In article ,
says...

On Tuesday, June 11, 2013 3:23:55 PM UTC-4, iBoaterer wrote:
In article ,

says...



On Tuesday, June 11, 2013 8:54:07 AM UTC-4, iBoaterer wrote:

In article ,



says...



There must be some reason that nearly every track record is held by a

4 wheeled vehicle.



Cite?



Barcelona 2005

Formula One - Fisicella's Renault - 1:15.641 fast lap.

MotoGP - Gibernau's Honda - 1:42.337 fast lap.



PHILLIP ISLAND GRAND PRIX CIRCUIT LAP RECORDS

OUTRIGHT SIMON WILLS REYNARD 94D 13/02/2000 1.24.2215

FORMULA 4000 SIMON WILLS REYNARD 94D 13/02/2000 1.24.2215



PHILLIP ISLAND GRAND PRIX CIRCUIT

MOTORCYCLE LAP RECORDS

MotoGP Marco Melandri (Ita) Honda RC211V 1:30.332 16-Oct-05

Pole : Nicky Hayden (USA) Honda RC211V 1:29.020 16-Sep-06



A couple of results from a quick google. You can do the rest of the work.



Pretty much the only tracks where you'll find faster times for bikes are the tracks specifically designed for bikes. Cars obviously enjoy enough of an advantage from their superior traction, brakes and downforce that it negates the bike's advantage of less mass and better power/weight ratio. Not by a lot, but 6 - 27 seconds (the diff in the examples above) is a lot on a track.



Have fun.



Let's see. All tracks made exclusively with cars in mind.

Tracks that have taken special pains to cater to a bikes special needs are faster for bike. Take that advantage away, and the car is faster.

Now, how about
REAL cites? How about the physics behind your ASSumptions? Superior

downforce??? You DO realize, don't you, that a motorcycle, when it leans

INTO the curve is keeping it's CG in line with the forces, while a car

isn't, correct?

You do realize that the CG of the bike, when leaning into a turn, is attempting to push the tire ACROSS the pavement at the angle of the lean? Meanwhile the car's down force is pushing the tire directly down into the pavement. Keeping the downforce perpendicular is a good thing.

Besides, the bike couldn't corner if it didn't lean to keep the CG in line with the cornering force... that's what keeps it from flipping over. That's also what causes the increase of slip angle and traction loss.

Okay, time for a simple physcis lesson, let's start with vector
mechanics, shall we? An object in motion tends to stay in motion AND
tends to stay in a straight line. For a simple demonstration of this,
take two strings, both say a foot long. Attach a one ounce ball to one
of them, and a 5 ounce ball to the other. Then swing them in a circle
and see which one takes the most effort on your part to hold on to. This
is an example of mass and velocity trying to keep those balls in a
straight line and you are having to restrain them from doing so by
holding the string. Force equals mass times acceleration. Simple as
that. What has more mass, the motorcycle or the car that weighs 4 times
as much? The car at the same speed has 4 times the force and this force
wants to stay in a straight line. So, it takes 4 times the resistance to
achieve this. The only thing affecting this is the tire coefficent of
friction. Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force. NOW we have the
problem of the car being at a much higher center of gravity vertically.
While the motorcycle's CG changes to be more inline with the force
vector, the cars remains unchanged and is not as near to in line with
the vector as the motorcycle's?

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car also shifts its CG to load up the outside tires in a turn, applying more down force to them. And, the car applies it's down force (traction) in a turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do you think is more efficient way to apply down force to resist that tendency? You're concentrating on one tiny little aspect of the issue. Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent weaknesses of bikes, cars almost always turn faster lap times. The ability to take the turns faster and better brakes more than makes up for the bike's better acceleration on most tracks.

In article ,
says...

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car also shifts its CG to load up the outside tires in a turn, applying more down force to them. And, the car applies it's down force (traction) in a turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do you think is more efficient way to apply down force to resist that tendency? You're concentrating on one tiny little aspect of the issue. Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent weaknesses of bikes, cars almost always turn faster lap times. The ability to take the turns faster and better brakes more than makes up for the bike's better acceleration on most tracks.

ME "open that basic physics book"??? Please, do tell me the physics
behind your allegations. Now, for your first question, like I stated
earlier, the "more efficient way" to apply down force is to have the CG
more in line with the vector, which is what a motorcycles does when it
corners! Thanks for making my point!

wrote in message
...

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car
also shifts its CG to load up the outside tires in a turn, applying
more down force to them. And, the car applies it's down force
(traction) in a turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do
you think is more efficient way to apply down force to resist that
tendency? You're concentrating on one tiny little aspect of the
issue. Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent
weaknesses of bikes, cars almost always turn faster lap times. The
ability to take the turns faster and better brakes more than makes up
for the bike's better acceleration on most tracks.

------------------------------------

As evidenced by virtually all real world tests done on the subject.
I had to let iBoaterer out of the Bozo bin to see what the heck he was
talking about.
His analysis on the subject is flawed. A car can overcome the
centrifugal forces (to a point) due to transferring them to the two
outside tires, allowing it to corner at faster speeds. If you could
measure the forces, they would be huge. A motorcycle rider can't
compensate enough by leaning at the same speed or even near the same
speed. He's relying on a "counterbalance" effect which can't be
nearly high enough. Now, if the motorcycle rider had outriggers that
he could climb out onto for additional mechanical advantage, he could
corner faster.

On 6/12/13 9:57 AM, Eisboch wrote:


wrote in message
...

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car also
shifts its CG to load up the outside tires in a turn, applying more down
force to them. And, the car applies it's down force (traction) in a
turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do you
think is more efficient way to apply down force to resist that
tendency? You're concentrating on one tiny little aspect of the issue.
Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent
weaknesses of bikes, cars almost always turn faster lap times. The
ability to take the turns faster and better brakes more than makes up
for the bike's better acceleration on most tracks.

------------------------------------

As evidenced by virtually all real world tests done on the subject. I
had to let iBoaterer out of the Bozo bin to see what the heck he was
talking about.
His analysis on the subject is flawed. A car can overcome the
centrifugal forces (to a point) due to transferring them to the two
outside tires, allowing it to corner at faster speeds. If you could
measure the forces, they would be huge. A motorcycle rider can't
compensate enough by leaning at the same speed or even near the same
speed. He's relying on a "counterbalance" effect which can't be nearly
high enough. Now, if the motorcycle rider had outriggers that he could
climb out onto for additional mechanical advantage, he could corner faster.



Every video of top drivers I've seen, one on a top of the line racing
bike and the other in a hot car, shows the bike typically
outaccelerating the car in the straights, and the car pretty close to
catching the bike in sharp turns, both because it has better brakes and
more ability to corner. In the end, though, in a "race" that
incorporates multiple circuits of the course, the bike typically "wins"
because of its acceleration.

In article ,
says...

On 6/12/13 9:57 AM, Eisboch wrote:


wrote in message
...

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car also
shifts its CG to load up the outside tires in a turn, applying more down
force to them. And, the car applies it's down force (traction) in a
turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do you
think is more efficient way to apply down force to resist that
tendency? You're concentrating on one tiny little aspect of the issue.
Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent
weaknesses of bikes, cars almost always turn faster lap times. The
ability to take the turns faster and better brakes more than makes up
for the bike's better acceleration on most tracks.

------------------------------------

As evidenced by virtually all real world tests done on the subject. I
had to let iBoaterer out of the Bozo bin to see what the heck he was
talking about.
His analysis on the subject is flawed. A car can overcome the
centrifugal forces (to a point) due to transferring them to the two
outside tires, allowing it to corner at faster speeds. If you could
measure the forces, they would be huge. A motorcycle rider can't
compensate enough by leaning at the same speed or even near the same
speed. He's relying on a "counterbalance" effect which can't be nearly
high enough. Now, if the motorcycle rider had outriggers that he could
climb out onto for additional mechanical advantage, he could corner faster.



Every video of top drivers I've seen, one on a top of the line racing
bike and the other in a hot car, shows the bike typically
outaccelerating the car in the straights, and the car pretty close to
catching the bike in sharp turns, both because it has better brakes and
more ability to corner. In the end, though, in a "race" that
incorporates multiple circuits of the course, the bike typically "wins"
because of its acceleration.

Well there you go, videos are the facts and physics is just hokum......

In article ,
says...

wrote in message
...

On Wednesday, June 12, 2013 8:36:01 AM UTC-4, iBoaterer wrote:

Okay, time for a simple physcis lesson...

Therefore the car has to have enough surface area, and
friction ability to to overcome 4 times the force.

The car has far more than 4 times the contact patch. And, the car
also shifts its CG to load up the outside tires in a turn, applying
more down force to them. And, the car applies it's down force
(traction) in a turn like this:

_|_

while the bike is like this:

_\_

If the desired result is to not slide across the pavement, which do
you think is more efficient way to apply down force to resist that
tendency? You're concentrating on one tiny little aspect of the
issue. Time to open your mind and that basic physics book!

Fact is, unless the track is specifically designed for the inherent
weaknesses of bikes, cars almost always turn faster lap times. The
ability to take the turns faster and better brakes more than makes up
for the bike's better acceleration on most tracks.

------------------------------------

As evidenced by virtually all real world tests done on the subject.
I had to let iBoaterer out of the Bozo bin to see what the heck he was
talking about.
His analysis on the subject is flawed. A car can overcome the
centrifugal forces (to a point) due to transferring them to the two
outside tires, allowing it to corner at faster speeds. If you could
measure the forces, they would be huge. A motorcycle rider can't
compensate enough by leaning at the same speed or even near the same
speed. He's relying on a "counterbalance" effect which can't be
nearly high enough. Now, if the motorcycle rider had outriggers that
he could climb out onto for additional mechanical advantage, he could
corner faster.

You are totally and conveniently forgetting the laws of physics. When
the car loads the two outside tires, thus transferring most of the force
to half of the contact area (since the two inside tires are doing
virtually no work) that in fact does right the opposite, less contact
area, less traction. The motorcycle, on the other hand because of fact
that it's CG is in line with the vector only causes more friction by
force. And while there is more friction by force on the car's two
outside wheels, there is also less friction by force on the inside
wheels. So, you now have a car with 4 times the mass using about the
same tire contact area as the motorcycle.

"iBoaterer" wrote in message
...


You are totally and conveniently forgetting the laws of physics. When
the car loads the two outside tires, thus transferring most of the
force
to half of the contact area (since the two inside tires are doing
virtually no work) that in fact does right the opposite, less contact
area, less traction. The motorcycle, on the other hand because of fact
that it's CG is in line with the vector only causes more friction by
force. And while there is more friction by force on the car's two
outside wheels, there is also less friction by force on the inside
wheels. So, you now have a car with 4 times the mass using about the
same tire contact area as the motorcycle.

----------------------------------------

You are talking two different things here. Stiction/Friction
(traction) is one thing. Centrifugal forces due to the turn is
another.
In the case of high speed motorcycle cornering the latter is the
governing issue, traction is secondary (until both the car and the
motorcycle exceeds the limit). The gyroscopic effect of the
motorcycle cannot be overcome by a weight shift by the rider
sufficiently to make a high speed turn as quickly as the car.
Granted, at parking lot speeds a motorcycle can turn faster than a
car, but that's due to it's much shorter turn radius and the absence
of any significant centrifugal force. But at high speeds,
centrifugal force becomes the deciding factor.

In article ,
says...

"iBoaterer" wrote in message
...


You are totally and conveniently forgetting the laws of physics. When
the car loads the two outside tires, thus transferring most of the
force
to half of the contact area (since the two inside tires are doing
virtually no work) that in fact does right the opposite, less contact
area, less traction. The motorcycle, on the other hand because of fact
that it's CG is in line with the vector only causes more friction by
force. And while there is more friction by force on the car's two
outside wheels, there is also less friction by force on the inside
wheels. So, you now have a car with 4 times the mass using about the
same tire contact area as the motorcycle.

----------------------------------------

You are talking two different things here. Stiction/Friction
(traction) is one thing. Centrifugal forces due to the turn is
another.
In the case of high speed motorcycle cornering the latter is the
governing issue, traction is secondary (until both the car and the
motorcycle exceeds the limit). The gyroscopic effect of the
motorcycle cannot be overcome by a weight shift by the rider
sufficiently to make a high speed turn as quickly as the car.
Granted, at parking lot speeds a motorcycle can turn faster than a
car, but that's due to it's much shorter turn radius and the absence
of any significant centrifugal force. But at high speeds,
centrifugal force becomes the deciding factor.

Yes, centrifugal force does indeed become the deciding factor. Which has
more centrifugal force if the radius of the turn is the same and the
speed is the same a 3200 pound car or a 600 pound motorcycle??

On Wednesday, June 12, 2013 10:04:24 AM UTC-4, iBoaterer wrote:

So, you now have a car with 4 times the mass using about the
same tire contact area as the motorcycle.

BS. Cite?

BTW... like many sports and race cars, my old Boxster's rear tires had a lot of camber to allow the tire to have better contact with the road when in a high speed turn. Wears out the inside edge quickly, but increases grip dramatically. That big, flat patch of rubber stays on the pavement.

Bikes can't have flat surfaced tires, so their contact patches are very small all the time.