Hard braking down hill blowouts

On Sun, 6 Apr 2008 21:23:43 +0000 (UTC),
(Luns Tee) wrote:

In article ,
wrote:
For a 40 mph coasting-speed hill, the peak of the braking power curve
was around 23 mph. Faster or slower than 23 mph meant less braking
effort.

The relationship between 40mph and 23mph comes from assuming the
only forces at play are gravity vs. wind and braking, with gravity and
wind being linear and cubic with respect to velocity. The braking power
is then:

P= k1*v - k2*v^3

This being zero at 40mph tells us that k1/k2=(40mph)^2. The
maximum power is when the derivative of this WRT the velocity is zero,
or:
k1 - 3*k2*v^2 = 0 - v=sqrt(k1/k2/3) = 40mph/sqrt(3) = 23mph.

It would take considerable (and impractical) calculations to
discover which two points on either side of the peak of the curve
correspond to each other.

For any given velocity v, if there's another velocity v2 with
the same braking power, then

k1*v - k2*v^3 = k1*v2 - k2*v2^3
k1*(v-v2) = k2*(v^3-v2^3)
k1/k2= v^2 + v*v2 + v2^2 = (40mph)^2 (unless v=v2)

This is a simple quadratic, with solution of:

v2= sqrt(40mph^2 - 3/4*v^2) - v/2


That is, for 15 mph on the hill, the same braking power is needed at
some speed between 23 mph and 40 mph, but the precise speed is not
practical.

sqrt(40^2 - 3/4 * 15^2) - 15/2 ~= 30.3mph

This is as precise as the underlying assumptions are.

-Luns

Dear Luns,

It may be too much trouble, but can that be used to produce the graph
of speed 0-40 versus braking power from the old post, whose link is
now dead?

I think that Jim Smith was using your approach:
http://groups.google.com/group/rec.b...705c3d6e475419

But his graph is gone:
http://groups.google.com/group/rec.b...c12e0e024dcf14

Cheers,

Carl Fogel