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...

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.

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.

In article ,
says...

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?

http://www.porsche.com/international/models/911/911-
carrera/featuresandspecs/
(140kg laden weight)

http://en.wikipedia.org/wiki/Ducati_848
(Dry weight 370 pounds, so if you take it's laden weight of say 600
pounds we're close)

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.

Oh, now you want to talk specialty cars, but street motorcycles!!!!!
Well, the contact are for a road course motorcycle is large as well for
just that purpose. Plus, you've forgotten that pesky physical fact that
you are trying to turn 4 or 5 times the mass of something that wants to
go straight.

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...

"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??