What stops faster "Car VS bike" - debate

What stops faster - modern car or moderne sportbike


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From that MCN chart, a 1098s stops from 60mph in 115 feet. A Sentra stops from 70mph in about 200 feet. But it really isn't a fair comparison. The Sentra would have tires that are designed far more for longevity than grip.

Actually, I looked up the Sentra SE-R numbers and it drops to 186 feet with gripier tiers, which Car and Driver thought was terrible due to a bad ABS controller.

The other important factor is the different level of skill involved. The motorcycle numbers can only be achieved by someone who is very skilled, while anyone can push the pedal to the floor on a modern car with ABS and get optimal braking.

So, my verdict would be on vehicles with similar performance envelopes (ss vs sports car, standard vs midsize car etc), the car will win unless the bike is being ridden by a highly skilled rider.
 
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All things aside, for the same compound: the bigger the surface area, the more traction, therefore more braking force.

not quite,
the traction force is a function of the weight of the vehicle and frictional properties of the tire and pavement. So when a vehicle weighs more it will also produce much more braking force which is why its possible for a heavier vehicle to out brake a lighter one.
 
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http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=4673

Anyone find a similar one for a car? Then we can plug numbers in all day and night for every model of bike and car possible.

Interesting that people are voting for ABS? With a skilled driver ABS takes longer to brake than non-ABS threshold braking, especially on dry pavement and a straight line.


What's even MORE interesting is the people voting for bikes are choosing non-abs, where as the people voting for cars are choosing abs. What does that tell you about the Car voters? :laughing8:
 
What's even MORE interesting is the people voting for bikes are choosing non-abs, where as the people voting for cars are choosing abs. What does that tell you about the Car voters? :laughing8:

cars have better abs systems? not too many cars even come without
 
star once did this test.
GSXR1000 vs porsche 911

car won stopping distance wheels like oil drums turned on their sides = many square inces of contact patch.

bike also won top speed in that test but the runway was not long enough
 
Rubber contact patch has NOTHING to do with braking. Frictional coefficent of a tire in NOT DEPENDANT OR RELATED TO AREA. The way your pads work is they convert forward movement into heat. The only time your tires come into play is in the rain, or if you lock them up, and if you are locking up your front tire frequently when you brake...you need to learn how to ride.

The motorcycle weighs 1/8 what a car does, so it has very little inertia (inertia=mass times velocity), so granted you can achieve a coeficient of friction that will allow your motorcycles brakes to absorb energy(and that coefficient of friction is NOT RELATED TO AREA), the determining factor is the amount of energy one's brakes can absorb in proportion to the mass of the car vs. bike, versus their respective velocities. I don't know about you, but the brakes on my R1 can absorb equal to or more than the brakes of your average Porsche...so since my mass is much less, and we are assuming our velocities are equal, my bike will brake better than any car short of an F1 car.
 


That whole article is bang on. Too bad James R. Davis didn't elaborate with respect to 4 wheeled vehicles. At least he unquestionably answeres the question regarding weight.

Let me point out a few things to start with. In order to determine the fastest possible stopping distance and time you need only know the bike's speed, the amount of grade, if any, the bike is riding on, the static coefficient of friction available to the tires by the roadway surface, and the efficiency of braking. You do NOT need to know WEIGHT!
 
Rubber contact patch has NOTHING to do with braking. Frictional coefficent of a tire in NOT DEPENDANT OR RELATED TO AREA. The way your pads work is they convert forward movement into heat. The only time your tires come into play is in the rain, or if you lock them up, and if you are locking up your front tire frequently when you brake...you need to learn how to ride.

The motorcycle weighs 1/8 what a car does, so it has very little inertia (inertia=mass times velocity), so granted you can achieve a coeficient of friction that will allow your motorcycles brakes to absorb energy(and that coefficient of friction is NOT RELATED TO AREA), the determining factor is the amount of energy one's brakes can absorb in proportion to the mass of the car vs. bike, versus their respective velocities. I don't know about you, but the brakes on my R1 can absorb equal to or more than the brakes of your average Porsche...so since my mass is much less, and we are assuming our velocities are equal, my bike will brake better than any car short of an F1 car.


LOL... please read - http://www.msgroup.org/Tip.aspx?Num=209&Set=199-231


Especially this part
Let me point out a few things to start with. In order to determine the fastest possible stopping distance and time you need only know the bike's speed, the amount of grade, if any, the bike is riding on, the static coefficient of friction available to the tires by the roadway surface, and the efficiency of braking. You do NOT need to know WEIGHT!

And this part
As to dual disc brakes ... your rate of deceleration is not limited by your brakes! (As we noted in the beginning, you can lock your wheels with the brakes you have so they are capable of stopping you as quickly as the environment will allow.) Your rate of deceleration is limited by how much of available traction you can use. If the amount of traction available supports a deceleration rate of, say, 1.1g's, then if your bike can do a stoppie you CANNOT stop that fast. So one or two discs is of no importance. (A second disc merely allows you to use your brakes longer before they overheat.)
 
not quite,
the traction force is a function of the weight of the vehicle and frictional properties of the tire and pavement. So when a vehicle weighs more it will also produce much more braking force which is why its possible for a heavier vehicle to out brake a lighter one.

wouldn't more weight = more momentum as well?

I'm just confused now. so ya.:D
 
wouldn't more weight = more momentum as well?

I'm just confused now. so ya.:D


Exactly.. more weight = more traction, and more momentum....
they cancel each other out 1 for 1...

weight is of no consequence, and therefore irrelevant for stopping distance calculations.
 
Here... http://www.nhtsa.dot.gov/Cars/rules/CAFE/NewPassengerCarFleet.htm

Gives an idea to the average curb weight of a passenger car, about 3000lbs it seems. A bike is about 400 + say 200lbs for the rider (OK I'm being generous).

There's a start, you have to take momentum into account first of all so pick some values and play with them.

My feeling...slightly scientific..is that the bike has the edge but the values are close as cars have 4 wheel braking to disperse the momentum whereas bikes have 90+% front with the remainder (if any) at the rear. So even though momentum is a lot less for the bike the car can disperse momentum over 4 wheels.

Then I think you have to look at the efficiency of the brakes/tires too obviously and again my feeling is that cars and bikes are pretty similar just due to rotor sizes given the average stock brakes but bikes would have the edge due to softer compound tires (on stock vehicles).

Last things to consider are aerodynamics and any engine braking.
 
i believe the reason we should use a sport bike and compare it to a family sedan (although i'd go ahead and pick sportier version of a family sedan) is because a large percentage of bikes out there are sport bikes, i'd sa 60 to 75% of bikes are sport bikes. That doesnt mean every rider can take it close, or even come to the limit of the bikes ability, but the truth is, a lot of bikes out there are sport bikes, super sports or sport touring, and thats why, the majority of bikes will be represented against majority of 4 wheeled vehicles, which are family sedans.
 
http://www.wheels.ca/Motorcycles/article/521012

"A few years ago, Wheels organized an informal grudge match at the old Picton airport. Car guru Jim Kenzie brought along a $200,000 Porsche 911 Turbo X50, while Wheels motorcycle racing columnist Larry Tate and I brought a Suzuki Hayabusa and a GSXR1000, both of which retailed for less than 13 grand. There were four events – a slalom, an approximate quarter-mile acceleration contest, braking from 160 km/h to zero, and whatever top speed we could get in the space available.

The bikes smoked the Porsche at every event except braking, and that was because we were deathly afraid of locking the front wheel while braking hard on the many pavement cracks, divots and tufts of grass growing up through the old runway. All Kenzie had to do was mash on the brake pedal and let the huge discs and wide, sticky rubber on the Porsche do their job"
 
Exactly.. more weight = more traction, and more momentum....
they cancel each other out 1 for 1...

weight is of no consequence, and therefore irrelevant for stopping distance calculations.

...under ideal conditions. But then again under ideal conditions you would have what is considered "perfect" dampers and springs as suspension components. Unless of course you elect to remove those elements completely in order to further simplify the corresponding equations, which would also allow you to disregard other phenomenon (such as weight shifting, skidding, etc), but would render this whole thought experiment moot as we are now dealing with unrealistic vehicles.

In order for any of us unqualified schlubs to make any kind of conclusive arguments, we would have to simplify these scenarios out so that they have very specific and clear boundaries. The more "real world" elements we throw in, the harder this becomes to analyze and comprehend....especially on an internet forum.
 
Exactly.. more weight = more traction, and more momentum....
they cancel each other out 1 for 1...

weight is of no consequence, and therefore irrelevant for stopping distance calculations.

wow
 

If you understand all the points here - http://www.msgroup.org/Tip.aspx?Num=209&Set=199-231



you would see that adding weight (in the right places) on a sportbike would decrease it's stopping distances- but overall weight has no affect.


In the examples given, they are able to increase the rate of deceleration by a measurable factor, by lowering the CG by only 0.7 inches...
 
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If you understand all the points here - http://www.msgroup.org/Tip.aspx?Num=209&Set=199-231



you would see that adding weight (in the right places) on a sportbike would decrease it's stopping distances..


In the examples given, they are able to increase the rate of deceleration by a measurable factor, by lowering the CG by only 0.7 inches...

did I say anything about adding weight to a bike? You said weight has no relevance in stopping distances!
 


Very interesting..

especially

The force on a vehicle during a stop is just the vehicle's mass times the (negative) acceleration, F = ma. That force has to be applied at the tires via their traction. The friction equation is F = μW (where W is the weight of the vehicle and μ is the Greek letter mu, the coefficient of friction — again see laws of friction for details). The weight of the vehicle is the mass m times the gravitational force g, so F = μmg. The maximum stopping force that can be applied is the maximum frictional force that the tires can sustain, so ma = μmg; and we can cancel the mass which appears on both sides to get the maximum deceleration possible:
a = μg

And the whole section dealing with - Can you brake faster than cars?



Can you brake faster than cars?

Good question. During our emergency braking exercises I always ask this question, and I get two answers, both wrong. One is "Certainly not; motorcycle tires have two small contact patches, car tires have four larger contact patches, so motorcycles have less traction." You can read my page on traction and contact patch area to see why contact patch area is irrelevant, so this answer is wrong. The other answer is "Certainly; you can accelerate faster than cars because you're lighter. You can decelerate faster than cars, because you're lighter." I subscribed to the second view until one of my students, Eli Baldwin, said that the physics in the two situations is entirely different. He's right. Motorcycles can accelerate faster than (most) cars because the ratio of power to weight is greater for motorcycles; there may be less horsepower but it's pushing a lot less weight. But for braking, the horsepower of the engine is irrelevant. To find out whether motorcycles brake better than cars we have to look deeper into the physics of braking. The force on a vehicle during a stop is just the vehicle's mass times the (negative) acceleration, F = ma. That force has to be applied at the tires via their traction. The friction equation is F = μW (where W is the weight of the vehicle and μ is the Greek letter mu, the coefficient of friction — again see laws of friction for details). The weight of the vehicle is the mass m times the gravitational force g, so F = μmg. The maximum stopping force that can be applied is the maximum frictional force that the tires can sustain, so ma = μmg; and we can cancel the mass which appears on both sides to get the maximum deceleration possible:
a = μg
Now before we go further, let's note some assumptions. One is that the downwards force in the friction equation is actually the weight of the vehicle. Race cars use airfoils to develop a downward force to improve their traction, so their stopping distances would be better than an unassisted vehicle (at speeds allowing the airfoil to work). If street cars ever begin to use this technology then the conclusions would have to change to take that into account.
Another assumption is that the limiting factor in stopping is traction, rather than the ability of the brakes to dissipate the energy. This is true of cars, as their brakes can overwhelm the traction of the tires, causing a slide. But it isn't true of large trucks. Their additional mass provides additional traction, as shown in F = μmg, but the additional energy is more than the brakes can deal with, resulting in longer stopping distances.
The limiting factor in a stop of a motorcycle may also not be the traction, but the stability of the vehicle. We've all seen sportbikes with the rear tire in the air in a stop. The front tire isn't sliding, so there may be still more traction available to slow, but any additional braking will just result in the motorcycle going over the front tire.
But with the assumption that stopping distance is limited by the traction of the tires, a = μg shows that the mass of the vehicle is not relevant; it does not enter into the equation. The only difference might be in the value of μ for car and bike tires. There is some evidence that bike tires have stickier rubber than auto tires: My motorcycle tires last between 10,000 miles (for the rear) and 20,000 miles (front), and the tires on my cars will go 40,000 or more. Softer rubber is stickier than harder rubber, so I think this indicates that motorcycle tires have better traction than auto tires, and thus that motorcycles will stop shorter.
But there is one final complication: Can you use all the superior traction of your motorcycle tires? If you get too hard on the brakes in your car, you slide, you let off the pedal to resume rolling, you get back on the brakes. When the same thing happens on your motorcycle, the slide generally results in a fall. Thus motorcyclists are reluctant to approach the limit of their braking, where the same isn't true of auto drivers.
So where does this leave us? Here's my rule: I figure the guy in front of me might be able to outbrake me, so I leave enough room between us, and look well ahead of him, so that I won't hit him. And I figure I can outbrake the guy behind me, especially if he's too close or not paying attention, so I leave even more room in front (to reduce the probability that I'll have to brake hard), and keep a close eye on my mirrors, and stay in the proper gear so that after braking hard I can escape if needed.
What do you think about antilock brakes?

The word from the experts at the motorcycle magazines, the people who do the 60-to-zero braking tests, is that under test conditions, they can outperform antilock brakes. That is, when their skills are already highly-developed, and if they have three or four tries to get tuned up, and when they know exactly when they're going to start braking, and if there are no traction surprises (like going over a sandy or painted patch of pavement), then they can outperform the machinery. But under real-world conditions, they say that antilock brakes win.
That's good enough endorsement for me. (Furthermore, my guess is that over the next few years even test conditions won't be enough to outperform antilock brakes.) Neither of my motorcycles has antilock brakes. My next motorcycle will have them.
 
did I say anything about adding weight to a bike? You said weight has no relevance in stopping distances!

Right- it doesn't, now you're getting it. But changing the center of gravity does.

Removing weight from below the center of gravity of a bike, would INCREASE stopping distances.

BECAUSE sportbikes typically can't utilize their available traction, since the bike will stoppie / flip over.
 
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