# More on turning and braking

I can't let it go. There is more here to explore. First, I can't believe that I looked at braking and then turning but not turning then braking. And what about braking while turning? I will model braking and then turning - but it won't stop here.

Consider a few motions. I could turn and then brake (which I am looking in this post). Another option would be to brake and then turn. I already showed that this takes a longer distance than just stopping. Braking and then turning in general won't work. Suppose I brake and slow down to a certain speed. Now I am a certain distance away from the wall going at a certain speed. In that previous post, I showed that it takes less distance to just stop rather than turn.

Turning first might be different. If the car turns a little bit, then there will be a greater distance that it can travel before it hits this really wide wall. So, here is what I am going to do. I will use python to model this motion and calculate the x-distance it takes to stop the car. For the first run, I will just go straight. Then I will turn just a tiny bit and then brake. I will keep increasing the amount the car turns before stopping and record the distance (in the direction of the wall).

Here is the data: First, a quick check. Does this data agree with my previous data? Using these same parameters (initial speed and coefficient of friction) I get the same stopping distance for no turn and 90 degree turn. So that is good.

I am not surprised that stopping without turning is still the shortest distance. But, I am surprised that turning 90 degrees is not the largest stopping distance. From the above data, it looks like the worse thing to do is to turn about 60 degrees and then stop. Whatever you do, don't do that.

One more thing - there is more to do. I have looked at:

• Braking
• Turning
• Braking then turning
• Turning then braking

But this is not every possibility. Is it possible you could turn right then left then brake? Maybe. There is a way to deal with these more generic functions - but I am not going to deal with this now.

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> Is it possible you could turn right then left then brake? Maybe.

But it will have a longer stopping distance than just braking. See my comment to your last post - you can show this is the best strategy without any calculation (assuming your starting velocity is normal to the wall and the wall is infinitely wide).

> But, I am surprised that turning 90 degrees is not the largest stopping distance.

It's an interesting result, but easily understood.

The more you've turned (i.e. the larger the angle between your velocity vector and the normal to the wall) the more "efficient" turning becomes: in the limit that you're moving parallel to the wall, 100% of the acceleration of turning is normal to the wall.

So if you're already at a large angle (having turned, say, 70%) it's now better to turn (where most of the acceleration is going into reducing the normal vector to the wall) than brake (where most of the acceleration is NOT reducing your velocity along the normal vector).

I tried a little bit of back-of-envelope math, and it put the "tradeoff angle" happening at the solution to the equation (1-sin(x)) = (cos(x))/2, which is around 37 degrees. This is the angle of travel at which it's better to turn rather than brake, and does NOT include the braking vs. the turning required to get to that angle (assuming starting normal to the wall), but with a little bit more work you should be able to turn this into your numerical result.

By Anonymous Coward (not verified) on 16 Aug 2010 #permalink

Ok, what about a 180 degree skid and accelerating, with smoking tires and a loud pounding bass line, and a supermodel riding shotgun, twins (with puppies) in the back seat, and a MacGuffin in the trunk? Does it make a difference if you are Bruce Willis?

By Anon (not verified) on 16 Aug 2010 #permalink

It's fairly obvious that the optimal solution is to always have 100% of your deceleration be normal to the wall. Let's break your force into two parts: Normal Force and Parallel Force, and similarly normal/parallel displacement and velocity. It's fairly simple to see that you will stop approaching the wall when (NF * ND) + 1/2 * m * NV^2 = 0; parallel force and motion is entirely irrelevant.

If you start moving straight towards the wall, 100% of force being away from the wall is deceleration. If you don't, it's turning, because the optimal solution doesn't involve any change in PV, and in fact your PV should remain constant over the entire period of the turn.

So, at any given angle, how much of your force will be turning, and how much will be braking? Well, the turn vector will be a force at 90 degrees from your current course, and the deceleration will be a force at 180 degrees from your current course, and you want them to add up to a force which is directed away from the wall, your braking winds up being sin(approach angle) * coefficient of friction, your turning winds up being cos(approach angle) * coefficient of friction.

By Anthony (not verified) on 16 Aug 2010 #permalink

Oh, and, since the normal force is constant, your path when using an optimal braking solution when coming in at an angle will be a segment of a parabola.

By Anthony (not verified) on 16 Aug 2010 #permalink

I'd agree with #3 that, in the case with a barrier of infinite width, the best case is the one in which all of the force is exerted normal to the wall.

Of course, in the more realistic case of an object with finite lateral extent, it may be better to swerve than attempt to stop. Of course, this depends heavily on the width of the object to be avoided. In the limit of an infinite object it should reduce back to the results you are getting here.

By Tercel (not verified) on 16 Aug 2010 #permalink

I suspect if you look hard enough an autos motions can be decomposed into a few basic types, just like a planes motions are in Stick and Rudder. A plane basically can fly level, climb, glide (decrease altitude) and turn, everything else is just a superpostion of these. With these motions you can get planes to slip and skid (rudder versus aileron motion I don't get the details) and stall and spin (bad news) Unless you assume an asymettric auto, then a car can accelerate, slow, or turn. Again its just a super position for the more complex motions. The symmetry assumption is to make left and right the same otherwise you take turns to left turns and right turns.

By Lyle (not verified) on 16 Aug 2010 #permalink

You know there is one other option no one mentioned: turn and accelerate. It is well known that you can get more centripetal force from the drive wheels when racing on surfaces like dirt and ice than you can from the lateral grip of even the most expensive tires.

Now in this case you don't turn the motion of the car, you turn the car relative to its motion so the nose is pointing somewhat towards the center of the turning circle followed by the center of mass. The advantage of this is tied into the fact that the coefficient of friction is not a constant independent of the motion of the tire relative to the surface. However, I believe most of the analysis remains unchanged. In this last case, the shortest stopping distance would be found by snapping it around in a 180 and flooring the gas rather than the brake.

By CCPhysicist (not verified) on 22 Aug 2010 #permalink