|
General » rec.autos.simulators » Re: Basics of a car simulator
| Re: Basics of a car simulator [message #495000] |
Mon, 31 October 2005 08:33 |
|
> There's plenty of engine data out there for whatever you want to do.
I wasn't sure on this. As posted, if you have the actual engine
data, use it. A lot of race engine manufacturers don't publish
their torque curves, but bike engines have similar torque curves
to racing engines.
>> > RPM is evolved over time as a function of flywheel torque
>> This is only good for hitting the throttle with the clutch in, or
>> in neutral.
> Depends how you model the drivetrain. Flywheel torque most certainly
> effects every drivetrain part and the tires in my model. As it does in
> reality.
I mistunderstood on this one. I though you meant flywheel inertia,
not engine torque at the flywheel.
> Rear wheel force is actually not a function of engine torque. Instead,
> a proper tire model is going to output the force as a function of slip
> ratio/load and so on. This force then DETERMINES the "rear wheel
> torque," not the other way around.
Agreed. One complication of wheel spinning is that engine acceleration
will depend on inertial momentum of the entire drive train from the
crank to the tires, in addition to the friction losses, and resistance
from the actual torque supplied by the tire.
>> Slip ratio is relative to rear wheel torque and the downforce on the tire,
>> not velocity. I'm not sure why so many texts use velocity based equations
>> since this is an effect, not a cause.
> No, the definition of slip ratio is as described in my first reply and
> it's written that way in texts because, well, that's the definition of
> it. :-) Torque and downforce are not in the equation. They end up
> being whatever they are because of slip ratio, not the other way
> around.
It's still my belief that the relative rotation rates are the effect, not
the cause of slip ratios. It's a way of defining slip ratio, and can be
used to measure slip ratio, but what's desired is to calculate slip ratio
based on the factors that cause slip ratio. These are longitudinal force,
normal force, and velocity (usually road velocity is used).
A lot of real tire testing varies longitudinal force, normal force, and
road velocity, to create a table of slip ratios. Given a 3 dimensional table
of sample points, the issue with low velocities is eliminated, since this is
just a table lookup with interpolation. Here is a link to one tire testing
program:
http://www.millikenresearch.com/fsaettc.html
I'm not sure where to find real world data for specific tires.
There's also the issue of jerk, or the transitions in force, I don't know
how much tire testing is done regarding jerk.
> Again, I recommend using the actual definition of slip ratio. That is
> not a function of torque or downforce at all. No such tables needed.
But the tables are more accurate, and don't have a low velocity issue.
>> Slip angles can occur without slippage. The contact patch is flexible, so
>> it's direction is not perpendicular to the tire's axis when there is
>> a side load. The smallest maximum slip angles occur on IRL type cars,
>> with a working slip angle of around 2%. Modern bias ply tires can go
>> over 5%, and the older tires were higher still.
> Slip angle is just what the term states it is. An "angle." Angles are
> expressed generally in degrees or radians, not as percentages.
Hmm, can't enter degrees as a symbol, sorry. I meant degrees.
> 95% of web sites are wrong about the radial tire force drop off and
> continue to propogate this myth. The feeling that a
> tire breaks away and loses grip is because the force merely stops
> rising at some point. I.e., it flattens off more suddenly generally
> with a radial than a bias tire.
I'm not sure that this is a myth. Here is a link to a site that claims
that the lateral force curve shape is affected by the normal force on
the tire. At low loads, there is no drop off, but at higher loads there
is somewhat of a drop off. The peak coefficient of friction also lower
at higher normal forces, again according to this web site.
http://www.smithees-racetech.com.au/ackerman.html
|
|
|
| Re: Basics of a car simulator [message #495001 ] |
Mon, 31 October 2005 10:03 |
|
Jeff Reid wrote:
<snip>
> >> Slip ratio is relative to rear wheel torque and the downforce on the tire,
> >> not velocity. I'm not sure why so many texts use velocity based equations
> >> since this is an effect, not a cause.
>
> > No, the definition of slip ratio is as described in my first reply and
> > it's written that way in texts because, well, that's the definition of
> > it. :-) Torque and downforce are not in the equation. They end up
> > being whatever they are because of slip ratio, not the other way
> > around.
>
> It's still my belief that the relative rotation rates are the effect, not
> the cause of slip ratios.
Relative rotation rate IS the SAE definition of slip ratio, period.
Nothing to argue there. How the slip ratio develops is another matter
of course. But to predict it in a sim such as the OP wants to create,
you need to have slip ratio as an input, not an output.
It's a way of defining slip ratio, and can be
> used to measure slip ratio, but what's desired is to calculate slip ratio
> based on the factors that cause slip ratio. These are longitudinal force,
> normal force, and velocity (usually road velocity is used).
>
In simulators you don't know what the longitudinal force is until you
know what the slip ratio is. Tire models use slip ratio and normal
force as inputs, then output the longitudinal force from that, not the
other way around.
> A lot of real tire testing varies longitudinal force, normal force, and
> road velocity, to create a table of slip ratios. Given a 3 dimensional table
> of sample points, the issue with low velocities is eliminated, since this is
> just a table lookup with interpolation. Here is a link to one tire testing
> program:
>
> http://www.millikenresearch.com/fsaettc.html
>
I'm familiar with the site. In fact, I'm meeting Doug Milliken, owner
of that company that actually runs those tire tests and does
simulation/vehicle dynamics engineering for the big toy boys, on
Tuesday to help assemble an RC tire testing machine that does the same
thing. Interestingly enough, our testing process will be the same as
what's done for big car tires.
Anyway, they are most certainly not "creating a table of slip ratios"
from longitudinal forces, I can assure you. We won't be doing it
either because there's no point in doing so. What you do is load the
tire up, then vary the slip ratio/angle while recording the
longitudinal/lateral forces that result from that. Slip ratio is the
independent variable; the INPUT. The forces are the OUTPUT.
> I'm not sure where to find real world data for specific tires.
>
> There's also the issue of jerk, or the transitions in force, I don't know
> how much tire testing is done regarding jerk.
>
Angular jerk is the derivitive of angular acceleration. This is sort
of modelled through the implementation of relaxation lengths, although
not really. This is sometimes measured by planting a tire at a slip
angle, then measuring the gradual force buildup as the tire is rolled
forward or the belt/drum is moved. However, pure angular jerk in
itself is not really a useful thing to bother modelling and to my
knowledge is never tested.
> > Again, I recommend using the actual definition of slip ratio. That is
> > not a function of torque or downforce at all. No such tables needed.
>
> But the tables are more accurate, and don't have a low velocity issue.
>
Where did you hear about these tables that "look up the slip ratio
given longitudinal force" and so on?
> >> Slip angles can occur without slippage. The contact patch is flexible, so
> >> it's direction is not perpendicular to the tire's axis when there is
> >> a side load. The smallest maximum slip angles occur on IRL type cars,
> >> with a working slip angle of around 2%. Modern bias ply tires can go
> >> over 5%, and the older tires were higher still.
>
> > Slip angle is just what the term states it is. An "angle." Angles are
> > expressed generally in degrees or radians, not as percentages.
>
> Hmm, can't enter degrees as a symbol, sorry. I meant degrees.
>
> > 95% of web sites are wrong about the radial tire force drop off and
> > continue to propogate this myth. The feeling that a
> > tire breaks away and loses grip is because the force merely stops
> > rising at some point. I.e., it flattens off more suddenly generally
> > with a radial than a bias tire.
>
> I'm not sure that this is a myth. Here is a link to a site that claims
> that the lateral force curve shape is affected by the normal force on
> the tire. At low loads, there is no drop off, but at higher loads there
> is somewhat of a drop off.
Well, I've got a book here that has quite a lot of measured tire force
data. In some cases, yes, there is a very, very slight drop off after
the peak in the dry. However, I have seen no difference between
radials and bias tires other than the shapes of the curves leading up
to the peak. Radials usually have a higher cornering stiffness and
peak at a lower slip angle, in general. They also tend to roll off
more quickly into the peak, again, in general, which makes a tire feel
more like it's losing grip suddenly. It's not, the force has just
stopped rising more suddenly, giving that impression. Lots of folks
swear that a car actually accelerates when it runs off into the grass.
Of course it's not, but it feels that way. Same principle.
These diagrams that you see in many books and web sites where the
radial tire hits a peak and then plummets off at a huge negative slope,
while a bias tire on the same graph stays pretty flat are just plain
wrong. Doug Milliken, owner of the company with the link you referred
me to there, is the one that pointed this out to me right here at ras a
few years ago. I couldn't verify it until only fairly recently, but it
appears that he's indeed right ;-)
Anyway, ask these people that post these goofy graphs where they got
the data for them. They won't be able to tell you other than it came
from another book where the data didn't come from an actual tire test
either.
Plenty of this stuff in one of these books shows many street tires,
both radials and bias tires alike, tested out to 20 degrees with the
force still gently climbing. One data extraction in another book
showed the force continuing to rise even at 28 degrees. What force
drop off? :-) I can't tell by looking at any of these graphs whether
the tires were bias or radial by simply looking at what happens after
the peak. Usually because even at 15 or 20 degrees most of them have
still not even peaked.
Now, if you're talking about butyl rubber or something else super
sticky, that's a little bit different story, but there's no reason why
NR and SBR compounds (typical for street tires) would drop off at
anything other than very extreme temperatures and really high speeds.
I.e., lock a wheel up and fry the contact patch or get it sideways at
150mph and then yes, you'll get some pretty significant force drop off,
but I've never seen any more than maybe 10%-15% or so under these
severe conditions. I've even seen data measured that showed the force
climbing right out past 60% slip ratio on at least one tire.
Anyway, to keep this all on point, my argument was that radials and
bias tires are not really noticably different in terms of what happens
after they peak. At least, not from any tests I've seen. I'll be
happy to scan an example for you if you'd like.
The peak coefficient of friction also lower
> at higher normal forces, again according to this web site.
>
Yes, tires exhibit what's called "load sensitivity." This is true of
all tires at anything but the very lightest loads.
|
|
|
| Re: Basics of a car simulator [message #495003 ] |
Mon, 31 October 2005 11:14 |
|
Jeff Reid wrote:>
> I'm not sure that this is a myth. Here is a link to a site that claims
> that the lateral force curve shape is affected by the normal force on
> the tire. At low loads, there is no drop off, but at higher loads there
> is somewhat of a drop off. The peak coefficient of friction also lower
> at higher normal forces, again according to this web site.
>
> http://www.smithees-racetech.com.au/ackerman.html
One more thing... Note that the peak effective friction coefficient of
the tire shown in the blue picture is astronomically high. This a kick
butt racing tire just about up to F1 grip levels. From your link:
"The 300lb blue curve might represent the inside tyre. It has a high
co-efficient of friction, 2. Thus maximum lateral force is 2 times
vertical load. The 900lb curve might represent the more heavily loaded
outside tyre. The co-efficient of friction is less at 1.6 and therefore
the maximum lateral force is only 1.6 times vertical load.
"
1.6 to 2? This is absolutely not a street tire rubber compound. The
grip is enormous, approaching twice what you get from a street tire
under similar loads, and yet the force curve doesn't drop off very much
at all, even at 1200 lb load (looks to be about a 5% drop, agree? Yes,
it will drop more as you go past the test slip angle range shown of
course). Take a look at the 300 lb load line, it's as flat as a
pancake after the peak. Even at 600 lb the drop off is barely
noticable, not even 2-3% probably. And these are the super sticky
tires that so many people insist should plummet after the peaks even
more so than "radials do."
Here's some more real, measured data, but this time from a few typical
street tires:
http://performancesimulations.com/files/tire1.JPG
Now, can you tell whether the top two tires are radials or bias ply?
Me neither... The bottom picture, however, shows an example of a
radial vs. bias. See how the radial rises more sharply (higher
cornering stiffness), but then suddenly changes its slope at one point
as it progresses out toward its peak? This is why a radial tire may
feel like it's loosing grip more suddenly than a bias tire and a driver
would call it unforgiving. It's not unforgiving and snappy because
it's lost grip after the peak, but rather that *the rate at which it
was increasing* suddenly changed somewhere on the way up, whereas the
bias tire rolls off more progressively. And again, this is taken out
well past 12 degrees and the forces are still clearly climbing on both
tires, just at different rates. Most of the other radial tire data
from this source looks more like the bias tire line does; there's not
another example that shows such a sudden change in slope part way up
the line as in this picture. I.e., in general the radials look very
much like bias tires.
The small picture at the top right is of two tires that have identical
tread widths, radii, and rubber compound. The only thing that's
different is the construction. I.e., cord angles, perhaps the number
and arrangement of the plies, and so on. Similar to what one might
expect when comparing a bias to a radial tire. The shapes are very
different indeed, but they aren't losing force anywhere out to 14
degrees slip angle. Again, once you get out there it's usually as flat
as a pancake, or perhaps there's a miniscule drop in grip way, way out
there. But it doesn't look much different between the two
constructions out beyond the peak.
Also, in the lower graph, note the dashed line for "limiting friction."
This seems to imply that the force assymtotically approaches that
line. I.e., it won't actually hit it until you're well off the right
side of the graph on either tire.
|
|
|
| Re: Basics of a car simulator [message #495014 ] |
Mon, 31 October 2005 11:25 |
|
> In simulators you don't know what the longitudinal force is until you
> know what the slip ratio is.
My fault, I stated the wrong terms, what I remember is that the inputs were
normal force, rear wheel torque (not longitudinal force), and velocity.
> How the slip ratio develops is another matter
> of course. But to predict it in a sim such as the OP wants to create,
> you need to have slip ratio as an input, not an output.
How do you determine the slip ratio then? It would seem that rear
wheel torque would be a factor here. If the tire started slipping, then
rotational acceleration would occur, with inertial momentum of the
entire drive train and tires, friction (including contact patch), and
throttle limited rpms of the engine. If the tires doesn't slip, then
there must be some relationship between applied rear wheel torque
and the actual longitudinal force applied to the road.
> These diagrams that you see in many books and web sites where the
> radial tire hits a peak and then plummets off at a huge negative slope,
> while a bias tire on the same graph stays pretty flat are just plain
> wrong.
Why does driver induced understeer work (at least on some cars)? Turn
the front tires inwards enough, and the result is understeer, and this
seems to occur at relatively small slip angles (less than 15 degrees)
on some cars. I've confirmed that this works on a Trans Am (front
heavy car) and a Caterham (rear heavy car). I've also received
emails from a couple of club racers that use a bit of induced understeer
to stabilize otherwise oversteery situations (like very high speed
turns where a lot of power is applied to overcome aerodynamic drag).
Is this because of caster / camber effects?
|
|
|
| Re: Basics of a car simulator [message #495015 ] |
Mon, 31 October 2005 11:25 |
|
> http://performancesimulations.com/files/tire1.JPG
BTW, thanks for taking the time to educate at least me, if not
others reading this thread.
One issue is that most drivers would like a somewhat consitent
repsonse to steering inputs. With a sharp bend in the curve, it
must be more difficult to drive at the limits.
|
|
|
| Re: Basics of a car simulator [message #497177 ] |
Mon, 31 October 2005 21:22 |
|
Jeff Reid wrote:
> > In simulators you don't know what the longitudinal force is until you
> > know what the slip ratio is.
>
> My fault, I stated the wrong terms, what I remember is that the inputs were
> normal force, rear wheel torque (not longitudinal force), and velocity.
>
> > How the slip ratio develops is another matter
> > of course. But to predict it in a sim such as the OP wants to create,
> > you need to have slip ratio as an input, not an output.
>
> How do you determine the slip ratio then? It would seem that rear
> wheel torque would be a factor here. If the tire started slipping, then
> rotational acceleration would occur, with inertial momentum of the
> entire drive train and tires, friction (including contact patch), and
> throttle limited rpms of the engine. If the tires doesn't slip, then
> there must be some relationship between applied rear wheel torque
> and the actual longitudinal force applied to the road.
>
In simulating a car, what you'll do is calculate all these things in an
order that lets you solve for everything. You would start off a
cycle/step using the car's position/orientation as it was at the end of
the last cycle/step. Since you know the position/orientation and the
race track's geometry, you can calculate what the suspension deflection
at each end of the car is. Given the deflection, you can calculate the
normal loads (either from the suspension geometry, like I do it, or
assuming a more simplified wheel rate exists). You also at this point
know the camber/steer angles at all four wheels (if you're modelling
camber).
Next, you'd calculate the slip ratio, which is just 1 - the ratio of
actual to free rolling velocities at each tire. You calculate the slip
angle as well.
The tire model will now PRODUCE the lateral/longitudinal forces from
these slip ratios/angles and the normal load that you got right at the
beginning of the step.
Next, you can add these forces to the car and wheels. You don't move
anything just yet, however.
Following that, you'd now find the torques at the wheels and so forth.
That slip ratio produced a longitudinal force, which will create a
torque at the driven wheels (or any braked ones; rolling resistance
torque can be added to this too). This is really a "road reaction
torque," i.e., it's the "other end" of the torque that is being fed
back through the drivetrain that's resists the engine speeding up, for
example.
Now, the total torque acting on the wheel is the sum of this "road
reaction torque" and any torque coming in from the engine/differential
or whatever you're modelling. Here's the interesting part:
Now that you have the total torque acting on the wheels, you know their
angular accelerations. You can now speed up or slow down the wheels'
rotational velocities based on their torques. We can now move the car,
spin the wheels a little bit, and repeat this whole process for the
next step. Every few cycles we render a graphics frame and voila, the
car moves.
On the very next step (step 2), the slip ratio is going to change just
as you described. I.e., the vertical load, longitudinal force, and
tire radius CONTROLLED the acceleration of the wheel, so the new slip
ratio that you calculate will indeed fall right where it would in
reality. So yes, the longitudinal and vertical forces, tire radius,
differential action and engine torque and so on are indeed controlling
how the slip ratio changes just as you described, but to really solve
for it properly you want to use slip ratio just as it's defined.
There's no need to precalculate anything and try to store slip ratios
as some sort of lookup value based on the other parameters. It all
just works out if you use slip ratio as an input into the tire model,
then let the forces accelerate the wheels however they want. On the
next step, the slip ratio will be right where it should be given all
those "other inputs."
> > These diagrams that you see in many books and web sites where the
> > radial tire hits a peak and then plummets off at a huge negative slope,
> > while a bias tire on the same graph stays pretty flat are just plain
> > wrong.
>
> Why does driver induced understeer work (at least on some cars)? Turn
> the front tires inwards enough, and the result is understeer, and this
> seems to occur at relatively small slip angles (less than 15 degrees)
> on some cars. I've confirmed that this works on a Trans Am (front
> heavy car) and a Caterham (rear heavy car). I've also received
> emails from a couple of club racers that use a bit of induced understeer
> to stabilize otherwise oversteery situations (like very high speed
> turns where a lot of power is applied to overcome aerodynamic drag).
>
> Is this because of caster / camber effects?
Caster/camber effects can effect it, but in general caster effects will
reduce the induced understeer you're referring to. Even without any
caster/camber effects you should still get induced understeer at some
point, especially if the center of gravity is low relative to the
wheelbase. I.e., a low slung open wheeler should have the effect more
pronounced than a regular road going car would.
The reason it is occurs is because even after you've passed the peak of
the lateral force curve, as you continue to dial in more steering, the
amount of side force in the car's coordinate system (pointing straight
out the left/right side of the car) is getting progressively smaller.
I.e., if you steered the front tires to 90 degrees, the tires are
making plenty of "side force," but only in the tire's frame of
reference. I.e., to the left/right of the tire plane there's a bunch
of force. However, relative to the car, all this force is now pointing
towards the rear bumper. Crank in 90 degrees of steering at the tires
and you'll stop the car in a real hurry as though you locked up the
front wheels.
The diagram below shows the same tire forces as the previous link
(tire1.JPG), but this time it shows it in the car's reference frame
instead of the tire's.
http://performancesimulations.com/files/tire2.JPG
The vertical axis is showing the amount of side force in the car's
reference plane instead of the tire's. As the tire is being steered
further and further the lateral force in the tire's frame of reference
is increasing (the length of the straight line going up to the u=0.91
point is growing. At some point it becomes constant (distance from the
origin to the line stays the same), but as you steer the tire further
and further that "side force" is pointing more towards the rear of the
car, leaving less for side force in the car's reference frame. As can
be seen, if you steered the tire right out to 90 degrees, you have no
side force in the car's reference frame at all.
Consequently, the horizontal component of the force in the diagram is
the induced drag and is what slows you down more and more rapidly as
you steer the tire further.
The point where the line is drawn up to u=0.91 is where you'll get the
least understeer (maximum side force in the car's reference frame).
Any more or less than this and you'll get understeer. Note that even
at this point the tires may not have even peaked yet, but you'll still
get the best performance on the skidpad at this slip angle. With
street tires that peak at huge slip angles it could be expected that
the quickest way around the race track would actually be to operate
quite a ways below the peak force slip angle! I covered that in an
LFSforum post recently. Maybe I'll dig that up and provide a link.
You may remember this diagram too:
http://performancesimulations.com/files/steering3.JPG
Going back to the polar diagram again (
http://performancesimulations.com/files/tire2.JPG ), the induced drag
(horizontal axis) will cause forward weight transfer, and this will get
larger and larger as you steer out towards 90 degrees slip angle at the
front. This will actually cause the entire plot to grow. I.e., if you
steered a little bit past the u=.91 point, since you're increasing the
forward weight transfer you will get even more lateral force in both
the tire and car reference frames. The higher the center of mass is in
relation to the wheelbase, the greater this effect will be. So with
cars that have a really high center of mass, you'll probably not get
the induced understeer until you've really got a whole bunch of
steering lock dialed in.
Also, it should be noted that even if the tire lateral force NEVER
peaked at all, but continued to grow and grow right out to 90 degrees
slip angle, you would still get understeer at some point, so the fact
that the tires peak or perhaps fall off after the peak does not in
itself cause the phenomenon. It will indeed effect when it happens,
but this tire behavior is not necessary for induced understeer to
occur.
|
|
|
Gehe zu:
aktuelle Zeit: Fri Jan 9 23:34:52 CET 2009
Insgesamt benötigte Zeit, um die Seite zu erzeugen: 0.07125 Sekunden |
