I understand that there's a language barrier, so I'll try my best to correct misconceptions where there may be any. You may already know some of what I said in Dutch, but just not expressed correctly in English; in which case, I apologize. Note that I am (and has been) very specifically speaking about oversteering not due to increasing engine output (putting your foot to the accelerator pedal).
If you have a car with 50/50 AWD, the force on the tires are distributed 25% on each of the 4 wheels, when a 50/50 AWD is driving in a straight line. With a FWD or a RWD, the forces on the tires are ditributed over only two wheels.
With "forces" I mean the force applied from the engine power.
With that last caveat, I guess that's a fair generalization. I think something like "engine forces" will be a better term because just "forces" is too general. Technically gravity is exerting a downward force and the road is exerting a normal force that is equal and opposite of gravity (assuming the road is flat). Just because the two forces don't cancel each other out does not mean that they don't exist. There are more forces acting upon a tire than just forces due to engine output.
As I said, the speed of a WRX in a corner is higher without losing grip than with a FWD/RWS.
Why? (Let me give you a hint, if two cars are identical in every way other than one is 2WD and the other is 4WD, then the maximum force that a tire can handle without sliding is the same)
In that case, because the speed is higher, the forces are higher and you will exceed the sideway forces a tire can handle faster than you think.
It's really not the speed that is higher that's causing the forces to be higher. It's that a higher speed (vector-less, direction-less) causes a higher
acceleration in order to maintain that speed, thus the higher
acceleration that is causing a higher force.
If you say that changing larger velocity causes a larger force, then that is correct, as the change in velocity (directioned vector) is acceleration.
Btw, physics wise, acceleration is a vector with a direction. If I'm traveling straight at 100kph, then turn left 90 degrees while maintaining 100kph, then keep traveling left at 100kph, I would have accelerated, and my velocity would have changed, but my speed wouldn't have.
However, with that said, the whole thing is based on the premise that the WRX will automatically be able to go through a corner faster, which isn't necessarily true.
Don't forget that in my previous posts, I mentioned shifting the weight of the car onto the front tires, which makes the rear tires lose grip even faster.
True.
So, you can definitely oversteer a WRX easily, as long as the speed is high enough (that was my argument in previous post ==> showing off to friend etc.... ).
Yup again. No argument about this.
Let's take one wheel of the front suspension for instance. Driving straight forward, there is only one force on the tire.
Only one
net force. There are multiple forces acting on the tire, which aren't too important when traveling at a constant velocity, but they do get very important very quickly as soon as you accelerate.
A force in a straight line. If you turn a secondary sideway force is applied to the tire. If the sideway force exceeds the straight forward force and the tire isn't designed to cope with this sideway force, this tire will lose grip and you have understeer.
I'm not 100% sure if you are saying what you said up there, but ummm...
So if you turn, a "sideway force" is indeed added, but it's the sum of the "sideway" and "straight forward" force that matters. If the sum of the two forces exceed the tire's maximum static friction (the tire's grip capabilities), then the tire will start to slip.
When presented with two forces that are at 90 degrees to each other, then you can sum them up by essentially drawing a diagonal across the rectangle that the force vectors makes. In real life, that's not necessarily the case as force vectors may not necessarily be at right angles to each other, but we'll simplify it so that they are.
If the forces are applied 25% on each wheel, it takes longer for the forces to exceed the forces a tire can handle thus making a 50/50 4WD more grippy than a FWD or RWD with all the engine power on only two tires instead of 4 tires.
It's not that a 4WD car is "grippier" than a 2WD car. The two have the same amount of grip at the tires. The maximum amount of static friction a tire has does not change whether it is 2WD or 4WD. It is that a 4WD car can put a greater amount of power to the ground when accelerating out of the corner without slipping than a 2WD car because as you said, the forces are distributed to four wheels instead of two.
But then this is untrue during lift off oversteer as there's no engine power at that time (technically there's engine braking, but let's ignore its effects for now).
The above example is theoretical and not entirely correct because I didn't mention the mechanics of a differential, the suspension in general etc... .
I don't know anything about the mechanic's and electronic's of a car.
Those don't matter. It only matters that we assume two cars are exactly identical except one is 2WD and the other is 4WD. This doesn't happen in the real world, so the practical effects may be slightly different, but as a thought process, one must assume that the only difference is the number of drive wheels, otherwise, one would be unable to attribute what caused the difference in vehicle behavior.
@Doog In the states do you have to be a legal adult to buy a car under your name, over here you don't.
Yes, in the US, you must be a legal adult (or be an emancipated minor) to own a car in your name, as with any other property. When you're not a legal adult, technically everything of yours is owned by your parent(s)/legal guardian(s).
I highly doubt that property can be owned by a non-legal adult in Canada.
