my simulator is almost complete, except these points:
1)what is the formula for rolling resistance? i think it woul dbe something
to do with (for each wheel): the weight on the wheel, the contact area on
the ground (flat tyre etc), the speed, and a coefficient. does anybody know
the real formula?
2) when accelerating, how does the wheelspin affect the acceleration? i
thought the tyre might push with it's maximum ability, and the rest of the
torque would go into spinning the wheels, however, this is wrong - you can
spin the wheels on the spot with enough power, or crappy tyres.
3) is air pressure = 1.3086 kg/metre cubed?
(0.7809x28.0)+(0.2095*32.0)+(0.0093*79.9)
------------------------------------------------ =
22.4litres
0.02931227kg / 0.022400metres cubed
is this correct?
thanks for any help here.
Nick.
Rolling resistance and torque split in a home-made sim
Nick G asked for rolling resistance formulas and how engine torque is distributed to driven wheels. Todd Wasson, Ruud van Gaal and others said rolling resistance is roughly coefficient times load and largely speed independent, and that an open differential sends equal torque to each wheel. Matt Jessick and Doug Milliken pointed to SAE papers, and posters corrected a claim that friction coefficients cannot exceed 1.
With Matt Jessick (Motorsims); Doug Milliken (Milliken Research Associates)
physics/realism debateRacer
Automated summary.
one other thing - how does the torque from the engine go to the wheels, is
it half and half to LR and RR (50-50 weight distribution) or 1/4 to each
wheel in 4WD cars? or is it same torque to all? i think it is 1/(no of
wheels) * transmission torque going to each wheel, but i want to check.
cheers, Nick.
Nick G <[email hidden]> wrote in message
news:[email hidden]...
> my simulator is almost complete, except these points:
I'll comment where I can:
> 2) when accelerating, how does the wheelspin affect theacceleration? i
> thought the tyre might push with it's maximum ability, and the restof the
> torque would go into spinning the wheels, however, this is wrong -you can
> spin the wheels on the spot with enough power, or crappy tyres.
Does Pacejka mean anything to you? If not, go looking...
> 3) is air pressure = 1.3086 kg/metre cubed?
Atmospheric air pressure? That's ~101.3 kPa AFAIK... or are you
talking about something else? (kPa = thousand newtons per *square*
metre)
Jonny
Also, I'd try the forums on the "Team Lightspeed" website. Address your
questions to Doug Arnao. He's an expert on that type of information.
Good Luck!! Mick
Nick G wrote:
>
> my simulator is almost complete, except these points:
>
> 1)what is the formula for rolling resistance?
>1)what is the formula for rolling resistance? i think it woul dbe something
>to do with (for each wheel): the weight on the wheel, the contact area on
>the ground (flat tyre etc), the speed, and a coefficient. does anybody know
>the real formula?
>
You want the formula that derives the rolling resistance? Or do you want the
formula that gives the force *from* rolling resistance? I believe that rolling
resistance coefficient can be used to get the force that slows the car by:
Rolling resistance coefficient = Roll_Res_Coeff
Vertical force = Load
Force that slows the car = Force
Force = Load * Roll_Res_Coef
I'm not 100% sure on that, but that's what it looks like the power equation
would brake down to.
>2) when accelerating, how does the wheelspin affect the acceleration? i
>thought the tyre might push with it's maximum ability, and the rest of the
>torque would go into spinning the wheels, however, this is wrong - you can
>spin the wheels on the spot with enough power, or crappy tyres.
>
First, how are you getting longitudinal force at the tires (the forward
force)?
Air pressure is not a "meters cubed" thing.. Perhaps you mean air density?
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
Quoted text (8 lines)
>3) is air pressure = 1.3086 kg/metre cubed? > >(0.7809x28.0)+(0.2095*32.0)+(0.0093*79.9) >------------------------------------------------ = > 22.4litres > >0.02931227kg / 0.022400metres cubed >
Nick G <[email hidden]> schreef in berichtnieuws
[email hidden]...
> my simulator is almost complete, except these points:
>
> 1)what is the formula for rolling resistance? i think it woul dbesomething
> to do with (for each wheel): the weight on the wheel, the contact area on
> the ground (flat tyre etc), the speed, and a coefficient. does anybodyknow
> the real formula?
The rolling resistance is almost independant of speed. Furthermore, I guess
it will mainly depend on how round your tyre is. And this depends on the
load on the tyre and the tyre pressure.
For a typical saloon car, the total roling resistance is in the order of 200
N and slightly DEcreasing as a function of the velocity (40 km/h: 220 N, 120
km/h: 190 N)
>
> 2) when accelerating, how does the wheelspin affect the acceleration? i
> thought the tyre might push with it's maximum ability, and the rest of the
> torque would go into spinning the wheels, however, this is wrong - you can
> spin the wheels on the spot with enough power, or crappy tyres.
Their is a maximum friction between tyre and road, when the contact patch is
not movint relatively to the road (which is the case if the vehicle is
moving without wheelspin).
When wheelspin starts, the friction force will be much less which gives more
wheelspin etc.
athmospheric pressure is 1013 hPa, about 10000 kg per square metre
furthermore: you could use the gas-laws (Gay-Lussac, Boyle etc).
if the absolute (Kelvin) temperature increases, the pressure increases
proportionally.
if the amount of air (moles) increases, the pressure increases
proportionally
if the volume increases, the pressure DEcreases proprortionally.
greetings,
hgrm
> thanks for any help here.
> Nick.
>
>
Quoted text (11 lines)
> > 3) is air pressure = 1.3086 kg/metre cubed? > > (0.7809x28.0)+(0.2095*32.0)+(0.0093*79.9) > ------------------------------------------------ = > 22.4litres > > 0.02931227kg / 0.022400metres cubed > > is this correct? >
Nick G <[email hidden]> schreef in berichtnieuws
[email hidden]...
> one other thing - how does the torque from the engine go to the wheels, is
> it half and half to LR and RR (50-50 weight distribution) or 1/4 to each
> wheel in 4WD cars? or is it same torque to all? i think it is 1/(no of
> wheels) * transmission torque going to each wheel, but i want to check.
>That depends on the transmission. Some differentials change the
torque-balance if one wheels start to spin.
Besides, in 4WD cars, the torque is not necessairily splitted 50/50 between
front and rear.
greetings,
hgrm
> cheers, Nick.
>
>
On Sun, 22 Jul 2001 15:26:06 +0100, "Nick G" <[email hidden]>
wrote:
>my simulator is almost complete, except these points:
>
>1)what is the formula for rolling resistance? i think it woul dbe something
>to do with (for each wheel): the weight on the wheel, the contact area on
>the ground (flat tyre etc), the speed, and a coefficient. does anybody know
>the real formula?
Marco's Monster page uses a formula that's based on velocity, but John
Dixon's book 'Suspension, Tires and Handling' uses just
F_rr=coeff*load. It seems mostly velocity independent from most of
what I heard.
>2) when accelerating, how does the wheelspin affect the acceleration? i
>thought the tyre might push with it's maximum ability, and the rest of the
>torque would go into spinning the wheels, however, this is wrong - you can
>spin the wheels on the spot with enough power, or crappy tyres.
Hm, when spinning the wheels enormously (like what Valentino Rossi
does after a win, keeping the front wheel locked and the rear
spinning), the friction goes down quite a lot, probably also due to
heat (it's melting/burning, from the looks of all that smoke).
With wheelspin the force just goes down. According to RCVD, just to
about 60% with reasonably severe spin (SR=1, so wheel rotating twice
as fast as the road below it).
>3) is air pressure = 1.3086 kg/metre cubed?
That's air density, I use 1.225, a standard value but a little high
perhaps for high-heat tracks in Spain for example.
Check out http://www.windpower.dk/stat/unitsw.htm
Lots of air densities there.
Cheers,
Ruud van Gaal, GPL Rank +53.25
Pencil art : http://www.marketgraph.nl/gallery/
Free car sim : http://www.marketgraph.nl/gallery/racer/
On Sun, 22 Jul 2001 15:30:06 +0100, "Nick G" <[email hidden]>
wrote:
>one other thing - how does the torque from the engine go to the wheels, is
>it half and half to LR and RR (50-50 weight distribution) or 1/4 to each
>wheel in 4WD cars? or is it same torque to all? i think it is 1/(no of
>wheels) * transmission torque going to each wheel, but i want to check.
Same torque to all. I had the same thought when applying suspension
spring force; half up and half down? But it's just the complete force
every which way. It's up to the tires whether they respond to the
torque.
Ruud van Gaal, GPL Rank +53.25
Pencil art : http://www.marketgraph.nl/gallery/
Free car sim : http://www.marketgraph.nl/gallery/racer/
Ruud van Gaal <[email hidden]> wrote in message
news:[email hidden]...
> On Sun, 22 Jul 2001 15:30:06 +0100, "Nick G" <[email hidden]>
> wrote:
>
> >one other thing - how does the torque from the engine go to thewheels, is
> >it half and half to LR and RR (50-50 weight distribution) or 1/4 toeach
> >wheel in 4WD cars? or is it same torque to all? i think it is 1/(noof
> >wheels) * transmission torque going to each wheel, but i want tocheck.
>
> Same torque to all. I had the same thought when applying suspension
> spring force; half up and half down? But it's just the completeforce
> every which way. It's up to the tires whether they respond to the
> torque.
... providing it's a totally 'open' diff. LSDs require slightly more
work!
Jonny
yeah, sorry, i meant air density.
thats what you get from typing so damn quickly coz of the phone bill (in the
UK)
Nick.
here's a new one:
one other thing i haven't implemented is engine braking.
i dont want to mess with my current formula coz it's so damn good:
i put in the gear ratios and a couple of other measurements from a Ford
Sierra RS Cosworth and a Ford Fiesta RS Turbo (two cars i know alot about,
and got within 1.6mph of their tested top speeds. and thats with a velocity
dependent rolling resistance with a coefficient of about 0.7 (which is a
good approx, but not true).
what i want is to be able to lift of the throttle, drop the revs and slow
the car down using engine braking and whatever. i wrote a side-program that
used the engine power output (in kW) and the power required to drive the
wheels to see if that was how it works, but i cant figure out a method that
works correctly (or as good as).
cheers, Nick.
Nick G <[email hidden]> wrote in message
news:[email hidden]...
> yeah, sorry, i meant air density.
>
> thats what you get from typing so damn quickly coz of the phone bill(in the
> UK)
What are you reading with? Outlook Express lets you compose
off-line... and ntl give you unlimited, 24-hour 'net access through a
dialer box on a BT line (which is what I'm using).
Jonny
Nick G wrote:
>
> my simulator is almost complete, except these points:
>
> 1)what is the formula for rolling resistance?
This paper discusses models used in research and the
limitation of various models as well:
"Modelling of Rolling Resistance Test Data,"
Parmeet S. Grover, SAE980251. SAE, 1998.
You could probably still download this paper online for
a small fee from www.sae.org.
Here is one model discussed in the paper that has/had a
particular use for comparing various tests:
RR = A0 + A1 L + A2 V + A3 (L^2 / P) + A4 L V + A5 V^2
where ^2 is a squaring function, L = vertical load,
P is pressure, V is speed. A0..5 are coeffficients from test data,
and RR is the rolling resistance. See paper for units, etc.
What you will find is that there are a lot of models
from a lot of tests for a lot of tires and no obvious
universal model that meets all needs.
- Matt
--
Matthew V. Jessick Motorsims
[email hidden] http://www.moto1.net
Vehicle Dynamics Engineer (972)910-8866 Ext.125, Fax: (972)910-8216
On Mon, 23 Jul 2001 18:28:54 +0100, "Jonny Hodgson"
<[email hidden]> wrote:
Ah damn, yes that's right ofcourse. Diffs, not too easy though.
Ruud van Gaal, GPL Rank +53.25
Pencil art : http://www.marketgraph.nl/gallery/
Free car sim : http://www.marketgraph.nl/gallery/racer/
Quoted text (10 lines)
>Ruud van Gaal <[email hidden]> wrote in message >news:[email hidden]... >> Same torque to all. I had the same thought when applying suspension >> spring force; half up and half down? But it's just the complete >force >> every which way. It's up to the tires whether they respond to the >> torque. > >... providing it's a totally 'open' diff. LSDs require slightly more >work!On Mon, 23 Jul 2001 18:41:53 +0100, "Nick G" <[email hidden]>
wrote:
...
>what i want is to be able to lift of the throttle, drop the revs and slow
>the car down using engine braking and whatever.
I use an upper torque limit, and a lower one. For the upper limit I
use a curve, since that's so characteristic for an engine. For the
lower limit, it seems a mostly linear relationship to RPM.
So:
throttleInput=0..1
maxTorque=curve(rpm)
minTorque=-engineBrakingCoeff*rpm
engineTorque=throttleInput*(maxTorque-minTorque)+minTorque
That's all folks. ;-)
Ruud van Gaal, GPL Rank +53.25
Pencil art : http://www.marketgraph.nl/gallery/
Free car sim : http://www.marketgraph.nl/gallery/racer/
On Mon, 23 Jul 2001, mjessick-Motorsims wrote:
Or SAE might make you buy SP-1338 (collection of tire/wheel papers from
1998). Just took a look, several other interesting papers in there.
-- Doug Milliken
Quoted text (11 lines)
> Nick G wrote: > > 1)what is the formula for rolling resistance? > This paper discusses models used in research and the > limitation of various models as well: > > "Modelling of Rolling Resistance Test Data," > Parmeet S. Grover, SAE980251. SAE, 1998. > > You could probably still download this paper online for > a small fee from www.sae.org.Since you've had heaps of answers on rolling resistance...
With any object, there is a coefficient of friction for sliding, and
one for static, when you are gripping. The friction force is pretty
simple, if you know the force from the tyre to the ground, it is
simply coeff x reaction force = lateral force. Once the lateral force
is exceeded, the coeff will change from the static one to the sliding
one.
So the amount of grip available drops once you spin the tyres, but
this will heat the tyres at the same time. So the "wheel spinning
grip" will rise with temp. to a point, then drop. Also the pressure
will rise, reducing the contact patch
What coefficients to use? Depends on tyres, heat, road surface, etc.
You'll need to get a data book on experimental values to be accurate.
Normal tyres are around 0.3-0.4 generally. I know that F1 cars develop
a coeff of around 3-3.5, mainly due to the grippy rubber and the huge
downforce at speed. Without downforce, it's impossible to go higher
than 1.0
Rafe Mc
Quoted text (18 lines)
>2) when accelerating, how does the wheelspin affect the acceleration? i >thought the tyre might push with it's maximum ability, and the rest of the >torque would go into spinning the wheels, however, this is wrong - you can >spin the wheels on the spot with enough power, or crappy tyres. > >3) is air pressure = 1.3086 kg/metre cubed? > >(0.7809x28.0)+(0.2095*32.0)+(0.0093*79.9) >------------------------------------------------ = > 22.4litres > >0.02931227kg / 0.022400metres cubed > >is this correct? > >thanks for any help here. >Nick. >Actually, that was indeed on of the 'rules' that were thought to be
quite true by physicists, but current tyre compounds proved the rule to
be false.
-Gregr
Rafe McAuliffe wrote:
> Without downforce, it's impossible to go higher
> than 1.0
>
> Rafe Mc
>
On Tue, 24 Jul 2001 10:37:18 GMT, [email hidden] (Rafe
McAuliffe) wrote:
>With any object, there is a coefficient of friction for sliding, and
>one for static, when you are gripping. The friction force is pretty
>simple, if you know the force from the tyre to the ground, it is
>simply coeff x reaction force = lateral force. Once the lateral force
>is exceeded, the coeff will change from the static one to the sliding
>one.
This is very much oversimplification for tires and I would advise
against your regular static vs. sliding friction model. The tire is a
flexing piece of rubber, and preferably has a slow transition from
sticking to sliding.
The 'coeff' you speak of can be taken from Pacejka curves, but that's
hardly trivial when you start with a sim. Transitions from static to
sliding coefficients are not like shoving around a cupboard.
>So the amount of grip available drops once you spin the tyres, but
>this will heat the tyres at the same time. So the "wheel spinning
>grip" will rise with temp. to a point, then drop. Also the pressure
>will rise, reducing the contact patch
It seems both raising and lowering pressure away from the optimal
point reduces grip btw. Wasn't somewhere like 155kPa appropriate for a
sports car tire?
Anybody now the relationship of tire heating vs. slip ratio for
example? Don't know whether it would be just linear or quadratic, but
it seems to me a simple temp system could consist of:
heat (in Joules for example) += slipRatio*timeStep*heatCoefficient
heat += slipAngle*timeStep*heatCoefficient(Lat)
heat -= disippationCoefficient*(tireTemp-airTemp)*timeStep
And derive pressure and things like that from that.
>What coefficients to use? Depends on tyres, heat, road surface, etc.
>You'll need to get a data book on experimental values to be accurate.
>Normal tyres are around 0.3-0.4 generally.
Actually, your own car probably has tires doing just over 1.0, in the
area of 1.1 somewhat. See Genta's book; a normal tire these days can
do about 1.1, the Ferrari tires are ~1.7 (the b2 Pacejka constant),
and I would assume therefore the F1 cars today will do about 2.0-2.4.
> I know that F1 cars develop
>a coeff of around 3-3.5, mainly due to the grippy rubber and the huge
>downforce at speed.
The friction coefficient probably doesn't get this high, but thanks to
downforce, you get more upforce, normally referred to as load, and as
F_long=someCoefficient*normalForce (normalForce=load), the added
downforce leads to more available longitudinal force. That's raising
the load, not the coefficient.
>Without downforce, it's impossible to go higher
>than 1.0
As Gregor pointed out, this is not true. Perhaps tire hysterisis or
effects like that (rubber gripping into the surface), but it can
easily outdo it's normal force. Extreme example; a pole of concrete
stucked into the ground. You surely have to apply more lateral force
to get it to move (if you succeed at all), than the force dragging it
down (gravity). Something sticky like that happens with rubber I
think, on a small scale.
But there I may be totally wrong, but still, friction coefficients are
generally 1.1-2.2 from road car to F1.
Ruud van Gaal, GPL Rank +53.25
Pencil art : http://www.marketgraph.nl/gallery/
Free car sim : http://www.marketgraph.nl/gallery/racer/
