Sorry it took me a little while to reply.
I'd be happy to expand a little on what I said. This isn't a really great way to have a back-and-forth conversation, but let me first ask:
Are you familiar with the "friction circle" concept?
If you are, here's a way of thinking about a couple of important cases for a race car in a corner:
First, consider the case of a simple circular track. If the car is following a circular path, then the combined vector of all of the forces from the tires must point towards the centre of the circle (centripetal force). That force will be both a combination of lateral force (in the car's frame of reference) and forward force (in the car's frame of reference) and the car will be yawed slightly so that the combined vector of those forces points at the centre. At maximum speed, that combined vector will of course be right at the outer edge of the car's friction circle.
OK, now let's examine a simple 90 degree corner, and let's assume that the car is at the perfect entry point for the corner at the maximum speed from which it can start. Let us further assume for this ideal case, that the friction circle concept is ideal and doesn't take account of the fact that there are times when the cars ability to accelerate (to generate force forward) is less than the friction the tires could deliver and instead, we are free to use the same maximum force in any direction with respect to the car's direction.
Now. Let's brake down the car's velocity vector in two components at 90 degrees to each other and with each vector pointing 45° from the car's direction of travel at the moment of corner entry. So there is one velocity vector parallel to the bisector of the corner, but pointed two the outside of the corner (call it "v"), and there another vector normal to that one, but pointed "across the corner" to where the car will end up (call it "u"). I don't want to draw diagrams at the moment, but can you envision it?
From the symmetry of the situation—with no adjustments for the difference between acceleration forces and braking forces, we can see that wherever the car is at the moment it begins cornering (after having braked in a straight line to get to the proper corner entry speed and location), it is going to be in the same position at the end of the corner; reflected across the bisector of the corner, but with the velocity vector changed only in the component parallel to the bisector. That component of velocity will be changed from pointing outside the corner to one pointing INSIDE. The other component—the component across the corner will be UNCHANGED.
So now we see for a 90° corner, we're changing one vector from v to -v, and therefore the force we need to put on the car to create that change (while never changing u at all) is going to be a force vector in the -u direction at all times. And since we're using an idealized friction circle, that force is going to be constant in magnitude at the outermost edge of that circle. All the while, the velocity, v, isn't changing.
In your mind's eye, imagine a corner coming down at a 45° diagonal from the upper left, then making a simple circular bend so that it exits at a 45° diagonal to the upper right.
Imagine a car at the entry point on the left. Moving across that corner horizontally at a constant velocity with its velocity in the vertical axis changing from straight down to straight up changing at a constant rate.
If the car really is capable of the same maximum friction force in any direction—a perfect friction circle, then the curve that describes the car's movement through the corner is a simple parabola.
