Question about the Physics Model

[Prologue]

Hey jlv, I've got a couple questions about the way you've modeled the physics. Basically a few people were discussing braking distances on the server and it got me thinking, 'I wonder how the game models it?'. So basically some people were claiming the stopping distance was shorter on the lighter bikes and the only way I could figure that would happen would be that the game considers air resistance.

MY conclusion is based on the assumption that:
1) The tires and brakes are the same for all bikes
2) and the aerodynamic cross section is the same for all bikes (if the aerodynamics come into play)

If aerodynamics were not coming into play then the only force that would be stopping the bike is ground friction. This friction varies linearly with the mass of the bike. So when the force needed to stop something goes up (because it is heavier), the ground friction goes up because the tire 'digs in' more which gives the additional needed stopping ability. And it is in the proper ratio to always end up with the same deceleration for a bike of any mass.

Some things not taken into account would be different moments of inertia and the irregularities in the ground, *this* could affect stopping ability as the suspension tries to react to the bumps. This brings up my next question. Is the suspension damping in the settings really only damping? Or is it perhaps a combo of damping and a change in spring rate? If it is not changing the spring rate then how is the spring determined for each bike? Varying these parameters would suggest that the bikes could have the potential to handle differently due to varying mass/spring combos.


Could you shed some light on this topic?

[Prologue]

The brakes don't have to be the same I suppose, they just have to be able to apply enough force to reach the point of barely locking up the wheel.

[Phathry25]

Your physics are flawed. Heavier bikes take longer to get stopped. Plain and simple.

The braking power is adjustable in the bike setup menu and can go higher than what a real bike/rider would be able to obtain.

[jlv]

I don't model wind resistance. At motocross speeds it wouldn't be a very large force.

Ground friction is downforce * the coefficient of friction and can also vary depending on slip rate now.

The suspension settings only change the damping.

[Prologue]

Your physics are flawed. Heavier bikes take longer to get stopped. Plain and simple.

The braking power is adjustable in the bike setup menu and can go higher than what a real bike/rider would be able to obtain.

An explanation of why I am wrong would be nice. I don't believe I am. Using the simple model where the friction force is proportional to the normal force you end up with an expression that looks like this

[Prologue]

Man, I really wish you could go back and edit posts on this forum! Sorry for the double post, I hit submit instead of preview.

My approach was to equate the kinetic energy with the energy expelled by the brakes (the work done by the brakes). Then solve for d, the stopping distance. Since the frictional force depends on mg (the weight), there is a coincidence within the equation that happens to cancel the mass, which means the mass doesn't matter. Of course, this is just a model (a pretty realistic one as low speeds) and the mass only doesn't matter within this model, but if jlv uses an equivalent technique then the result still works.

[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.

It's been a while since I looked at equations like that, but it looks like your final equation "d" is solving for the friction force.

It's fine that that happens to not include mass.

However, that friction force still has to be applied to the kinetic energy of the bike, which still includes mass as a factor.

[Prologue]

If the braking force (F in the equations) is constant, and d is the distance it moves then the work done is Fd. You can look at it two ways, maybe you apply an equal force but is not caused by the brakes, but is the same magnitude. If you start at zero velocity and then apply that same force over the distance d, then you will end up with a kinetic energy exactly equal to 1/2mv^2. Of course this isn't what is happening but doing the opposite thing (starting with 1/2mv^2 and braking down to zero) doesn't change the result. I just think that the 'speeding up and slowing down way' of looking at it is beneficial to visualize what the *work* means. It is one of those things that is always baffling. You've got this heavy thing and of course it takes more force to slow it down in the same distance as something with a smaller mass. But since it weighs more, the tires can dig in more, and that allows additional braking force. And it ends up providing the perfect amount.

[Prologue]

Sorry for the double post. I just realized I forgot a factor of a half in the last two equations, just forgot to insert them when I generated the images. So here are the corrected bottom two:




Still, it is independent of mass. I just had some time to rifle around on some websites and found these links

http://hyperphysics.phy-astr.gsu.edu/HBASE/crstp.html
http://www.physicsforums.com/showthread.php?t=236295
http://www.msgroup.org/forums/mtt/topic ... PIC_ID=119

For the most part my claim is substantiated in those pages for a real life scenario.


Now for the game. JLV, thanks for giving the info that 'Ground friction is downforce * the coefficient of friction and can also vary depending on slip rate now.'. I think this fits in with the model I was asking about as long as the wheels aren't slipping. But also, it probably fits if the tires are slipping the same amount too.

[jlv]

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.

[Phathry25]

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?

[DJ99X]

Most intelligent thread ever posted on this forum. I likes it.

I think his point is that Prologues maths is correct, but only in those conditions

[Phathry25]

I meant of the thread. Logic will tell me that a heavier bike takes longer to stop.

[jlv]

The same logic would tell you weighting the front wheel would make it more likely to slide. The truth is it doesn't matter much with dry friction.

[Phathry25]

You guys are all way over my head. So tell me why does it take longer for a heavier bike/car to stop?