-=> On 09-05-08 05:41 TIM RICHARDSON wrote to JIM HOLSONBACK <=-


 RS>> First, we drove about 60 miles, between 65 and 70 mph.

 RS>> We refilled the gas tank, and drove the same distance, at
 RS>> 55 mph.

 RS>> It was at 55 we saved all the gas.

 bk>    F=MV^2

 RS>I'm afraid I'm really lousy at math, what does this mean?

 TR> F (Force) = M (Mass) times V (Velocity)squared!

 JH>Not only does BK have the formula wrong, it is not applicable to the
 JH>energy required to keep a vehicle is traveling at constant speed.  See
 JH>my post to Frank in this packet.

 JH>Reminder: the correct version of Klahn's formula is
 JH>Work = F x s = 1/2 mv^2 = kinetic energy

 TR> I realize that. The formula above is for a body moving at a constant
 TR> velocity, isn't it?

 JH>It is, but it is also correct for a given mass at a given instantaneous
 JH>velocity while it accelerates or decelerates.

 TR> By `formula above' I was refering to the formula Klahn tossed out there
 TR> in the beginning.

No, Klahn's formula isn't correct for _any_ case.  He did some googling
and found an article which seemed to agree that his formula was valid in
the case of an impact or collision, but the publishers later admitted
the formula was incorrect in that it left out the "1/2" factor.  By
reply to him, I showed how to correctly calculate the "force of impact"
using the correct formula and other formulas straight from Newton's laws
of motion.  Klahn's formula cannot be correct because the units don't
match.  For his formula, the left side of the equation is Force (pounds
in English units), and the right side works out to be a Force times a
distance (foot-pounds in English units).

 TR> The situation Sauer is talking about involves a time factor, and a non-
 TR> constant velocity.

 JH>To simplify the situation he presented, I think it is fair to assume
 JH>that the vehicle accelerated to highway speed, and pretty much held that
 JH>speed for the duration of the trip.  i.e. highway driving vs city
 JH>driving.

 TR> If his `friend' were using a `cruise control', the results might be
 TR> more acceptable.

The basic unacceptability of Sauer's conclusion was that they didn't
measure miles traveled vs. fuel used between fill-ups.  They just relied
on their observations of the fuel gauge in the vehicle, and fuel gauges
can be very unreliable, especially for a short trip just after a
fill-up.

 TR> I think the above formula has to do with kinetic energy or some such.

 JH>Yes, the 1/2 mv^2 side of the equation is the kinetic energy of a mass
 JH>moving at a given speed.

 TR> And the speed is `constant'.

No, as I tried to indicate above, that side of the equation is
instaneously valid as it passes thru a given speed all during periods of
acceleration and deceleration.

 TR> Nothing to do with fuel savings at varying speeds in limited time
 TR> periods.

 JH>Keeping it simple, best assume the trip was basically all at highway
 JH>speeds.

 TR> Given a distance traveled, a steady velocity of travel, and ignoring up
 TR> and down hill motion.

Again, no need over-complicate the situation.  I dunno how hilly it is
between Sheboygan and Green Bay.  In the general case of climbing a
hill, extra energy is required for that, and the extra energy ends up as
potential energy, which is recovered during an economical downhill run.

This can be looked at in terms of the left side of the equation,
(F (x) s). For a simple example, moving a 3300# vehicle to the top of a
10 ft hill adds 3300x10 = 33000 ft-pounds of potential energy.  Since 1
horsepower = 33000 ft-pounds per minute, it would use 1 extra HP to lift
the vehicle the 10 ft in one minute, and 20 extra HP to lift it 10 ft
in 3 seconds. At a speed of 100 fps (~68mph) for 3 seconds, the vehicle
would travel 300 feet, and the corresponding grade would be 10ft/300ft,
i.e. a 3.3% grade.

An exception to the recovery of that potential energy on the downhill
run is where downhill grades are so long and steep that the driver must
brake in order to reduce the speed of the vehicle.  In that case, the
potential  energy being recovered is lost as heat.

 JH>But Sauer's assertion was nut-numbingly dumb when you look at it a bit.
 JH>See the made-up scenario I posted-  With a 16 gal tank and 30mpg at 55
 JH>mph, the vehicle would only get 10mpg at 65-70mpg in order to save 1/4
 JH>a tank of gas.  (using 6 gals one way and 2 gals the other). My point
 JH>was that _nobody_ believes such a vehicle exists.

 TR> *Nor* does this latest of Sauer's *friends*!

It's hard to tell what Klahn believes in these cases, since he simply
loves to argue.  In the thread in Debate, he is now reduced to
what I'll refer to as full Klahnian debate mode, consisting
entirely of niggling, quibbling and nit-picking, he being unable to
refute what I have written.

- - -  JimH.



... "Jim, how do you spell 'argumentum ad nauseum'? - - Bubba
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