I had mistakenly set the engine braking
coefficient to 6.24 which means that at 7000 rpm you get;
(7000 / 60) = 116.66
6.24 * 116.66 = 727.95
727Nm of engine braking torque!!! If the accelerator is not being
pressed then all of this -ve torque is being sent to the driving
wheels and hence producing some seriously -ve slipRatios, hence the
oversteer, all of the lateral grip from the traction circle is being
taken away by these -ve slip ratios.
This braking coefficient has now been reduced to a more sensible 1.0
and that is the setting in raceDemo 0.13.
Colin
On 7 Feb 2002 16:04:17 -0800, [email hidden] (Colin Reed) wrote:
>> Great demo! What's the bug? I haven't noticed it.
>...
...
>This braking coefficient has now been reduced to a more sensible 1.0
>and that is the setting in raceDemo 0.13.
I once plucked that coefficient from an Arrows F1 car, which was 0.74.
Seems street cars have much more engine braking, like 1.5 (that's just
an educated guess though). Figured that out once I redid my drivetrain
to include the full analog clutch and the revving of the engine was
funny.
Loved the demo btw; looks great and drives pretty well (although I
only tested with the keyboard). Too bad you're going to have time off
to move.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (7 lines)
>Ever since the first raceDemo I had mistakenly set the engine braking >coefficient to 6.24 which means that at 7000 rpm you get; > >(7000 / 60) = 116.66 >6.24 * 116.66 = 727.95 > >727Nm of engine braking torque!!!
rrevved <[email hidden]> wrote in message news:[email hidden]...
> I hope you're not allowing the lowered aerodynamic drag
> of a F1 car vs. a street car to cloud your calculations of
> 'engine braking'.
>
> Any thoughts on this?
*Increased* drag, surely... F1s are optimised for downforce,
not drag.
Jonny
On Fri, 08 Feb 2002 06:46:02 -0600, rrevved <[email hidden]> wrote:
First, there was some talk about this not too long ago:
http://dynamic2.gamespy.com/~hg/forums/showthread.php?s=&threadid=34167
(probably wraps, so paste it together in your browser).
Second, I may indeed be wrong, but here's my view: an F1 car runs at
much higher revs than a street car. Having MORE engine braking than a
street car (at the same rpm) would give the racecar a disadvantage
since you have to pull the engine through all that torque trying to
brake the engine. I'd say an F1 car engine would be built more
efficient in that area than a street car (although other goals might
contradict this goal).
There are 2 parameters that decide the speed of revving up and down
(in neutral); the engine torque (which includes engine braking and
positive torque when the throttle pedal is pressed), and the engine
inertia.
A big flywheel gives more inertia, and that's the parameter I know
little about. Haven't got too many clues what a street car engine has
for inertia vs. an F1 car.
Still, I'd think an F1 car will use a much lower inertia (less
weighing flywheel) to be more responsive. Although ofcourse a higher
inertia gives you a more stable accelerating/braking car.
>My Lotus Eleven LeMans racecar had a 3 pound flywheel
>in 1956.
Did you drive it at LeMans back then? Would be way cool. :)
...
>ie: I would think that engine braking on a modern F1 car
>is MASSIVE. Very light driveline components, high
>compression, ratios, higher-revs/lower gearing etc.
>would make them slow down REALLY fast when the throttle
>is closed.
I think that's more to do with the low INERTIA of the components,
rather than the high braking coefficient (which just gives you the
torque with which the engine is trying to slow down, not the actual
speed at which this finally happens, since that is related to inertia
as well).
F1 engines must be powerful to push the cars through the aero drag,
even with high wing settings. So a massive engine braking contradicts
that goal, since at high RPM the engine braking would really fight the
engine from generating any torque to push the car forward through the
air. So I'd think striving for less engine braking would be beneficial
for an F1 car; at high speeds you depend on the aero downforce and
brakes to get good braking, instead of using the engine as a brake.
All this is just how I see it; perhaps someone with more knowledge
(Doug?) can shed a more practical light on this.
> You can hear this by listening to the revving
>of their engines while in the paddock. When the throttle
>is opened and then closed, they slow down REAL quick.
>Much quicker than ... a Toyota Celica.
Right, but I think that's to do with the inertia of the Toyota engine
being far higher than the F1's. As the speed of engine rpm change is
like:
engineAcceleration=engineTorque/engineInertia
I think the product at the right side here is higher on an F1 car just
because the inertia is much lower, although the engine braking torque
is a little lower.
>I hope you're not allowing the lowered aerodynamic drag
>of a F1 car vs. a street car to cloud your calculations of
>'engine braking'.
Nope, they're completely separate. I define the engine braking
coefficient as being a measure of compression braking, and any
friction in the engine. I don't try to include any other implicit
effects. Couldn't do that even, because that would make it harder to
later implement strong winds.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (9 lines)
>On Fri, 08 Feb 2002 12:23:50 GMT, [email hidden] (Ruud van Gaal) wrote: > >>I once plucked that coefficient from an Arrows F1 car, which was 0.74. >>Seems street cars have much more engine braking, like 1.5 (that's just >>an educated guess though) > >Ruud, I'm certainly no physics expert, but how do you >conclude that a typical street car has more engine braking >than a purpose-built racecar?[email hidden] (Ruud van Gaal) writes:
> There are 2 parameters that decide the speed of revving up and down
> (in neutral); the engine torque (which includes engine braking and
> positive torque when the throttle pedal is pressed), and the engine
> inertia.
hm. I always thought engine output and engine braking were completely
separate. engine braking is, like you say, just mass, while power
output decreases with increasing mass.
> A big flywheel gives more inertia,
actually, a heavy flywheel does, but let's not be picky. anyway;
I think there's a lot of magic around the flywheel. I seem to recall
something about fitting a heavier flywheel might increase max power,
since the ratio of rotating mass to, uh, "mass that moves up and
down" [1] (i.e the pistons and con-rods) gets bigger. or smaller. [2]
not that any of this matters, as I'll make a helpful comment in a
little while.
> and that's the parameter I know
> little about. Haven't got too many clues what a street car engine has
> for inertia vs. an F1 car.
you're probably right [3], but [4] I'm pretty sure F1 cars have
back-torque limiters, which would mean you can pretty much just make
something up, or even make it user configurable, if you're a really
nice guy.
> Still, I'd think an F1 car will use a much lower inertia (less
> weighing flywheel) to be more responsive. Although ofcourse a higher
> inertia gives you a more stable accelerating/braking car.
pah! stability schmability! :)
> F1 engines must be powerful to push the cars through the aero drag,
> even with high wing settings. So a massive engine braking contradicts
> that goal,
oh ok. you knew that.
is it bedtime yet?
[1] forgive me, I didn't get any sleep last night. :)
[2] see [1]
[3] I mean, who has ever heard of a heavy part for an F1 car.
[4] this is the helpful comment.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
FJ�SE!
On Fri, 08 Feb 2002 06:46:02 -0600, rrevved <[email hidden]> wrote:
Yeah. When Tiff Needell tested the Jaguar R1 he said(paraphrased) that
letting off the throttle was like slamming the brakes on any normal
car, slamming the brakes on the F1 was like hitting a brick wall. :)
Quoted text (8 lines)
>ie: I would think that engine braking on a modern F1 car >is MASSIVE. Very light driveline components, high >compression, ratios, higher-revs/lower gearing etc. >would make them slow down REALLY fast when the throttle >is closed. You can hear this by listening to the revving >of their engines while in the paddock. When the throttle >is opened and then closed, they slow down REAL quick. >Much quicker than ... a Toyota Celica.na_biker <[email hidden]> writes:
well, yeah. I read somewhere that an F1 car at 200 kph will
decellerate at about 1 G if you just press the clutch.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
silence is FOO!
Quoted text (14 lines)
> On Fri, 08 Feb 2002 06:46:02 -0600, rrevved <[email hidden]> wrote: > > >ie: I would think that engine braking on a modern F1 car > >is MASSIVE. Very light driveline components, high > >compression, ratios, higher-revs/lower gearing etc. > >would make them slow down REALLY fast when the throttle > >is closed. You can hear this by listening to the revving > >of their engines while in the paddock. When the throttle > >is opened and then closed, they slow down REAL quick. > >Much quicker than ... a Toyota Celica. > > Yeah. When Tiff Needell tested the Jaguar R1 he said(paraphrased) that > letting off the throttle was like slamming the brakes on any normal > car, slamming the brakes on the F1 was like hitting a brick wall. :)On 08 Feb 2002 18:59:34 +0100, Gunnar Horrigmo <[email hidden]>
wrote:
Uhm, engine braking is not mass, but torque, as I see it. The net
torque an engine can produce is the sum of the part that make it want
to accelerate (fuel burning), and the part that wants to brake it
(pistons/air being compressed).
To simulate this, I have 2 'curves'; one for the upper engine torque
limit (when throttle=100%) and one lower curve, which states the
torque at 0% throttle. Games like F1-2001 do it like that; a minimal
and maximum torque. As the minimum is almost linear, I use just a
single coefficient for that, so engineBrakingTorque=coeff*rps.
The torque being sent out by the engine then is just linearly
interpolated from the lower to upper curve. I know, this isn't really
linear in real life, but for F1 cars, anything's possible (as it's fly
by wire), and linear seems quite ok (your controller isn't linear
anyway).
>> A big flywheel gives more inertia,
>
>actually, a heavy flywheel does, but let's not be picky.
Ok. :) I indeed meant 'heavy'.
...
>you're probably right [3], but [4] I'm pretty sure F1 cars have
>back-torque limiters, which would mean you can pretty much just make
>something up, or even make it user configurable, if you're a really
>nice guy.
Hm, interesting, never heard of them. I'll have to google that.
...
>is it bedtime yet?
Dinnertime here, lol. Must... get... food...
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (10 lines)
>[email hidden] (Ruud van Gaal) writes: > >> There are 2 parameters that decide the speed of revving up and down >> (in neutral); the engine torque (which includes engine braking and >> positive torque when the throttle pedal is pressed), and the engine >> inertia. > >hm. I always thought engine output and engine braking were completely >separate. engine braking is, like you say, just mass, while power >output decreases with increasing mass.
rrevved <[email hidden]> wrote in message news:[email hidden]...
> >Uhm, engine braking is not mass, but torque, as I see it.
>
> Huh?
>
> Engine braking is the drag of the engine/driveline parts, plus
> the momentum of those parts, Ruud.
Yup: the engine's internal friction creates a torque which opposes
the main, driving torque. Close the throttle, and the friction (drag)
torque becomes greater than the driving torque, hence the *total*
torque is negative or braking.
> Higher mass = Lower engine braking
> Lower mass = Higher engine braking.
Actually, no. The mass is irrelevant to the steady-state torque
produced by the engine, be that driving or braking torque. What
increased mass *does* do is make the car effectively heavier, so
it requires a greater braking torque to achieve the same deceleration.
There's an interesting concept called 'equivalent mass', which
includes the rotating parts in the complete vehicle inertia. An
odd side effect is that the car is actually heavier in lower gears
- something I've noticed with my model race cars, which accelerate
better than I'd expect when I fit silly tall gearing for a big
outdoor track.
> Torque has -nothing- to do with it. Why would you think that
> 'torque' would have anything to do with engine braking when
> the closed-throttle engine is producing almost NO torque?
Because applying torque to the drivetrain is the *only* way the
engine can affect anything...
> Engine 'braking' doesn't occur when the engine is not idling.
No, because then the overall torque is positive.
Jonny
rrevved <[email hidden]> wrote in message news:[email hidden]...
Are you going to actually counter any of my points, or just
laugh like a hyena smoking something illegal?
Jonny
Quoted text (9 lines)
> Really? > LOL!! > LOL!! #2 > LOL!! #3 > Bye Jonny.
rrevved <[email hidden]> wrote in message news:[email hidden]...
> On Fri, 8 Feb 2002 13:37:01 -0000, "Jonny Hodgson" <[email hidden]> wrote:
> >*Increased* drag, surely... F1s are optimised for downforce,
> >not drag.
>
> Jonny, are you really this thick?
No. Are you?
> Try this:
>
> Stand in front of an F1 car and then stand in front
> of a Toyota Celica. Look closely at each of them.
>
> Tell me Jonny, which is optimised for lower drag?
The one with*out* the exposed wheels and big wings.
The one which *doesn't* generate more than its own weight in
downforce at 100mph.
Hint: there's a thing called a 'lift:drag ratio'. Basically
it means that if you want downforce, it'll cost you drag. Racing
cars before the 70s were optimised for low drag. Then some bloke
called Chapman discovered downforce, and things changed a little.
Second hint: a Celica has perhaps 100 hp, and a top speed of
perhaps 120 mph. The F1 has 800 hp and a top end of 210 mph.
Aero power requirements rise with the cube of speed.
(120^3 / 100^3)^0.33 = 1.2 (~= mph / hp)
(210^3 / 800^3)^0.33 = 0.26. IOW, the Celica is getting far
more maximum speed per horsepower than the F1 car - which suggests
it has just a tiny bit less drag.
Jonny
Jonny Hodgson <[email hidden]> wrote in message news:oZY88.11029$[email hidden]...
Okay, I'll try that again without cubing the power. (See, Todd,
I told you you should never trust my equations - especially this
late at night!)
P = k1 * drag * speed^3
drag = k2 * P / speed^3
For the Celica, that gives drag of 100 / 120^3 = 0.000 057 8 [units]
For the F1, it's 800 / 210^3 = 0.000 086 4 [units].
Not a huge difference - the F1 is smaller but has a higher Cd
- but still the same way round.
Jonny
proving that you *can* get a decent degree without being able
to get the right answer first time...
Quoted text (10 lines)
> Second hint: a Celica has perhaps 100 hp, and a top speed of > perhaps 120 mph. The F1 has 800 hp and a top end of 210 mph. > > Aero power requirements rise with the cube of speed. > > (120^3 / 100^3)^0.33 = 1.2 (~= mph / hp) > > (210^3 / 800^3)^0.33 = 0.26. IOW, the Celica is getting far > more maximum speed per horsepower than the F1 car - which suggests > it has just a tiny bit less drag.On Fri, 08 Feb 2002 14:35:14 -0600, rrevved <[email hidden]> wrote:
>On Fri, 08 Feb 2002 20:08:01 GMT, [email hidden] (Ruud van Gaal) wrote:
>...
>>Uhm, engine braking is not mass, but torque, as I see it.
>
>Huh?
>
>Engine braking is the drag of the engine/driveline parts, plus
>the momentum of those parts, Ruud.
Right, and drag is the engine producing torque to counter the
rotation. The momentum just decides how fast the reaction will be
(plus the momentum of the rest of the drivetrain).
Drag is a force, not a mass. It couldn't be, since mass stays equal
all the time. You can *express* it as a virtual mass, for d'Alembert
type physics, but when you're separating these things, drag is a
force.
Too bad btw that the US uses lbs, which is dependent on gravity, which
is a force, so you might swap weight for force thinking it's the same
thing. Mass has no direct relation to force on the other hand (in the
SI system).
>Higher mass = Lower engine braking
>Lower mass = Higher engine braking.
We're probably talking about the same things here, though in a
different context or perspective, I think.
If we take:
engineAcceleration=engineTorque/engineInertia
you can indeed see that higher (flywheel+engine parts) mass, or rather
inertia means engineInertia goes up, so engineAcceleration goes down.
Which is lower engine braking.
No argument there.
>Torque has -nothing- to do with it. Why would you think that
>'torque' would have anything to do with engine braking when
>the closed-throttle engine is producing almost NO torque?
The way I use 'engine braking' in my sim is the torque the engine
generates (at the gearbox input) when no throttle is applied at a
certain rpm. If you're running 10,000 rpm and close the throttle,
surely the engine generates a negative torque to slow the drivetrain
down; I mean, pressing the clutch will let the car coast on while the
engine slows down faster.
So, in that perspective, at closed-throttle at 10,000 rpm, the engine
is generating a lot of negative torque which slows down the
drivetrain. This is what I call 'engine braking'; the amount of torque
that the engine generates at closed-throttle under a certain rpm.
Seems like this can't be far from the truth (it's working properly in
my sim).
The flywheel just constitutes more inertia, and I add the flywheel
inertia to the engine inertia to get a total inertia (at one end of
the clutch).
>Engine 'braking' doesn't occur when the engine is not idling.
You'd say it does, otherwise the engine would accelerate?! If fuel is
burning, and no friction/braking occurs, the engine would accelerate
(if you applied the clutch, otherwise it would be dependent on the
wheels etc).
Idling is a balance between engine friction+braking and applying fuel.
It must be that we're using a different term of 'engine braking'. If
nothing would counter the engine from rotating, you would have an
accelerating engine. The fuel gives the power to rotate the engine
uptill the engine fights back just hard enough so that rpm neither
goes down, or up.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (11 lines)
>Jonny, are you really this thick? > >Try this: > >Stand in front of an F1 car and then stand in front >of a Toyota Celica. Look closely at each of them. > >Tell me Jonny, which is optimised for lower drag? > >Let us know what you find out. >>Okay, I'll try that again without cubing the power. (See, Todd,
>I told you you should never trust my equations - especially this
>late at night!)
>
Hehe.. I'll be watching you more closely from now on :0)
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm
On Fri, 08 Feb 2002 14:20:01 -0600, rrevved <[email hidden]> wrote:
>On Fri, 08 Feb 2002 14:25:57 GMT, [email hidden] (Ruud van Gaal) wrote:...
>>Still, I'd think an F1 car will use a much lower inertia (less
>>weighing flywheel) to be more responsive. Although ofcourse a higher
>>inertia gives you a more stable accelerating/braking car.
>
>Correct. And that's why engine braking is MUCH higher on any racing
>car than a street car. Have you ever driven a purpose built racecar?
Nope, sorry to say that I haven't.
>In case you haven't, when you close the throttle into a corner,
>you get much more natural (engine) braking than a street car.
>A LOT more. It changes the handling of the car dramatically.
Ok, so then it seems both the engine inertia is lower, AND the engine
is fighting (giving negative torque) more than in a streetcar.
>The reason is the lightened driveline components as well as the
>generally lower gearing of the gearbox.
The lower gearing is a secondary effect though and has nothing to do
with the engine wanting to brake itself. The thing is I'm not looking
for a net effect, since if some mysterious power was rotating the
wheels, the engine wouldn't brake anyway, and you'd think engine
braking torque was 0, which it isn't; it's just counterbalanced.
So forget everything behind the engine and flywheel. Uptill there it's
an autonomous system and should be viewed as such to look at 'engine
braking'. Keeping the car in neutral or with the clutch applied, for
example.
> But forget the gearbox
>for a second.
Oh, yes, right. :)
> That racecar will slow very quickly, without any
>brakes whatsoever versus a street car. I'm talking about a
>purpose-built race car here. F1/F3/Sprint Cars/Nascar/ CART, etc..
>They all use lightened flywheels. They all will slow down much
>quicker than a street car.
Ok, then probably, with your experience, given a 0.74 coefficient for
an F1 car, a street car might end up more like half of that (instead
of being more).
>As I said, just listen to the cars in the paddock. That should be
>enough proof that the engines slow down very quickly when the
>throttle is suddenly closed after being blipped. There are no
>aerodynamics or torque or wings or anything else when they
>do that. Just engine and flywheel.
That's true.
>>Did you drive it at LeMans back then? Would be way cool. :)
>
>Graham Hill drove it at LeMans in 1956, I got it in about 1970
>and raced SCCA with it. Chassis number 247.
Must have been awesome. :)
>I sold it in about 1975 because I'm stupid ..:)
Ouch! ;-)
Not at all; inertia is a fixed property of an object, just like mass.
Nothing 'causes' inertia; it's there because the thing has
molecules/mass.
Like F=m*a, T=I*w where T=torque, I=inertia, w=angular acceleration.
Mass is a constant property of an object, just like inertia (rather
the inertia matrix).
> If the object is lightweight, there
>isn't as much inertia as with a heavy object.
Right, but the inertia of the object remains fixed; it's the torque
that changes (the engine pushing or pulling rotationally), and the
angular acceleration is a result of that (w=T/I).
Surely the engine is pushing the drivetrain in the reverse direction
of which it (the engine/drivetrain) is rotating in. Otherwise it
wouldn't decelerate. So the engine is torqueing the drivetrain, no
doubt in my mind.
> Drag and inertia decide what happens, not power.
Power is just a measurement of energy and is just a nice statistic,
drag and inertia are inputs, yes.
Didn't know about the braking improvements, but that surely does make
sense. It seems though the downforce and brakes are most important,
since the downforce generates implicitly the maximum longitudinal
braking force you can apply to the car without locking the wheels. I'd
say the brakes of an F1 car are powerful enough at any speed to lock
the wheels, so in that sense, engine braking is of no importance for
the total amount of braking you can apply to the tarmac; the limit is
easily reached by the brakes alone, so engine braking just gets you
there faster.
But I may be completely off there, and F1 brakes may be impossible to
lock at high speeds (but I don't think so).
>With your theory, a racecar would be better off with
>heavy driveline components, which of course is not the
>way it is done.
Hm, it's getting late. Have to rethink that. Indeed, less
flywheel/engine inertia gives better deceleration. In my perspective,
it's probably that the brakes can overcome the flywheel inertia as
part of their job, so at a certain point you don't need a lighter
flywheel.
But I guess then maybe the wheel brakes aren't capable of locking the
drivetrain at 300 km/h? Otherwise they could brake both the wheels and
the engine, so the flywheel wanting to spin on wouldn't make that much
of a difference.
>> engineAcceleration=engineTorque/engineInertia
>
>Ruud, guess what causes that inertia? One -major- thing is the
>flywheel.
Right, that and the engine parts themselves.
It's in there; my engine inertia includes both flywheel and engine
parts themselves (they are summed). The rest of the drivetrain
ofcourse also has inertia and can be tweaked in every which way.
The inertia just doesn't come into play until there's a torque from
the engine and you want to know how fast all the parts in the
drivetrain are going to accelerate (which is done in 2 parts in my
sim; preclutch and postclutch; they can rotate at different speeds,
which I need because I have an analog clutch simulated).
Interesting chat, haven't written this much stuff in a post for some
time. :)
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (9 lines)
>>I think that's more to do with the low INERTIA of the components, >>rather than the high braking coefficient (which just gives you the >>torque with which the engine is trying to slow down, not the actual >>speed at which this finally happens, since that is related to inertia >>as well). > >Correct. Guess what causes inertia? You need an object >which has weight and a force which puts it in motion. The result >is a certain amount of inertia.Quoted text (8 lines)
>>F1 engines must be powerful to push the cars through the aero drag, >>even with high wing settings. So a massive engine braking contradicts >>that goal, since at high RPM the engine braking would really fight the >>engine from generating any torque to push the car forward through the >>air. > >When the throttle is closed on any car, it isn't 'pushing' >anything.Quoted text (7 lines)
>>So I'd think striving for less engine braking would be beneficial >>for an F1 car; at high speeds you depend on the aero downforce and >>brakes to get good braking, instead of using the engine as a brake. > >Nope. The purpose of low inertia driveline components >is for weight reduction, acceleration enhancement and braking >improvements. I'm surprised you don't know this.Quoted text (10 lines)
>>Nope, they're completely separate. I define the engine braking >>coefficient as being a measure of compression braking, and any >>friction in the engine. I don't try to include any other implicit >>effects. > >You should. > >Lightened flywheel/driveline is a MAJOR performance >consideration in a racecar. They actually drive totally >differently.>hm. I always thought engine output and engine braking were completely
>separate. engine braking is, like you say, just mass, while power
>output decreases with increasing mass.
>
>
"Engine braking" in the context Ruud, Jonny, and myself are speaking of is a
torque, nothing else.
The "rate at which the engine slows down" from this engine braking torque is
dependent on the engine braking torque and the mass combined (more
specifically, the polar moment of inertia of the engine/flywheel, which is
basically how massive it is with weight and size combined.)
As Ruud said:
Engine_Deceleration = Torque / Rotation_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
Quoted text (7 lines)
>>The mass is irrelevant to the steady-state torque >>produced by the engine, be that driving or braking torque. What >>increased mass *does* do is make the car effectively heavier, so >>it requires a greater braking torque to achieve the same deceleration. >> > >LOL!!Quoted text (7 lines)
>> I've noticed with my model race cars, which accelerate >>better than I'd expect when I fit silly tall gearing for a big >>outdoor track. >> > >LOL!! #2 >
I go away for a day and miss a physics slug fest!!
Box Score -- Modelers 1 (Ruud, Jonny), ex-racers 0 (rrevved), as I see it.
But no one yet has mentioned the magic words -- motoring torque. It's hard
data to find, I don't have any that I can share for F1 or other racing
engines, sorry.
Steady-state motoring torque must be the sum of the torque due to friction
and pumping losses (actually "vacuum pumping", since the pistons are trying
to suck air through a closed throttle.) And a few other losses that could
be broken out like running the accessories (water/oil pumps, alternator,
etc.)
Note that race engines have a lot of valve overlap (assuming no variable
valve timing is used?), so at low rpm they are crappy pumps.
Motoring torque is measured on a powered dyno, where the engine under
test can be spun up to any rpm. Them measure the torque required. This
is most often done (afaik) when developing an automatic transmission
for a specific vehicle.
-- Doug Milliken
On Sat, 9 Feb 2002, Ruud van Gaal wrote:
Quoted text (182 lines)
> On Fri, 08 Feb 2002 14:20:01 -0600, rrevved <[email hidden]> wrote: > > >On Fri, 08 Feb 2002 14:25:57 GMT, [email hidden] (Ruud van Gaal) wrote: > ... > >>Still, I'd think an F1 car will use a much lower inertia (less > >>weighing flywheel) to be more responsive. Although ofcourse a higher > >>inertia gives you a more stable accelerating/braking car. > > > >Correct. And that's why engine braking is MUCH higher on any racing > >car than a street car. Have you ever driven a purpose built racecar? > > Nope, sorry to say that I haven't. > > >In case you haven't, when you close the throttle into a corner, > >you get much more natural (engine) braking than a street car. > >A LOT more. It changes the handling of the car dramatically. > > Ok, so then it seems both the engine inertia is lower, AND the engine > is fighting (giving negative torque) more than in a streetcar. > > >The reason is the lightened driveline components as well as the > >generally lower gearing of the gearbox. > > The lower gearing is a secondary effect though and has nothing to do > with the engine wanting to brake itself. The thing is I'm not looking > for a net effect, since if some mysterious power was rotating the > wheels, the engine wouldn't brake anyway, and you'd think engine > braking torque was 0, which it isn't; it's just counterbalanced. > > So forget everything behind the engine and flywheel. Uptill there it's > an autonomous system and should be viewed as such to look at 'engine > braking'. Keeping the car in neutral or with the clutch applied, for > example. > > > But forget the gearbox > >for a second. > > Oh, yes, right. :) > > > That racecar will slow very quickly, without any > >brakes whatsoever versus a street car. I'm talking about a > >purpose-built race car here. F1/F3/Sprint Cars/Nascar/ CART, etc.. > >They all use lightened flywheels. They all will slow down much > >quicker than a street car. > > Ok, then probably, with your experience, given a 0.74 coefficient for > an F1 car, a street car might end up more like half of that (instead > of being more). > > >As I said, just listen to the cars in the paddock. That should be > >enough proof that the engines slow down very quickly when the > >throttle is suddenly closed after being blipped. There are no > >aerodynamics or torque or wings or anything else when they > >do that. Just engine and flywheel. > > That's true. > > >>Did you drive it at LeMans back then? Would be way cool. :) > > > >Graham Hill drove it at LeMans in 1956, I got it in about 1970 > >and raced SCCA with it. Chassis number 247. > > Must have been awesome. :) > > >I sold it in about 1975 because I'm stupid ..:) > > Ouch! ;-) > > >>I think that's more to do with the low INERTIA of the components, > >>rather than the high braking coefficient (which just gives you the > >>torque with which the engine is trying to slow down, not the actual > >>speed at which this finally happens, since that is related to inertia > >>as well). > > > >Correct. Guess what causes inertia? You need an object > >which has weight and a force which puts it in motion. The result > >is a certain amount of inertia. > > Not at all; inertia is a fixed property of an object, just like mass. > Nothing 'causes' inertia; it's there because the thing has > molecules/mass. > Like F=m*a, T=I*w where T=torque, I=inertia, w=angular acceleration. > Mass is a constant property of an object, just like inertia (rather > the inertia matrix). > > > If the object is lightweight, there > >isn't as much inertia as with a heavy object. > > Right, but the inertia of the object remains fixed; it's the torque > that changes (the engine pushing or pulling rotationally), and the > angular acceleration is a result of that (w=T/I). > > >>F1 engines must be powerful to push the cars through the aero drag, > >>even with high wing settings. So a massive engine braking contradicts > >>that goal, since at high RPM the engine braking would really fight the > >>engine from generating any torque to push the car forward through the > >>air. > > > >When the throttle is closed on any car, it isn't 'pushing' > >anything. > > Surely the engine is pushing the drivetrain in the reverse direction > of which it (the engine/drivetrain) is rotating in. Otherwise it > wouldn't decelerate. So the engine is torqueing the drivetrain, no > doubt in my mind. > > > Drag and inertia decide what happens, not power. > > Power is just a measurement of energy and is just a nice statistic, > drag and inertia are inputs, yes. > > >>So I'd think striving for less engine braking would be beneficial > >>for an F1 car; at high speeds you depend on the aero downforce and > >>brakes to get good braking, instead of using the engine as a brake. > > > >Nope. The purpose of low inertia driveline components > >is for weight reduction, acceleration enhancement and braking > >improvements. I'm surprised you don't know this. > > Didn't know about the braking improvements, but that surely does make > sense. It seems though the downforce and brakes are most important, > since the downforce generates implicitly the maximum longitudinal > braking force you can apply to the car without locking the wheels. I'd > say the brakes of an F1 car are powerful enough at any speed to lock > the wheels, so in that sense, engine braking is of no importance for > the total amount of braking you can apply to the tarmac; the limit is > easily reached by the brakes alone, so engine braking just gets you > there faster. > But I may be completely off there, and F1 brakes may be impossible to > lock at high speeds (but I don't think so). > > >With your theory, a racecar would be better off with > >heavy driveline components, which of course is not the > >way it is done. > > Hm, it's getting late. Have to rethink that. Indeed, less > flywheel/engine inertia gives better deceleration. In my perspective, > it's probably that the brakes can overcome the flywheel inertia as > part of their job, so at a certain point you don't need a lighter > flywheel. > But I guess then maybe the wheel brakes aren't capable of locking the > drivetrain at 300 km/h? Otherwise they could brake both the wheels and > the engine, so the flywheel wanting to spin on wouldn't make that much > of a difference. > > >> engineAcceleration=engineTorque/engineInertia > > > >Ruud, guess what causes that inertia? One -major- thing is the > >flywheel. > > Right, that and the engine parts themselves. > > >>Nope, they're completely separate. I define the engine braking > >>coefficient as being a measure of compression braking, and any > >>friction in the engine. I don't try to include any other implicit > >>effects. > > > >You should. > > > >Lightened flywheel/driveline is a MAJOR performance > >consideration in a racecar. They actually drive totally > >differently. > > It's in there; my engine inertia includes both flywheel and engine > parts themselves (they are summed). The rest of the drivetrain > ofcourse also has inertia and can be tweaked in every which way. > > The inertia just doesn't come into play until there's a torque from > the engine and you want to know how fast all the parts in the > drivetrain are going to accelerate (which is done in 2 parts in my > sim; preclutch and postclutch; they can rotate at different speeds, > which I need because I have an analog clutch simulated). > > Interesting chat, haven't written this much stuff in a post for some > time. :) > > > Ruud van Gaal > Free car sim: http://www.racer.nl/ > Pencil art : http://www.marketgraph.nl/gallery/ > > StopIT! You guys are making my head hurt just BYPASSING these threads....
dave henrie
"Doug Milliken" <[email hidden]> wrote in message
news:[email hidden]...
>
> I go away for a day and miss a physics slug fest!!
> Box Score -- Modelers 1 (Ruud, Jonny), ex-racers 0 (rrevved), as I see it.
> w=angular acceleration.
[email hidden] (Ruud van Gaal) writes:
of course engine braking is torque, but I always thought that almost
_all_ of this was from inertia. take two-strokes as an example, they
have insane compression ratios and virtually zero engine braking.
> >you're probably right [3], but [4] I'm pretty sure F1 cars have
> >back-torque limiters, which would mean you can pretty much just make
> >something up, or even make it user configurable, if you're a really
> >nice guy.
>
> Hm, interesting, never heard of them. I'll have to google that.
it's basically one of those simplex-clutches I can't remember what are
called. very common on bikes.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
gitaren er en sjingke
Quoted text (17 lines)
> On 08 Feb 2002 18:59:34 +0100, Gunnar Horrigmo <[email hidden]> > wrote: > >[email hidden] (Ruud van Gaal) writes: > > > >> There are 2 parameters that decide the speed of revving up and down > >> (in neutral); the engine torque (which includes engine braking and > >> positive torque when the throttle pedal is pressed), and the engine > >> inertia. > > > >hm. I always thought engine output and engine braking were completely > >separate. engine braking is, like you say, just mass, while power > >output decreases with increasing mass. > > Uhm, engine braking is not mass, but torque, as I see it. The net > torque an engine can produce is the sum of the part that make it want > to accelerate (fuel burning), and the part that wants to brake it > (pistons/air being compressed).[email hidden] (J. Todd Wasson) writes:
> I'm afraid he's exactly right, rrevved. I think you may not be aware that
> Jonny is speaking of two different things seperately here, where it sounds like
> you might be combining them in one concept, although mathematically, they are
> seperate things. The steady state torque the engine produces has nothing to do
> with the mass of the engine.
so a (lot) heavier flywheel will _not_ increase engine braking?
anyone know of any (simple) texts on this subject? engine mechanics
for dummies or somesuch? :)
hm. ok. that helped somewhat.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
"a poster is a human being or the software equivalent"
Quoted text (9 lines)
> Increasing the mass of the engine causes the engine to accelerate/decelerate > more slowly: > > Rotational_Acceleration = Torque / PM_Inertia > > The *torque* in this equation, is not dependent on the engine's mass.. But > rather, the acceleration is dependent on both the torque and the mass. I > suspect you're referring to "engine braking" as Rotational_Acceleration, > whereas Jonny, Ruud, and Collin are referring to engine braking as Torque.[email hidden] (Ruud van Gaal) writes:
> you can indeed see that higher (flywheel+engine parts) mass, or rather
> inertia means engineInertia goes up, so engineAcceleration goes down.
> Which is lower engine braking.
I'm having serious problems with this. a heavier flywheel makes for
lower engine braking? that doesn't make sense to me...
> >Engine 'braking' doesn't occur when the engine is not idling.
>
> You'd say it does, otherwise the engine would accelerate?! If fuel is
> burning, and no friction/braking occurs, the engine would accelerate
fuel is burning even when the throttle is closed.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
gnus don't kill people. gnus-summary-lower-score kills people.
rrevved <[email hidden]> writes:
maybe, but in that case whoever wrote it, should do a write up, as he
offered aerodynamic drag as the explanation. :)
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
"a poster is a human being or the software equivalent"
Quoted text (25 lines)
> On 08 Feb 2002 20:03:47 +0100, Gunnar Horrigmo <[email hidden]> wrote: > >na_biker <[email hidden]> writes: > >> On Fri, 08 Feb 2002 06:46:02 -0600, rrevved <[email hidden]> wrote: > >> > >> >ie: I would think that engine braking on a modern F1 car > >> >is MASSIVE. Very light driveline components, high > >> >compression, ratios, higher-revs/lower gearing etc. > >> >would make them slow down REALLY fast when the throttle > >> >is closed. You can hear this by listening to the revving > >> >of their engines while in the paddock. When the throttle > >> >is opened and then closed, they slow down REAL quick. > >> >Much quicker than ... a Toyota Celica. > >> > >> Yeah. When Tiff Needell tested the Jaguar R1 he said(paraphrased) that > >> letting off the throttle was like slamming the brakes on any normal > >> car, slamming the brakes on the F1 was like hitting a brick wall. :) > > > > Yep to the above. Hell it was even strong in my old Lotus. > > >well, yeah. I read somewhere that an F1 car at 200 kph will > >decellerate at about 1 G if you just press the clutch. > > I bet if you read that again, it would say that an F1 car > will decelerate at +1G if you- close the throttle- at 200K[email hidden] (J. Todd Wasson) writes:
oh aha. I get it. in my head "engine braking" has always been "the
rate at which the vehicle slows down" minus brakes and aero drag. oh well.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
silence is FOO!
Quoted text (9 lines)
> >hm. I always thought engine output and engine braking were completely > >separate. engine braking is, like you say, just mass, while power > >output decreases with increasing mass. > > "Engine braking" in the context Ruud, Jonny, and myself are speaking of is a > torque, nothing else. > > The "rate at which the engine slows down" from this engine braking > torqueOn 09 Feb 2002 14:14:50 +0100, Gunnar Horrigmo <[email hidden]>
wrote:
Think of it this way:
The [heavier]flywheel means more inertia, more inertia means that the
engine is less "willing" to rev down, i.e. it "wants" to keep it's
speed.
If you had a 500kg flywheel like in some old steamengines all braking
power would go to braking the _flywheel_....
Quoted text (8 lines)
>[email hidden] (Ruud van Gaal) writes: > >> you can indeed see that higher (flywheel+engine parts) mass, or rather >> inertia means engineInertia goes up, so engineAcceleration goes down. >> Which is lower engine braking. > >I'm having serious problems with this. a heavier flywheel makes for >lower engine braking? that doesn't make sense to me...;-) My celica has 393hp @ 5500rpm, a 'Colin Tune Special' but even so,
it should not have been producing -700Nm of -ve engine torque at
7000rpm ;-)
Colin
na_biker <[email hidden]> writes:
but that _is_ the way I'm thinking. and since it "wants" to keep it's
speed it has _more_ engine braking.
> If you had a 500kg flywheel like in some old steamengines all braking
> power would go to braking the _flywheel_....
where I come from, the flywheel is an engine part. :)
anyway; I've figured it out. you guys are just talking about the
engine, while I was thinking of the vehicle as a whole.
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
"Det er nok ingen ovn, men fartsm�leren v�r."
Quoted text (17 lines)
> > On 09 Feb 2002 14:14:50 +0100, Gunnar Horrigmo <[email hidden]> > wrote: > >[email hidden] (Ruud van Gaal) writes: > > > >> you can indeed see that higher (flywheel+engine parts) mass, or rather > >> inertia means engineInertia goes up, so engineAcceleration goes down. > >> Which is lower engine braking. > > > >I'm having serious problems with this. a heavier flywheel makes for > >lower engine braking? that doesn't make sense to me... > > Think of it this way: > > The [heavier]flywheel means more inertia, more inertia means that the > engine is less "willing" to rev down, i.e. it "wants" to keep it's > speed.Gunnar Horrigmo wrote:
Just the opposite! In fact, as far as engine "braking" is concerned,
the engine sees no difference between the mass of itself (it's internal
parts, it's external parts eg - flywheel) or the mass of the car,
wheels, etc. It's all just one big mass!
F = MA
Force = Braking, Mass = Mass of (car + wheel inertia + engine +
inertia of engine bits)
Question is, what causes acceleration/decelleration ie change in
speed/rpm?
Answer = friction + (compression/heat loss and pressure drops)
friction is obvious and insignificant?
Looking at 1 cylinder and it's 4 strokes -
Intake = low pressure (Closed throttle - Vacuum) ie small force
on top of cylinder (very small +)
Exhaust = Virtually 1 atmosphere greater than intake (-)
Compression = Obvious + we have heat loss we do not get back on
combustion stroke ie loss of energy from a non elastic event. (-)
Combustion - finally the engine momentum gets some substantial help
:)) (+)
Now lets get right down into the nitty gritty! LOL
As rpm increases Intake = virtual Vacuum ie zero effect
Exhaust = greater piston speed -> greater pressure
Compression - same as exhaust
Combustion - Hmmm, let's see. Can the gasses, ignited ATDC?, ever
catch up with a piston accelerating away at maybe 10,000 + RPM? ie
diminishing effect in forward direction.
Conclusion - as we raise rpm at idle throttle settings we increase
"engine braking"! Commonly referred to on the streets as "compression
lockup"!
Glad I got that off my chest. Now I'm back off to the nut house for
another couple of years :))
Quoted text (13 lines)
> > [email hidden] (J. Todd Wasson) writes: > > > I'm afraid he's exactly right, rrevved. I think you may not be aware that > > Jonny is speaking of two different things seperately here, where it sounds like > > you might be combining them in one concept, although mathematically, they are > > seperate things. The steady state torque the engine produces has nothing to do > > with the mass of the engine. > > so a (lot) heavier flywheel will _not_ increase engine braking? > anyone know of any (simple) texts on this subject? engine mechanics > for dummies or somesuch? :) > Well, you are/were actually right and wrong both times :)
Drag is proportional to speed under slipstreamed conditions.
Drag is proportional to the square of speed in turbulent conditions.
The speed at which slipstream airflow degenerates into turbulent
airflow will vary for all the different bits of said car!
Jonny Hodgson wrote:
Quoted text (30 lines)
> > Jonny Hodgson <[email hidden]> wrote in message news:oZY88.11029$[email hidden]... > > > Second hint: a Celica has perhaps 100 hp, and a top speed of > > perhaps 120 mph. The F1 has 800 hp and a top end of 210 mph. > > > > Aero power requirements rise with the cube of speed. > > > > (120^3 / 100^3)^0.33 = 1.2 (~= mph / hp) > > > > (210^3 / 800^3)^0.33 = 0.26. IOW, the Celica is getting far > > more maximum speed per horsepower than the F1 car - which suggests > > it has just a tiny bit less drag. > > Okay, I'll try that again without cubing the power. (See, Todd, > I told you you should never trust my equations - especially this > late at night!) > > P = k1 * drag * speed^3 > drag = k2 * P / speed^3 > > For the Celica, that gives drag of 100 / 120^3 = 0.000 057 8 [units] > For the F1, it's 800 / 210^3 = 0.000 086 4 [units]. > > Not a huge difference - the F1 is smaller but has a higher Cd > - but still the same way round. > > Jonny > proving that you *can* get a decent degree without being able > to get the right answer first time...
Byron Forbes wrote:
>
> Glad I got that off my chest. Now I'm back off to the nut house for
> another couple of years :))
Not a single personal insult to be found anywhere, I'm slightly
disappointed, impressed in one way, but also disappointed, you haven't
gotten old on us have you Byron ?
Beers and cheers
(uncle) Goy
"The Pits" http://www.theuspits.com/
* Spam is for losers who can't get business any other way *
"Spamkiller" http://www.spamkiller.com
Getting warmed up! :)
Goy Larsen wrote:
Quoted text (17 lines)
> > Byron Forbes wrote: > > > > Glad I got that off my chest. Now I'm back off to the nut house for > > another couple of years :)) > > Not a single personal insult to be found anywhere, I'm slightly > disappointed, impressed in one way, but also disappointed, you haven't > gotten old on us have you Byron ? > > Beers and cheers > (uncle) Goy > > "The Pits" http://www.theuspits.com/ > > * Spam is for losers who can't get business any other way * > "Spamkiller" http://www.spamkiller.comByron Forbes wrote:
>
> Getting warmed up! :)
Ah, just getting the feel of the place again are you ?
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--
>;-) My celica has 393hp @ 5500rpm, a 'Colin Tune Special' but even so,
>it should not have been producing -700Nm of -ve engine torque at
>7000rpm ;-)
>
Hehe... Probably not!
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm
Quoted text (11 lines)
>>I'm having serious problems with this. a heavier flywheel makes for >>lower engine braking? that doesn't make sense to me... > >Think of it this way: > >The [heavier]flywheel means more inertia, more inertia means that the >engine is less "willing" to rev down, i.e. it "wants" to keep it's >speed. > >If you had a 500kg flywheel like in some old steamengines all braking >power would go to braking the _flywheel_....Quoted text (7 lines)
>> The "rate at which the engine slows down" from this engine braking >> torque > >oh aha. I get it. in my head "engine braking" has always been "the >rate at which the vehicle slows down" minus brakes and aero drag. oh well. > >Quoted text (13 lines)
>> Jonny is speaking of two different things seperately here, where it sounds >like >> you might be combining them in one concept, although mathematically, they >are >> seperate things. The steady state torque the engine produces has nothing >to do >> with the mass of the engine. > >so a (lot) heavier flywheel will _not_ increase engine braking? >anyone know of any (simple) texts on this subject? engine mechanics >for dummies or somesuch? :) >On 09 Feb 2002 22:39:47 GMT, [email hidden] (J. Todd Wasson) wrote:
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.
Quoted text (20 lines)
>>>I'm having serious problems with this. a heavier flywheel makes for >>>lower engine braking? that doesn't make sense to me... >> >>Think of it this way: >> >>The [heavier]flywheel means more inertia, more inertia means that the >>engine is less "willing" to rev down, i.e. it "wants" to keep it's >>speed. >> >>If you had a 500kg flywheel like in some old steamengines all braking >>power would go to braking the _flywheel_.... > > > 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...Byron Forbes <[email hidden]> writes:
Hmmm, but since as long as there is any skin friction
there will be turbulence on (parts of) any profile, and
since the pressure drag totally dominates the skin friction
drag the resulting total drag will be proportional to the
square of the speed?
_
Mats Lofkvist
[email hidden]
Quoted text (8 lines)
> Well, you are/were actually right and wrong both times :) > > Drag is proportional to speed under slipstreamed conditions. > > Drag is proportional to the square of speed in turbulent conditions. > > The speed at which slipstream airflow degenerates into turbulent > airflow will vary for all the different bits of said car!rrevved <[email hidden]> writes:
uh. I don't know how I was thinking there. anyway; thanks for the
help everyone. :)
--
Gunnar
#31 SUCKS#015 Tupperware MC#002 DoD#0x1B DoDRT#003 DoD:CT#4,8 Kibo�: 2
gitaren er en sjingke
Quoted text (26 lines)
> On 09 Feb 2002 17:22:14 +0100, Gunnar Horrigmo <[email hidden]> wrote: > >na_biker <[email hidden]> writes: > >> On 09 Feb 2002 14:14:50 +0100, Gunnar Horrigmo <[email hidden]> > >> wrote: > >> >[email hidden] (Ruud van Gaal) writes: > >> > > >> >> you can indeed see that higher (flywheel+engine parts) mass, or rather > >> >> inertia means engineInertia goes up, so engineAcceleration goes down. > >> >> Which is lower engine braking. > >> > > >> >I'm having serious problems with this. a heavier flywheel makes for > >> >lower engine braking? that doesn't make sense to me... > >> > >> Think of it this way: > >> > >> The [heavier]flywheel means more inertia, more inertia means that the > >> engine is less "willing" to rev down, i.e. it "wants" to keep it's > >> speed. > > > >but that _is_ the way I'm thinking. and since it "wants" to keep it's > >speed it has _more_ engine braking. > > > > Keeping speed is not 'braking'. > > Slowing speed is 'braking.
Byron Forbes <[email hidden]> wrote in message news:[email hidden]...
> Well, you are/were actually right and wrong both times :)
>
> Drag is proportional to speed under slipstreamed conditions.
>
> Drag is proportional to the square of speed in turbulent conditions.
Mmm-hmm <nod>, but that's force.
(Drag) force (proportional to) speed^2
Power = force x speed
Power (proportional to) speed^3
Jonny
rrevved <[email hidden]> wrote in message news:[email hidden]...
> He's wrong. He has -never- driven a true purpose-built race car.
> I know he hasn't, because he would never be saying the things
> he does.
Okay, you're right there - I've never driven a 1:1 racecar.
Some day... :-)
> Not only does he not believe that lightened driveline, including
> flywheel, will dramatically increase acceleration and 'braking'
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.
> (LOL!!) , he probably doesn't even realize that flywheel weight
> can even be a -critical- handling consideration. ie: If you use
> a lightened (chopped/aluminum/carbon fibre) flywheel, your
> car will tend to oversteer from the increased 'braking' at
> the rear of the car. Assuming RWD, here.
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?
> I've raced (a lot). I've raced in (many) cars which have had
> light flywheels. I've driven them before the mods and after.
> I know what I'm talking about, but I'm finished trying to
> convince him. He is free to believe what he wants.
<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
Quoted text (9 lines)
>In a lightweight racecar, such as an F1, or a midget, bike, or probably >a kart, the light flywheel/engine braking effect will be much greater than in >say, a Cadillac, or a Toyota. In my judgement, this is due to the overall >mass >of the vehicle 'drowning-out' any advantage in braking power that a light >flywheel provides. Also, dirt track people are -very- sensitive to flywheel >weight in their sprints. It can alter the handling big-time with the >additional >braking at the rear.Quoted text (19 lines)
>>> I'm afraid he's exactly right, rrevved. I think you may not be aware >that >>> Jonny is speaking of two different things seperately here, where it sounds >like >>> you might be combining them in one concept, although mathematically, they >are >>> seperate things. The steady state torque the engine produces has nothing >to do >>> with the mass of the engine. >> >>so a (lot) heavier flywheel will _not_ increase engine braking? >>anyone know of any (simple) texts on this subject? engine mechanics >>for dummies or somesuch? :) >> >> > >He's wrong. He has -never- driven a true purpose-built race car. >I know he hasn't, because he would never be saying the things >he does.Quoted text (9 lines)
>a lighter flywheel means that when the throttle is twisted more power >is immediately available to accelerate the motorcycle. Releasing the >throttle, of course, reveals the other side: Having less stored >energy, the engine slows down more quickly. In other words, there is >an increase in engine braking. > >http://www.tripointengineering.com/Catalog_pages/mechanical.htm > >Quoted text (9 lines)
>> Sounds like a real handful. I'm jealous :-) Why in the world did you >sell >>it ? > >I had to sell it or lose my wife. I broke a half-shaft at IRP and >had a major spin-a-thon. She was there and got a little .. scared. > >:) >On 09 Feb 2002 00:12:14 GMT, [email hidden] (J. Todd Wasson) wrote:
To backup this, someone suggested me to link drag to downforce
coefficient by a factor, instead of specifying the coefficient for
drag and downforce totally independent.
And that was based on actual F1 aero results. So that would back up
Jonny's statement that with extra downforce comes extra drag. That's
also why BHP is so important in F1; it's there to push the car through
all that extra generated drag.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (17 lines)
>>Jonny, are you really this thick? >> >>Try this: >> >>Stand in front of an F1 car and then stand in front >>of a Toyota Celica. Look closely at each of them. >> >>Tell me Jonny, which is optimised for lower drag? >> >>Let us know what you find out. >> > > In defense of Jonny, the coefficient of drag of modern F-1 cars is usually at >or above 1.. Several times higher than a Toyota Celica. The frontal area is >lower, however, but if an F-1 car was optimized for low drag rather than >downforce, it would certainly not have a drag coefficient at least twice as >high as any road going car, would it?On Fri, 08 Feb 2002 16:55:17 -0600, rrevved <[email hidden]> wrote:
>On Fri, 8 Feb 2002 22:02:55 -0000, "Jonny Hodgson" <[email hidden]> wrote:
>>
>>rrevved <[email hidden]> wrote in message news:[email hidden]......
Really, revved, this is uncalled for; learn your Newton physics and
you'll see that Jonny is right here.
Mass=inertia=resistance to acceleration. Take F=m*a and you'll see
that to get the same 'a' with a higher 'm'(ass) you get a greater F
(force/torque) (just poetic freedom while mixing force & torque here,
which are basically the same).
...
>>> Engine 'braking' doesn't occur when the engine is not idling.
>>
>>No, because then the overall torque is positive.
>
>LOL!! #3
Well, the overall net torque is approx. 0, but still, engine braking
occurs but is countered by the fuel burning. Like I said, a balance.
And why is it that I go away for the weekend, have some social stuff,
watch some Olympics, and EVERYBODY else seemed to be chatting here!
;-)
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (12 lines)
>>> Higher mass = Lower engine braking >>> Lower mass = Higher engine braking. >> >>Actually, no. > >Really? > >> What >>increased mass *does* do is make the car effectively heavier, so >>it requires a greater braking torque to achieve the same deceleration. > >LOL!!On 09 Feb 2002 00:36:31 GMT, [email hidden] (J. Todd Wasson) wrote:
...
>Rotational_Acceleration = Torque / PM_Inertia
>
> The *torque* in this equation, is not dependent on the engine's mass.. But
>rather, the acceleration is dependent on both the torque and the mass. I
>suspect you're referring to "engine braking" as Rotational_Acceleration,
>whereas Jonny, Ruud, and Collin are referring to engine braking as Torque.
I think so too; revved talks from his racing perspective, and adds up
a couple of things while the sim modelers split that into the
Newtonian parts. At that moment units become important.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
On 09 Feb 2002 14:07:46 +0100, Gunnar Horrigmo <[email hidden]>
wrote:
I think this depends on the wheel brakes. Without wheels, a heavier
flywheel will cause less engine braking, since the flywheel wants to
continue spinning (has more inertia).
With good enough brakes though, the wheel brakes can brake both the
flywheel and the wheels (with their reactive forces of the road), and
so it would not make a whole lot of difference.
I think F1 brakes would be able to lock even at top speed, so the
flywheel becomes irrelevant there (you just need bigger brakes), but
ofcourse the flywheel has other effects (stabilizes things a bit and
enables idle throttling).
>anyone know of any (simple) texts on this subject? engine mechanics
>for dummies or somesuch? :)
Hehe, I can read RCVD for a 100 times and still learn something new
each time I read it, simply because my understanding of it all grows.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (9 lines)
>[email hidden] (J. Todd Wasson) writes: > >> I'm afraid he's exactly right, rrevved. I think you may not be aware that >> Jonny is speaking of two different things seperately here, where it sounds like >> you might be combining them in one concept, although mathematically, they are >> seperate things. The steady state torque the engine produces has nothing to do >> with the mass of the engine. > >so a (lot) heavier flywheel will _not_ increase engine braking?On 09 Feb 2002 22:39:47 GMT, [email hidden] (J. Todd Wasson) wrote:
...
> 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..
Right; that gets back to the original thing; with a lighter flywheel,
it still depends on the amount of engine braking *torque* what happens
with the net *acceleration* of the car.
Even if an F1 car has a lighter flywheel (which we know it does), we
can still have more or less acceleration of a car when closing the
throttle, depending on how much negative (braking) torque the engine
at that point generates.
The problem is thus; suppose we have an F1 engine and a Toyota engine,
both with exactly the same flywheel and nothing attached to it.
They're both running at say 10,000 rpm (the Toyota engine just not
blasting ;-) ), which engine would slow down quicker? (i.e. gives the
most negative torque).
The original proposition I gave was that the Toyota engine then slows
down faster (given the same flywheel). Ofcourse, that statement can be
shot, but hopefully with experimental results.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
On Sun, 10 Feb 2002 05:14:27 +0100, na_biker <[email hidden]>
wrote:
...
Lol; but when you're simulating an engine in a computer, you must
separate these things to get a good simulation. That's the nice thing
about it; you put in the low level effects, and the driver running the
program says 'hey, it's got great oversteer' while you never actually
put anything like that directly into the program. :)
And for understanding WHY your car oversteers, I'd say it is pretty
good for a race driver to know exactly how a car works so that
solutions can be found quicker and more to the point. Just 'this car
oversteers too much' isn't that helpful.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (7 lines)
>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. ;-)On 09 Feb 2002 14:01:23 +0100, Gunnar Horrigmo <[email hidden]>
wrote:
Inertia is just the resistance to get up (or down) to speed. Having
much inertia means you're going to spend a lot of energy getting to a
speed, but also means it won't fall down again that fast.
>> >back-torque limiters,...
>> Hm, interesting, never heard of them. I'll have to google that.
>
>it's basically one of those simplex-clutches I can't remember what are
>called. very common on bikes.
Yep, I thought about that one; didn't know such things were used in
cars themselves. Could be pretty relevant.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (22 lines)
>[email hidden] (Ruud van Gaal) writes: >> On 08 Feb 2002 18:59:34 +0100, Gunnar Horrigmo <[email hidden]> >> wrote: >> >[email hidden] (Ruud van Gaal) writes: >> > >> >> There are 2 parameters that decide the speed of revving up and down >> >> (in neutral); the engine torque (which includes engine braking and >> >> positive torque when the throttle pedal is pressed), and the engine >> >> inertia. >> > >> >hm. I always thought engine output and engine braking were completely >> >separate. engine braking is, like you say, just mass, while power >> >output decreases with increasing mass. >> >> Uhm, engine braking is not mass, but torque, as I see it. The net >> torque an engine can produce is the sum of the part that make it want >> to accelerate (fuel burning), and the part that wants to brake it >> (pistons/air being compressed). > >of course engine braking is torque, but I always thought that almost >_all_ of this was from inertia. take two-strokes as an example, they >have insane compression ratios and virtually zero engine braking.On Sat, 09 Feb 2002 15:09:29 -0600, rrevved <[email hidden]> wrote:
>On Fri, 08 Feb 2002 14:25:57 GMT, [email hidden] (Ruud van Gaal) wrote:
>
>>Having MORE engine braking than a
>>street car (at the same rpm) would give the racecar a disadvantage
>
>Better engine response and braking is a disadvantage in F1 racing?
Better engine response is an advantage; more (better is subjective and
can mean more or less) engine braking certainly is a disadvantage.
That's because we have a fundamental difference on the presence of
engine braking; to me, it's always there, always trying to brake the
engine (which in engineering terms it is; the fuel must overcome the
engine's intrinsic braking to get positive torque and accelerate,
instead of decelerate).
So, more engine braking means that at a given rpm, the engine is also
braking harder (more friction etc); so you need MORE power to get it
to accelerate (overcome the desire of the engine to decelerate). With
less engine braking, you could get more power on the track, instead of
spending it to fight the engine's will to decelerate.
Again, it boils down to you using 'engine braking' as an acceleration
term (acceleration is also deceleration in mathematical terms), so
you're using the net effect, while I define it as the desire (torque)
of the engine wanting to slow down.
In your terms, more engine braking would indeed be better; in my
terms, more engine braking would be worse (as it subtracts from the
engine's power output, or rather torque output to put it cleanly).
The missing factor, the coefficient (0.74 for the Arrows F1 car), is
something that you don't calculate with (but mix in with your 'engine
braking' experience). The problem is; does an F1 engine, given the
same flywheel, slow down faster (picture the engine hanging there
without a car around it) than a street car's engine?
You can go one way and say the F1 engine has bigger compression ratios
probably, so it slows down faster. On the other hand, the street car
engine might have more friction and slow down faster because of it.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
On 09 Feb 2002 14:26:12 +0100, Gunnar Horrigmo <[email hidden]>
wrote:
Right; but to simulate such things you need to break things apart. I
got pointed to this when adding an analog clutch; you then need to
simulate what happens if the engine is fully detached from the rest of
the car. For that situation, the car doesn't even need to be present,
so you can't simulate just by looking at what the whole car would do.
That just comes out of the equations in the end, implicitly.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (14 lines)
>[email hidden] (J. Todd Wasson) writes: > >> >hm. I always thought engine output and engine braking were completely >> >separate. engine braking is, like you say, just mass, while power >> >output decreases with increasing mass. >> >> "Engine braking" in the context Ruud, Jonny, and myself are speaking of is a >> torque, nothing else. >> >> The "rate at which the engine slows down" from this engine braking >> torque > >oh aha. I get it. in my head "engine braking" has always been "the >rate at which the vehicle slows down" minus brakes and aero drag. oh well.On 11 Feb 2002 05:39:20 GMT, [email hidden] (J. Todd Wasson) wrote:
Right, no car is worth losing a wife you love for. Well, that's my
immaterialistic view of things. :)
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (7 lines)
>>I had to sell it or lose my wife. I broke a half-shaft at IRP and >>had a major spin-a-thon. She was there and got a little .. scared. >> >>:) >> > > Youch.. Good choice though... I guess ;-) Oops. Right you are! Pretty much. :)
Point was (small as it is) that "power" is not proportional to
speed^2 or ^3. It's somewhere in between. Closer to ^2 at low speeds and
^3 at higher speeds, of course :)
Jonny Hodgson wrote:
Quoted text (17 lines)
> > Byron Forbes <[email hidden]> wrote in message news:[email hidden]... > > Well, you are/were actually right and wrong both times :) > > > > Drag is proportional to speed under slipstreamed conditions. > > > > Drag is proportional to the square of speed in turbulent conditions. > > Mmm-hmm <nod>, but that's force. > > (Drag) force (proportional to) speed^2 > > Power = force x speed > > Power (proportional to) speed^3 > > Jonny Not quite sure what your point is. Last time I looked, streamlined
flow was defined as flow without "eddys". Once these "eddys" come into
effect, this is turbulent flow. Without eddys you have a steady gradient
of relative speed of airflow from the "skin" outwards. As speed
increases we have lower and lower pressure areas demonstrating
turbulence known as eddys (swirling little whirlwinds) . As soon as the
first one of these appears around a wing we have "Turbulent Airflow"!
Mats Lofkvist wrote:
Quoted text (21 lines)
> > Byron Forbes <[email hidden]> writes: > > > Well, you are/were actually right and wrong both times :) > > > > Drag is proportional to speed under slipstreamed conditions. > > > > Drag is proportional to the square of speed in turbulent conditions. > > > > The speed at which slipstream airflow degenerates into turbulent > > airflow will vary for all the different bits of said car! > > Hmmm, but since as long as there is any skin friction > there will be turbulence on (parts of) any profile, and > since the pressure drag totally dominates the skin friction > drag the resulting total drag will be proportional to the > square of the speed? > > _ > Mats Lofkvist > [email hidden]On Tue, 12 Feb 2002 00:35:54 +1100, Byron Forbes
<[email hidden]> wrote:
> Not quite sure what your point is. Last time I looked, streamlined
>flow was defined as flow without "eddys".
I think Mats means that in most real situations, you'll have
turbulence at least somewhere, so as this is much more important for
the aero effects, you can just estimate the entire thing using just
the squared speed rule. IOW real slipstream situations hardly ever
occur on cars? Just a hunch. Mats can probably clarify, I shouldn't
try and read his mind. :)
>Mats Lofkvist wrote:...
>> Hmmm, but since as long as there is any skin friction
>> there will be turbulence on (parts of) any profile, and
>> since the pressure drag totally dominates the skin friction
>> drag the resulting total drag will be proportional to the
>> square of the speed?
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (23 lines)
>> 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.. > >Right; that gets back to the original thing; with a lighter flywheel, >it still depends on the amount of engine braking *torque* what happens >with the net *acceleration* of the car. >Even if an F1 car has a lighter flywheel (which we know it does), we >can still have more or less acceleration of a car when closing the >throttle, depending on how much negative (braking) torque the engine >at that point generates. >The problem is thus; suppose we have an F1 engine and a Toyota engine, >both with exactly the same flywheel and nothing attached to it. >They're both running at say 10,000 rpm (the Toyota engine just not >blasting ;-) ), which engine would slow down quicker? (i.e. gives the >most negative torque). > >The original proposition I gave was that the Toyota engine then slows >down faster (given the same flywheel). Ofcourse, that statement can be >shot, but hopefully with experimental results.[email hidden] (Ruud van Gaal) writes:
You succeeded quite well at reading my mind, Ruud :-)
From what I have understood (?) of aerodynamics, the only
way to get drag proportional to less than the square of
the speed is to ignore friction / viscous flow.
(Actually, this assumtion leads to _zero_ drag, but
I guessed the friction was the source of the 'linear
in speed' part.)
_
Mats Lofkvist
[email hidden]
Quoted text (12 lines)
> On Tue, 12 Feb 2002 00:35:54 +1100, Byron Forbes > <[email hidden]> wrote: > > > Not quite sure what your point is. Last time I looked, streamlined > >flow was defined as flow without "eddys". > > I think Mats means that in most real situations, you'll have > turbulence at least somewhere, so as this is much more important for > the aero effects, you can just estimate the entire thing using just > the squared speed rule. IOW real slipstream situations hardly ever > occur on cars? Just a hunch. Mats can probably clarify, I shouldn't > try and read his mind. :)Quoted text (8 lines)
> > >Mats Lofkvist wrote: > ... > >> Hmmm, but since as long as there is any skin friction > >> there will be turbulence on (parts of) any profile, and > >> since the pressure drag totally dominates the skin friction > >> drag the resulting total drag will be proportional to the > >> square of the speed?
rrevved <[email hidden]> wrote in message news:[email hidden]...
I've re-read it, and I still don't see what I've written which is
"ludicrous". If you're going to disagree with someone, please do
so by stating exactly what you disagree with, and explaining what
you *do* believe instead.
"Really?", "LOL", and "Bye" simply don't carry enough meaning
for me to figure out what you don't like about my post.
-----------------------------
> :: Higher mass = Lower engine braking
> :: Lower mass = Higher engine braking.
> :
> :Actually, no.
>
> Really?
Your statement here, which Ruud(?) disgreed with, is very loosely
correct in terms of effect, but is about as helpful (and accurate)
as saying that a road car with four passengers plus luggage has
"less braking" than one with just a driver.
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> :The mass is irrelevant to the steady-state torque
> :produced by the engine, be that driving or braking torque. What
> :increased mass *does* do is make the car effectively heavier, so
> :it requires a greater braking torque to achieve the same deceleration.
>
> LOL!!
<frowns> Please re-read my paragraph, word by word. Especially
"steady-state", and my use of "torque" vs "deceleration".
You agreed with
> >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.
so I don't see your problem with the above, which is the same thing
stated the other way around.
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> : I've noticed with my model race cars, which accelerate
> :better than I'd expect when I fit silly tall gearing for a big
> :outdoor track.
>
> LOL!! #2
Again, this was an entirely valid and truthful statement. In a
1200g (about 2.5 lbs) car, the motor (running to about 20 000 rpm
at 40 mph outdoors) forms a significant part of the total inertia.
Increasing the gearing by 50% (which happens when going from indoor
to outdoor tracks) makes a noticeable difference, which illustrates
my point:
> >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.
which you then agreed with.
-----------------------------
> :: Engine 'braking' doesn't occur when the engine is not idling.
> :
> :No, because then the overall torque is positive.
>
> LOL!! #3
Again: *what is wrong with my statement?* The total torque applied
by the engine to the clutch can be positive, in which case the car
accelerates; or negative, in which case it slows down. In the latter
case we call it "engine braking".
-----------------------------
> Bye Jonny.
> --
> "Good job. You made me even less moderate than I was when I woke
> up this morning--AGAIN" - rotflmao7d1 in alt.security.terrorism
Yeeees....
Carry on being this un-moderate, and I won't say goodbye - I'll
just add you to my killfile and get on with participating in
discussions here without your interruption.
I'm more than happy to listen to alternative viewpoints and see
if I learn something which adds to my knowledge, but you aren't
going to "convince me" of /anything/ by laughing hysterically in
response to my posts - since I then don't know /what/ you're trying
to convince me of.
Jonny
Quoted text (10 lines)
> >Now, where do you disagree in that lot? > I agree with everything you said in this post, so far. > >Do you disagree with that? > Nope. I agree. > Jonny, why did you not mention your prior post in which you > claimed some ludicrous things? > > Here it is again, with my original commentsOn 11 Feb 2002 19:43:35 +0100, Mats Lofkvist <[email hidden]> wrote:
Hah, I'm beginning to feel psychic! :) (almost an anagram of physics,
hm, lol).
>From what I have understood (?) of aerodynamics, the only
>way to get drag proportional to less than the square of
>the speed is to ignore friction / viscous flow.
Weird, drag sounds like friction by definition. But perhaps that's why
it's mostly turbulent and thus proportional to the square.
Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art : http://www.marketgraph.nl/gallery/
Quoted text (16 lines)
>[email hidden] (Ruud van Gaal) writes: > >> On Tue, 12 Feb 2002 00:35:54 +1100, Byron Forbes >> <[email hidden]> wrote: >> >> > Not quite sure what your point is. Last time I looked, streamlined >> >flow was defined as flow without "eddys". >> >> I think Mats means that in most real situations, you'll have >> turbulence at least somewhere, so as this is much more important for >> the aero effects, you can just estimate the entire thing using just >> the squared speed rule. IOW real slipstream situations hardly ever >> occur on cars? Just a hunch. Mats can probably clarify, I shouldn't >> try and read his mind. :) > >You succeeded quite well at reading my mind, Ruud :-)As it happens, lift is proportional to velocity squared
even with assumption of inviscid (frictionless) flow.
The first velocity comes in from mass flow across the wing:
The faster you fly the more air mass you are effecting per second.
The second velocity term arrives because the overall lift
effect of the wing on the airflow is deflecting some
of it downward. So the force is proportional to the
downwash velocity component which is proportional to
the airspeed. (So for a particular shape that rotates
the velocity a particular angle downward, the lift force
is proportional to airspeed squared)
The above might be one way to motivate the correctness of
aerodynamic forces being proportional to velocity squared.
(Without mentioning vorticity once, mostly ;)
The usual way is straight "dimensional analysis".
It is pretty clear the force is proportional to area,
both experimentally and philosophically.
This means that the rest of the equation has to have the
units of force/area. Since this part is also proportional
to density, the remaining part pretty much has to be
velocity squared because it has to have units of
length squared per time squared.
Ruud van Gaal wrote:
> Weird, drag sounds like friction by definition. But perhaps that's why
> it's mostly turbulent and thus proportional to the square.
Sir Isaac Newton screwed that up too
(the question of linear or quadratic),
doing thought experiments.
Then people did flawed physical
experiments and mistakenly confirmed his screwed up theory. ;)
(If you rotate a wing on the end of a stick perpendicular
to a rotating shaft and measure the forces you can get
approximately linear force with speed.
But that is mostly because you spin up the air in the
test area as you do this kind of test.)
--
Matthew V. Jessick Motorsims
[email hidden] http://www.moto1.net
Vehicle Dynamics Engineer (972)910-8866 Ext.125, Fax: (972)910-8216
