energy-conservationeveryday-lifenewtonian-mechanicsrotational-kinematicssimulations
I am writing simple car simulation.
Assume non friction, then in straight line the car doesn't lose speed. But what if the car is turning, there should be some kinetic energy loses to change the direction of the car, so I should subtract some speed based on changing direction, isn't it right?
Deeper explanation is welcome.
It's hard to make the wheels spin at high speeds because you're in a higher gear, so the torque at the wheels is less. So I assume you are only asking about wheel spin in first gear i.e. it's quite easy to spin the wheels when pulling away in first gear but much harder if e.g. you're travelling at 10 mph in first gear.
The reason is that if you're stationary and drop the clutch the angular momentum of the engine contributes to the torque. That is, the torque at the wheels is the torque from the engine plus the torque from angular momentum stored in the flywheel, crankshaft etc. This happens because the engine is spinning faster that it would if the clutch were engaged, so engaging the clutch slows the engine speed. The extra torque is given by:
$$ \tau = I\frac{d\omega}{dt} $$
where $I$ is the moment of inertia of the spinning bits of the engine and $\omega$ is the engine speed, so $d\omega/dt$ is the rate of change of engine speed. If you drop the clutch the engine speed changes rapidly so $d\omega/dt$ is large and the extra torque is large. If you ease the clutch out $d\omega/dt$ is small so the extra torque is small and the wheels won't spin.
When you're driving at (e.g.) a steady 10 mph the engine speed matches the wheel speed, so if you now suddenly stamp on the accelerator it's only the torque from the engine that's available to spin the wheels. You don't get the contribution from $d\omega/dt$.
To see this try driving at 5 mph, then disengage the clutch, rev the engine and drop the clutch. As the clutch bites the wheels will spin just as they do when the car is stationary.
It's worth noting that a powerful car can spin the wheels in first gear even without playing with the clutch. In fact an old sports car I had many years ago would spin the wheels in second gear in the dry and in third gear if the road was wet!
The whole point of braking is to dissipate kinetic energy. Not the kinetic energy of the wheel as you said, but the kinetic energy of the car, even though you may do that through transmission to the wheel. Some trucks or busses actually brake by transforming part of their KE into electricity, which may sometime be reused, or is dissipated into heat as eddy (or Foucault) currents.
However, the most common way to dissipate kinetic energy is friction. In the case of cars there are two possible frictions : bretween the brake and the wheel (not the rubber itself hopefully) and between the rubber and the road.
But there is energy dissipation only if there is motion with (kinetic) friction creating a resisting force (in the case of friction braking). The word kinetic is in parentheses, because it may require some further precision (see below).
When the car is rolling normally, there is no (or marginal) kinetic friction because the wheel is at rest relative to the road in the contact part. If you brake, this may no longer be true, because the wheel may not turn fast enough. On some surfaces, like a wet road (but apparently not all surfaces) the friction is more important if the speed of the wheel part in road contact is not too important relative to the road. Beyond a certain speed, the tire can even sort of surf on a thin layer of water, and the friction goes down, thus dissipating less energy. This happens much faster if you block the brakes.
So, with the brakes blocked, there is no energy dissipated by friction in the brakes, and the wheels may be skidding too fast to dissipate energy efficiently. Hence, it take a longer time to dissipate, meaning a longer time to stop.
The ideal situation is dissipating energy both in the brakes and in the rubber. But that is not easy to attain, because the static friction coefficient is usually greater than the dynamic coefficient. As soon as the wheel starts slipping, the friction reaction force of the wheel that preserved some motion in the brakes may become too low for the brakes to allow for motion, and the brakes block, no longer providing any dissipation, and increasing further the skidding speed of the wheel.
ABS prevents blocking the brakes by removing briefly the friction, and allows the wheels to turn some, so that the relative speed of their contact with the road does not get too high.
But why should it work on a dry road ? According to Wikipedia, there is another phenomenon to be considered. The transition from static to dynamic friction coefficient is not a discontinuous phenomenon. Apparently the "maximum braking force is obtained when there is approximately 10%-20% slippage between the braked wheel's rotational speed and the road surface", beyond which "rolling grip diminishes rapidly" to kinetic friction. So that is where the heat dissipation is at its maximum, since maximum dissipation requires maximum motion with the greatest motion compatible friction (actually, it is the product that is to be maximized). The role of ABS will be to let go when the slippage becomes too important so that the slippage remains in the optimal range (in addition to above issues).
But apparently some surfaces behave differently, and ABS may actually brake more slowly. I would guess that this is due to the specific properties of the function that relates the friction force and the slippage speed for that kind of surface in contact with rubber wheels. But on such surface, the advantage of keeping better control of the car, by slipping less, is also an issue.
Another role of ABS systems is to distribute the braking effort between front and rear wheels. Front and rear wheels have different internal pressure, thus different contact surface with the road. They are also subjected to different forces as the car is braking (more force in the front), so that the friction coefficient acts more effectively where the force is greater. Hence slippage control has to differ in the front and in the back. It may also balance left and right if for some reason the two sides behave differenlty.
A last issue was actually raised by @tohecz. Where should the energy be dissipated, or according to what ratio between brakes and rubber-road? His opinion is that it should be in the braking system, not in the wheel-road contact. I did not find any information stating that, if there is a choice, it should one more than the other, but it may indeed be preferable to spare the tires (I do not really know). It is however worth considering the issue and the degree of freedom of choice.
We can analyze somewhat this ratio by considering extreme cases. If you block the wheels (assuming no ABS), no energy is dissipated in the brakes. Thus it is all dissipated in the rubber-road contact. On the other hand, if you brake slowly, the wheels surface remains in static contact with the road (no ABS needed) and all the energy is dissipated in the brakes. This runs contrary to some belief that violent braking could heat the braking system: frequent and slow braking will, while violent braking without ABS will heat and wear the rubber.
So the question of the ratio, with an ABS system, occurs really only when you brake strongly enough so that wheel slippage will occur and the ABS can be used to control it. Here a proper analysis would really require working on actual figures, as there are many possible scenarii.
It should be the case that optimal braking, with fastest energy dissipation, will impose a precise pressure on the brakes resulting in a precise dissipation ratio between brakes and rubber. However, given the hiccup behavior of ABS system, this corresponds probably to an unstable setting requiring a dynamic control of the pressure so as not to leave the optimal dissipation zone. I did not find any information about this ratio.
If the pressure on the brake pedal does not indicate urgency for fast braking, the ABS system can probably choose, according to its programming, what amount of pressure to apply, and when, so as to determine where most of the energy will be dissipated, between brakes and rubber. But there does not seem to be much public information on that.
A last remark is that the choice of optimal pressure for whatever result is desired should also depend on the current speed of the car. It is probably hard to get any slippage from a very slow car. Hence the process has to be dynamically controlled for that reason too.
Note: In this analysis of ABS braking, the careful reader will have noticed that I talk of forces, when actually it should be torques in many cases. My reasons for doing this are the following:
the main issue is friction and friction forces, which become torques because of the structure of the devices considered;
talking of torque would necessarily require the description to introduce size considerations (wheel and brakes radius), which would complicate the analysis without bringing in any essential insight regarding ABS;
this is just a qualitative analysis, without using any actual figures. Developing complete formulae would of course require to bring in size issues, and to consider torques. But I deemed it simpler not to do that here.
To dissipate any misunderstanding and any heat that could result from it, I should make it clear that this looked to me like an interesting problem to work on, but that I have no particular expertise, and I did what I could with the information I could find. Comments and criticisms are welcome.
Best Answer
Assuming non friction, we must assume the car turns because it is held by a rope or chain to a point. The turning itself will not take energy from the system, what you have now is a constant speed that is tangent to the circle the car is following at the current car position.