Conference rusure::math

>    Consider a car coming to a stop sign: it slows, stops, then accelerates.
>    Now consider a convoy of cars, with each car maintaining a distance from
>    the preceding car which is one car length (say 15 feet) for every 10 mph
>    velocity of the car.
    
>    Can the convoy of cars legally negotiate the stop sign?  How?  Can this
>    be solved so that each car follows the same 'curve', just time-displaced?

1) Most modern cars are actually 18-20 feet long. I once had reason to 
measure my Chevy station wagon and found [to my surprise] that it was
almost exactly 20' long, and on the same wheelbase as my sedan. 

2) I think what you want to know is how fast the convoy can be moving and 
still clear the intersection without creating a queue. I.e. how far behind
the second car must be to avoid hitting the first. For that, the second car
should arrive near the tail of the first, decellerating, as the first pulls
away. Relevant data includes stopping curves, starting curves, and how long
it takes to go from convoy speed to 0 within 20 feet. That last can be found
from Consumer Reports! 

A complication where I live is VA law that the minimum duration of the stop
at a sign is 3 seconds. [Nobody observes it, but wotthe'ell] 

    Just yesterday, a friend of mine who is a professor of mechanical
    engineering mentioned this problem and how he uses it in class. 
    The rate of decelleration is greater for each car whose driver's
    reaction time exceeds the time necessary to cover the distance to the
    next car.  Obviously the decelleration must get everyone to a velocity
    of 0 at the same time to avoid collisions.  
    
    fwiw,
    
    Dick

>    ... Obviously the decelleration must get everyone to a velocity
>    of 0 at the same time to avoid collisions.  

Sounds like a somewhat different problem. I assumed the problem was for 
each car to reduce speed *only* for the intersection, not for other cars 
stopping ahead of it. My interpretation requires the cars to be separated 
enough so the behavior of the first car has no effect on the second, etc.

What say you, Peter?

    I worked a problem similar to this in grad school.  Turned out that the
    most important factor was driver reaction time (i.e., how long it took
    for each driver to witness the deceleration of the vehicle in front & 
    begin to decelerate.  (We weren't working with convoys)
    
    One of the most interesting scenarios we modeled involved accidents - 
    when a "lead" car came to an abrupt stop, the second and third cars 
    could successfully avoid an accident but the fourth and fifth would
    collide.  Note:  The magnitude of the deceleration was a function of
    the distance between the driver and the car in front of him.

I used to wonder why the "stop line" was so far back from the intersection.
To be completely legal AND safe, it required me to stop at the stop line, then
crawl to the actual intersection and check for traffic, then go.

But coming out of the DSG plant here in Westford, Mass. I've made an interesting
observation.

If I stop at the line, and look for traffic coming from my left, I can only
see so far, before the road curves.  If I crawl to the intersection, I can
see farther, and check a greater distance for on-coming traffic.  All that
is not the interesting part.

Oh, I need to mention that this stop sign is at a T intersection, and you can
only go left or right, and I always turn right on my way home.

The interesting observation I made is that if I ONLY stop at the stop line,
and don't see any oncoming traffic from the left, and then I go and DON'T
stop at the intersection again, although I wasn't able to check farther down
for traffic, I don't need to, since my added velocity as I enter the
intersection decreases the required safe distance to the nearest oncoming
traffic !

So there's a tradeoff here, and an interesting one.  There's a class of
intersections at which stopping at the official stop line decreases your
vision distance, but that's o.k. because you can enter the intersection with
enough velocity so that any cars that were blocked from your vision are
too far away to endanger you anyway.

Putting it another way, there's a set of distant traffic that will hinder me
if I stop at the intersection but will not hinder me if I only stop at
the stop line and not again at the actual intersection.

    What I was hoping for was something like this:
    
    Let f (t) be the position of the lead car as a function of time, f (t)
         1							      2
    is the the position of the second car, and so on.  So,
    
    	f (t) + K�f'(t) + l = f (t)
    	 2	   2	       1
    
    f (t) = position of the second car (where its front bumper is)
     2
    
    f'(t) = velocity of second car
     2
    
    K = 'safe distance' factor (the units are seconds, fwiw)
    
    l = length of the first car (and/or any car)
    
    f (t) = position of the first car (where its front bumper is)
     1
    
    Furthermore, both cars must stop at a stop sign (the same stop sign).
    Assume without loss of generality that the stop sign is at position 0,
    so that:
    
    	f'(t) = 0 when f (t) = 0; and f'(t) = 0 when f (t) = 0.
    	 1		1	       2	      2
    
    Finally, can a third car join the convoy?  Do the above conditions cause
    a fender-bender (if l > 0)?
    
    
    It'd be pretty (but not necessary) for the second and following cars to
    travel the same route as the first car, except with a time delay.  Thus,
    
    	f (t+j) = f (t)
    	 2	   1