Newsgroups: rec.autos.driving
Subject: Braking Distances
From: cal@codewks.nacjack.gen.nz (Cal Gulliver)
Date: Sat, 21 Nov 92 02:51:19 NZST
Organization: The Code Works Limited, PO Box 10 155, Auckland, New Zealand

As Paul Stafford said to All on 11-12-92:
 
PS>FROM pauls@hpnmdla.sr.hp.com
PS>TO All
PS>
PS> Well, here is another try. Anybody find anything wrong?
PS>  there are TWO independant equations here.
PS>
PS>      S = V0t + 1/2at**2
PS>
PS>and V = V0 + at
PS>
PS>  s     distance (feet)
PS>  V0    initial speed ( in feet per second)
PS>  t     time  ( seconds)
PS>  a = acceleration ( in this case, deceleration) (feet per second 
PS>squared)
PS>
PS>  we assume braking to zero velocity, so V = 0
PS>
PS>    solving for t = V0/a ( assuming constant braking force,
PS>                            i.e. a != f(t)   )
                   ^^^^^^^^  here's the mistake:
                             should be  t = -V0/a
 
PS>
PS>      s = V0*V0/a + 1/2 a *( (V0 * V0)/ (a*a))
PS>        =  1.5*(V0*V0)/a
             ^^^^^^^^^^^^^^
should be:   -0.5 V0*V0/a   (remember, acceleration is negative,
                             so this will yield a positive distance)
 
PS>
PS>  or in other words, proportional to velocity squared, and 
PS>inversly proportional to deceleration force- what we might expect.
 
True. But ...
 
PS>
PS>     the deceleration "a" = weight of car * coefficient of
PS>                                            friction
PS>                          = mgu
 
No. That gives you the decelerating force. Divide the force by the
mass of the car to get the deceleration rate, i.e.
 
                     a  = - (m * Cf * g / m)
 
Mass cancels out, so we get:
 
                     a  = -Cf x g                    ...1
and                  s  = -V0 x V0 / 2 x a           ...2
Substitute 1 in 2:   s  =  V0 x V0 / 2 x Cf x g
Rearranging,         Cf =  V0 x V0 / 2 x s x g
 
Substituting your typical stopping distance of 150' (45.7m)
from 60mph (26.8m/s), and g (acceleration due to gravity)
of 9.8 m/s**2,

we get:
                     Cf = 0.80
 
which is exactly what you'd expect on a dry asphalt road.
 
Check out some AA road tests of new cars sometime. You'll see
stopping distances that yield coefficients of friction of
(usually) 0.8 to 0.95. But the maximum possible coefficient is 1.0,
and the motorcyclist who stops from 60 in 114' (I think I have the
figure right) does slightly better than that! But the following
factors can get in the way of accurate measurement:
      Air resistance adds to deceleration;
      Aerodynamic downforce increases braking force;
      Many speedometers read fast;
      It's hard to tell exactly where you started braking,
           or to brake at a particular point - and if you
           measure the length of a skid mark, some braking
           will have been achieved before the wheels locked.
 
And yes, the biggest truck can stop in the same distance as the most 
British of sports cars, all else being equal (i.e. air resistance, 
aerodynamics, road surface and weather, brake and tyre condition).
 
 
 
As Andy Lowry said to All on 11-13-92:
 
 
AL>In article <1992Nov13.070620.19475@nntp.hut.fi> 
AL>jaa@hutcs.cs.hut.fi (Jari Alasuvanto) writes:
AL> > Sorry, in a message I posted earlier I stated that the
AL> >coefficient for friction is around 0.3 for dry road and good
AL> >tires.
AL>
AL>It seems to me that friction between tires and road is the wrong 
AL>thing to be looking at here, in trying to determine the
AL>deceleration during effective braking.  For that to be a factor,
AL>we're assuming the wheels are locked and the car is sliding along
AL>the surface.  Effective braking would instead have the tires
AL>rolling along the surface and the brake pads sliding along the
AL>disk or drum, right?  So isn't the friction we're concerned
AL>with the friction encountered by the brake pad?
 
Braking is indeed more effective if ALL wheels are just on the verge
of locking. When the wheels are locked, melting rubber and bitumin
form a lubricant, reducing the coefficient of friction. Typical
coefficient of sliding friction is 0.6, compared with 0.8 as above.
 
Unfortunately, weight distribution changes in braking - the front
wheels carry more weight than the back. The braking power (the
force exerted on the discs or drums) needs to be unevenly distributed
to match, in order that all wheels can be simultaneously about to
lock. However, the bias needed varies with the type of road surface
and the optimal adjustment for a hard, dry surface will be very
wrong for a wet or gravel road.
 
Jearl Walker, Scientific American's Amateur Scientist, goes into this
question very thoroughly in the Feb., 1989 issue. He concludes:
 
JW> ... locking all four wheels and sliding may be the best strategy
JW> in some emergency stops if the idea is to minimize the stopping
JW> distance.
JW> ... one serious flaw, however: if the car is fully sliding, you
JW> have virtually no control ...
 
 
AL>  ... (Of course, to determine the deceleration caused by that
AL>friction, we need to know how hard we can brake without causing
AL>the tires to lose their traction, and that *is* related to the
AL>friction between the tires and the road surface.  But assuming
AL>we're just teetering below that threshold, braking distance is
AL>determined by the deceleration caused by friction experienced
AL>by the brakes themselves, not the tires.)
 
You're right: the kinetic energy can be converted to heat by friction
at the disc/pad surface or the tyre/road surface, or a mixture of
both; the maximum retarding force at each surface is its coefficient
of friction multiplied by the force with which the two surfaces are
pushed together. 
With a strong leg and/or power brakes (and even little cars have 
vacuum-assisted brakes these days) you can always lock the brakes,
so the upper limit on braking performance is the tyre/road friction.
 
 
By far the best answer, of course, is ABS (Anti-lock Braking System).

--
cal@codewks.nacjack.gen.nz (Cal Gulliver)
The BBS Works -- +64 9 630 7739 NZL	New Zealand's Oldest BBS
Newsgroups: rec.autos.driving