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I take it the angle of the steering wheel in a car is roughly proportional to the curvature, and therefore inversely proportional to the turning radius.

When I am driving slowly and want to turn the car, I have to turn the steering wheel at a very large angle to get it to turn in a reasonable amount of time, sometimes turning the wheel to the ends of travel to get it to keep to a small turning radius.

If I am driving fast, I barely have to turn the wheel at all to get the car to turn. Which makes sense, because the turning radius is much larger.

So here's my question: what happens if I were to drive down the highway at 65MPH and turn the steering wheel as hard as I could?

  • would the steering wheel reach the end of travel at a smaller angle?
  • would the steering wheel be too hard to move + thus prevent me from doing something stupid?
  • would the car flip over?
  • would the car go into a skid?

For obvious reasons, I don't want to try this myself.

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  • $\begingroup$ Depends entirely on the car. A car with anti-skid system will lock your steering, a car without such a system with low center of mass will skid and an SUV, truck or trailer with high center of mass will flip over. Pick your favorite scenario. Any of that can happen. $\endgroup$ – CuriousOne Oct 4 '14 at 5:12
  • $\begingroup$ You know about lateral acceleration $a_L = \frac{v^2}{r}$? $\endgroup$ – ja72 Oct 4 '14 at 22:39
  • $\begingroup$ of course I do... (aside from the fact that t.c. mentioned it) What I don't know is what happens in the steering mechanism and with the vehicle dynamics, where when you are steering the wheels turn at slightly different rates, so I don't know what's going to happen first, whether I'm going to lose traction, or whether something else is going to happen. $\endgroup$ – Jason S Oct 4 '14 at 23:36
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Regardless of whether you are driving at 65mph or 1mph, as long as your steering angle is the same, you will travel the same path. (Of course in real life understeer or oversteer has to be considered)

However, due to the much greater centripetal acceleration experienced, where it is proportional to the square of the velocity,

$$a_c = \frac{v^2}{r}$$

You will find that the centripetal force will be much greater.

The maximum speed at which the car can negotiate the turn is $v_{max}^2 ={\mu{}gr}$, where $\mu$ is the coefficient of friction.

If you are travelling at 65MPH with the wheel locked to one side, your $v$ would probably exceed $v_{max}$, and your car will no longer have enough traction to negotiate the turn, and skid off the turn.

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    $\begingroup$ Good start. You might want to expand your answer to discuss under what circumstances the car would flip over rather than skid (which was part of the question) - and maybe include why the Ford Explorer was prone to this when the tire pressure recommendation was lower than it is today... (not in the question but an interesting point to ponder when you think of what it takes to flip a car). $\endgroup$ – Floris Oct 4 '14 at 5:12
  • $\begingroup$ You can take the braking distance to calculate the coefficient of friction $\endgroup$ – Paul Young Jul 30 '19 at 19:40
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This is an actual test that's done by auto manufacturers and tire companies. It's call the "step steer" test. But it's not normally done to full lock (the maximum steering angle). Normally, temporary mechanical stops are installed in the steering system to stop the steering angle at a desired point that's determined by the engineers. (Source: I'm engineering testing manager for an auto company.)

I honestly don't know what would happen if you went to full lock at high speed but I strongly recommend against trying it! It's fun to speculate, though. First, yes, you absolutely would be able to turn the steering wheel all the way. But on the highway you'd already be off the road before you turned the wheel that far, unless you were able to turn it very fast. If we assume that there's enough room for you to stay on the pavement (i.e., more like a giant skid pad than a highway) then it would depend on how fast you turned the steering wheel. If you turned it rapidly enough you would most likely go into a skid, in most cars. There would certainly be lots of tire squealing. If you turned the wheel slowly enough you would, in most passenger cars, enter a turn in a severe understeer condition--i.e., the front wheels would be cocked hard over, squealing like heck, and the rear wheels would be just kind of following them around the curve. In more sporting cars (rear-engine Porsche, for example) or many racing cars you would likely end up in a spin even if you turned the steering wheel slowly.

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would the steering wheel reach the end of travel at a smaller angle?
NO. the range of travel of the wheel is set by mechanical parts in the chain. They don't know how fast you are going. There has been talk of "steer by wire" where a computer reads the angle of the steering wheel and decides how much to move the front wheels. As opposed to the throttle, I am not aware of a car that does this, but there could be a few. would the steering wheel be too hard to move + thus prevent me from doing something stupid? NO. the force on the wheel is quite light. would the car flip over? would the car go into a skid?
It is not a bad approximation to assume the tires can make a certain $g$ level in whatever direction you want (at low speed limited by the steering travel). On flat ground if the CG is high enough and the $g$ level is high enough, you will roll. If the CG is low or the $g$ level is low, you will slide. If you drive over a ramp with the inside tires you can increase the probability to roll. Before tires loose all traction they operate at a high angle of attack, where they apply force perpendicular to the orientation, but move outside the orientation. You may have read about oversteer and understeer. Oversteer happens when the front tires hold better than the rear tires (operating at a smaller angle of attack). The car tends to point into the turn and the driver must correct for that.

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