rec.autos.simulators

I've recently began writing a vehicle simulator, and I currently trying to
get to grips with engine/wheel stuff. I think I have a model which is
correct, but I just wanted to run it by you guys to see if I'm completely on
the wrong track. Here's the way I think things work, feel free to poke holes
in it :)

I have an engine which for a given RPM produces an amount of Torque Te (for
now I'm ignoring resistive/frictional forces). For a given gear, we can
calculate the torque on the axle Ta by:

Ta = Te*Rg*Rd

where Rg is the gear ratio and Rd is the differential ratio.

Now here's the bit I'm not convinced about. The torque on the axle(s) is
used to accelerate the rotation of the wheels, ie. from the torque, the
angular acceleration of each wheel can be calculated by

Aw = Ta / Iw

where Iw is the MOI for the wheel. This angular acceleration can then be
integrated to give a new angular velocity for the wheel.

Once we have an angular velocity for each wheel, the slip ratios can be
calculated. These can then be used to calculate a traction force using the
longitudinal version of the magic formula. This traction force (ignoring
friction limits) then accelerates the car forwards, as well as again
applying a torque to the wheel in the opposite direction (by Newton's 3rd
law).

OK, so thats what I think happens so far. The thing I'm not very clear on is
rolling resistance. In the litrature I've seen, rolling resistance is always
treated as a force which is applied to the vehicle as a whole. To my mind,
rolling resistance should be applied to the wheels' rotation, which would
indirectly cause the vehicle to slow down using the above. Am I right? and
if so how do I calculate rolling resistance. What I've read seems to be very
vague on this, and just says it's some constant multiplied by the velocity
of the vehicle. This seems like a cheap approximation to me ...

Any help appreciated ...

Cheers,

Mike.

Would love to help Mike... but this just isnt my field.

ed_

I am in very much the same situation. What I was taught in A-Level mechanics
is that the friction of a sliding object is = uR, where u is the coefficient
of friction 0<=u<=1, and R is the reaction force from the ground. The
reaction force from the ground is equal to the weight of the car on that
suspension point + the weight of the wheel, brake and suspension units. The
weight on each suspension unit can be calculated from the Centre of Mass
using simple moments.

We always assumed rolling resistance to be negligible when compared to air
resistance, and treated air resistance as the total car resistance. The
formula for air resistance (from a technical aeroplane book) is proportional
to the surface area of the object, the velocity of the object and the
density of the air. There is a great explanation in the Air Pilot's Manual
Vol 4: The Aeroplane: Technical available at
http://www.amazon.co.uk/exec/obidos/ASIN/1840371552/qid%3D995399866/2...
782-1771049 for 16.

However, I cannot help you in your quest for the rolling resistance.

Nick.

The main reason I'm interested in rolling resistance is to slow down the
rotation of the wheels. If you have a vehicle suspended in the air with it's
wheels rotating, the wheels will gradual slow down then stop. This is the
behaviour that I want to model. This will also make a contribution to
slowing the vehicle down if the wheel is in contact with the ground.

Thanks for the link, I'm sure it will become useful soon.

Mike.

http://www.racesimcentral.net/%3D995399866/2...

It was from that web page that I found out about this newsgroup :)

At the moment I'm just trying to learn this stuff, and I don't expect the
first iteration to be very useful other than as a learning aid to me. Once
I'm sure I have a suitable model, then I'll start to write a sim for real,
and the physics part will probably be open source. I'll post a link to the
group when that happens.

  Right.  If you decide later to model a differential gear itself, this would
be torque applied to the ring gear.

  >Now here's the bit I'm not convinced about. The torque on the axle(s) is

  Yes, although if you want to go a bit further, remember that (at least) the
half-shaft MOI also should be included with this.  This will help stability in
your system somewhat, as well as keep overall acceleration closer to real life
results.  

  >Once we have an angular velocity for each wheel, the slip ratios can be

  That's how I do it too.

    >The thing I'm not very clear on is

  Currently, I'm using a cheap force approximation.  However, looking through
Milliken's "Race Car Vehicle Dynamics" on pages 71 and 72 (tire behavior),
there is a good treatment on rolling resistance.  If you don't have the book
yet, get it :-)  Hopefully, Doug himself will pop in here with the best link to
use.

  The section describes how, due to the tire elements' compression and
re-extension as they move through the contact patch, there is an offset where
this force is applied (rather than being dead-center in the patch.)  The
example shows a .22 inch offset for one particular tire, so that might be a
good ballpark figure.  I'm not sure if this stays constant or not, but is
probably a decent approximation for starters.  Given the effective tire radius
(squashed tire from spring type behavior), vertical load, and the offset
described above, the rolling resistance force is given in the example by:

  rolling_resistance_force = offset * vertical_load / effective_tire_radius

  In the example, it was:

  rolling_resistance_force = .22inch * 1200lbs / 13.7inches
                                     = 19.2 lbs

  Anyway, the way I was originally handling rolling resistance was by adding a
braking torque to the tire directly that would give the same results as the
approximation you spoke of in the last post.   However, this doesn't appear to
be the correct way to handle it.  Since the rolling resistance force is applied
in the direction opposite to vehicle travel (if wheels are free rolling), the
resulting torque would actually speed up the tire and slow down the car
simultaneously, rather than slow down both at the same time as my braking
torque idea would.  

  In short, it looks as though the original approximation you talked about is
probably right.  Now, try finding the resulting torque on the tire and add it
in while also applying the force directly to the car.  The wheels should
accelerate slightly (keeping non-negative slip ratios, unlike a braking method
would produce), yet the car should decelerate, *I think* :-)

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

  Oh, that's not rolling resistance, it's bearing friction combined with
compression losses with closed throttles (if the clutch is engaged.)  To do
this, add a negative torque to the driveline/wheels, as though a small brake
was at work somewhere.  Not sure on real values for this though.  The
Straightline Acceleration Simulator program at my site works well with real
life values of about 75 lbs-feet for driveline loss.  That seems like an awful
lot of friction to me, but the results are consistent with real car tests so
far.  

  Rolling resistance is a seperate effect, and is due to tire flex in the
contact patch.  If the car's suspended in mid air, there is 0 "rolling
resistance," if my definition is correct.  
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://www.racesimcentral.net/

   Lowering flame retardent face shield...
    Charlie

  Horsepower = ((Coefficient_rolling_resistance * Weight +
air_density*frontal_area*miles_per_hour^2) * miles_per_hour) / 375

  With a 3500 lb car, using a supposedly high typical rolling resistance
coefficient of .015, a frontal area of 24 sq feet, and a coefficient of drag
(the wind resistance part) of .42, to travel at these speeds would require this
much power: (air density is .0025 in this example)
              Rolling resistance  --Wind resistance   -Both effects  -% rolling
resistance
                    only                         only
  10 mph--->  1.4 hp                     .06 hp                1.46 hp      
96%
  20 mph--->  2.8 hp                     .53 hp                3.33 hp      
84%
  40 mph--->  5.6 hp                    4.30  p                9.90 hp      56%
  45 mph--->  6.3 hp                    6.12 h               12.42 hp      50%

  At 45 mph (for this particular car), the wind and rolling resistance both
contribute about the same amount.  Continuing on, the wind resistance really
takes over:

              Rolling resistance  --Wind resistance   -Both effects  -% rolling
resistance
                    only                         only
  60 mph--->      8.4 hp              14.51 hp                22.91 hp    
36.6%
  80 mph--->    11.2 hp              34.40 hp                45.60 hp     24.6%
100 mph--->    14.0 hp              67.20 hp                81.20 hp     17.2%
120 mph--->    16.8 hp            116.12 hp               132.92 hp    12.6%
140 mph--->    19.6 hp            184.39 hp               204.00 hp      9.6%
160 mph--->    22.4 hp            275.25 hp               297.65 hp      7.5%
180 mph--->    25.2 hp            391.91 hp               417.11 hp      6.0%
200 mph--->    28.0 hp            537.60 hp               565.60 hp      4.9%
220 mph--->    30.8 hp            715.00 hp               746.34 hp      4.1%
240 mph--->    33.6 hp            928.97 hp               962.57 hp      3.5%
260 mph--->    36.4 hp           1181.11 hp              1217.50 hp    3.0%
280 mph--->    39.2 hp           1475.17 hp              1514.37 hp    2.6%
300 mph--->    42.0 hp           1814.40 hp              1856.40 hp    2.3%

  I've never layed this out this way before, but it looks as though rolling
resistance is a larger factor than I previously believed, but yeah, it does
look like it becomes somewhat negligable after a point.  The percentage column
is "rolling resistance only" divided by "both effects", giving the percentage
contribution of rolling resistance.

  The interesting part (to me, anyway :-)) here is, it looks as though the
rolling resistance force is constant and independent of speed, while the wind
resistance force (not power, like shown above) quadruples when the speed
doubles (squared relationship.)  Power is cubed, of course.

  Note that racing bicycles with hard, skinny tires should have far, far less
rolling resistance than this car example does.  They're designed probably with
that sole purpose in mind, so the percentages would drop off extremely quickly,
leaving aerodynamics (even at typical bicycle racing speeds) as the ***
force.  Well, that's what I'd bet anyway :-)

  > I know almost nothing in this field but I would like to make a small

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://www.racesimcentral.net/

  One more thing.  If you do seperate the driveline/bearing and engine friction
from true rolling resistance at the tires, note that the rolling resistance
force pushes in the opposite direction of the car itself.  This would give the
same negative acceleration to the car as it would if you applied the force
directly to the center of mass of the car, so the way you've read is correct.  

   The thing that intrigues me though, is that since the force is at the
contact patch of the tire, it will torque the body (pitch) as well as torque
the tire itself, and unless I'm missing something here, this torque would
actually serve to *accelerate* the tire, while simultaneously slowing the
vehicle.  Kinda wierd.  Anyone know for sure if this is true, or have I been
drinking too much lately?  :-)
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

Get your bathroom scales and hold them vertically up against the back of
your car. Make sure it's in neutral, handbrake off, on flat ground, and then
shove the car along with the scales jammed between your palms and the back
of the car. That way you can measure the force you're applying. It doesn't
matter if you're moving at a slow walk or an all-out sprint, the force is
the same. At least, it was for my Vauxhall Nova down our drive.... by the
way it's a good idea to have someone in the front seat who can apply the
brakes when you're done :0)

As for aerodynamic drag, it's 0.5 * Cd * rho * A * V^2 where Cd is the drag
coefficient (approx 0.3 to 0.5 for road cars), rho is density of air, A is
frontal area, and V is the velocity w.r.t the air.

Not sure if any of this stuff helps in your conversation as I sort of lost
my way in your maths, but hopefully it's slightly relevant!

Cheerio
Jim Seamus

Have you read the physics of racing series by Brian Beckman?  Its all great
stuff, but the earlier parts are keyed towards the race-driver wanting to
know "how things work", it is great background reading.  The later parts are
just awesome for the budding Sim-writer talking about the pacekja "magic"
formula as a way to get the Force generated by a particular tyre at a given
slip-ratio and slip-angle.  He is currently writing part 26 and 27, which
cover slip-ratio near zero (a problem with the normal formulas for
calculating slip ratio).

Sorry thats a bit illegible, I haven't woken up properly yet :)

Oh, here is the link -
http://www.esbconsult.com.au/ogden/locost/phors/phors.htm

All the best,
Ash - budding sim writer member #131151232 ;)

the question is really interesting. If you only applied the rolling
resistance as a force pointing backwards to the direction of travel at
the contact point of the wheel, when the system settles a slip ratio
develops with another force being produced that exactly counters this
force, and, as amazing as it may seem, there would be absolutely no
braking effect from such a force.

The rolling resistance, however, is defined as happening in the already
settled situation. The total torque on the wheel is actually very close
to 0 in this situation (*), but a net braking force exists. The
explanation for this is that there is an additional torque produced by
the vertical force due to the offset in vertical load that you
mentioned, and in a settled situation (*) this exactly cancels the
torque on the wheel as produced by the rolling resistance. The overall
result is that rolling resistance as a total force must effectively be
applied at the center of the wheel (*).

Of course, a very good question remains whether slip ratio is nonzero at
this point or not (meaning that the tyre may indeed rotate faster or
slower). I do think that the effect should be pretty negligible, though,
at least with low rolling resistance.

-Gregor

(*) as long as the effective MOI/R^2 is much lower than the mass of the
car

That's where I've been getting confused. The stuff I've read tends to group
all these effects together under the name rolling resistance. Thanks for
clearing that up.