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Re: Driving Pet peeves
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Subject: Re: Driving Pet peeves
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From: Michael Johnson <MICHAEL@MAINE.MAINE.EDU>
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Date: Wed, 21 Jun 1995 15:12:16 EDT
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Newsgroups: rec.autos.driving
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Organization: University of Maine System
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References: <3rabr4$fv@noc2.drexel.edu> <3raq3l$1v1@news.primenet.com> <3rc11m$1e6o@news.doit.wisc.edu> <3rhkeh$d67@dunx1.ocs.drexel.edu> <3ri1ga$k63@martha.utk.edu> <3rids6$1gdi@usenetw1.news.prodigy.com> <3rk853$ef@vixen.cso.uiuc.edu> <3rltq7$r73@acmex.gatech.edu> <95170.122611MICHAEL@MAINE.MAINE.EDU> <95172.074955MICHAEL@MAINE.MAINE.EDU>
In article <95172.074955MICHAEL@MAINE.MAINE.EDU>, I said:
>
> course, it is not going to be. The rate of velocity change is actually a
> parabolic curve, so that the velocity curve is a third-order curve and
> the actual distance traveled is a fourth-order curve.
Actually, upon reflection, the rate of velocity change (e.g. deceleration)
is a constant (A=F/M) so the velocity curve is a second-order curve and the
distance curve is a third-order curve.
There will be a little bit of deviation from constant deceleration due to
rolling resistance and wind resistance, but that only comes into play at
high speeds and favors the trailing car anyway.
Further on I wrote:
>3. I assumed that car 1 stops braking when it has changed speed by 10 mph.
> Usually sudden braking in highway situations will be like this. However,
> it is possible that someone might be making a panic stop, in which case
> their velocity continues to drop, at an increasing rate, while the car
> behind is losing velocity, but always at a lesser rate.
Actually, once car 2 gets on the brakes, the two cars will once again have a
constant (but non-zero) relative velocity.
> In this case, it DOES make a difference how fast you are going, because
> the time it takes to slow from 90 mph to 0 mph is greater than the time it
> takes to slow from 50 mph to 0 mph, during all of which time car 2 is
> closing on car 1 at an INCREASING rate, up until the time that car 1 comes
> to a stop, at which point car 2 is closing at a DECREASING rate.
The closure rate is actually constant, until car 1 comes to a stop, then the
closure rate starts dropping.
The rest of what I wrote still holds true. The only consequence of my
error (i.e. assuming that deceleration was parabolic) is that the necessary
following distance to avoid a collision decreases.
The time to go from speed to zero increases linearly with speed, since
the rate of velocity change is linear with respect to time. So moving at
90 mph means it takes twice as long to stop as it does from 45 mph. Assuming
a velocity drop of 10 mph before car 2 gets on the brakes, that means that
car 2 is closing on car 1 at 10 mph for the entire time they are both slowing
down. So for a speed of 90 mph, car 2 should be twice as far back from car 1
as for 45 mph in order to avoid a collision. This is why the 2-second
following rule works.
Note that this does not change the case where both cars decelerate by an
equal amount (e.g. 10 mph) from a given speed (e.g. 90 mph or 50 mph). The
only reason why following distances need to be longer for 90 mph than for
45 mph is that the total time before car 2 comes to rest relative to car 1
differs in panic stops from those two speeds.
Again, if car 2 has better brakes than car 1, then the safe following
distance decreases by some factor.
Pardon my errata, it's been a number of years since I did a physics lab.
Michael Johnson, Relay Technology, Inc.
michael@maine.maine.edu, michaelj@relay.relay.com
"I will choose a path that's clear. I will choose Free Will." -- Neil Peart
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