Question about the Physics Model

[jlv]

It shouldn't take longer on dry concrete.

[Prologue]

Prologue is correct. Given perfect dry sliding friction and strong enough brakes, the stopping distance will be the same.

Neither of which we'll ever have in real life. So what's the point here?

Good question, my answer would be, you gotta start somewhere! Also, this *is* a game and not real life, you have to attempt to model what happens in real life. You do that with mathematical models. So, when someone says, 'why is a lighter bike better?', I would like to give them a concrete answer, not just what feels right. Feelings can be very deceptive.

On an even more precise, for the most part (with adequate braking systems) cars/motorcycles have nearly identical braking distances when using tires of similar compound - in REAL life. So all this math stuff really pans out in real life. Again this is all on flat ground and without suspension effects. But, the analysis can be extended to slopes, and once that is done I think the case is pretty much sealed. I just did the slope case and it turns out to be independent of mass again (no surprise because without air resistance all things fall at the same rate). The math looks like this.






And again, the stopping distance does not depend on mass. You can do another analysis on it using other techniques to see that the time it takes to stop is the same too.

[Prologue]

Sorry to double post but I just found a picture that gives an idea of where I got the angles from and such. They use different names for the variables though. In their graph mentally replace s with θ, and w with mg. With those substitutions, the picture describes what I am saying (at least somewhat!).

[Garasaki]

How could the mass not matter?

Your are talking about disappaiting (sp?) the kinetic energy of the bike when you apply brakes.

The kinetic energy of the bike is based on the speed it's moving and the MASS of the bike.

Prologue is correct. Given perfect dry sliding friction and strong enough brakes, the stopping distance will be the same. The extra energy of the more massive bike goes into heating the brakes since the more massive bike required more pressure on the brake pads proportional to its mass to exploit the traction from the downforce the extra mass provided. Hope that sentence made sense.

Strong enough breaks? Dosen't that mean increased braking force?

I'm sorry bout I'm not following this arguement either.

Given 2 bodies travelling at the same velocity, assume 1 has x mass and the other has 2x mass, then the kinetic energy of the second body will be twice that of the first.

You guys are arguing that the braking force does not change based on mass (right?).

I agree that's true in the simplified example (and generally speaking in real life).

However that does not mean the braking distance is the same. The braking distance is (in my head) the the amount of work done by said force (area under the curve). This amount of work must equal the kinetic energy of the body for the body to come to a complete stop.

If the force is the same amoung our 2 examples, but the beginning kinetic energy is different (in this case, twice as much), then the braking distance will be different.

Note that kinetic energy is based on the square of the velocity. And we are talking about small (5% ish, or 1.0 vs 1.05) differences in weight (errrrrrr mass!). So as speed increases, the difference between the 2 bodies kinetic energy becomes smaller. So the difference in braking distance resulting from increased mass would be much more apparent at much lower starting speeds, and become exponentially smaller as the speeds increase.

So in other words I can't follow your math cause it's been 8 years since I "cos"ed anything, and i got 2 kids (1 is 3 months old) so I'm a little foggy in the brain bucket, but I dont see how you could possibly present a situation where 2 bodies with different starting kinetic energies have the same braking distance (although the real world differences may be negligable).

And yes, all the energy the brakes absorb is turned into heat. Which is neither here nor there.

[Garasaki]

ut the beginning kinetic energy is different (in this case, twice as much), then the braking distance will be different.

Note because the assumption here is that both bodies start at the same velocity but their masses, therefore KE, are different

[DJ99X]

Even though the heavier bike will have a greater kinetic energy, it also produces a greater stopping force, due to the greater frictional force (how he replaced Fd with (mew)mgd)

[jlv]

You guys are arguing that the braking force does not change based on mass (right?).

No. The braking force required is higher for the more massive bike. We were using the assumption that traction was the limiting factor, not braking force. If the brakes are too weak to lock the wheels up the more massive bike will have a longer stopping distance.

[Prologue]

Yeah what they said, we are saying that the bigger bike is able to produce a higher braking force (before the tire slips) because of the added weight on the tire. The key is that this is all before the tire slips. The tires on the heavier bike (although identical) produce more grip because of the added weight of the bike. Of course the brakes could in theory be as powerful as you want but there is a point where the tires will just skid and beyond that point you have a diminished stopping force.

So the key is to keep the tire just on the edge of slipping, and we have assumed this is what is happening in the analysis. Both bikes are on the edge of locking the tires. But since the heavier one makes the tires 'dig in' more, it has the ability to sustain a higher braking force before the tires break loose.

[jlv]

Even if you lock everything up it should still be the same as long as the static friction coefficient is the same as the sliding coefficient.

[Prologue]

Even if you lock everything up it should still be the same as long as the static friction coefficient is the same as the sliding coefficient.

Yeah, definitely. I was just talking about the shortest braking distance but you are right, lock em up and it is still the same.

[Prologue]

Sorry to double post. But jlv, the static doesn't have to be the same as the dynamic, the static on one bike (tire) just has to equal the static on the other and the same for the dynamic (sliding).

[Garasaki]

You guys are giving me a headache.

An interesting set of assumptions indeed. The arguement seems a bit circular to me. Assume traction remains equal then make the braking force heavier on the bigger bike - because it can get more traction because it has more mass...etc

We'll just say the heavier bike takes longer to stop then??

I imagine the effects of the suspension and overall geometry of the bike make calculating the true friction force quite hard. Nevermind the interaction of knobbies on dirt. Would be interesting to see that attempt though.

I still feel quite strongly that the actually added weight would have no real impact on the friction force (I'm having trouble expressing this elequently nor do I have mathmatical evidence, but I think the center of gravity and the suspension design has more influence on the weight pushing down (normal) on the front wheel then does the actual mass)

Furthermore, the beauty of MX is that the rider:

1) Has a mass approximately equal to that of the bike
2) Has the ability to reposition that mass, greatly affecting the center of gravity

But anyway, I'm going to try my best to not participate further - for my own sanity

[Voutare]

I can stop faster in my empty pickup than I can when the 29' camper is behind it.

[jlv]

That's because it isn't adding downforce when you tow it. If you crushed the camper and put it in the bed it would have about the same stopping distance as unloaded.

[yomo]

I know this isn't exactly the right place to add this...

But with the antilock brakes in game, do they calculate the perfect about of braking force before the tires slip? Or is it better to use analog braking and get it perfect that way (although it is hard to do).

Also if the theoretical distance for two cars to stop on a flat dry surface is only dependant on the amount of friction which doesn't matter on the mass thus any two cars will have the same theoretical stopping distance. Why can a formula 1 car out-brake a standard car? Is the only reason because formula one cars have brakes that can apply more braking force (And of course using aerodynamics and suspension to its advantage)