Tow Vehicle Fuel Consumption - Page 3 - Fiberglass RV

 Fiberglass RV Tow Vehicle Fuel Consumption

 03-28-2008, 01:47 PM #29 Senior Member   Name: Per Trailer: 2000 Burro 17 ft Widebody towed by Touareg TDI Oregon Posts: 863 Registry My ancient '99 Odyssey has a computer which automatically changes the timing to take advantage of higher octane. Probably in conjunction with a knock sensor (device used by earlier SAABs and others I seem to recall). Now if I weren't so cheap I would already have tested this out. __________________
 03-28-2008, 07:03 PM #30 Senior Member     Name: Ches Trailer: 1992 Kustom Koach 17 FT British Columbia Posts: 4,897 My truck which i have at present time is a Diesel 6.7liter. On my way to Chilliwack going empty i averaged 20 mpg Imperial. That was a distance of 1164KM. Most of this was highway driving and some town driving. On my way back i took the long hilly way and got 16 mpg Imperial pulling trailer. Trip back was 1429.8 Kms Lots of my route was in mountains and at times very slow. I never drove over 100kms/60 mph. Total fuel bill was $483.52 CDN. Total Kms was 2593.8 Kms Cost/Km is about$0.18 __________________ __________________ Retired Underground Coal Miner. Served in Canadian Army (1PPCLI)
 03-28-2008, 08:49 PM #31 Senior Member     Name: Clive Trailer: Boler American #3104 Florida Posts: 554 Registry [quote] Attachment 12305 Nice memories, huh? \$5 for 3 Gallons !!
 03-28-2008, 10:00 PM #32 Senior Member     Trailer: Scamp Posts: 3,072 That was taken in August 2001, on my way back to Washington from Alaska. People were complaining loudly about the cost of fuel at that time, they are complaining now and they will be complaining in the future.
 03-29-2008, 02:56 PM #34 Senior Member     Trailer: Scamp Posts: 3,072 If the bearings are a problem they should be heating up and be detectable. Umm, something wrong with your numbers -- They don't match your conclusion about slowing down, which of course, they should. From WiKi: QUOTE Power The power required to overcome the aerodynamic drag is given by: P_d = \mathbf{F}_d \cdot \mathbf{v} = {1 \over 2} \rho v^3 A C_d. Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power. It should be emphasized here that the drag equation is an approximation, and does not necessarily give a close approximation in every instance. Thus one should be careful when making assumptions using these equations. END QUOTE
 03-29-2008, 03:02 PM #35 Senior Member     Name: Ches Trailer: 1992 Kustom Koach 17 FT British Columbia Posts: 4,897 OK Pete---Put that in lay mans terms __________________ Retired Underground Coal Miner. Served in Canadian Army (1PPCLI)
 03-29-2008, 03:17 PM #36 Senior Member     Trailer: Scamp Posts: 3,072 Buried in the middle of that is: "the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW)." Double speed, increase power need by eight times! However, that relationship is less true for lower speeds -- I recall reading somewhere that the air drag factor really starts to come into play at something like 45 mph. Maybe someone with a consumption readout in their buggy can look at 45 mph compared to 90 mph and give us some real-world readings (without trailer, of course). A lot might depend on the speed at which the car may have been designed to give the least drag.
03-29-2008, 04:32 PM   #38
Senior Member

Name: Peter
Trailer: 2005 19 ft Scamp 19 ft 5th Wheel
Oregon
Posts: 1,519
Registry
Quote:
 Umm, something wrong with your numbers -- They don't match your conclusion about slowing down, which of course, they should.
I read your post three or four times before I realized where you thought my numbers were off, then I realized: You were assuming I drove the trailer both ways, so . . .
Leg 1: I drove the Santa Fe northward without a trailer at 65mph (average) and got 24 mpg
Leg 2: I drove the Santa Fe southward with the trailer at 55-60mph and got 17.5 mpg

The loss in gas mileage was not because I slowed down, it was due to hauling 2000 lbs of trailer behind me.

--Peter

03-29-2008, 04:38 PM   #39
Senior Member

Name: Peter
Trailer: 2005 19 ft Scamp 19 ft 5th Wheel
Oregon
Posts: 1,519
Registry
Quote:
Thank you Steve . . . You need to teach a class in Trailer Tires for us!

03-29-2008, 11:22 PM   #40
Senior Member

Trailer: Scamp
Posts: 3,072
Quote:
 I read your post three or four times before I realized where you thought my numbers were off, then I realized: You were assuming I drove the trailer both ways, so . . . Leg 1: I drove the Santa Fe northward without a trailer at 65mph (average) and got 24 mpg Leg 2: I drove the Santa Fe southward with the trailer at 55-60mph and got 17.5 mpg The loss in gas mileage was not because I slowed down, it was due to hauling 2000 lbs of trailer behind me. --Peter
OK! Got it!

03-30-2008, 12:17 AM   #41
Senior Member

Name: Mike
Trailer: Bigfoot 25 ft
Posts: 7,317
Quote:
 I know that (some of the rest) of the moderators frown at excessively technical posts but I'm not above tweaking things once and awhile.
I think technical info is a good thing.

03-30-2008, 12:24 AM   #42
Senior Member

Name: Peter
Trailer: 2005 19 ft Scamp 19 ft 5th Wheel
Oregon
Posts: 1,519
Registry
Quote:
 The power required to overcome the aerodynamic drag is given by: P_d = \mathbf{F}_d \cdot \mathbf{v} = {1 \over 2} \rho v^3 A C_d.
Converted to layman's terms

There are amount of power (energy) needed to push your car through the air is

For 60mph:
It takes 27% more energy to push the air aside when you're driving 65 instead of 60
It takes 59% more energy to push the air aside when you're driving 70 instead of 60
It takes 95% more energy to push the air aside when you're driving 75 instead of 60

And 65:
It takes 25% more energy to push the air aside when you're driving 70 instead of 65
It takes 54% more energy to push the air aside when you're driving 75 instead of 65

So the short rule is: At freeway speeds increasing your speed by 15 miles an hour roughly doubles the amount of gasoline required to overcome wind resistance. You can save a whole mess of gasoline by slowing down five or ten miles an hour.

Another factoid: Most cars get their best gas mileage at speeds between 47 and 54 miles per hour.

So, here's the table comparing efficiencies at speeds over 55mph:
It takes 30% more energy to push the air aside when you're driving 60 instead of 55
It takes 65% more energy to push the air aside when you're driving 65 instead of 55
It takes 100% more energy to push the air aside when you're driving 70 instead of 55
It takes 154% more energy to push the air aside when you're driving 75 instead of 55

And, even more impressive, for 50 mph:
It takes 33% more energy to push the air aside when you're driving 55 instead of 50
It takes 73% more energy to push the air aside when you're driving 60 instead of 50
It takes 220% more energy to push the air aside when you're driving 65 instead of 50
It takes 275% more energy to push the air aside when you're driving 70 instead of 50
It takes 336% more energy to push the air aside when you're driving 75 instead of 50

See the [b]BIG jump at 65 mph? We, as a country, could save a whole mess of gasoline and reduce our greenhouse gas emissions by a [b]HUGE amount by setting our metropolitan area (where the highest number of freeway miles are driven) speed limit to 50 or 55 mph and our rural area speed limits to 60 mph.

Science stuff:

Spelled out mathematically, the amount of energy required to push the air aside is:
1/2 the thickness/pressure of the air
multiplied by your car's windface (the total area of your car or trailer facing into the wind)
multiplied by the amount of aerodynamic drag for your car/trailer (more aerodynamic = smaller number)
multiplied by your car's speed again
multiplied by your car's speed a third time

Shortened even more, the amount of energy needed to push your car through the air is
(a bunch of stuff you can't control)

--Peter
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