Winter Riding Motorcycle harassment

[soto]

I think I'd prefer one of those hovering speeders from star wars for winter.

[caboose483]

I love when I can rely on my personal experience and not opinions in left field. If you are concerned about your braking, don't tailgate, drive slower, and don't speed up when someone goes to pass. That's just being rude.

So everyone else in the world is subject to certain laws of physics but you are able to use a different set? That is what you have discovered in your personal experience? Do i understand you correctly?

[slogan]

I think winter riding would make a lot of fun on trails or so. But never in the city with a traffic.

[Baggsy]

There's a picture floating around the internet of a rider who may have read one of those 'Bikes can outbrake cars" bits of nonsense.
His head is embedded in the back of a tractor trailer, and the pictures are pretty graphic.

I have yet to see where mass fits into any braking formula. Please do the math and post your formula for motorcycle braking distances.

It's possible that since bikes are routinely 10% slower than the speedometer indicates, that someone might think that they are stopping faster than they actually are.

A real world test such as caboose483 has suggested would be the best way to prove to you once and for all that it isn't true.

What scares me is that some young people will read drivel posted on the web, believe it to be true, and lose their lives over it, because someone has been to lazy to do a little bit of basic research before posting.

Be careful about what you post.

[soto]

I think he's saying that he can anticipate ALL traffic situations and ride in a way that his brakes will always stop fast enough regardless of conditions.

OP is either very young, inexperienced, or worse.

I'm not saying never to ride in winter. I do it myself. I'm saying it's important to understand the additional risk involved and make informed decisions rather than deluding oneself.

[ItIsWhatItIs]

There's a picture floating around the internet of a rider who may have read one of those 'Bikes can outbrake cars" bits of nonsense.
His head is embedded in the back of a tractor trailer, and the pictures are pretty graphic.

I have yet to see where mass fits into any braking formula. Please do the math and post your formula for motorcycle braking distances.

It's possible that since bikes are routinely 10% slower than the speedometer indicates, that someone might think that they are stopping faster than they actually are.

A real world test such as caboose483 has suggested would be the best way to prove to you once and for all that it isn't true.

What scares me is that some young people will read drivel posted on the web, believe it to be true, and lose their lives over it, because someone has been to lazy to do a little bit of basic research before posting.

Be careful about what you post.

And movie producers should watch what they put in their movies out of concern someone will be inspired? Watch the movie. They are riding on city streets with traffic. The roads were icy. As for braking, stop tail gating and their won't be a problem.

http://tutor4physics.com/formulas.htm

Newton's laws of motionThrough Newton's second law, which states: The acceleration of a body is directly proportional to the net unbalanced force and inversely proportional to the body's mass, a relationship is established betweenForce (F), Mass (m) and acceleration (a). This is of course a wonderful relation and of immense usefulness.
F = m x a click for calculator
Knowing any two of the quantities automatically gives you the third !!

---

MomentumMomentum (p) is the quantity of motion in a body. A heavy body moving at a fast velocity is difficult to stop. A light body at a slow speed, on the other hand can be stopped easily. So momentum has to do with both mass and velocity.
p = mv click for calculator
Often physics problems deal with momentum before and after a collision. In such cases the total momentum of the bodies before collision is taken as equal to the total momentum of the bodies after collision. That is to say: momentum is conserved.

[turbodish]

OP is either very young, inexperienced, or worse.

I was thinking excess drug or alcohol consumption myself, combined with a good dose of excess ego.

I'm not saying never to ride in winter. I do it myself. I'm saying it's important to understand the additional risk involved and make informed decisions rather than deluding oneself.

Ditto. It can be done and I've done it when the roads are decent in winter, but I do recall a few rides coming home in a storm where I wished that I was anywhere else but there at that moment. It's a whole lot easier to do it on 4 wheels than on two wheels once the roads get wet/slushy/snowing/drifted. It also hurts less when you don't get it right in marginal conditions.

[caboose483]

And movie producers should watch what they put in their movies out of concern someone will be inspired? Watch the movie. They are riding on city streets with traffic. The roads were icy. As for braking, stop tail gating and their won't be a problem.

http://tutor4physics.com/formulas.htm

You forgot the other half of the equation accounts for the braking force in terms of the coefficient of static friction and the normal force and thus cancels out mass. Of course that's all based on assumption that gravitational mass and inertial mass are equivalent.

But i digress..... to my point that you don't understand the laws of physics.

[dru9]

Someone on the boards (can't remember offhand) uses one of his "brilliant" quotes as his sig. Very funny, and shows the same ignorance of physics.

[ItIsWhatItIs]

You forgot the other half of the equation accounts for the braking force in terms of the coefficient of static friction and the normal force and thus cancels out mass. Of course that's all based on assumption that gravitational mass and inertial mass are equivalent.

But i digress..... to my point that you don't understand the laws of physics.

Stop confusing meta physics with physics.

[ItIsWhatItIs]

Someone on the boards (can't remember offhand) uses one of his "brilliant" quotes as his sig. Very funny, and shows the same ignorance of physics.

Is that you Caboose?

Someone with next to no posts go out and starts attacking. Very troll like. What did you do opened up another email account created a new id to start trolling.

[caboose483]

Stop confusing meta physics with physics.

I'm not the one who is confused.

[dru9]

Lol. No we are not the same person - and more than a few have rebutted you, so why point out just one?

[Flywheel]

http://www.imdb.com/name/nm0000553/news?year=2000

Seems like Liam has some firsthand experience on how well a motorcycle stops.

[ZX600]

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 and motorcycles at normal speeds, 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. As speeds increase, even cars and motorcycles may become limited by the ability of the brakes to deal with the energy (which increases as the square of the speed). If this point is reached then motorcycles may have an advantage in stopping distance, as motorcycles generally weigh a quarter or less of a car's weight, so the kinetic energy which must be converted to heat is also a quarter or less. I don't know where the tipping point is between tire traction and braking energy as the limiting factor, the difference in stopping distances between smaller passenger cars and larger SUVs and pickup trucks widens dramatically from 60mph to 80mph, suggesting that energy becomes a factor even at those speeds. Brake design plays a role. Drum brakes don't deal with energy as well as modern disk brakes. Two large brake disks on a sportbike will move more energy than a single small disk on a cruiser. More pad area pushed by more pistons will reduce lever effort on the part of the rider, making it easier to achieve the maximum braking force. The same factors in auto brakes will affect which auto can outbrake which motorcycle. 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. I had speculated that motorcycle tires have stickier rubber than auto tires, because of the difference in tire life. Softer rubber being stickier than harder rubber, and having a shorter life, it might follow that motorcycle tires are stickier than auto tires. But a reader, Blane Baysinger, pointed me to a Society of Automotive Engineers article comparing motorcycle and auto tires. This article indicates that the coefficient of friction of both auto and motorcycle tires is about 1.2 on dry surfaces (declining to .7 to .9 when skidding). The difference in longevity appears to be due to the greater amount of rubber on the auto tires; and it appears that if you can stop faster than a car, it's because you're better at using the brakes, not because of any inherent superiority in the braking capability of a motorcycle. And there is one final complication: Can you use all the 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.


..and here is the other side of the equation

[ItIsWhatItIs]

I'm not the one who is confused.

Does your braking also include when the brakes fail and the vehicle start sliding. How does your meta physics apply to that?

[Flashmonkey]

We're getting trolled guys.

This. Come on guys

[ItIsWhatItIs]

Another assumption is that the limiting factor in stopping is traction, rather than the ability of the brakes to dissipate the energy

car

48651.48 N
1814 kg
26.82 m/s 60 mph

bike

3647.62 N
136 kg
26.82 m/s

A fraction of the energy to disipate by the brakes and the weight of the rider ought to exert a downward pressure on the tires to increase friction.

[ZX600]

car

48651.48 N
1814 kg
26.82 m/s 60 mph

bike

3647.62 N
136 kg
26.82 m/s

A fraction of the energy to disipate by the brakes and the weight of the rider ought to exert a downward pressure on the tires to increase friction.

I would recommend you read and understand my post before you end up as a sticker in the back of a truck, if you decide to not understand the whole picture and just pick and choose the info that will out of context help your point then there is nothing I can do about it.

peace out

[caboose483]

Does your braking also include when the brakes fail and the vehicle start sliding. How does your meta physics apply to that?

It does not apply at all.

But my boring old regular physics tells me that you apply the coefficient of dynamic friction instead of static friction. Assuming you have really good balance and can keep the bike upright with the front tire locked.

car

48651.48 N
1814 kg
26.82 m/s 60 mph

bike

3647.62 N
136 kg
26.82 m/s

A fraction of the energy to disipate by the brakes and the weight of the rider ought to exert a downward pressure on the tires to increase friction.

And the weight of the car itself doesn't exert more downward pressure on the tires to increase friction?