G-forces when maximum braking?
Dave
What are the g-forces upon the vehicle and objects fastened to a
vehicle when undergoing maximum braking at 100mph under ideal
conditions?
Does ABS provide the maximum/quickest deceleration possible?
What are the g-forces upon the vehicle and objects fastened to a
vehicle when undergoing a sudden stop (i.e. hitting a brick wall) from
100mph?
If you can't provide these answers do you know where I could locate
them?
Thanks,
Dave d...@mindspring.com
bo...@smart.net
I can answer part of your question:
a=(v^2 - vo^2)/2s where:
a - acceleration
vo - original velocity
v - final velocity
s - distance
the g-force will equal a/32.2
So, 100mph = 147 ft/sec.
Assuming 5 feet stopping distance (as seen by the passenger)
as the car hits a brick wall, you have an acceleration of (0
-147^2)/(2*5) = -2160 ft/sec^2 = 67g. (splat)
I'm not an automotive engineer, but I assume that the g-forces
will be pretty constant with maximum braking, regardless of the
initial velocity. You will only be subjected to them for a longer
period of time.
Also, since ABS prevents skidding, and as a result stops the
car in the shortest distance possible, you would experience the
highest deceleration rates with it.
Offhand, I would think the max braking would be in the order
of about 1g. At 100mph, that would mean a stopping distance of approx
336 ft. By using the formula v=at, where t = time, then it would take
about 4.6 seconds to bring the car to a stop.
Regards,
Bob Van Jones, PE
Kas R. Osterbuhr
> Offhand, I would think the max braking would be in the order
> of about 1g. At 100mph, that would mean a stopping distance of approx
> 336 ft. By using the formula v=at, where t = time, then it would take
> about 4.6 seconds to bring the car to a stop.
A top fuel dragster accelerates 0-100 in about a second. Equivalent to
about 3g's. I always thought that rubber, and almost everything, had no
more that coef. of friction of 1, but if that is so, then a dragster is
not following the laws of physics. I've though about this one quite a
bit. I know a consumer car is no dragster, but I always wondered how a
car could pull 3 g's acceleration without adhesion to the surface....
Anyone here an expert on rubber tires? I assume that there is adhesion,
and compliance of the rubber to the asphalt, in effect letting it act
sort of like velcro (with non horizontal contacts to create forward
motion).
--
Kas Osterbuhr
Metallurgist/CAE Specialist
Sherpa Unlimited
Computer Products Division
Denver, CO
email: she...@plinet.com
cc: cla...@midusa.net
phone: 1.303.987.2139
Doug Milliken
On Sat, 1 Mar 1997, Kas R. Osterbuhr wrote:
> A top fuel dragster accelerates 0-100 in about a second. Equivalent to
> about 3g's. I always thought that rubber, and almost everything, had no
> more that coef. of friction of 1, but if that is so, then a dragster is
> not following the laws of physics. I've though about this one quite a
> bit. I know a consumer car is no dragster, but I always wondered how a
> car could pull 3 g's acceleration without adhesion to the surface....
>
> Anyone here an expert on rubber tires? I assume that there is adhesion,
> and compliance of the rubber to the asphalt, in effect letting it act
> sort of like velcro (with non horizontal contacts to create forward
> motion).
In the course of our vehicle dynamics (handling) work we often measure tire
force and moment properties, using the large flat belt tire tester at
Calspan SRL Corp in Buffalo, NY. Race tires usually deliver a "coefficient
of friction" (horizontal force / load) of well over 1.0, and often over 2.0
-- so this cannot be called "classic friction". It is pretty obvious that
the tire is both "sticking/adhering" to the road/belt surface and also
"mechanically gearing" itself to the bumps in the road surface.
Since I'm not a rubber chemist or a researcher on surface effects, I can't
explain what is happening in detail, but race cars regularly accelerate,
brake and corner at well over 1.0 g.
Of course with the addition of "downforce" from a banked turn and/or
from aerodynamics, race cars are capable of even higher "effective
coefficients". I believe that max braking for a Formula 1 car from
high speed can be 4 or 5 g's.
Sincerely,
-- Doug
Douglas Milliken <bd...@freenet.buffalo.edu>
Milliken Research Associates Inc.
Jeffrey...@gat.com
>> Offhand, I would think the max braking would be in the order
>> of about 1g. At 100mph, that would mean a stopping distance of approx
>> 336 ft. By using the formula v=at, where t = time, then it would take
>> about 4.6 seconds to bring the car to a stop.
>
>A top fuel dragster accelerates 0-100 in about a second. Equivalent to
>about 3g's. I always thought that rubber, and almost everything, had no
>more that coef. of friction of 1, but if that is so, then a dragster is
>not following the laws of physics. I've though about this one quite a
>bit. I know a consumer car is no dragster, but I always wondered how a
>car could pull 3 g's acceleration without adhesion to the surface....
>
>Anyone here an expert on rubber tires? I assume that there is adhesion,
>and compliance of the rubber to the asphalt, in effect letting it act
>sort of like velcro (with non horizontal contacts to create forward
>motion).
>
>
>
>--
>Kas Osterbuhr
>Metallurgist/CAE Specialist
>
>Sherpa Unlimited
>Computer Products Division
>Denver, CO
>
>email: she...@plinet.com
>cc: cla...@midusa.net
>phone: 1.303.987.2139
Tires do not follow the basic laws of friction when it comes to
friction. The tire can get non-friction based adhesion to the
irregularities in the road surface. (note that by basic friction
theory, tire width would have no bearing on adhesion) "Sticky" tires
are those that have a soft enough compound to engage more of the road
features. The price you pay is in tire wear. Dragster tires are the
ultimate trade where they only have to last for 5 seconds so they can
have REALLY STICKY TIRES. Road tires must last just a little longer
and typically are capable of coefficients of friction ranging from .7
(50,000 mile warranty) to .9+ (how often do you replace the tires on
your corvette?) Also, dragsters and other race cars take advantage of
downforce from the aerodynamic aids allowing them to increase normal
foce and therfore accelerations. I think a top fuel dragster is
actually accelerating faster at the end of the quarter mile than at
the beginning. So, in answer to the original question...
A passenger car will not be capable of anything approaching 1g due to
brake bias difficiencies and tire limitations. Something more like
8g would be quite ambitious. As for ABS, this is not the best way to
stop a car. If you look at rec.auto.sport.tech I believe there may be
a thread about this. The best type of braking is called threshold
braking where the wheels are turning just slightly slower than that
for rolling. (i.e. about 20% slip) An ABS system cannot do this
without some sort of road speed measurement. What ABS does is pulse
between locked and unlocked brakes which gives you an average stopping
foce which is less than true threshold braking. BTW race drivers use
threshold braking all day long.
If you hit a wall the g's experienced will be a function of the
stiffness of the wall/object and the crush distance. Only tests for
the actual vehicle can give the real number especially when spikes are
considered. The numbers however are in the tens of g's typically.
Hope this helps.
Mike McDermott
"Classic" coefficient of friction is often defined as the tan of
the angle of an inclined plane when a body starts to slip down
it. Tan can take any value from zero to infinity. It's a myth to
say it can't exceed 1.0.
Mike McDermott
Paul Skoczylas
bo...@smart.net wrote:
> Also, since ABS prevents skidding, and as a result stops the
> car in the shortest distance possible, you would experience the
> highest deceleration rates with it.
While ABS prevents skidding, it does not do so in the most efficient way
possible. A professional (or even a well trained amateur) driver can
use "threshold braking" to stop a car on ice in a shorter distance than
ABS. I seem to remember in the one test I saw on TV recently (this was
not a rigorous test, but a demonstration by the local chapter of the
CAA) the car with ABS actually took a longer distance to stop on ice
than the same type of car without ABS, locking its wheels. The main
difference is that if you have ABS, you can maintain control of the
vehicle, while if your wheels lock, you can't. (BTW, in the same test,
the car stopped with threshold braking stopped in a shorter distance
than either of the other two, and maintained control.)
-Paul
Carl P Baker
Jeffrey...@gat.com wrote:
>
>
> Tires do not follow the basic laws of friction when it comes to
> friction.
This is probably true, but it's almost certainly a second order
effect in this case. I find it difficult to believe that
dragster tires have a coefficient of 3.0. I think that the
real answer is _aerodynamics_. Look at these cars - they
have lifting surfaces all over the place. As the car picks
up speed, the aerodynamic down force supplies additional
traction.
--
carl
Carl P Baker <cp_b...@pnl.gov>
Tom Burton
On Sat, 01 Mar 1997 22:34:30 -0700, "Kas R. Osterbuhr" <she...@plinet.com> wrote:
>A top fuel dragster accelerates 0-100 in about a second. Equivalent to
>about 3g's. I always thought that rubber, and almost everything, had no
>more that coef. of friction of 1, but if that is so, then a dragster is
>not following the laws of physics. I've though about this one quite a
>bit. I know a consumer car is no dragster, but I always wondered how a
>car could pull 3 g's acceleration without adhesion to the surface....
>
A dragster typically has a spoiler that increases its weight w at speed significantly, but increases its mass m only slightly, thereby increasing the acceleration a=fw/m, where f is the coefficient of friction (which may exceed one, as others have pointed out).
Tom Burton Brahea Consulting 616/436-7436
Joseph (Joe) R. Steinman
Braking force is a function of several physical phenomenon:
First is the coefficient of friction (u)(acting horizontally).
Secondly is the combined normal force (N)(ie vertical forces due to the
vehicle weight, any down force created by aerodynamic aids, and weight
transfer onto the front of the vehicle.
Physical laws state that the force causing deceleration (F)(braking) is
equal to the product of the coefficient of friction (u) and the normal
force (N). (ie F=u.N)
Lets assume that a 2000lb race car can decelerate at 1g (32.8 ft/s^2) and
assuming no down force and weight transfer is negligable. Since F=u.N
then u=F/N. => u=32.8(ft/s^2)/2000lb => u=.0164 s^2/ft.lb which is much
less than 1 !
Hope this helps
--
Joseph (Joe) R. Steinman
Wild Horses Motorsports
Dallas Texas
Fast Cars And Wild Horses - No Place But Texas!
Joseph (Joe) R. Steinman
A correction to my previous statement:-(
Joseph (Joe) R. Steinman wrote:
>
> Braking force is a function of several physical phenomenon:
>
> First is the coefficient of friction (u)(acting horizontally).
> Secondly is the combined normal force (N)(ie vertical forces due to the
> vehicle weight, any down force created by aerodynamic aids, and weight
> transfer onto the front of the vehicle.
>
> Physical laws state that the force causing deceleration (F)(braking) is
> equal to the product of the coefficient of friction (u) and the normal
> force (N). (ie F=u.N)
The deceleration force F is a product of the vehicle mass and the deceleration (assume 1g)
F=mg
> Lets assume that a 2000lb race car can decelerate at 1g (32.8 ft/s^2) and
> assuming no down force and weight transfer is negligable. Since F=u.N
> then u=F/N. => u=32.8(ft/s^2)/2000lb => u=.0164 s^2/ft.lb which is much
> less than 1 !
F=m.a=u.N from the assumptions above m=N. Then a=u (the deceleration = the coefficient of
friction)
When you apply the other factors to the normal force N=W+downforce+wt_xfr. Then
u=m.a/(W+downforce+wt_xfr) which says that u may be less than 1 for decelerations greater
than 1g !
sorry for the error.
Doug Milliken
On Wed, 5 Mar 1997, Carl P Baker wrote:
> Jeffrey...@gat.com wrote:
> > Tires do not follow the basic laws of friction when it comes to
> > friction.
>
> This is probably true, but it's almost certainly a second order
> effect in this case. I find it difficult to believe that
> dragster tires have a coefficient of 3.0. I think that the
Check again, the drag strips give out a time slip with several
times-to-distance, not just the full quarter mile. I think that they time
0 to 100 feet (? or thereabouts) -- short enough that aero downforce is
only a minor contributor. I think that 3 g's off the line is about right
for the top classes. That rubber is _sticky_ when it is heated up
to operating temperature.
Hugh Hunkeler
Doug Milliken wrote:
>Kas R. Osterbuhr wrote:
> > A top fuel dragster accelerates 0-100 in about a second. Equivalent to
> > about 3g's. I always thought that rubber, and almost everything, had no
> > more that coef. of friction of 1, but if that is so, then a dragster is
> > not following the laws of physics.
>
> force and moment properties, using the large flat belt tire tester at
> Calspan SRL Corp in Buffalo, NY. Race tires usually deliver a "coefficient
> of friction" (horizontal force / load) of well over 1.0, and often over 2.0
> -- so this cannot be called "classic friction".
> ...but race cars regularly accelerate,
> brake and corner at well over 1.0 g.
I assure you that dragsters and other racing cars DO obey the laws of
physics. There is nothing magic about the number "1" as a coefficient of
friction. Remember that the coefficient of friction is the tangent of
the angle of the ramp where an object would just begin sliding, so a CF
of 1.0 would mean that the object would slide down a 45 degree ramp but
not a 44 degree ramp. (Arctan(1.0)=45 degrees.) By using very sticky
rubber compounds, a CF of greater than 1.0 can be achieved on dry
pavement. Of course, the wear characteristics may be bad, but the
trade-offs are different than in passenger cars, where you have to
contend with snow, ice, rain, and longer life expectancy. Aerodynamics
will also come into play. Some race cars can generate two to three times
their weight in down force by aerodynamic design. The acceleration force
(or braking or cornering force) each tire can provide depends on the
down force (weight + aerodynamic + load shift) at each tire. Finally,
recall what the road surface is at the start of a drag strip--it's not
just pavement, but a lot of rubber left by many tires of previous races.
Rubber-on-rubber evidently gives good traction.
George Jefferson
:Tires do not follow the basic laws of friction when it comes to
:friction. The tire can get non-friction based adhesion to the
:irregularities in the road surface.
Much good stuff said about tire adhesion..I have to cringe at the above
terminology though. "classic" high school physics friction
is called coulomb friction. Something deviating from that might
be called non-coulomb or maybe non-linear friction.
Anyway a point on coulomb friction to help understand why some systems
dont follow it..
The reason you have no contact area effect with classic friction is
that due to surface roughness the true contact area is *very* small
compared to the "apparent" contact area. As you increase normal loading
the true contact area increases and the available friction increases along
with it. There is no fundamental reason to expect "u=1" to be a bound on
this phenomenon, that is just an observation for many common systems.
Coulomb friction does not hold where the true contact area is not
small to begin with..Tires are a good example of this, the rubber is soft
so it easily conforms to any irregularities in the road surface.
Another interesting "anomoly" is if you have two highly polished surfaces
then you get very high friction under low applied normal loading.
Unknown
On Sat, 01 Mar 1997 17:58:06 GMT, bo...@smart.net wrote:
>I can answer part of your question:
> a=(v^2 - vo^2)/2s where:
> a - acceleration
> vo - original velocity
> v - final velocity
> s - distance
>
> the g-force will equal a/32.2
>
> So, 100mph = 147 ft/sec.
>
> Assuming 5 feet stopping distance (as seen by the passenger)
>as the car hits a brick wall, you have an acceleration of (0
>-147^2)/(2*5) = -2160 ft/sec^2 = 67g. (splat)
>
(snip)
>
>Regards,
>Bob Van Jones, PE
>
>
Not quite. While the braking forces will be nearly constant (ignoring
brake fade, etc.) the stopping forces when hitting a wall will be far
from constant over the stopping distance. In fact, when the bumper
first contacts the wall, the stopping force will be zero. Next,
moving the bumper back a few inches doesn't take much force at all (I
know this from experience :-) ) Actually, the further you crush, the
more things start to pile up in the engine compartment and the more
the forces rise. A more reasonable assumption would be that force
rises sort of linearly from zero at impact to a maximum just prior to
stopping. The easiest way to calculate the peak force is using energy
methods. The car has 1/2*m*v^2 of energy. For simplicity, lets
assume the car weighs 2000 lb. That means that at 100 mph it would
have 668,047 ft-lb of energy. The area of the force*distance energy
triangle must now equal that number. If the base of the triangle is 5
ft. and the area is 668,047 ft-lb, then the peak force must be 267,218
lb. This force will decelerate our 2000 lb car at 133.6 g's.
_Double_Splat!
Note that our assumption of 2000 lb is irrelevant to the deceleration
number for our other assumptions, it just eases the calculation.
Also note that this 133.6 g's is equivalent to a deceleration
efficiency (defined as the ratio of average force to peak force for
the same stopping distance) of about 50%. This isn't actually all
that bad. The best spring/damper systems can only achieve about 85%
for the same type problem.
Chris Hall
In <E6LzG...@freenet.buffalo.edu>, Doug Milliken <bd...@freenet.buffalo.edu> writes:
>On Wed, 5 Mar 1997, Carl P Baker wrote:
>
>> Jeffrey...@gat.com wrote:
>> > friction.
>>
>> effect in this case. I find it difficult to believe that
>> dragster tires have a coefficient of 3.0. I think that the
>
>Check again, the drag strips give out a time slip with several
>times-to-distance, not just the full quarter mile. I think that they time
>0 to 100 feet (? or thereabouts) -- short enough that aero downforce is
>only a minor contributor. I think that 3 g's off the line is about right
>for the top classes. That rubber is _sticky_ when it is heated up
>to operating temperature.
>
>-- Doug
> Douglas Milliken <bd...@freenet.buffalo.edu>
> Milliken Research Associates Inc.
Drag strip surfaces don't rely on friction to give the grip. Firstly,
a good track will use a concrete base planed to near optically
flat. After that you drag second-hand slick tyres up and down the
track (weighted down by heavy sleds) for hours on end to grind a
layer of rubber into the surface. Finally you spray it with a traction
compound (similar stuff to a contact adhesive), and let it dry in the
sun. A well prepared start line will remove the shoes from your feet
unless they're laced up REAL tight.
When you run your car/bike up the track, you perform a "burn-out"
to warm the tyres until they smoke. This mobilises the aromatic
compounds in the rubber to make it extremly sticky. Then you put
the tyres on a contact adhesive surface....
The end result is grip of profoundly unbelievable proportions. The
onboard datalogger on my motorcycle has shown peak acceleration
of 3.1g at a point 0.8 seconds into the run. Top fuel cars are seeing
in excess of 4.5g peaks.
We get timing information at 60, 330, 660, 1000 and
1320 feet. My bike runs 60 foot times in the 1.1 second bracket, fuel
cars are getting into the low 0.8's (already close to 90mph).
But this performance ain't coming from "friction", it's coming from
"glue".
Chris Hall,
Funnybike racer, computer programmer and ex-ME