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Old 05-19-2008, 01:24 PM   #1
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Rogers' Transportation Law

Hello, folks,

Here is a tool to help people save fuel while driving--

ROGERS TRANSPORTATION LAW

The following law came down to me just before I dozed off last night?

Δ gallons = - 1/30 (mph /mpg) (Δ minutes)

This simple relationship tells you how fuel consumed on a trip changes if you change your speed. If you adjust your driving to change the travel time by ?Δ minutes?, then the amount of fuel used will change by ?Δ gallons ?.

In simpler terms, suppose you want to get to your destination 30 minutes sooner. The amount of extra fuel used will be your travel speed divided by your miles-per-gallon. Conversely, if you lengthen the travel time by 30 minutes, the gallons of fuel saved will be the speed divided by your mpg. If you want to save more (or less) time, the amount of extra fuel will be proportionately more (or less).

Here is the way the formula looks for Europeans?

Δ litres = - 1/3000 (kph) (litres /100 km) (Δ minutes)

The formula is strictly true for small changes, and for a car where fuel consumption varies as the square of velocity. Notice that the formula does not depend on the length or duration of the trip. (Or when you initiate the action.)

Obviously, ?your mileage may vary.? This is a rule-of-thumb kind of formula, as anything applying to a complicated transportation system must be. Most people only have a rough guess for the ?mpg? anyway. Some may ask which speed to put into the formula. Probably the higher speed is best since most people will guess a little high on the mpg number.

The formula can be generalized to work for almost any mode of transportation. In which case, it might look like this?

ΔFuel = - n V (Fuel /distance) Δt

where fuel consumption varies as the nth power of V, and the formula is strictly true for small changes. An empirical choice for n might work best. Such could also be the case for an automobile, and then you would change the 1/30 constant.

Let?s say you travel about 70 mph, your car gets 25 mpg, and you slow down, lengthening the trip by 10 minutes. Here is your fuel saving?

Δ gallons = (10 /30) x (70 /25) = 0.93 gallons

So, how much were you paid for your time, at $3.75 per gallon?

0.93 x 60/10 x 3.75 = $21 /hour

And, you didn?t even have to work for it!
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Old 05-20-2008, 05:52 AM   #2
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Originally Posted by Ernie Rogers View Post
ROGERS TRANSPORTATION LAW

Δ gallons = - 1/30 (mph /mpg) (Δ minutes)
Not E=MC^2, but as simple at first glance, and with undoubtedly as many unexpected implications. I hope this one is more easily put into practice than striving for 99.99% the speed of light!
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Old 05-20-2008, 06:23 AM   #3
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Is there a significance to the "-" before the 1/30 constant? You include it in both the US and metric formulae, but it isn't used in the example.
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Old 05-20-2008, 07:03 AM   #4
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Originally Posted by Lug_Nut View Post
Not E=MC^2, but as simple at first glance, and with undoubtedly as many unexpected implications. I hope this one is more easily put into practice than striving for 99.99% the speed of light!
Yeah, but how much fuel will you use at that speed? I suggest no more than Plaid Speed, or if you're in a hurry, Ludicrous Speed.
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Old 05-20-2008, 09:09 AM   #5
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Exclamation

I've been playing with the formula and plugging in some long term fuel mpg and speed numbers I have consistently gotten over the years.
I get, consistently and repeatedly, 50 mpg at 60 mph. Thats 1.2 gallons of fuel used for a 60 mile trip in one hour. If I slow down to 45 mph, increasing my trip time for that 60 miles by 20 minutes the calculation becomes: [(20/30) x (60/50)], or [.6667 x 1.2], or .8 fewer gallons of fuel for that 60 mile trip.
Since I would use 1.2 gallons at 60 mph, and the estimate at the 45 mph speed is .8 fewer gallons, the fuel estimated to be needed is .4 gallon for the 60 mile trip, or 150 mpg. I know that to not be valid since steady, as near to constant as possible operation, for hundreds of miles, at 45 mph returns barely 80 mpg.
This margin of error is too great for me to continue to seriously consider this approximation formula a tool I can use, even with the caveat about only being an effective estimation when applied to small changes.
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Old 05-21-2008, 05:38 PM   #6
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I knew that was too complicated--

Hello, Lug Nut,

Nice to see you are still out causing trouble.

Oh, I knew that was going to happen. You didn't "follow the rules." Instead of worrying about that, let's just say I think I have fixed the problem, and added some seasoning. I will post that under a new thread after dinner.

Ernie Rogers

Quote:
Originally Posted by Lug_Nut View Post
I've been playing with the formula and plugging in some long term fuel mpg and speed numbers I have consistently gotten over the years.
I get, consistently and repeatedly, 50 mpg at 60 mph. Thats 1.2 gallons of fuel used for a 60 mile trip in one hour. If I slow down to 45 mph, increasing my trip time for that 60 miles by 20 minutes the calculation becomes: [(20/30) x (60/50)], or [.6667 x 1.2], or .8 fewer gallons of fuel for that 60 mile trip.
Since I would use 1.2 gallons at 60 mph, and the estimate at the 45 mph speed is .8 fewer gallons, the fuel estimated to be needed is .4 gallon for the 60 mile trip, or 150 mpg. I know that to not be valid since steady, as near to constant as possible operation, for hundreds of miles, at 45 mph returns barely 80 mpg.
This margin of error is too great for me to continue to seriously consider this approximation formula a tool I can use, even with the caveat about only being an effective estimation when applied to small changes.
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Old 05-21-2008, 06:54 PM   #7
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Is there a significance to the "-" before the 1/30 constant? You include it in both the US and metric formulae, but it isn't used in the example.
I think the minus signs in formulas are supposed to be ignored. Here, the minus is to remind us that when the trip time increases (delta t is positive) then the fuel change is negative. In the revised formula, I get rid of the mumbo jumbo (but it remains implicitly).

The new thread will be--

"Three Laws of Car Fuel Economy"

Ernie
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