The Physics of Drag Racing: A question
Jim Nelson
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
> We got to discussing this question, but were not able to come up with a
> scientific explanation - intuitively, it seems that wider rear tires would
> increase traction, but that's not a function of friction, is it? If a
> narrow tire were used, the force vector perpendicular to the ground would
> be increased because point of contact was smaller (higher pounds per square
> inch).
>
> My guess is that the answer is more mundane - namely, that the pressure
> on a narrow tire, combined with the friction, would make blow-outs more
> likely. But could you use a thick-walled, narrow tired and achieve the
> same effects as a wide, but thinner tire?
>
> If someone could help explain this mathematically and provide the relevant
> physical concepts, we would greatly appreciate it (because we'd be able to
> get back to our work).
>
> Thanks!
>
> -- Sam Blackman
Boy--wait 'til you get to wrinkle-walls. *That* will really
make things interesting in your department.
>
Balaguru Kalundavelu Sambandam
: The contact patch does not remain constant. As the car leaves, the load is
: transferred to the rear tires and causes them to flatten out thus creating a
: larger area for friction to occur. The applied force is not entirely static.
: Mathematically it would take me a while to figure it out but trust me, wider
: tires do promote better traction.
: Joe
I could be wrong, but I thought that wider tires yeild better grip, because as
the tire is loaded it deforms, and the rubber fills the pockets in the
pavement. The wider the tire -- the more number of pockets are filled. It's
like holding yourself up by one finger vs four fingers. But that's just my
humble theory...seems to also explain why softer tires have more grip.
-Bala
Jim Carr
Nicely crossposted. That explains the comment about wrinkle walls!
I love the pictures showing the wheel rotating before the car moves,
which must be critical to exceeding 3 g at launch when the clutch
is dropped.
sc...@midway.uchicago.edu (Samuel C. Blackman) writes:
>
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>We got to discussing this question, but were not able to come up with a
>scientific explanation
The answer is: your "if" is wrong. The coefficient of friction is
not a constant independent of the loads placed on the material.
The simplest way to understand this from the physics side is to
actually do some measurements. Experiments answer lots of questions.
The next best way is to note how the tire slips. On a good surface,
rubber from the tire stays stuck to the road but is ripped from the
tire itself. The limit to the grip that can be supplied by a bit of
rubber on the tire is related to the molecular forces that hold the
tire together and the rubber to the rubber bonded to the track and
the bond between the rubber and the pavement. This is a force that
cannot be helped much by increasing the normal force on the tire
and thus determines the maximum friction per unit area a given tire
compound can provide.
> intuitively, it seems that wider rear tires would
>increase traction, but that's not a function of friction, is it?
Traction is used many ways by many people, often to make a
distinction between static friction and tread effects, but also
to include dynamic effects such as deformation of the tire.
--
James A. Carr <j...@scri.fsu.edu> | In its first 24 hours MSNBC "news"
http://www.scri.fsu.edu/~jac/ | misrepresented the hardware and O/S
Supercomputer Computations Res. Inst. | used in Independence Day to promote
Florida State, Tallahassee FL 32306 | a MS product. Gates uber alles!
Shawn P. Church
sc...@midway.uchicago.edu (Samuel C. Blackman) wrote:
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
<snip>
While I'm sure that there are other reasons involved, you must
remember that the traction available is not dictated simply by
the coeffient of friction. There is also the mechanical
interaction between the rubber layed down on the burnout and the
sticky rubber of the slicks. An appropriate analogy _might_ be
to consider gluing two objects together. Would you rather use a
dab of glue, or a large amount? I believe that the effective
coefficient of friction for a Top Fuel dragster is in the 4.0
range.
Oh yes, there is also something about the way the slicks tend to
_grow_ in height as the car accelerates. This provides for a
longer thinner contact patch which is considered important
(versus a wider one for lateral acceleration).
Rgds,
Shawn
Detroitjoe
Well Samuel, pretty big can of worms you opened here. I like good
questions. The coefficient of rubber on cement is very high compared to
many other materials rubbing on the same patch of cement. The coefficient
of a material is based on the same size contact patch of each of the given
materials. With this in mind, the coefficient of friction will tell you
how a material will resist sliding on other materials. It is a ratio
basically. If a rubber on cement has a coefficient of say 1.5, and rubber
on rubber has a coefficient of 3.0, then a tire 5" wide riding on a track
coated with rubber would produce about the same traction as a 10" tire on
cement. Sound reasonable?
I tried........
Michael Weiner
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
> increase traction, but that's not a function of friction, is it? If a
> narrow tire were used, the force vector perpendicular to the ground would
> be increased because point of contact was smaller (higher pounds per square
> inch).
>
> My guess is that the answer is more mundane - namely, that the pressure
> on a narrow tire, combined with the friction, would make blow-outs more
> likely. But could you use a thick-walled, narrow tired and achieve the
> same effects as a wide, but thinner tire?
>
> If someone could help explain this mathematically and provide the relevant
> physical concepts, we would greatly appreciate it (because we'd be able to
> get back to our work).
>
> Thanks!
>
> -- Sam Blackman
>
> Samuel C. Blackman ! InterNet : blac...@tigger.uic.edu
> MD/PhD Student (3/7) ! Disclaimer : Who cares what I say, I'm a student!
> Univ. of Ill. at Chicago ! Quote : "Quandro potro io finir di stupire?"
> Dept. of Pharmacology ! Phone : 312/996-4983 (lab) Fax: 312/996-1225
I'm new to this whole newsgroup thing, so this is a response as well as a
test to see if my mail is even getting through.
I haven't done any research in this area, but with the speed and force
being transmitted through those tires, I'd bet that a skinny tire would
just melt into a puddle of muck. (I could be wrong though)
**********************************************************
Michael Weiner
Graduate Student
Colorado State University
'95 - '96 Formula SAE Team
Joe Wiles
sc...@midway.uchicago.edu (Samuel C. Blackman) wrote:
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>We got to discussing this question, but were not able to come up with a
>scientific explanation - intuitively, it seems that wider rear tires would
>increase traction, but that's not a function of friction, is it? If a
>narrow tire were used, the force vector perpendicular to the ground would
>be increased because point of contact was smaller (higher pounds per square
>inch).
>
>My guess is that the answer is more mundane - namely, that the pressure
>on a narrow tire, combined with the friction, would make blow-outs more
>likely. But could you use a thick-walled, narrow tired and achieve the
>same effects as a wide, but thinner tire?
>
>If someone could help explain this mathematically and provide the relevant
>physical concepts, we would greatly appreciate it (because we'd be able to
>get back to our work).
>
>Thanks!
>
>-- Sam Blackman
>
>
>--
The contact patch does not remain constant. As the car leaves, the load is
Samuel C. Blackman
My Ph.D. advisor and I are trying to find the answer to the following
question, wholly unrelated to our field (pharmacology):
"Why do drag racers use wide rear tires, if the coefficient of friction
is independant of surface area?"
We got to discussing this question, but were not able to come up with a
scientific explanation - intuitively, it seems that wider rear tires would
increase traction, but that's not a function of friction, is it? If a
narrow tire were used, the force vector perpendicular to the ground would
be increased because point of contact was smaller (higher pounds per square
inch).
My guess is that the answer is more mundane - namely, that the pressure
on a narrow tire, combined with the friction, would make blow-outs more
likely. But could you use a thick-walled, narrow tired and achieve the
same effects as a wide, but thinner tire?
If someone could help explain this mathematically and provide the relevant
physical concepts, we would greatly appreciate it (because we'd be able to
get back to our work).
Thanks!
-- Sam Blackman
--
Roger W. Faulkner
Balaguru Kalundavelu Sambandam wrote:
> I thought that wider tires yeild better grip, because as
> the tire is loaded it deforms, and the rubber fills the pockets in the
> pavement. The wider the tire -- the more number of pockets are filled.
> -Bala
I think that is right as far as it goes. Racing tires are much softer than
ordinary tires, and often have no tread. The fact that no tread produces better
traction on dry roads is compatible with your explaination...except rather than
pockets, I think it is the tops of rocks sticking up above the road plain that
are most effective.
Also, the tires are soft enough that there is a degree of tack which temporarily
bonds the tires to the surface. That is why race drivers often burn a little
rubber to break in new tires...the sticky, reverted surface actually gives more
grip.
...Roger Faulkner
aaron charles bernstein
In article <DupBy...@midway.uchicago.edu>,
Newsgroups: sci.physics,rec.autos.sport.tech,rec.autos.sport.misc
Subject: Re: The Physics of Drag Racing: A question
Summary:
Expires:
References: <DupBy...@midway.uchicago.edu>
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Cc: me
Samuel C. Blackman <sc...@midway.uchicago.edu> wrote:
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>... could you use a thick-walled, narrow tired and achieve the
>same effects as a wide, but thinner tire?
The big tires are to dissipate heat. Think of the car with a
final velocity at the end of the track. A certain amount of energy must
go into translation of the car to get it to that velocity. Now, for
narrower, thicker tires which could possibly dissipate the same amount of
heat as the wider, thinner ones (say they have the same mass), then more
energy would have to be spent in just *rotating* the tires. Less would be
left for translation of the car.
The concept of different shapes of wheels requiring different
amounts of energy to rotate them is demonstrated by a simple experiment:
two wheels with equal mass but different shape (radius) may roll down an
incline at different rates of acceleration. The distribution of mass
within the wheel as well as the wheel's radius determines whether this
occurs.
So why not just make the wheels have a really really small radius
to allow for ease of rotation? Because in heat-dissipation, consideration
must be given to how many times a part of the tire hits the pavement per
second. Consider a large-radius tire rolling along at the same velocity
as a small-radius tire. A given part of the small-radius tire will strike
the pavement and experience the heat-producing friction many more times
per second than the large-radius tire.
So, given basic material properties of the rubber and pavement,
there is the balancing of these two main effects in designing the tire in
order to keep it at optimum "pavement gripping" temperature: 1) energy it
takes to rotate a tire and 2) rotation frequency.
Aaron Bernstein
Leland Lewis
Jim Nelson wrote:
>
> Samuel C. Blackman wrote:
> >
> > My Ph.D. advisor and I are trying to find the answer to the following
> > question, wholly unrelated to our field (pharmacology):
> >
> > "Why do drag racers use wide rear tires, if the coefficient of friction
> > is independant of surface area?"
> >
> > We got to discussing this question, but were not able to come up with a
> > scientific explanation - intuitively, it seems that wider rear tires would
> > increase traction, but that's not a function of friction, is it? If a
> > narrow tire were used, the force vector perpendicular to the ground would
> > be increased because point of contact was smaller (higher pounds per square
> > inch).
> >
> > My guess is that the answer is more mundane - namely, that the pressure
> > on a narrow tire, combined with the friction, would make blow-outs more
> > likely. But could you use a thick-walled, narrow tired and achieve the
> > same effects as a wide, but thinner tire?
> >
> > If someone could help explain this mathematically and provide the relevant
> > physical concepts, we would greatly appreciate it (because we'd be able to
> > get back to our work).
> >
> > Thanks!
> >
> > -- Sam Blackman
>
> make things interesting in your department.
> >
I believe your answer may lay in the laws of resistance (or lack of?).
I don't have the answer in terms of physics, but a something to look at
is
the distortion of a tire under load and at speed. The tire flattens on
initial
launch due to weight transfer giving traction then grows in diameter due
to
centrifigal force giving more top end speed by dynamically changing the
gear
ratio. The tire patch on the pavement narrows giving, we are told, less
rolling resistance.
In the sixties they used your "narrow tire, thick sidewall" theory and
with around
one third of the power of today's cars spun the tires most if not all
the way
down the 1/4 mile. One man, Flamin' Frank Pedregon, used to actually
ignite the
tires.
To add to the discussion, why is it that a load is easier to move (or
seems so)
if it rests on narrow runners rather than resting flat on the ground, if
the
coefficient of friction is the same?
Lee
Naruhisa Takashima
Could it be that friction is limited by the shear stress on the
tire surface?
For a given amount of friction, increasing the contact area
reduces the amount of shear stress on the tire which ultimately limits
the amount of friction.
Does this sound plausible?
All we really need is an engineer from a tire company to answer
this question...
Naru Takashima
--
College Park Bicycle Club: http://www.glue.umd.edu/~naru/cpbc.html
.. __o __o __o __o __o __o __o __o ..
.. -\<, -\<, -\<, -\<, -\<, -\<, -\<, -\<, ..
..(_)/(_)(_)/(_)(_)/(_)(_)/(_)(_)/(_)(_)/(_)(_)/(_)(_)/(_)..
Chuck Tomlinson
Samuel C. Blackman said...
>
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
The coefficient of friction for pneumatic tires is dependent on loading
(pounds per square inch). As the vertical loading increases, the
coefficient of friction will decrease. This is true even for street cars.
IMO, drag tires have adhesion working for them, as well as "normal"
friction. In other words, a hot drag slick wants to stick to the pavement,
and can generate some traction even if the normal load is zero. With
adhesive tires, you want as much consistent surface area as possible.
But even with (non-adhesive) cold street tires, wider tires generate more
traction. For example, Grand Sport Corvettes require more rpm for optimium
launches, compared to the "regular" LT4 Vette. The drivetrains in both cars
are identical except for the wider rear tires on the Grand Sport.
--
__
___| |____ Chuck Tomlinson <toml...@ix.netcom.com>
/___LT-1___/ Mouse Power!
|__| '94 Vette Z07/ZF6
Kevin Anthony Scaldeferri
Who maintains the rec.auto.* FAQ anyways. I've pointed out several
times when this has come up that you don't even have to examine the
special behavior of tires to get a mu of greater than 1. For some
reason there is this common misconception (even among PhD physicists)
that this limit exists when, in fact, it does not. Now, there aren't
many substances that exhibit this sort of behavior, but they do exist.
Kevin Scaldeferri
In article <4slrt2$3...@skipper.balltown.cma.com>,
richard welty <we...@skipper.balltown.cma.com> wrote:
>
>
>here is the pertinent section from the rec.autos.* FAQ:
>
>
>Q: According to my physics class, you can't ever accelerate or corner at
> more than 1.0G. Is this true?
>
>
>
>A: the equations for friction used in freshman physics textbooks presume
> that the surfaces are smooth, dry and non-deformable, none of which
> are the case with tires & pavement.
>
> Pavement is _never_ smooth; it is always irregular to a greater or lesser
> extent. Tires, which are not really dry and solid (as rubber is a
> substance which in its natural form is liquid, and which has only been
> coerced into a semblance of solidity by chemical magic), deform to match
> the surface of the pavement which a vehicle is traveling over. The
> mechanisms by which tires grip pavement are not well understood, but
> are clearly not those described by the "Introduction to Physics"
> formula.
>
> A better model for tire grip is offered in _Race Car Vehicle Dynamics_,
> William and Douglas Milliken, Society of Automotive Engineers, 1995.
> In this model, mu, the constant coefficient of friction, is replaced
> by a lateral (xy plane) force coefficient, whose value may vary depending
> on the vertical (z axis) force being applied, and on other conditions.
> For a modern Formula 1 race tire, the lateral force coefficient may
> be as high as 1.8 for light loading in the z axis. See chapter two
> of this fine book for an exhaustive discussion of tire behaviour
> in extreme conditions.
>
>cheers,
> richard
>--
> Not Insane! in '96
> George Leroy Tirebiter for President
> One Organism: One Vote
Jim Carr
kfis...@iglou.com (Ken Fischer) writes:
>
> Right before the race starts they "burn rubber" to
>heat it up so it is sticky, then they back up and start
>the race.
There are other ways to preheat tires, but it does help.
Note the care taken to put the car over the nice fresh rubber they
just laid down. The main reason for a burnout can be found in the
CRC "rubber handbook" for Physics: the coefficient of static friction
for rubber on clean surfaces varies from 1 to 4 and, for a wide
range of materials, the coefficient is quite large between two
surfaces made of the same stuff (adhesion effects). It is better
to race on rubber bonded to pavement than on bare pavement.
Ken Fischer
Samuel C. Blackman (sc...@midway.uchicago.edu) wrote:
: My Ph.D. advisor and I are trying to find the answer to the following
: question, wholly unrelated to our field (pharmacology):
: "Why do drag racers use wide rear tires, if the coefficient of friction
: is independant of surface area?"
Right before the race starts they "burn rubber" to
heat it up so it is sticky, then they back up and start
the race. The more sticky area, the more friction,
but if they "break friction" (the wheels spin too fast)
they will lose the race. Math doesn't do much for this
problem.
Ken Fischer
Nitro
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
why???because skinny tires look stupid....and besides...nitro and rubber smell so
nice...
Elaine DiRico
In article <DupBy...@midway.uchicago.edu>, sc...@midway.uchicago.edu says...
>
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>scientific explanation - intuitively, it seems that wider rear tires would
>increase traction, but that's not a function of friction, is it? If a
>narrow tire were used, the force vector perpendicular to the ground would
>be increased because point of contact was smaller (higher pounds per square
>inch).
>
>My guess is that the answer is more mundane - namely, that the pressure
>on a narrow tire, combined with the friction, would make blow-outs more
>likely. But could you use a thick-walled, narrow tired and achieve the
>same effects as a wide, but thinner tire?
>
>If someone could help explain this mathematically and provide the relevant
>physical concepts, we would greatly appreciate it (because we'd be able to
>get back to our work).
>
>Thanks!
>
>-- Sam Blackman
>
>
>Samuel C. Blackman ! InterNet : blac...@tigger.uic.edu
>MD/PhD Student (3/7) ! Disclaimer : Who cares what I say, I'm a
student!
>Univ. of Ill. at Chicago ! Quote : "Quandro potro io finir di stupire?"
>Dept. of Pharmacology ! Phone : 312/996-4983 (lab) Fax: 312/996-1225
This is a question that my nephew--who has had considerable success racing
sprint and other oval track cars tried to debate in junior high. He knew for
a fact that wider tires DO increase traction. The reason why the increase in
surface area is not canceled out by the decrease in unit pressure is because
THE COEFFICIENT OF FRICTION IS NOT ACTUALLY A CONSTANT. Due to shearing off
of rubber particles or whatever else is going on at the interface of the
asphalt and rubber or dirt and rubber, the coefficient of friction goes down
as the unit pressure goes up. When you plug that into the equation, wider
tires, besides providing known better traction, in no way contradict the laws
of physics.
Don from Don's Automotive in Austin Texas
(Also a stock car racer myself)
Bob Gilbert
In article <DupBy...@midway.uchicago.edu>, sc...@midway.uchicago.edu (Samuel C. Blackman) writes:
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
> We got to discussing this question, but were not able to come up with a
> scientific explanation - intuitively, it seems that wider rear tires would
> increase traction, but that's not a function of friction, is it?
Bingo! Friction is usefull for modeling the forces between two relatively
hard and non-deforming surfaces. Tires (especially those used by drag
racers) are neither hard or non-deforming. Further there are the affects
of adhesion (stickyness) that add into the total "traction" formula.
Friction just does not accurately model the forces between tire and road,
although friction is a component of the total traction.
-Bob
me...@cars3.uchicago.edu
In article <4sjlfu$l...@opus.cs.utexas.edu>, bal...@cs.utexas.edu (Balaguru Kalundavelu Sambandam) writes:
>
>: The contact patch does not remain constant. As the car leaves, the load is
>: transferred to the rear tires and causes them to flatten out thus creating a
>: larger area for friction to occur. The applied force is not entirely static.
>: Mathematically it would take me a while to figure it out but trust me, wider
>: tires do promote better traction.
>
>: Joe
>
>like holding yourself up by one finger vs four fingers. But that's just my
>humble theory...seems to also explain why softer tires have more grip.
>
did so far, I feel obliged. The person who brought up the question
took as a starting point the claim that the friction force doesn't
depend on the contact area (which is, to first order, OK). However,
the shear stresses in the tire (tangential to the surface), while
acelerating, depend very much on the contact area (to first order
being inversely proportional to it). Once these stressees exceed the
yield stress of the tire material, an attempt to accelerate faster
results in the outer layer of the tire being peeled of while the wheel
is spinning in place. So, for any given load and desired top
acceleration there is a minimal tire size necessery if you don't want
your tires to self destruct.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Jim Carr
we...@skipper.balltown.cma.com (richard welty) writes:
>
Hey, Richard, nice to see you in a physics group.
>here is the pertinent section from the rec.autos.* FAQ:
The first paragraph could use a minor adjustment or two.
>Q: According to my physics class, you can't ever accelerate or corner at
> more than 1.0G. Is this true?
>
>A: the equations for friction used in freshman physics textbooks presume
> that the surfaces are smooth, dry and non-deformable, none of which
> are the case with tires & pavement.
I won't try to rewrite the FAQ, but this misconception is spread by
teachers, not textbooks. For example, Halliday and Resnick states
very clearly (over more than a page) that friction equations are
approximations (for the reasons mentioned above, and others) that
replace a complex phenomenon with an empirical constant.
H&R also say explicitly that the coefficient can be bigger than 1.
I know one book that includes a table of kinetic coefficients, one
of which (rubber on dry "cement") is bigger than 1.
Thus the student might be advised to read more than just the final
equation used for some idealized cases. After all, a more careful
analysis such as the one you cite still uses mu*N, but makes mu
depend on some key variables so it models a real tire. In addition,
note that the CRC Handbook lists several materials, one of them being
rubber, that has a coefficient larger than 1.
The question above is an urban legend spread by profs of physics and
engineering who have not read the book they are teaching out of.
richard welty
In article <4sjlfu$l...@opus.cs.utexas.edu>,
Balaguru Kalundavelu Sambandam <bal...@cs.utexas.edu> wrote:
>: The contact patch does not remain constant. As the car leaves, the load is
>: transferred to the rear tires and causes them to flatten out thus creating a
>: larger area for friction to occur. The applied force is not entirely static.
>: Mathematically it would take me a while to figure it out but trust me, wider
>: tires do promote better traction.
>I could be wrong, but I thought that wider tires yeild better grip, because as
>the tire is loaded it deforms, and the rubber fills the pockets in the
>pavement. The wider the tire -- the more number of pockets are filled. It's
>like holding yourself up by one finger vs four fingers. But that's just my
>humble theory...seems to also explain why softer tires have more grip.
as rough explanations go, it's not bad.
the fundamental flaw in the initial premise is assuming that the standard
formulas in physics for friction apply; they do not. the coefficient of
friction is not a constant when we're dealing with soft tires on pavement.
here is the pertinent section from the rec.autos.* FAQ:
Q: According to my physics class, you can't ever accelerate or corner at
more than 1.0G. Is this true?
A: the equations for friction used in freshman physics textbooks presume
that the surfaces are smooth, dry and non-deformable, none of which
are the case with tires & pavement.
Pavement is _never_ smooth; it is always irregular to a greater or lesser
Balaji Srinivasa
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
> We got to discussing this question, but were not able to come up with a
> scientific explanation - intuitively, it seems that wider rear tires would
> narrow tire were used, the force vector perpendicular to the ground would
> be increased because point of contact was smaller (higher pounds per square
> inch).
>
> My guess is that the answer is more mundane - namely, that the pressure
> on a narrow tire, combined with the friction, would make blow-outs more
> likely. But could you use a thick-walled, narrow tired and achieve the
> same effects as a wide, but thinner tire?
>
> If someone could help explain this mathematically and provide the relevant
> physical concepts, we would greatly appreciate it (because we'd be able to
> get back to our work).
>
> Thanks!
>
> -- Sam Blackman
>
> --
> Samuel C. Blackman ! InterNet : blac...@tigger.uic.edu
> MD/PhD Student (3/7) ! Disclaimer : Who cares what I say, I'm a student!
> Univ. of Ill. at Chicago ! Quote : "Quandro potro io finir di stupire?"
> Dept. of Pharmacology ! Phone : 312/996-4983 (lab) Fax: 312/996-1225
it has all the physics for Racing (not necessarily Drag).
it applies in principle to drag too!
http://reality.sgi.com/employees/rck/PhOR/
--
_______________________________________________________________________________
;-Balaji Srinivasa - Platinum Technology/Aston-Brooke Lab - bal...@platinum.com
Bruce Augenstein
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
Well, Mr. Blackman, you've got two or three reponses in this thread that
get to one of the more important aspects of traction: Namely, the idea
that the ultimate coefficient of friction is not constant when the unit
loading changes, because the tensile strength of rubber is relatively
low, so it tears away more readily under high unit loadings.
Of course, that's only part of the picture, because putting wide tires on
a vehicle does nothing to put more rubber on the road. It simply changes
the shape of the identical-sized contact patch (assuming the same
inflation pressure). Contact patch area is determined solely by load on
that wheel, divided by inflation pressure, assuming similar tire
construction (meaning, sidewall and tread flexibility/rigidity).
One of the things a larger tire does is allow you to run lower inflation
pressures, thus increasing the size of the contact patch, which does
nothing to increase the static coefficient of friction, but it *does*
lower the unit loading, so you don't have rubber tearing away as readily
as you do with a higher unit loading. Thus, the ultimate dynamic
coefficient of friction is raised. Of course, you can lower the inflation
pressures with smaller tires, as well, but, you run into sidewall/contact
patch distortion earlier (meaning, the contact patch on a smaller
tire starts deforming at relatively higher pressures), so portions of the
contact patch have higher unit pressures, begin tearing away, and supply
ready-made ball bearings to interfere with those sections of the contact
patch that have lower unit loadings, destroying the ultimate dynamic
coefficient of friction.
************************************************************************Note that I have only mentioned "larger" tires, as opposed to wider.
Wider tires with the same height are worth amazingly little, unless they
are made of higher traction rubber.
************************************************************************
The second advantage that larger tires have is taller sidewalls assumimg
they use the same wheel size. Taller sidewalls allow more "give" on hard
takeoffs or during shifts, so torque spikes at the contact patch are
softened. You've already read a couple of replies that speak of
"wrinklewall" slicks, which in effect are so compliant as to allow the
contact patch to rotate soewhat in relation to the wheel, giving you the
ultimate in torque spike cushioning.
The gentleman who wrote about the Corvette Gran Sport LT4 seemed to make
the point that the wider tires on that model improved straight line
traction. In fact, that is *not* the case, since those tires are the same
height as the tires on the standard Vettes, and therefore have
relatively shorter sidewalls (meaning, sidwall height vs tread width),
and thus are very hard to launch with. The relatively more rigid
sidewalls are extremely unforgiving, faithfully transmitting any and all
torque spikes in undiminished form, thus breaking the tires loose. Any
ZR-1 driver can tell you these tires are a bear to work with at the drag
strip. On the other hand, they're very good at cornering, where the
relatively more rigid sidewalls keep the contact patch more firmly and
evenly planted under high G conditions.
So, bigger tires allow for lower tire pressures, increased contact patch
area, and lower unit loadings without the same level of distortion
experienced with smaller tires, which would lead to uneven contact patch
loading, and thus "little rubber ball bearings" tearing away from the
more highly loaded sections. However, even at identical inflation
pressures (and thus identical contact patch area and unit loading), the
larger tire's taller, more compliant sidewalls will allow harder launches
and shifts while still maintaining traction, because they'll more
effectively damp torque spikes before they get to the contact patch.
Side note: A tire, any tire, *must* slip in order to generate traction.
As the tire begins to slip, the coefficient of friction begins to *rise*.
It peaks at some level of slip, and then begins to fall off, more and
more rapidly as slip increases. Modern radials tend to generate peak
traction at around eight to ten percent slip, and begin to fall off
rapidly after that, which is why they're so hard to work with at the drag
strip. Drag racers say that radials don't "recover", meaning once you've
broken them loose, they'll stay broken loose.
Slicks, on the other hand, have a more gradual curve, peaking at
something above 35% slip, and tapering off much more gradually after
that, which is why they have a fair amount traction even when broken
loose, and thus recover well when you feather the throttle.
No math here, but I hope that helps.
Bruce
Joe Wiles
Nitro <mro...@mts.net> wrote:
>Samuel C. Blackman wrote:
>>
>> My Ph.D. advisor and I are trying to find the answer to the following
>> question, wholly unrelated to our field (pharmacology):
>>
>> "Why do drag racers use wide rear tires, if the coefficient of friction
>> is independant of surface area?"
>
>
>
>nice...
THAT'S IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Steve Hess
Samuel C. Blackman wrote:
>
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>
Because it works!
Chuck Tomlinson
Bruce Augenstein said...
>
>************************************************************************
>Note that I have only mentioned "larger" tires,
>as opposed to wider.
>Wider tires with the same height are worth amazingly little, unless they
>are made of higher traction rubber.
>************************************************************************
I agree with most of your post, but I have to address a couple of items.
IMHO, wider tires with the same height should allow lower tire pressures
while maintaining acceptable contact patch distortion (and pressure
distribution). This will increase the contact patch area and available
traction.
>The second advantage that larger tires have is taller sidewalls assumimg
>they use the same wheel size. Taller sidewalls allow more "give" on hard
>takeoffs or during shifts, so torque spikes at the contact patch are
>softened. You've already read a couple of replies that speak of
>"wrinklewall" slicks, which in effect are so compliant as to allow the
>contact patch to rotate soewhat in relation to the wheel, giving you the
>ultimate in torque spike cushioning.
>
>The gentleman who wrote about the Corvette Gran Sport LT4 seemed to make
>the point that the wider tires on that model improved straight line
>traction. In fact, that is *not* the case, since those tires are the same
>height as the tires on the standard Vettes, and therefore have
>relatively shorter sidewalls (meaning, sidwall height vs tread width),
>and thus are very hard to launch with. The relatively more rigid
>sidewalls are extremely unforgiving, faithfully transmitting any and all
>torque spikes in undiminished form, thus breaking the tires loose. Any
>ZR-1 driver can tell you these tires are a bear to work with at the drag
>strip. On the other hand, they're very good at cornering, where the
>relatively more rigid sidewalls keep the contact patch more firmly and
>evenly planted under high G conditions.
I'm the one who mentioned the Grand Sport. I understand you point that
they may make the car less forgiving of torque spikes (i.e. abrupt clutch
engagement). But I believe that once the clutch is engaged, the bigger
tires offer more available traction.
Regarding sidewall stiffness, the sidewalls on Goodyear EMTs are about the
stiffest around. AFAIK, EMTs do not corner as well as normal GS-Cs of the
same size. So the cornering advantage of the 315/35's comes from extra
width.
jc
In article <4sjh1h$5...@kocrsv08.delcoelect.com>,
Joe Wiles <jjw...@koasm010.delcoelect.com> wrote:
>sc...@midway.uchicago.edu (Samuel C. Blackman) wrote:
>>My Ph.D. advisor and I are trying to find the answer to the
following
>>question, wholly unrelated to our field (pharmacology):
>>
>>"Why do drag racers use wide rear tires, if the coefficient of
friction
>> is independant of surface area?"
>>
load is
>transferred to the rear tires and causes them to flatten out thus
creating a
>larger area for friction to occur. The applied force is not entirely
static.
>Mathematically it would take me a while to figure it out but trust
me, wider
>tires do promote better traction.
>
>
You can show mathematically that a drag racer should not be able to
accelerate at more than 1g; above 1g the tires have no traction. The
reason they CAN accelerate at more than 1 g is that their tires
shred in the process; wider tires provide more shredding.
Kevin Anthony Scaldeferri
In article <31f13...@news.sisna.com>, jc <ix...@sisna.com> wrote:
>
> You can show mathematically that a drag racer should not be able to
>accelerate at more than 1g; above 1g the tires have no traction. The
>reason they CAN accelerate at more than 1 g is that their tires
>shred in the process; wider tires provide more shredding.
Please enlighten us by showing mathematically why a drag racer
shouldn't be able to accelerate at more than 1g. This I'd like to
see.
Kevin Scaldeferri
Jim Carr
ix...@sisna.com (jc) writes:
>
> You can show mathematically that a drag racer should not be able to
>accelerate at more than 1g; above 1g the tires have no traction.
What does math have to do with it?
Physics says the force is mu*N, and mu can be anything, certainly
it can be bigger than one. Just go out and measure it. Usually
a nice fresh eraser, what the brits call a rubber, will do it.
>The
>reason they CAN accelerate at more than 1 g is that their tires
>shred in the process; wider tires provide more shredding.
Interesting hypothesis. You thing you can get 2 more g that way?
I suggest you get up close and personal with a racing tire some
day and see just how sticky those rubber compounds are.
Spencer Stevens
Bruce Augenstein <bruce.au...@mro.mts.dec.com> writes
>
>Side note: A tire, any tire, *must* slip in order to generate traction.
>As the tire begins to slip, the coefficient of friction begins to *rise*.
>It peaks at some level of slip, and then begins to fall off, more and
>more rapidly as slip increases. Modern radials tend to generate peak
>traction at around eight to ten percent slip, and begin to fall off
>rapidly after that, which is why they're so hard to work with at the drag
>strip. Drag racers say that radials don't "recover", meaning once you've
>broken them loose, they'll stay broken loose.
>
>Slicks, on the other hand, have a more gradual curve, peaking at
>something above 35% slip, and tapering off much more gradually after
>that, which is why they have a fair amount traction even when broken
>loose, and thus recover well when you feather the throttle.
>
On reading your post, I think your point comes across slightly
confused: you talk about radials and then write 'slicks, on the
other hand...' which gives the impression slicks and radials are
somehow opposites. I'm sure this wasn't your intention! Of course
radial tyres refer to tyre construction not to treads (for the
benefit of others reading this post).
I'm not so clued up on drag racing so I won't try to make any
statements about grip and slip on acceleration, but certainly in
cornering slick tyres grip well to a given amount of slip (cornering
slip being slightly different and expressed in terms of angle), but
that drops off very suddenly which is hard to control. Treaded
tyres drop off much more suddenly thanks to the additional give
due to block movement (of course the total available grip is less
as well). This is why it is more important for drivers with
slick, grippy tyres to drive smoothly than those with treaded
tyres.
--
Spencer Stevens
jc
In article <4smaor$n...@ds8.scri.fsu.edu>,
j...@ds8.scri.fsu.edu (Jim Carr) wrote:
>we...@skipper.balltown.cma.com (richard welty) writes:
>>
> Hey, Richard, nice to see you in a physics group.
>>
>
>
corner at
>> more than 1.0G. Is this true?
>>
>>A: the equations for friction used in freshman physics textbooks
presume
>> that the surfaces are smooth, dry and non-deformable, none of
which
>> are the case with tires & pavement.
>
> teachers, not textbooks. For example, Halliday and Resnick states
> very clearly (over more than a page) that friction equations are
> approximations (for the reasons mentioned above, and others) that
> replace a complex phenomenon with an empirical constant.
>
> H&R also say explicitly that the coefficient can be bigger than 1.
> I know one book that includes a table of kinetic coefficients, one
> of which (rubber on dry "cement") is bigger than 1.
>
> Thus the student might be advised to read more than just the final
> equation used for some idealized cases. After all, a more careful
> analysis such as the one you cite still uses mu*N, but makes mu
> depend on some key variables so it models a real tire. In addition,
> note that the CRC Handbook lists several materials, one of them
being
> rubber, that has a coefficient larger than 1.
>
> The question above is an urban legend spread by profs of physics and
> engineering who have not read the book they are teaching out of.
>
What is true is that a wheel provides as much lift as it does
forward motion, in terms of g's. That holds even after you adjust for
the friction of the tires. That means that, theoretically, wheels are
unable to accelerate at more than 1g since 1g completely reduces
traction to zero, regardless of the friction of the tires. This can
be checked by working out the g's involved in the better E.T.'s; no
substance in the tires has the coefficient of friction required to
achieve them.
However, autos can accelerate faster than 1 g
and this is accomplished with soft tire formulations; the tires have
to shred for this to work. This also explains the use of wide tires -
more shredding. Any other explanation involving only friction
increases the lifting force of the rotating wheel, reducing traction,
and therefore is not an adequate explanation.
Kevin Anthony Scaldeferri
In article <31f27...@news.sisna.com>, jc <ix...@sisna.com> wrote:
>
> What is true is that a wheel provides as much lift as it does
>forward motion, in terms of g's. That holds even after you adjust for
>the friction of the tires. That means that, theoretically, wheels are
>unable to accelerate at more than 1g since 1g completely reduces
>traction to zero, regardless of the friction of the tires. This can
>be checked by working out the g's involved in the better E.T.'s; no
>substance in the tires has the coefficient of friction required to
>achieve them.
This is utter bollocks. The normal force on a tire is, to a first
approximation, always equal to the weight of the car. It can only be
altered by aerodynamic forces on the car. But, most cars, and race
cars in particular, generate downward aerodynamic forces. There is no
theoretical limit on the coefficient of friction. Substances with
CoF's greater than 1 are easily found in the CRC, and include most
rubbers.
On a side note, downforces can allow a car to pull more g's than the
CoF of the tires. But, the CoF can certainly be over 1 to begin with.
Kevin Scaldeferri
Eric Sillman
Laura Baehner wrote:
>
> Makes sense to me... But I don't think it's as technical as the guy's
> looking for....
Laura, the mail I've read so far is focused on the first 60 feet ot so of
a drag race. I can assure you that what happens during the remainder of
the 1320 feet has some bearing on ET.
There are many variables that determine what size tire is needed. It is
not always true that wider tires give better ETs. For example, the
larger tires produce more rolling drag than smaller tires so having big
tires when you don't need them can make your car go slower. Basically,
you want to transfer as much force to the tires as you can with as little
slip as possible to get the most forward force. This should provide the
greatest acceleration. As the car begins to move, the size of the
contact patch shrinks, but less force is required per sq in because in
inertia forces are being overcome. Somewhere down the track, the rolling
resistance increases as tire rpm increases and that starts to consume
more and more energy.
Bottom line is that I try an run the smallist tire I can get away with in
width and as tall a tire as possible (tall being diameter) as long as
wheelspin at the start is not excessive ( one or two turns max). This
gives me the best reaction and initial acceleration without having to
drag a huge tire the full length of the track.
With a properly set up car, you can turn 10 sec runs on 9" wide tires.
And thats in a 3000 pound full steel body. Check out the stock
eliminator records. They all run 9" tires. Oh yes, I have a 67 Plymouth
Barracuda with a 340 that I drag race here in Arizona. It is street
legal and turns high 10 sec. runs on pump gas. I use 10" wide 30" tall
slicks with about 6.5 PSIG of pressure in them. Hope this helps.
Eric Sillman (sil...@azstarnet.com)
Chuck Tomlinson
jc said...
>
> What is true is that a wheel provides as much lift as it does
>forward motion, in terms of g's. That holds even after you adjust for
>the friction of the tires. That means that, theoretically, wheels are
>unable to accelerate at more than 1g since 1g completely reduces
>traction to zero, regardless of the friction of the tires. This can
>be checked by working out the g's involved in the better E.T.'s; no
>substance in the tires has the coefficient of friction required to
>achieve them.
Wherever you got this theory about lift (!!??), it has no basis in
reality. Try to find someone who has a firm grasp of basic physics,
and have them show you why your "lift theory" is um... unfounded.
--
Chuck Tomlinson
Joe Wiles
>This is utter bollocks. The normal force on a tire is, to a first
>approximation, always equal to the weight of the car. It can only be
>altered by aerodynamic forces on the car. But, most cars, and race
Don't forget that the load on the tire is altered under acceleration because
of the old proof that a body at rest tends to stay at rest unlest a force acts
upon it. Think of the body and the engine and such as components of a moment
arm acting upon the rear tire.
WILDMAN
jc
In article <4sutb0$7...@dfw-ixnews9.ix.netcom.com>,
ground. Try pushing on the ground with your legs at 1.1g and tell me
how much traction you (momentarily) have, regardless of the soles of
your shoes.
Fred Farzanegan P185
In article <4srivf$p...@mulligan.eng.umd.edu>, cool...@Glue.umd.edu (Kevin Anthony Scaldeferri) writes:
|> In article <31f13...@news.sisna.com>, jc <ix...@sisna.com> wrote:
|> >
|> > You can show mathematically that a drag racer should not be able to
|> >accelerate at more than 1g; above 1g the tires have no traction. The
|> >reason they CAN accelerate at more than 1 g is that their tires
|> >shred in the process; wider tires provide more shredding.
|>
|> Please enlighten us by showing mathematically why a drag racer
|> shouldn't be able to accelerate at more than 1g. This I'd like to
|> see.
Coefficient of friction: force needed to move the object laterally/gravity
Using the most simple assumptions, the downward force on a tire is
equal to gravity * mass. Traction is the coefficient of friction over the
surface area * gravity. Simplistically, the coefficient of friction
is 1 or less, therefore, you shouldn't be able to accellerate faster than
gravity. The coefficient of friction is the 'normal' friction between
two objects. It is derived from the force needed to move an object
over a surface. A coefficient of 1 means that the force needed to
move the object is equal to gravity. In theory, this means that it doesn't
matter how wide your tires are, because wider tires mean more surface
area, but the mass of the vehicle is distributed over a greater area,
giving less force/unit.
However, lets look at an example of a block of wood sitting on a piece
of glass. We measure how much force it takes to move the block laterally
and find that its less than 1. Now we take that block of wood and glue
it to the glass. The force needed to overcome the friction between the
block and glass is >>1. This is what happens with rubber on asphalt.
The rubber deforms and grips greater than 'normal'. Hot sticky rubber
is like the glue in the above example. More glue can be obtained by
a wider tire...
Now a *real* question is 'can a single-track vehicle (motorcycle)
corner at >1g'.
-fred
(who can also prove that a bullet can't penetrate a piece of paper:
measure the distance between bullet and paper every time the distance
between them is halved: 10 ft, 5 ft, 2.5 ft, 1.25 ft,..... 0.0000000001 ft, etc.)
Jim Carr
fr...@bnr.ca (Fred Farzanegan P185) writes:
>
> .... Simplistically, the coefficient of friction is 1 or less, ...
For wood, maybe, but not for rubber.
Try an experiment with your drag slicks and report back to us. The
Physics of Racing pages describe an easy one.
>However, lets look at an example of a block of wood sitting on a piece
>of glass. We measure how much force it takes to move the block laterally
>and find that its less than 1. Now we take that block of wood and glue
>it to the glass. The force needed to overcome the friction between the
>block and glass is >>1. This is what happens with rubber on asphalt.
Good point, since it also illustrates the tradeoff between traction and
drag. Your wheel glued to the table will not roll very well after it
initially accelerates. Force is required to 'shred' it so it can roll
forward, and that retards the vehicle. The small % of slip at optimal
grip must help with this, but even at zero slip mu is > 1.
Bruce Augenstein
Spencer Stevens wrote:
>
> Bruce Augenstein <bruce.au...@mro.mts.dec.com> writes
cuts......
> >Slicks, on the other hand, have a more gradual curve, peaking at
> >something above 35% slip, and tapering off much more gradually after
> >that, which is why they have a fair amount traction even when broken
> >loose, and thus recover well when you feather the throttle.
> >
>
> On reading your post, I think your point comes across slightly
> confused: you talk about radials and then write 'slicks, on the
> other hand...' which gives the impression slicks and radials are
> somehow opposites. I'm sure this wasn't your intention! Of course
> radial tyres refer to tyre construction not to treads (for the
> benefit of others reading this post).
Good point, but I kinda meant what I said......sort of. :-)
I should have said drag slicks. Drag slicks are the amoebas of the tire
world, with soft, compliant sidewalls *and* tread infrastructure. You can
drop the pressure all the way down to 'bout here, and the contact patch
doesn't distort worth a damn. Race slicks such as those for Indycars and
Formula 1 aren't anything like drag slicks.
B.F. Goodrich announced a radial drag slick recently, mainly for street
cars, and I believe it's the first.
Bruce
nko...@rosenet.net
Hi all, I've just joined here (sci.physics), so forgive me if this answer has
been presented before.
> You can show mathematically that a drag racer should not be able to
> accelerate at more than 1g; above 1g the tires have no traction. The
> reason they CAN accelerate at more than 1 g is that their tires
> shred in the process; wider tires provide more shredding.
First off, the coefficient of friction is not limited to 1.0. We have some
measuring spoons with plastic-coated handles, and when these are placed on a
clean glass surface, the coefficient of friction is at least 100.
Ordinarily the coefficient of friction is independent of the pressure of
contact, and this is why we can have a such thing as a "coefficient of
friction" in the first place.
In the case of drag tires, the surface becomes heated, and this makes the
surface sort of gummy or sticky like hot-melt glue. If the tire was narrower,
the increased shear force (per unit area) would simply remove the sticky
surface too soon. If the tire were wider there would be insufficient heating
to produce the sticky surface.
Neil.
Jim Carr
toml...@ix.netcom.com (Chuck Tomlinson) writes:
>
>Regarding sidewall stiffness, the sidewalls on Goodyear EMTs are about the
>stiffest around.
Stiffer than my BFG R1's? They come with a warning to check inflation
because you can't tell one is "flat" until it comes off the rim. Which
brings up a question I have wondered about. Drag tires are run at low
inflation pressure. What keeps the bead seated under such extreme loads?
Jeffrey Bakal
i
On 20 Jul 1996, Jim Carr wrote:
> ix...@sisna.com (jc) writes:
> >
> > You can show mathematically that a drag racer should not be able to
> >accelerate at more than 1g; above 1g the tires have no traction.
>
> What does math have to do with it?
>
> Physics says the force is mu*N, and mu can be anything, certainly
> it can be bigger than one. Just go out and measure it. Usually
> a nice fresh eraser, what the brits call a rubber, will do it.
>
> >The
> >reason they CAN accelerate at more than 1 g is that their tires
> >shred in the process; wider tires provide more shredding.
>
> Interesting hypothesis. You thing you can get 2 more g that way?
> I suggest you get up close and personal with a racing tire some
> day and see just how sticky those rubber compounds are.
>
The average F1 car corners at over 4.5g.. so there
Jim Carr
Is it quite that high? I thought it was more like 3.5 g but they
do brake at 4 g. I would love to see a g-analyst readout.
Anyway, they get some of that from downforce increasing the normal force
rather than just the tires. There was a time when the downforce in F1
was more than enough for them to have N = mg while driving on the ceiling.
I don't think they can come close to the 3+ g that T/F dragsters get from
a standing start where only the rubber matters.
Jim Carr
In article <31f27...@news.sisna.com> ix...@sisna.com (jc) writes:
>
> What is true is that a wheel provides as much lift as it does
>forward motion, in terms of g's. That holds even after you adjust for
>the friction of the tires.
Lift? There is no aerodynamic lift on the tires at launch. Also
no aerodynamic downforce. Just the weight of the car and various
dynamical effects related to lash and tire distortion when you
apply 6000+ ft-lbs of torgue to the drive train.
You must be confusing something with the effect of that torgue on
lifting the front end, but that helps traction, and is controlled
by design and weight distribution.
>That means that, theoretically, wheels are
>unable to accelerate at more than 1g since 1g completely reduces
>traction to zero, regardless of the friction of the tires.
You can prove otherwise by measuring the coefficient of friction
for some sticky rubber, like a drag tire or some other racing tire
not designed for longterm road use. (My autocross tires claim an
050 treadwear rating, but are really only good for an hour or so of
'proper' use. More agressive tires are almost tacky to the touch.)
>This can
>be checked by working out the g's involved in the better E.T.'s; no
>substance in the tires has the coefficient of friction required to
>achieve them.
Rubber has a static coefficient of as much as 4. As someone else
noted, "traction" effects actually increase the friction coefficient
with small slip angles before it falls off to the sliding value.
> However, autos can accelerate faster than 1 g
>and this is accomplished with soft tire formulations; the tires have
>to shred for this to work.
Sticky. And they have to slip a little bit. Shred is not a word
I like to use in this context. (Flatspots are a bad thing.) You
leave a little bit of rubber behind, yes, but so does a street tire.
>This also explains the use of wide tires - more shredding.
Better to say optimal slip.
More shredding was the way they did in the good old days, when the
idea of a 4.59 ET was "impossible". Lots of smoke, signifying nothing.
Mark Preston
physics. See my program is the compuserve racing library called
Dynamic Response which simulates a racing car going round a
racing track. We can simulate laps times to within 2% with simple
coefficient of friction algorithms!!
Mark Preston
>> PRESSPLEY
Kevin Anthony Scaldeferri
In article <4svsc1$5...@kocrsv08.delcoelect.com>,
Aack, mea culpa. Of course the normal force on the rears can be
increased via weight transfer during acceleration. What I meant is
that the total normal force (all four tires) is constant. And since
the allegation to which I was replying was that the normal force was
_reduced_ under acceleration, it doesn't really change the fact that
whoever said that was full of it.
Kevin Scaldeferri
Kevin Anthony Scaldeferri
In article <31f38...@news.sisna.com>, jc <ix...@sisna.com> wrote:
> Wheels transfer rotational force to forward force by pushing on the
>ground. Try pushing on the ground with your legs at 1.1g and tell me
>how much traction you (momentarily) have, regardless of the soles of
>your shoes.
As mentioned in another thread, it's important to think about these
things in terms of vectors. Certainly a 1.1g push in an _upward_
direction would seperate me from the ground and reduce traction to
zero while I wasn't in contact with the ground. However, were I
wearing my rock climbing shoes, I wouldn't have much trouble with a
1.1g sideways acceleration.
Kevin Scaldeferri
Spencer Stevens
Jim Carr <j...@ds8.scri.fsu.edu> writes
>Jeffrey Bakal <jba...@acs.ucalgary.ca> writes:
>>
>>The average F1 car corners at over 4.5g.. so there
>
>Is it quite that high? I thought it was more like 3.5 g but they
>do brake at 4 g. I would love to see a g-analyst readout.
>
A typical F1 car corners at around 3.5g maximum now. This is at
high speed with high downforce. Without downforce you could expect
an F1 car to corner in the 2.0g region. Under braking the balance
is easier to maintain and the tyres grip in a slightly different mode
so slightly higher g forces are possible, around 4g as you say at
high speed.
Back in the 'good ol' days' of full ground effect and sliding
skirts (many years ago...) F1 cars could corner at approaching
5g when the downforce came into good effect. If these things were
still allowed and developed to the state of current F1 technology
the drivers may well have been the limiting factor of the cars
cornering ability...
--
Spencer Stevens
Chuck Tomlinson
Spencer Stevens said...
>
>A typical F1 car corners at around 3.5g maximum now. This is at
>high speed with high downforce. Without downforce you could expect
>an F1 car to corner in the 2.0g region. Under braking the balance
>is easier to maintain and the tyres grip in a slightly different mode
>so slightly higher g forces are possible, around 4g as you say at
>high speed.
I believe F1 braking accel is higher in magnitude than cornering accel
because the high speed aero drag forces on an F1 car are high enough
to generate close to 1g decel near top speed, without any assistance
from the brakes or engine. During braking, the aero drag forces are
superimposed on the braking forces.
--
Chuck Tomlinson
funkraum
>sc...@midway.uchicago.edu (Samuel C. Blackman) wrote:
>My Ph.D. advisor and I are trying to find the answer to the following
>question, wholly unrelated to our field (pharmacology):
>
>"Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
>We got to discussing this question, but were not able to come up with a
>scientific explanation - intuitively, it seems that wider rear tires would
>increase traction, but that's not a function of friction, is it? If a
>narrow tire were used, the force vector perpendicular to the ground would
>be increased because point of contact was smaller (higher pounds per square
>inch).
[...]
As I understand it, the force is not transmitted through friction but
through grip.
Drag cars are able to pull an instantaneous 8g off the line and 6g
shortly thereafter (probably more nowadays). It would not be possible
to pull more than one g if 'friction' were the sole method of
transmission of force.
The friction model takes place in a 'noiseless' model environment with
perfectly flat surfaces, etc. Grip is the result of the rubber
deforming around the asperities and interstices of the asphalt which,
I imagine, are tricky to model.
Brent Mazur
On 22 Jul 1996, Jim Carr wrote:
> I don't think they can come close to the 3+ g that T/F dragsters get from
> a standing start where only the rubber matters.
Initially, T/F dragsters pull over 5g. It's not for long, but it happens.
_ _
|_)(_`|\/| Brent Mazur | America borders on the magnificent...
|_)._)| | Winnipeg, Manitoba | CANADA!!!
~RW~
Chuck Tomlinson
Jim Carr said...
>
>toml...@ix.netcom.com (Chuck Tomlinson) writes:
>>
>>Regarding sidewall stiffness, the sidewalls on Goodyear EMTs are about the
>>stiffest around.
>
>Stiffer than my BFG R1's? They come with a warning to check inflation
>because you can't tell one is "flat" until it comes off the rim. Which
>brings up a question I have wondered about. Drag tires are run at low
>inflation pressure. What keeps the bead seated under such extreme loads?
I don't know how stiff BFG R1 sidewalls are, but Goodyear GS-C EMTs
(Extended Mobility Tires a.k.a "run-flats") will let you drive 200 miles
(nominally) with the tires at 0.0 psig. They are available as options on
the Corvette, but you also have to order the expensive Low Tire Pressure
warning system.
I saw a Goodyear promo video that showed a Vette being driven very
aggressively. All four EMTs on the Vette had 2" holes cut in the center of
the tire tread. It was not obvious at a glance that the tires were totally
deflated, and they stayed on the rims :-)
FWIW, I believe the vertical spring rate for EMTs at normal inflation
pressures is on the order of 450-500 N/mm. That's about twice as stiff as a
normal car tire, and at least 50% stiffer than the normal GS-Cs.
I don't know exactly how drag slicks are prevented from rotating relative to
the wheel. I've seen drag car wheels with screw heads spaced around the
wheel. I'm guessing that the screws "bite" into the bead to resist
slippage. I've also seen ads for wheels with "bead locks", but I don't know
how they work. Hopefully, a drag racer can help here.
--
__
___| |____ Chuck Tomlinson <toml...@ix.netcom.com>
/___LT-1___/ Mouse Power!
|__| '94 Vette Z07/ZF6, '89 Mustang LX 5.0/T-5
Kevin Anthony Scaldeferri
In article <4sv2bn$9...@brtph500.bnr.ca>,
Fred Farzanegan P185 <fr...@bnr.ca> wrote:
>
>In article <4srivf$p...@mulligan.eng.umd.edu>, cool...@Glue.umd.edu (Kevin Anthony Scaldeferri) writes:
>|> Please enlighten us by showing mathematically why a drag racer
>|> shouldn't be able to accelerate at more than 1g. This I'd like to
>|> see.
>Coefficient of friction: force needed to move the object laterally/gravity
>
>Using the most simple assumptions, the downward force on a tire is
>equal to gravity * mass. Traction is the coefficient of friction over the
>surface area * gravity. Simplistically, the coefficient of friction
>is 1 or less, therefore, you shouldn't be able to accellerate faster than
>gravity.
Faulty assumption. CoF can certainly be greater than 1.
<snip>
>
>Now a *real* question is 'can a single-track vehicle (motorcycle)
>corner at >1g'.
>
Hmm...this is a more difficult question because you have to consider
things like ground clearance for the foot pegs and the effects of the
rider leaning into the curve. Assuming the pegs don't hit the ground,
it should be possible. However, I don't have any information on
maximum cornering of real live motorcycles availible to me.
Kevin Scaldeferri
Jim Carr
I noticed that the newsgroup line on this article had been trimmed
before I posted to sci.physics. Followups are set to all groups.
ix...@sisna.com (jc) writes:
>
> 1.1 to 1.2 g's are entirely possible,
They are common place. What is rare is to have the powerplant required
to achieve that acceleration in the forward direction.
You can walk into an automobile showroom and purchase a number of cars
that will corner at over 0.9 and brake at close to 1.0 on OEM street
tires designed to last more than 15,000 miles on the highway. If you
put street legal racing tires on that same car, you can brake and turn
in the 1.2 g territory, but the tires may only last an hour or so. You
can do even better (more traction, less lifetime) with racing slicks.
Why? Because if you measure the friction coefficient you will find
that it is bigger than 1, perhaps as much as 1.6 at the optimum slip
angle. Why? Because they are soft and sticky. We autocross on old
asphalt, so it is not unusual to find pea gravel stuck to (or in) tires.
However, the car is unlikely to have the power to use all of that traction
to launch it in the forward direction. Even if you can afford an F50
you will not have the power to accelerate that much.
>but you won't get anywhere
>near the E.T.'s which drag racers achieve with them. As I said,
>friction alone is not going to cut it.
Friction (all of its elements, including adhesion and mechanical grip)
only serves to put power to the pavement. It will not compensate for
a small engine or poor brakes. If you do have the power, say 6000+ HP,
it is friction that transfers power from the motor to the road.
msjo...@ks.symbios.com
>Anyway, they get some of that from downforce increasing the normal force
>rather than just the tires. There was a time when the downforce in F1
>was more than enough for them to have N = mg while driving on the ceiling.
>I don't think they can come close to the 3+ g that T/F dragsters get from
>a standing start where only the rubber matters.
Your numbers are a mite old...T/F cars pull about 5 G off the start line and
are still running about 2.5 in the top end. Average acceleration is now in the
3+ G range for the entire pass (312 mph in 4.59 seconds comes out to just a
bit more than 100 ft/sec/sec).
Another aspect not mentioned in this thread to any great extent is that the
larger contact patch of wide, low pressure tires reduces the shearing force
applied on an areal basis.
Jim Carr
msjo...@KS.Symbios.COM writes:
>
>Your numbers are a mite old...T/F cars pull about 5 G off the start line and
>are still running about 2.5 in the top end. Average acceleration is now in the
>3+ G range for the entire pass (312 mph in 4.59 seconds comes out to just a
>bit more than 100 ft/sec/sec).
Thanks for the details. I have never seen a g-analyst type graph for T/F
except on one of the tech bits during TV coverage. I should have qualified
my answer as an average acceleration. If you compare E/T and speed you
know that a is not constant, and that aerodynamic drag causes the
acceleration to fall off at the end ... explaining why funny cars are
almost in the same top speed territory while having longer times. They
are probably cleaner than an open wheel dragster.
>Another aspect not mentioned in this thread to any great extent is that the
>larger contact patch of wide, low pressure tires reduces the shearing force
>applied on an areal basis.
It was earlier, but that says it about as succinctly as possible. That
should be the way it is put in any FAQ. Short and sweet.
Dan Dunhem
Why is it that you both talk of "centrifugal force?" There is no such
force. It is a catch phrase for the normal force component vector of
the force that is being exerted on an object at any instant in time.
In other words, spin the bucket around, let go, and you can observe
the bucket fly in the direction of the tangental force component
vector. The force that is perpendicular to this in three space is the
normal force vector, which you refer to as the "centrifugal force."
C'mon guys, these are simple freshman physics concepts that we're
talking about.
Dan
j...@ds8.scri.fsu.edu (Jim Carr) wrote:
>Original only posted to sci.physics; followups set to sci.physics.
>At launch, the centrifugal force on a drag tire increases its diameter.
>This increases the final drive ratio, raises the rear of the car (is
>this what the poster meant by "lift"?), which increases momentarily
>the normal force on the rear tires. Probably also helps counter the
>tendency to do a wheelstand, as does flex in the frame.
>>That force is acceleration toward the center of the tire.
>A force is not an acceleration.
>A centrifugal force is directed outward.
>> This isn't that complicated for anyone who would stop for a
>>minute and think about spin and centrigual force
>I suggest you think for a few more minutes.
>> So thanks to those of you who tried to think about this
>>problem sensibly without diatribe.
>You ain't seen diatribe, kid.
>--
> James A. Carr <j...@scri.fsu.edu> | In its first 24 hours MSNBC "news"
> http://www.scri.fsu.edu/~jac/ | misrepresented the hardware and O/S
> Supercomputer Computations Res. Inst. | used in Independence Day to promote
> Florida State, Tallahassee FL 32306 | a MS product. Gates uber alles!
GO HOKIES!!!
Sugarbowl Champions!
Dan Dunhem
Huh?@vt.edu
http://sfbox.vt.edu:10021/D/ddunhem/
http://ddunhem.async.vt.edu/
Virginia Tech
Department of Chemical Engineering
We're not in hell... but we can see it from here!
msjo...@ks.symbios.com
In article <FHp8cPA3...@upthorpe.demon.co.uk> Oz <O...@upthorpe.demon.co.uk> writes:
>>Initially, T/F dragsters pull over 5g. It's not for long, but it happens.
>Must be kinda exciting the first few times.
>Hmmm, and the stopping too, I would imagine.
Yep...in fact, the reason "Big Daddy" Don Garlits finally had to retire the
last time was because he suffered detached retinas while experimenting with a
new parachute rig that delivered -7 G when deployed. His doctor told him to
quit that stuff or he'd go blind. [Gee, sounds like your mother :-)]
+5 to -5 G within the space of 5 seconds...hell of a shock. The other fun
comment from racer and racing school owner Frank Hawley is that most everybody
drives the last third of the track with their hand on the parachute lever -
ever thought of driving 300 mph one-handed? Hawley figures it this way - if
your control of the vehicle is such that you need both hands in the lights,
you ought to have the chutes out anyway.
Oz
In article <Pine.SOL.3.91.960723162542.25613C-
100...@winnie.freenet.mb.ca>, Brent Mazur <axq...@freenet.mb.ca> writes
>On 22 Jul 1996, Jim Carr wrote:
>
>> a standing start where only the rubber matters.
>
Must be kinda exciting the first few times.
Hmmm, and the stopping too, I would imagine.
-------------------------------
'Oz "When I knew little, all was certain. The more I learnt,
the less sure I was. Is this the uncertainty principle?"
David Wilkie
Fred Farzanegan P185 wrote:
> |> >Now a *real* question is 'can a single-track vehicle (motorcycle)
> |> >corner at >1g'.
>
> a single-tracked vehicle depends on gravity to accellerate it into the
> direction of lean. The max accelleration could only be 1g.
> Footpegs, clearance, traction, are irrelevant.
>
Fred,
Your statement is wrong. A motorcycle going round a corner needs
to lean into the corner to balance centripetal and gravitational
forces to prevent the bike flipping outward. Your are forgetting
that the friction of the tyre on the road provides the
centripetal force, and leaning simply balances this with the
gravitational force. It is perfectly possible for an idealized
planar motorcycle of unit mass to corner at >1g:
(I used tabs in my drawing; I hope your news reader interprets them
properly) Assume mass=1 (all motorcycle/rider combinations are like
this)
\
\
\
| \
| \ g
| \ /|\ No net vertical force
\|/ \ |
_________________g___a(\________________________________
f=v*v/R <-------
1) f=v*v/R A motorcycle travelling 50mph (22m/s) around
a 20m radius curve will have f=24m/s, 2.4g
The angle of lean a to balance all this is found from the relationship
Gcosa=fsina
tana=G/f
a=atan(GR/v^2)
In the above example a=22 degrees.
The limitation in this example is the frictional force that the
tyres can provide, which must be sufficient to prevent sliding.
I have no idea wether or not this example is possible. However,
as an avid motorcycle race viewer I can say that motorcycles
corner greater than 1g (a=45 degrees for a 1g turn.)
I apologise for not being able to explain it better without a
drawing.
David Wilkie
--
Lance Ferguson
>
> An interesting bit of info: Back a few years a top fuel dragster was
> heavily instrumented and careful high speed photography taken. Analysis
> of this data found that the car had actually reach 60mph BEFORE the
> rear wheels had made a complete revolution. This means that there can
> exist momentary acceleration peaks as high as 8 g's off the line.
>
> -Bob
I'm sure that eight G's is correct.
Something being missed by some in this thread is the enormous vertical
loading on T/F cars during launch. T/F Dragster chassis design ( a true
black art) yields a car that bows upward severly at launch. 18" of arch
would not suprise me.
It is true that on "burnouts" the tires do almost immediately begin to
grow in diameter, producing a lifting effect on the rear of the car that
is quite visible. HOWEVER, this effect is induced with water unnder the
contact patch as well as 'easing into' the clutch. During actual launch
the tires are 'shocked' by a downforce that mashes them damn near flat.
Just what is going on in the chassis as far as a moment of inertia etc.I'm
quite sure nobody really knows. Whatever it is produces thousands pounds
of downforce for a few tenths of a second.
I've talked to two men in the last few months who have seperately build
dozens of very successful T/F cars over a perion of decades. Both told me
they knew what worked but not why!
Another thing; This thread has promise of running longer than the infamous
"Torque" discussion. (If you don"t know about that one, don't ask.)
-Lance
--
Life is fatal...but not serious. O.Wilde
Jim Carr
>maximum cornering of real live motorcycles availible to me.
Doesn't cornering at more than a 45 degree angle offer a clue?
--
James A. Carr <j...@scri.fsu.edu> | "The half of knowledge is knowing
http://www.scri.fsu.edu/~jac/ | where to find knowledge" - Anon.
Supercomputer Computations Res. Inst. | Motto over the entrance to Dodd
Florida State, Tallahassee FL 32306 | Hall, former library at FSCW.
Jeffrey Bakal
On 22 Jul 1996, Jim Carr wrote:
> Jeffrey Bakal <jba...@acs.ucalgary.ca> writes:
> >
> >The average F1 car corners at over 4.5g.. so there
>
> Is it quite that high? I thought it was more like 3.5 g but they
> do brake at 4 g. I would love to see a g-analyst readout.
>
> Anyway, they get some of that from downforce increasing the normal force
> rather than just the tires. There was a time when the downforce in F1
> was more than enough for them to have N = mg while driving on the ceiling.
> I don't think they can come close to the 3+ g that T/F dragsters get from
> a standing start where only the rubber matters.
Well the point to end this belief that you couldn't accelerate in any
direction at more than 1g and further the question was posed of weather
you could hit 2g. so I couldn't stand it anymore anyway point is that as
anyone who has fallen off a motorcycle or bicycle or any moving thing
knows that pavement is not smooth and neither are slick tyres. therefore
highschool dry friction laws just don't cut it.
Jeff" there ain't no such thing as too much tyre" Bakal
Kevin Anthony Scaldeferri
In article <4t5diu$4...@brtph500.bnr.ca>,
>|> In article <4sv2bn$9...@brtph500.bnr.ca>,
>|> >
>|> >Now a *real* question is 'can a single-track vehicle (motorcycle)
>|> >corner at >1g'.
>|> >
>|>
>|> things like ground clearance for the foot pegs and the effects of the
>|> rider leaning into the curve. Assuming the pegs don't hit the ground,
>|> maximum cornering of real live motorcycles availible to me.
>
>a single-tracked vehicle depends on gravity to accellerate it into the
>direction of lean. The max accelleration could only be 1g.
>Footpegs, clearance, traction, are irrelevant.
>
I'm not sure I agree. It is more enlightening to state that a bike
depends on gravity to balance the 'centrifugal force' (yeah, yeah,
you know what I mean) which would like to highside you. However, at a
lean angle of more than 45 degrees, the sideways component of the
force should exceed the downward component. Whether you can achieve a
45 degree lean is dependent on footpeg clearance, etc, as well as the
CoF of your tires.
Kevin Scaldeferri
Bob Gilbert
Of course the tires used by many drag racers are designed to increase
substantially in circumfrence as they spin up, thus lifting the rear
of the car higher, and thus momentarily exerting additional normal
force at the rear.
Fred Farzanegan P185
In article <4t3f89$3...@latte.eng.umd.edu>, cool...@Glue.umd.edu (Kevin Anthony Scaldeferri) writes:
|> In article <4sv2bn$9...@brtph500.bnr.ca>,
|> Fred Farzanegan P185 <fr...@bnr.ca> wrote:
|> >
|> >In article <4srivf$p...@mulligan.eng.umd.edu>, cool...@Glue.umd.edu (Kevin Anthony Scaldeferri) writes:
|> >|> Please enlighten us by showing mathematically why a drag racer
|> >|> shouldn't be able to accelerate at more than 1g. This I'd like to
|> >|> see.
|> >Coefficient of friction: force needed to move the object laterally/gravity
|> >
|> >Using the most simple assumptions, the downward force on a tire is
|> >equal to gravity * mass. Traction is the coefficient of friction over the
|> >surface area * gravity. Simplistically, the coefficient of friction
|> >is 1 or less, therefore, you shouldn't be able to accellerate faster than
|> >gravity.
|>
|> Faulty assumption. CoF can certainly be greater than 1.
|>
|> <snip>
Thanks a lot for the snip, putz. You know, the part where I show the coef to be
greater than one?! Did you miss the part of 'most simple assumptions' in the
paragraph you did leave?
|> >Now a *real* question is 'can a single-track vehicle (motorcycle)
|> >corner at >1g'.
|> >
|>
|> Hmm...this is a more difficult question because you have to consider
|> things like ground clearance for the foot pegs and the effects of the
|> rider leaning into the curve. Assuming the pegs don't hit the ground,
|> it should be possible. However, I don't have any information on
|> maximum cornering of real live motorcycles availible to me.
Trick question. It cannot corner greater than 1g because cornering on
a single-tracked vehicle depends on gravity to accellerate it into the
direction of lean. The max accelleration could only be 1g.
Footpegs, clearance, traction, are irrelevant.
-fred
Leland Lewis
Just figured I'd throw another non-scientific point in here. Does anyone
remember what the g-force is of a fighter launching from the catapult
on an aircraft carrier?
Reason I ask is back when top fuel engines were around 3500HP Don
Garlits
out accelarated one in a short run.
Lee
Oz
In article <4t6k42$1...@ds8.scri.fsu.edu>, Jim Carr
<j...@ds8.scri.fsu.edu> writes>> ... However, I don't have any information on
>>maximum cornering of real live motorcycles availible to me.
>
>
Dunno about the USA, but in the UK most corners are slightly banked. I
guess this would help somewhat. :-)
Mark A. Friesel
Two (bad) things might happen when cornering a motorcycle: a tire might
slip, or the cycle might flip sideways. If the bike slips without
flipping it is because the squared instantaneous linear speed of the bike
times the bike's mass divided by the turn radius (to the CM) exceeds the
maximum frictional force between the tire and the roadway.
Since the contact patch of the tire is at rest wrt the roadway when there
is no slip, it's seductive to represent the maximum frictional force as
a coefficient of static friction times the normal force exerted by the
bike, i.e. its weight. In the case of a block on an inclined plane the
maximal acceleration of the block is g, but for some materials it's
necessary that the slope be great enough that the block fall off the
surface, and by using glue or rough surfaces the additional force
necessary to make the block slide off even a vertical slope can be many
times that exerted by gravity. Representing this 'frictional' force as
some f= kN therefore means that k can be much greater than 1 and there is
really no reason why a bike cannot corner at greater than 1 g from this
standpoint. It's a materials problem.
In the other case the clearance of the bike parts and roadway is
important since to corner without flipping, the torques about the contact
point of the tires must be zero (or cause net rotation towards the ground
which would eventually cause the footrest to contact the ground, and also
causes the vertical weight on the tires to decrease which can contribute
to slipping). The torgue exerted by gravity is mgRsinB where B is the lean
angle and R the distance from the center of mass of the bike to the
tire/road contact 'point'. The frictional force is centripetal and causes
the bike to turn without slipping, but it can't counter the torque about
the contact point which causes the bike to flip since the moment arm of
the frictional force is zero. This torque must equal m(v*v/r)RcosB where v
is the instantaneous speed of the bike and r the turn radius. To not flip
over (outwards) v*v/r <= gtanB, and as v gets higher for a given r, the
angle B must get closer to pi/2. If Bmax is pi/4 = 45 degress due to
clearance concerns, then the centripetal acceleration represented by v*v/r
must be less than or equal to g.
The same thing applies to cars, except that they can't lean very well and
the mass is distributed radially w/o leaning.
On 24 Jul 1996, Kevin Anthony Scaldeferri wrote:
> In article <4t5diu$4...@brtph500.bnr.ca>,
> Fred Farzanegan P185 <fr...@bnr.ca> wrote:
> >
> >In article <4t3f89$3...@latte.eng.umd.edu>, cool...@Glue.umd.edu (Kevin Anthony Scaldeferri) writes:
> >|> In article <4sv2bn$9...@brtph500.bnr.ca>,
> >|> Fred Farzanegan P185 <fr...@bnr.ca> wrote:
> >|> >
> >|> >Now a *real* question is 'can a single-track vehicle (motorcycle)
> >|> >corner at >1g'.
> >|> >
> >|>
> >|> Hmm...this is a more difficult question because you have to consider
> >|> things like ground clearance for the foot pegs and the effects of the
> >|> rider leaning into the curve. Assuming the pegs don't hit the ground,
> >|> it should be possible. However, I don't have any information on
> >|> maximum cornering of real live motorcycles availible to me.
> >
> >Trick question. It cannot corner greater than 1g because cornering on
> >a single-tracked vehicle depends on gravity to accellerate it into the
> >direction of lean. The max accelleration could only be 1g.
> >Footpegs, clearance, traction, are irrelevant.
> >
>
Anthony Potts
On 23 Jul 1996, Kevin Anthony Scaldeferri wrote:
> In article <4sv2bn$9...@brtph500.bnr.ca>,
> >
> >Now a *real* question is 'can a single-track vehicle (motorcycle)
> >corner at >1g'.
> >
>
> Hmm...this is a more difficult question because you have to consider
> things like ground clearance for the foot pegs and the effects of the
> rider leaning into the curve. Assuming the pegs don't hit the ground,
> it should be possible. However, I don't have any information on
> maximum cornering of real live motorcycles availible to me.
>
>
> Kevin Scaldeferri
>
>
>
Ha, motorcycles again. Back into my territory here. If a bike was going
to corner at 1g lateral acceleration, it would have to lean over to 45
degrees from the vertical. Any sportsbike can do this with ease. If I
look at my bike at rest, I can lean it to about 30 degrees from the
horizontal, which is a fair way past 1g equivalent. Of course, in motion
the suspension would be compressed, so the angle would be a bit less, but
I would also be hanging off the side of the bike, compensating slightly
for it.
So, 1g is actually pretty pedestrian in the sportsbike stakes, but I
can't tell you what the ultimate number is.
Cheers,
Anthony Potts
Anthony Potts
On Wed, 24 Jul 1996 msjo...@KS.Symbios.COM wrote:
> In article <FHp8cPA3...@upthorpe.demon.co.uk> Oz <O...@upthorpe.demon.co.uk> writes:
> >>Initially, T/F dragsters pull over 5g. It's not for long, but it happens.
>
> >Must be kinda exciting the first few times.
>
> >Hmmm, and the stopping too, I would imagine.
>
> Yep...in fact, the reason "Big Daddy" Don Garlits finally had to retire the
> last time was because he suffered detached retinas while experimenting with a
> new parachute rig that delivered -7 G when deployed. His doctor told him to
> quit that stuff or he'd go blind. [Gee, sounds like your mother :-)]
>
> +5 to -5 G within the space of 5 seconds...hell of a shock. The other fun
> comment from racer and racing school owner Frank Hawley is that most everybody
> drives the last third of the track with their hand on the parachute lever -
> ever thought of driving 300 mph one-handed? Hawley figures it this way - if
> your control of the vehicle is such that you need both hands in the lights,
> you ought to have the chutes out anyway.
>
>
Why do they need to stop so quickly? Is it just because the tracks don't
run on for long enough, or is it just in case something goes wrong, and
you want to pull up short?
Also, do today's drg cars have manual gears, or any gears at all? Do they
even have a throttle pedal, or is it just a big red button?
I can't really imagine changing gears manually whilst undergoing 5g, but
I suppose it is possible.
Cheers,
Anthony Potts
Anthony Potts
On Wed, 17 Jul 1996, Samuel C. Blackman wrote:
> My Ph.D. advisor and I are trying to find the answer to the following
> question, wholly unrelated to our field (pharmacology):
>
> "Why do drag racers use wide rear tires, if the coefficient of friction
> is independant of surface area?"
>
The answer is quite simple. When you skid, it is not thet tyre losing
contact with the road, but the rubber in the tyre breaking down under the
shear forces applied. A bigger tyre means that there is more rubber, and
so you have to push it harder to tear off a layer of it, so you don't
skid as easily.
Cheers,
Anthony Potts
msjo...@ks.symbios.com
>Why do they need to stop so quickly? Is it just because the tracks don't
>run on for long enough, or is it just in case something goes wrong, and
>you want to pull up short?
Most tracks used for fuel cars are around 3500 to 4500 feet in total length.
In addition to the quarter mile, another 150-200 feet is used up in staging
area, water box for burnouts (has to be well behind the line so that no water
gets carried on tires up to the launch pad), etc. That leaves 2500 to 3000 feet
to get stopped from 300-315 mph. Mostly, they try to get stopped by the second
turnoff point (with about 800-1000 feet left over) under "normal"
circumstances. The big-time tracks have sand pits, arresting gear, or some
such to stop runaways, which happen every now and again. Also, continuing to
accelerate past the finish line is a double whammy. I did some photo-analysis
of the video of a recent record setting run, in which Blaine Johnson drove a
good 150 feet beyond the finish line under full power. I figure he had to be
going about 330 mph by the time he got off the pedal.
>Also, do today's drg cars have manual gears, or any gears at all? Do they
>even have a throttle pedal, or is it just a big red button?
Depends on the class. The nitro-burners (Top Fuel and Funny Car) run a
pneumatically-controlled multistage slipping clutch that is actuated by timers
started when the pedal is floored - the initial stage is centrifugal. In
general, there will be five or six "steps" in the automatic clutch, with full
lock-up to direct drive occurring at about the 1000-foot mark (yep, only 300
feet from the finish line). They have a clutch pedal to allow them to
positively disengage during startup and while shifting to reverse after a
burnout. The supercharged methanol classes (Top Alcohol) typically have a
two-stage centrifugal clutch ahead of a three-speed planetary transmission.
Those guys generally shift manually, with two gear changes in about 5.6 to 5.9
seconds. Most use an air-shifter with a separate button for each shift. The
slower cars use conventional automatic or manual transmissions. Automatics are
favored because they are more consistent and put less strain on drivetrains at
the green light.
>I can't really imagine changing gears manually whilst undergoing 5g, but
>I suppose it is possible.
Sure. It isn't any hairier than reaching for the parachute lever while going
300 mph, leaving you driving at that speed one-handed. Fighter pilots have to
be able to manipulate controls while in 5 to 8 G turns all the time.
Chuck Tomlinson
On Thu, 25 Jul 1996 11:47:24 GMT, Anthony Potts wrote...
The answer is not nearly as simple as you think it is. At constant
inflation pressure, the contact patch area is pretty much independent of
the tire width (over a reasonable range), so wider tires do *not* have more
rubber in contact with the road.
And from your explanation, the path to increased traction would be to make
the tire out of a harder rubber compound that can support higher shear
stress without failure. Obviously, this does not agree with reality.
--
__
___| |____ Chuck Tomlinson <toml...@ix.netcom.com>
/___LT-1___/ Mouse Power!
|__| '94 Z07/ZF6, '89 LX5.0L/T5
Kevin Anthony Scaldeferri
In article <4t6k42$1...@ds8.scri.fsu.edu>,>> ... However, I don't have any information on
>>maximum cornering of real live motorcycles availible to me.
>
>
Yes, but I said that because I personally have never taken a corner
with that much lean. I've definately seen photos of races that seem
to have that much lean, but camera angles and such make it hard to
tell for sure. I'm betting that it can be done -- just not by me.
Kevin Scaldeferri
Mark A. Friesel
On 25 Jul 1996, Chuck Tomlinson wrote:
> On Thu, 25 Jul 1996 11:47:24 GMT, Anthony Potts wrote...
> >
> >On Wed, 17 Jul 1996, Samuel C. Blackman wrote:
> >
> >> "Why do drag racers use wide rear tires, if the coefficient of friction
> >> is independant of surface area?"
> >>
> >The answer is quite simple. When you skid, it is not thet tyre losing
> >contact with the road, but the rubber in the tyre breaking down under the
> >shear forces applied. A bigger tyre means that there is more rubber, and
> >so you have to push it harder to tear off a layer of it, so you don't
> >skid as easily.
>
> The answer is not nearly as simple as you think it is. At constant
> inflation pressure, the contact patch area is pretty much independent of
> the tire width (over a reasonable range), so wider tires do *not* have more
> rubber in contact with the road.
Since the tires only deform enough to support their own weight and the
weight of the car, it seems that the area of the contact patch should
be pretty independent of tire width, all else being equal. Certainly a
wider tire, under these conditions, will initially be rounder, so the
wider of two 'equal radius' tires will have a slightly larger effective
initial radius. If you want to keep your initial air pressure as low as
possible a wider tire may be preferrable especially since the tire
surface must distort circumferentially at least if it's going to flatten
out. This would mean that you're getting somewhat flakier contact over the
patch area than you'd get if you used a wider tire. The center region of
the wider tire will also tend to deform more at speed, bringing the
effective tire width closer to that of the narrower tire and slightly
increasing its effective radius. Interesting.... Wider tires may also
give a more stable ride.
>
> And from your explanation, the path to increased traction would be to make
> the tire out of a harder rubber compound that can support higher shear
> stress without failure. Obviously, this does not agree with reality.
Right - soft compounds increase the mechanical contact with the road
surface by allowing the tire surface to conform to all the little dips and
rises. Since the forces applied to the surface are tangential this can
help to increase traction, but it seems like a tough job to match
elasticity with strength. An iron tire (or a harder rubber one) will make
less contact with the road surface since it will tend to ride on the high
points and there will be reduced resistance to tangential forces. What
might be ideal is a tire which is vertically soft and circumferentially
stiffer and very tough - a very high shear strength - but this could
really shred the road surface. Another materials problem.
Bruce Augenstein
Leland Lewis wrote:
cuts......
> Just figured I'd throw another non-scientific point in here. Does anyone
> remember what the g-force is of a fighter launching from the catapult
> on an aircraft carrier?
> Reason I ask is back when top fuel engines were around 3500HP Don
> Garlits
> out accelarated one in a short run.
It depends on the aircraft, or course (i.e. - appropriate takeoff speed).
F-14s and F/A-18s pull about 4 Gs at launch, but they go to some trouble
with the catapult design to *not* instantly hit 4 Gs. Rather, they ease
into it, so to speak, for the first tenth second or two.
The current Road & Track has info on this, as a sidebar in their
"0-100-0" comparison.
My memory might be faulty, but I think that the Garlits race you
mentioned was a paper race only.
As an aside, one of my sons drives an F/A-18 for a living, and he says
the catapult shot feels like God's own shove :-). Not really brutal, but
patently irresistible.
Bruce
jc
In article <4t3f89$3...@latte.eng.umd.edu>,>>In article <4srivf$p...@mulligan.eng.umd.edu>, cool...@Glue.umd.edu
(Kevin Anthony Scaldeferri) writes:
>>surface area * gravity. Simplistically, the coefficient of
friction -snip-
>>is 1 or less, therefore, you shouldn't be able to accellerate faster
than
>>gravity.
>
>Faulty assumption. CoF can certainly be greater than 1.
>
>
CoF can be greater than 1, but it is unlikely to be high enough to
explain the g's which drag racers achieve. Try out your 1/2at^2=d,
and solve for a, and you'll see what I mean.
<snip>
>
>
>Kevin Scaldeferri
>
Leland Lewis
I have a magazine (somewhere in a box) that shows the car and plane side
by side
on the carrier but I don't remember if they are moving. Supposedly they
did
race though.
Lee
Mark T.
In article <4t8924$f...@sjx-ixn3.ix.netcom.com>,
Chuck Tomlinson <toml...@ix.netcom.com> wrote:
>On Thu, 25 Jul 1996 11:47:24 GMT, Anthony Potts wrote...
>>
>>On Wed, 17 Jul 1996, Samuel C. Blackman wrote:
>>
>>> "Why do drag racers use wide rear tires, if the coefficient of friction
>>> is independant of surface area?"
>>>
>>The answer is quite simple. When you skid, it is not thet tyre losing
>>contact with the road, but the rubber in the tyre breaking down under the
>>shear forces applied. A bigger tyre means that there is more rubber, and
>>so you have to push it harder to tear off a layer of it, so you don't
>>skid as easily.
>
>The answer is not nearly as simple as you think it is. At constant
>inflation pressure, the contact patch area is pretty much independent of
>the tire width (over a reasonable range), so wider tires do *not* have more
>rubber in contact with the road.
What kind of myth is this? Wider tires do not have more contact area?
This isnt right. Your contact patch grows wider as your tire grows wider
and grows longer as your tire gets taller, given the same pressure on your
slicks. The reasons for low tire pressure and soft sidwalls are to put
even more rubber on the ground during launch, the sidewalls wrinkle up and
the contact patch grows even longer. But I'm pretty sure a 10" wide slick
has 20% more contact patch then the same diameter 8" slick.
Take a trip to your local drag strip, and watch closely as a typical bracket
car launches. Its fun to watch how drag slicks react.
>
>And from your explanation, the path to increased traction would be to make
>the tire out of a harder rubber compound that can support higher shear
>stress without failure. Obviously, this does not agree with reality.
>
>--
> __
> ___| |____ Chuck Tomlinson <toml...@ix.netcom.com>
>/___LT-1___/ Mouse Power!
> |__| '94 Z07/ZF6, '89 LX5.0L/T5
>
Mark Tapper KJ7GK
ma...@halcyon.com
Anthony Potts
On Thu, 25 Jul 1996, Chuck Tomlinson wrote:
>
> The answer is not nearly as simple as you think it is. At constant
> inflation pressure, the contact patch area is pretty much independent of
> the tire width (over a reasonable range), so wider tires do *not* have more
> rubber in contact with the road.
>
ut you don't have constant inflation pressure. Fatter tyres allow you to
run lower inflation pressures. If you do this, you DO have more rubber on
the ground.
> And from your explanation, the path to increased traction would be to make
> the tire out of a harder rubber compound that can support higher shear
> stress without failure. Obviously, this does not agree with reality.
>
Not at all. There comes a point where the harder rubber slips on the
tarmac. It is a compromise between shredding the rubber and having it
soft enough to grip the tarmac. With this in mind, rather soft tyres, fat
enough not to shred is the way to go. This is borne out by watching the
actual tyre choice of drag racers and other racers (e.g. Formula 1).
Basically, fatter tyres are in general better, up to a limit.
Anthony Potts
CERN
Jeffrey Bakal
>
>
> > And from your explanation, the path to increased traction would be to make
> > the tire out of a harder rubber compound that can support higher shear
> > stress without failure. Obviously, this does not agree with reality.
> >
>
> Not at all. There comes a point where the harder rubber slips on the
> tarmac. It is a compromise between shredding the rubber and having it
> soft enough to grip the tarmac. With this in mind, rather soft tyres, fat
> enough not to shred is the way to go. This is borne out by watching the
> actual tyre choice of drag racers and other racers (e.g. Formula 1).
> Basically, fatter tyres are in general better, up to a limit.
to add my 2 cents as they say, softer tyres also go in between the "holes"
or cracks in the pavement providing increased surface area and a more
mechanical type grip, and provide more of a glue effect as was
mentioned earlier.
>
F1 as well as most other road and oval classes limit the tyre size and
you will generally see all of them running at the upper end of the limit.
these rules are generally in place to keep the costs and speed down on the
cars. though I don't know for sire but I would expect it may also be
limited by rules in drag racing. don't you think NASCAR would like a
little more tyre?
that is all.
Jeff
Brent Mazur
On Thu, 25 Jul 1996 msjo...@KS.Symbios.COM wrote:
> started when the pedal is floored - the initial stage is centrifugal. In
> general, there will be five or six "steps" in the automatic clutch, with full
> lock-up to direct drive occurring at about the 1000-foot mark (yep, only 300
> feet from the finish line). They have a clutch pedal to allow them to
I believe that the clutch is locked up before half track or 660 feet.
_ _
|_)(_`|\/| Brent Mazur | America borders on the magnificent...
|_)._)| | Winnipeg, Manitoba | CANADA!!!
~RW~
Kevin Anthony Scaldeferri
Why do you insist on imposing your preconcieved notion of what CoFs
can be achieved. While I am sure that increasingly higher CoF's
require more sophisticated/exotic materials, I see no reason for a
ceiling on it. Keep in mind that drag slicks are not the $30
Bridgestones you buy at Sears. Those things are expensive and
basically last about one day at the track.
People have pointed out that drag racers do indeed achieve about 4-5
g's max. But what's wrong with that? The fact that they are able to
do it is evidence that materials exist with CoF's in that neighborhood
since the definition of the CoF is the ratio of tangential to normal
force. Questions of weight transfer and downforces mean that you may
not need a CoF of 5, but maybe only 3 or so, which is perfectly
reasonable for a high performance rubber compound, go look at the CRC.
Kevin Scaldeferri
Carl Porter
Kevin Anthony Scaldeferri wrote:
>
> In article <4t6k42$1...@ds8.scri.fsu.edu>,
> Jim Carr <j...@ds8.scri.fsu.edu> wrote:
> >>maximum cornering of real live motorcycles availible to me.
> >
> >Doesn't cornering at more than a 45 degree angle offer a clue?
> >
>
> Yes, but I said that because I personally have never taken a corner
> with that much lean. I've definately seen photos of races that seem
> to have that much lean, but camera angles and such make it hard to
> tell for sure. I'm betting that it can be done -- just not by me.
>
> Kevin Scaldeferri
I believe the people at Motor Trend (possibly Car & Driver) got a
motorcycle jockey to try a skid pad not too long ago. If I remember right the
gentleman was limited to .75g. I think it was is an issue in which they got the
same guy to road race a ZR-1 Corvette, must of been Car & Driver. I'll see if I can
find it.
Carl Porter
Mike Kohlbrenner
Mark T. wrote:
>
> In article <4t8924$f...@sjx-ixn3.ix.netcom.com>,
> Chuck Tomlinson <toml...@ix.netcom.com> wrote:
> >inflation pressure, the contact patch area is pretty much independent of
> >the tire width (over a reasonable range), so wider tires do *not* have more
> >rubber in contact with the road.
>
> This isnt right. Your contact patch grows wider as your tire grows wider
> and grows longer as your tire gets taller, given the same pressure on your
> slicks. The reasons for low tire pressure and soft sidwalls are to put
> even more rubber on the ground during launch, the sidewalls wrinkle up and
> the contact patch grows even longer. But I'm pretty sure a 10" wide slick
> has 20% more contact patch then the same diameter 8" slick.
>
I have had this exact discussion with folks on the net. I would love
a definitive answer to this question -- especially if it shows my view
to be correct! ;-/ I'm pretty sure that while the contact patch of a
10" tire might be 25% WIDER than an 8" slick, it is somewhat shorter
in the direction perpendicular to that width such that the AREA of the
patch has not grown significantly. The SHAPE changes, but the AREA
does not change proportionately, as you are claiming.
Here's how I look at it...
Tires are flexible. They are filled with air pressure. When you rest
them on a surface and add some weight, they flatten out to some extent,
creating the familiar "contact patch". This much I hope no one disputes.
Their behavior in this respect should fall somewhere between the "soft-
sided balloon" and the "rigid-sided balloon".
A soft sided balloon is one where it cannot stretch easily, but can flex
easily. At rest, floating through the air, it is perfectly round due to
the equalization of the internal pressure with the tension in the skin.
When pushed against a surface, the required force balance creates a
"contact patch" whose AREA is mostly a function of the air pressure inside
and the force pushing against the surface.
A rigid sided balloon is one where it can neither stretch nor flex
easily. It would create a contact patch of AREA near zero, regardless of
the force applied, since the skin cannot flex -- it merely experiences
more bending stress to balance the applied force.
Of course both of these are idealized, but I hope everyone is still with
me.
Now, lets take the soft-sided balloon, and instead of the familiar round
shape, we get one of those oblong shapes, and set it down on its side and
apply the same force as before. The AREA of the contact patch will be
essentially the same as before, given the same inside pressure, correct?
The SHAPE of the contact patch, however will be very different.
This same comparison with differently shaped rigid sided balloons is
useless, correct?
So, let's go to tires. Tires do indeed have a significantly sized contact
patch AREA. The contact patch AREA of tires will indeed change with changes
in either loading or pressure. So, it's pretty clear that they are NOT
like the rigid sided balloons at all.
Are tires like soft-sided balloons? That is really what is meant when
someone claims that the contact patch is solely a function of the pressure.
Tires, however, have a pretty stiff carcass (some more than others...). So,
they really aren't EXACTLY like soft-sided balloons.
MY belief (which, of course, could be wrong...) is that tires are on the
"soft-sided balloon" side of the fence. This is easily supported merely by
looking at the changes in contact patch sizes as you pump up a flat tire.
Clearly, very stiff racing tires are further away from the purely soft
side of things, but I still BELIEVE that they are closer to that than the
rigid side.
As I said, I would LOVE to see some kind of REAL proof one way or the other
to put this to rest.
I am sure, however, that neither of the following are true:
* The contact patch AREA is solely a function of air pressure.
* The contact patch AREA is proportional to tire diameter and/or width.
It is somewhere in between. The question is: where?
Mike Kohlbrenner
<kohl...@an.hp.com>
Paul Resch
I remember that most measurements of CofF are pretty idealised. i.e.
They assume perfectly flat samples with no lubrication in either a
vacuum or at least quite standard test conditions.
Since the syrface of the track is definately rough (at least at the
microscopic scale) and the tire compound is soft, significant 'keying'
takes place and the CofF is less important that you'd think.
Consider this experiment.
Equip a dragster (on a dragstrip that will never be used again or that
needs repaving) with tires that had hardened steel surfaces and that
could 'bite' into the track. Even though the CofF would be low for
smooth steel on asphalt, it could probably move quite well.
I seem to remember that the sand drag racers 'paddle' tires that fling
sand to the rear and that really grab the sand are also capable of
pretty good acceleration, even though sand has a low CofF due to the
rolling effect of the loose grains. The Egyptians used sledges to some
extent after all to move heavy stones.
My point is that the dragstrip is a finely tuned relation of tire
'stick', track surface and allpication of huge amounts of torque, so
the normal textbook sort of calculations while useful, really do not
apply regarding limits.
This is similar to the often quoted statement that according to
aeronautical engineers, a bumblebee cannot fly, having insufficent
power to weight. Except that it does, and it does because all the
aeronautical calculations deal with rigid and not flexible and
constantly changing wing surfaces in level flight.
PAUL
Kevin Anthony Scaldeferri
I'll start by saying, what the hell is up with repling by email _and_
posting to the newsgroup? I've had a bunch of people do this. Is
there some messed up newsreader that does this automatically?
Now then, as I replied in email to this, the question is in which way
was he limited to .75g? Tire friction, footpeg clearance or nerves of
the rider? Keep in mind that willingly riding a motorcycle in a
circle faster and faster until you lose traction is not something most
riders will volunteer to do. On the other hand, on a realistic curve,
I think you will see a rider push it much harder for a short distance.
Kevin Scaldeferri
Rick Baartman
Mark T. wrote:
>
> In article <4t8924$f...@sjx-ixn3.ix.netcom.com>,
> Chuck Tomlinson <toml...@ix.netcom.com> wrote:
> >>
> >>On Wed, 17 Jul 1996, Samuel C. Blackman wrote:
> >>
> >>> "Why do drag racers use wide rear tires, if the coefficient of friction
> >>> is independant of surface area?"
> >>>
> >>The answer is quite simple. When you skid, it is not thet tyre losing
> >>contact with the road, but the rubber in the tyre breaking down under the
> >>shear forces applied. A bigger tyre means that there is more rubber, and
> >>so you have to push it harder to tear off a layer of it, so you don't
> >>skid as easily.
> >
> >inflation pressure, the contact patch area is pretty much independent of
> >the tire width (over a reasonable range), so wider tires do *not* have more
> >rubber in contact with the road.
>
> What kind of myth is this? Wider tires do not have more contact area?
> This isnt right. Your contact patch grows wider as your tire grows wider
It's not a myth. If you have 20 psi supporting a 1000 lb weight, it will spread effectively
over 50 square inches. If you have a 10 inch wide tire, the contact patch will be 5 inches
long. If you widen the tire to 50 inches, still using 20 psi, the contact patch will shorten
to 1 inch. To take advantage of the extra width, you need to drop the tire pressure to 4 psi.
--rick
Oz
In article <31FE5F...@an.hp.com>, Mike Kohlbrenner
<kohl...@an.hp.com> writes
>
>I think you missed my point. I guess I strayed from the drag racing topic
>a little to a more general issue encompassing "regular" tires. And by
>your last sentence above, it seems as though you agree with me, i.e. that
>real tires are at neither of the two extremes represented by my idealized
>models. I would hazard a guess, though that drag slicks are a lot closer
>to the idealized "soft-sided balloon" than any other tires made since they
>are purposely made very flexible and run at very low pressures.
Dragsters:
Exactly how much HP is being delivered through each tyre?
What is the maximum torque being applied through the sidewall?
Whilst there is a lot of rubber area in the sidewall if it's too soft it
will have trouble transmitting the forces appropriately. Of course the
sidewalls do not have to worry about cornering, so they might appear
soft compared to a regular tyre where lateral support whilst cornering
is of some importance.
Carl Porter
Anthony Potts wrote:
>
> On Wed, 24 Jul 1996 msjo...@KS.Symbios.COM wrote:
>
> > In article <FHp8cPA3...@upthorpe.demon.co.uk> Oz <O...@upthorpe.demon.co.uk> writes:
> > >>Initially, T/F dragsters pull over 5g. It's not for long, but it happens.
> >
> > >Must be kinda exciting the first few times.
> >
> > >Hmmm, and the stopping too, I would imagine.
> >
> > Yep...in fact, the reason "Big Daddy" Don Garlits finally had to retire the
> > last time was because he suffered detached retinas while experimenting with a
> > new parachute rig that delivered -7 G when deployed. His doctor told him to
> > quit that stuff or he'd go blind. [Gee, sounds like your mother :-)]
> >
> > +5 to -5 G within the space of 5 seconds...hell of a shock. The other fun
> > comment from racer and racing school owner Frank Hawley is that most everybody
> > drives the last third of the track with their hand on the parachute lever -
> > ever thought of driving 300 mph one-handed? Hawley figures it this way - if
> > your control of the vehicle is such that you need both hands in the lights,
> > you ought to have the chutes out anyway.
> >
> >
> run on for long enough, or is it just in case something goes wrong, and
> you want to pull up short?
>
> even have a throttle pedal, or is it just a big red button?
>
> I suppose it is possible.
>
>
> Anthony Potts
They used to have gears and air operated shifters. Now with the staged clutches
I don't know. They definitely have throttle pedals, they need them if the get sideways
or if the tires smoke so they get off the power ease it back on. Just exactly how you
ease 5000+ horse power is beyond me.
Regards,
Carl Porter
Carl Porter
Kevin Anthony Scaldeferri wrote:
>
> Aack, mea culpa. Of course the normal force on the rears can be
> increased via weight transfer during acceleration. What I meant is
> that the total normal force (all four tires) is constant. And since
> the allegation to which I was replying was that the normal force was
> _reduced_ under acceleration, it doesn't really change the fact that
> whoever said that was full of it.
>
> Kevin Scaldeferri
I was under the impression that any time there was a weight transfer
that total traction decreased as the total normal forces didn't equalize due to
roll couple and such.
Carl Porter
Kevin Anthony Scaldeferri
In article <31FE63...@MSG.TI.COM>, Carl Porter <3C...@MSG.TI.COM> wrote:
>Kevin Anthony Scaldeferri wrote:
>
>>
>> Aack, mea culpa. Of course the normal force on the rears can be
>> increased via weight transfer during acceleration. What I meant is
>> that the total normal force (all four tires) is constant. And since
>> the allegation to which I was replying was that the normal force was
>> _reduced_ under acceleration, it doesn't really change the fact that
>> whoever said that was full of it.
>>
>
>that total traction decreased as the total normal forces didn't equalize due to
>roll couple and such.
>
Umm...it's hard to see where any of the normal force would go. The
total normal force is a sum of gravity (const) and aerodynamic
downforce (increases normal force). At any rate, the traction of
interest in a drag race is at the rear wheels and the weight tranfer
greatly increases the normal force at the rear wheels.
Kevin Scaldeferri
Oz
In article <31FD58...@an.hp.com>, Mike Kohlbrenner
<kohl...@an.hp.com> writes
>
>
>As I said, I would LOVE to see some kind of REAL proof one way or the other
>to put this to rest.
>
>I am sure, however, that neither of the following are true:
>
> * The contact patch AREA is solely a function of air pressure.
>
> * The contact patch AREA is proportional to tire diameter and/or width.
>
>It is somewhere in between. The question is: where?
I think you may be looking at this backward. In order to maximise grip
and minimise tyre consumption the dragster wishes to have the maximum
contact area on the ground. To do this he requires minimum tyre pressure
and the softest sidewall. The contact area is clearly a function of
both, assuming the tarmac is uncompressible.
However there are problems of flexing and heating of the sidewall if the
deflection of the tyre is very great. Also the forces transmitted
through the sidewall to the road are very significant and he would like
to have a thick, strong, sidewall to cope with these. The engine power
is transmitted through the sidewall.
The effect (I calculated this once, long ago, but this is from old
memory) of the sidewall (for a given tyre construction) on the contact
area depends on it's vertical deflection. If the deflection is great (ie
a narrow small diameter wheel) then the effect of the sidewall stiffness
is greater. If small then it's smaller. So a large diameter tyre has
less defection and the effect of sidewall stiffness on the contact area
is smaller. So if you want a large area in contact with the ground and a
strong sidewall to take the forces, then you would tend to a large
diameter tyre inflated to a very low inflation pressure. Since there is
usually a maximum realistic diameter for other reasons, you then balance
up with the appropriate width for the application. Of course increasing
the width of the tyre also reduces the deflection and thus the effect of
the sidewall stiffness.
I don't think anyone has said what airpressures these drag tyres are
used at but I would guess under 10psi, probably 2 or 3?
Mike Kohlbrenner
Oz wrote:
>
> In article <31FD58...@an.hp.com>, Mike Kohlbrenner writes
> >
> >As I said, I would LOVE to see some kind of REAL proof one way or the other
> >to put this to rest.
> >
> >I am sure, however, that neither of the following are true:
> >
> > * The contact patch AREA is solely a function of air pressure.
> >
> > * The contact patch AREA is proportional to tire diameter and/or width.
> >
> >It is somewhere in between. The question is: where?
>
> I think you may be looking at this backward. In order to maximise grip
> and minimise tyre consumption the dragster wishes to have the maximum
> contact area on the ground. To do this he requires minimum tyre pressure
> and the softest sidewall. The contact area is clearly a function of
> both, assuming the tarmac is uncompressible.
I think you missed my point. I guess I strayed from the drag racing topic
a little to a more general issue encompassing "regular" tires. And by
your last sentence above, it seems as though you agree with me, i.e. that
real tires are at neither of the two extremes represented by my idealized
models. I would hazard a guess, though that drag slicks are a lot closer
to the idealized "soft-sided balloon" than any other tires made since they
are purposely made very flexible and run at very low pressures.
Mike Kohlbrenner
<kohl...@an.hp.com>
Shawn A (is for Amazing) Jefferds
I see all the quibbling over spuonds per squre inch. What I do not see
is any mention of the fact that once the driver jam the peddle to the
floor, suddenly there is a vastly larger ammount of weight transfered
to those square inches. This is not a static load that is applied,
it's dynamic! Ive never seen anything statically lift a set of front
tires straight up in the air.
Mark A. Friesel
It's called 'torque'.
Jonathan Barnes
In article <4tlqua$q...@mocha.eng.umd.edu>, Kevin Anthony Scaldeferri (cool...@Glue.umd.edu) writes:
>In article <31FE63...@MSG.TI.COM>, Carl Porter <3C...@MSG.TI.COM> wrote:
>>Kevin Anthony Scaldeferri wrote:
>>
>>>
>>> Aack, mea culpa. Of course the normal force on the rears can be
>>> increased via weight transfer during acceleration. What I meant is
>>> that the total normal force (all four tires) is constant.
>>
>> I was under the impression that any time there was a weight transfer
>>that total traction decreased as the total normal forces didn't equalize due to
>>roll couple and such.
>>
>
>Umm...it's hard to see where any of the normal force would go. The
>total normal force is a sum of gravity (const) and aerodynamic
>downforce (increases normal force). At any rate, the traction of
>interest in a drag race is at the rear wheels and the weight tranfer
>greatly increases the normal force at the rear wheels.
>
accelerated upwards. this is pased the point of 100% weight transfer. of
corce this is a situation that can't cary on for long as the car will be
fliping over :-).
Jonathan. | Barnes's Theorum: For every foolproof device |
| there exists a fool greater than the proof |
Jeff Tuckey
>>>>> "Oz" == Oz <O...@upthorpe.demon.co.uk> writes:
In article <4197SiBvck$xE...@upthorpe.demon.co.uk> Oz <O...@upthorpe.demon.co.uk> writes:
Oz> In article <31FE5F...@an.hp.com>, Mike Kohlbrenner
Oz> <kohl...@an.hp.com> writes
Oz> Dragsters:
Oz> Exactly how much HP is being delivered through each tyre?
Oz> What is the maximum torque being applied through the sidewall?
Oz> Whilst there is a lot of rubber area in the sidewall if it's
Oz> too soft it will have trouble transmitting the forces
Oz> appropriately. Of course the sidewalls do not have to worry
Oz> about cornering, so they might appear soft compared to a
Oz> regular tyre where lateral support whilst cornering is of some
Oz> importance.
i'm no dragster, but i did see the start of a drag race in slow motion
recently (in a tv commercial)...
the thing that impressed me was that the sidewalls did literally buckle, wrinkle
and twist as the power of the engine was initially applied. it looked
like the wheel rotated several inches (maybe half a foot ?) before the
tire was sufficiently deformed to actually begin the forward motion of
the vehicle...
really interesting to watch.
/j
--
Jeff Tuckey ext. 3739