Craig wrote:
>I've been trying to figure something out about slaloms
>and getting myself confused.
Oh man, you shoulda been here around four years ago. Right,
Byron?
>Problem is this: At the junctions between arcs, the
>car would have to have an instantaneous change in
>lateral acceleration from full-left to full-right in
>order to make this work. That obviously isn't
>possible.
Yep.
> The only way to get smooth changes in
>lateral acceleration would be to have the car take a
>sinusoidal path through the slalom, which looks a bit
>different.
Yep.
> Is this a better model of the car's path?
Much.
>Reason I'm asking is, I'm attempting to work out a
>physical explanation for whether a narrow car is
>faster through a slalom than a wide one.
No biggie. I'm gonna (roughly) repeat the model I posted several
years ago.
First, the assumptions: 1) the car follows a sinusoidal path, 2)
cones are equally spaced (you can account for varying spacing,
but it's better to start w/equal), 3) the component of the car's
velocity along the straight line connecting the cones is
constant, 4) lateral acceleration may be approximated as
occurring at right angles to the "cone line." Assumption 3) makes
the math much simpler, but, once you've acquired the necessary
insights, you'll see that it doesn't reduce the generality of the
results. 4) is only exactly correct at the instant the car passes
a cone, but that is also the peak value, which turns out to be an
important parameter.
The car moves along the x-axis:
y = Asin(omega*x),
where omega can be replaced by 2*pi/2*D = pi/D, where D = cone
spacing. A (amplitude) is the maximum deviation from the line
between the cones.
But x = V*t, where V = component of velocity along the "cone
line," and t = time. So
y = A sin[(pi/D)*V*t]
And
dy/dt = A*V(pi/D)*cos[(pi/D)*V*t]
Further
d^2y/dt^2 = - A*V^2(pi^2/D^2)sin[(pi/D)*V*t], which is our
"lateral acceleration" approximation. See assumption 4) above.
Of interest is the maximum value for lateral acceleration, which
is A*V^2(pi^2/D^2) and occurs at the point of maximum
displacement from the "cone line," which, if you're going the
quickest way, will occur just as you pass each cone. Given a
car's peak lateral acceleration capability P and a value for A,
you very easily come up with a maximum realizable value for V,
your velocity along the cone line:
V = (D/pi)*sqrt(P*32.2/A)
For example, if your car can corner at 1g (about 32.2 ft/s^2)
and it moves 10ft to either side of the cones in a 60 ft. slalom,
(ain't simplicity wonderful?), then, using 32.2ft/s^2 for the
acceleration of gravity, we get
V = 34.2 ft/s,
which is about 23 mph. Now, consider what happens if we bring A
to a minimum. If you car is, say, six feet wide, and the base of
a cone is 1 ft (for simplicity, remember?), then the absolute
minimum you can move laterally and not hit the cone is 3 /12 ft.
In that case,
V = 58 ft/s,
or about 39 mph. Keep in mind that's the same car and the same
cornering capability, just a reduction in the lateral
displacement.
Another item of interest: the time from cone to cone will be D/V
= (1/pi)*sqrt(A/P), which is a constant independent of D. Ergo,
cone-cone time for a given car will be the same for an open
slalom as it is for a tight one. The increased velocity just
makes up for the longer distance traveled in the more open
slalom.
>Craig "yeah, I KNOW I should get out more" Blome
Well, there's a little geek in lots of us. ;<)
Jay
|