Full RPM or high idle? Bush hogging fuel efficiency

[Redneck in training]

This one has more incorrect science than I care to correct, but I'll try to get through the first sentence or two:

"Power generated by the engine is always equal to power required to perform the task."

Just plain and simple wrong . Power generated by the engine is equal to
power required to perform the task plus many, many other losses. Ever notice your tractor warms up as you use it?

You are confusing power available in the fuel with power produced by the engine after all losses are acounted for.

"Fuel consumption is proportional to power required multiplied by time. "

Nope, same flawed science as before. You haven't factored in all the variable losses. You also don't account for engine/speed efficiency.

I might give you this one. Yes fuel consumption would be greater than proportionality would suggest

"Mowing at lower rpm will not save fuel because it takes longer to perform the task"

That's a total false premise. Nobody said the ground speed wouldn't be maintained.
If you mow the same amount of grass in the same time you will use the same amount of fuel (provided that efficiency is the same at either rpm)

Let me know if you want me to continue. Please say no.

No. I don't want to continue. My answers are in italics above.

 

[DarkBlack]

Your example is about the car is flawed due to aerodynamic drag difference at 30 and 60 mph. Aerodynamic drag increases with square of speed while rolling drag increases linearly. Therefore doubling the speed of the car will require eight times as much power. .

No, no no.. your'e trying to slither away again. Right here. This one above. Show your math for the above.
I'm not interested in you again changing it, and I don't want you to copy and paste other people's text.

 

[DarkBlack]

Actually, redneck, don't bother. I like passing on wisdom to those who want to learn, but I just ran out of ambition with you.

 

[Redneck in training]

look in post #100
I go for dinner. See you tomorrow. Bye.

 

[Redneck in training]

Actually, redneck, don't bother. I like passing on wisdom to those who want to learn, but I just ran out of ambition with you.

See. You can be civil. We might not agree but we don't need to insult each other.

 

[JoeBuyer]

I have about 60acres left to cut, more than welcome to come do the test at my place.

Sure. Stand outside and hold the gate open, I'll be right there. Actually it may take less time to mow your 60 acres than it takes this thread to give the OP a conclusive answer.

 

[284 International]

Domush, in case you haven't found it yet, there actually IS some scientific data for your machine, in Nebraska Tractor Test #1444. While it doesn't show reduced PTO horsepower settings, they do a reduced RPM 50% drawbar load along with a full RPM, 50% drawbar pull. Both are virtually the same horsepower, but the reduced RPM setting burns about 0.2 gallons per hour less. You can find a link to the test in .pdf form here.

As to the other comments, aerodynamic drag increases as the square of speed. In other words, double the speed, quadruple the aerodynamic drag. However, remember that power is work divided by time. We are doing 4 times the work to move an object the same distance twice as fast, but doing it in half of the time. 4 divided by 0.5 is 8. This is where we find the "8 times as much power to go twice as fast" principle.

 

[patrick_g]

I mow 160 acres and I do it with a dull 6 ft brush hog. Except for "eating" trees (up to 4 inches in diameter or so) I rarely run at the rated PTO speed as I can git 'er done at a more reasonable engine speed. Well yahh, if'n you are running a generator off the PTO or otherwise need 540 RPM I can do that but it is at about 2650 RPM on my 39HP Kubota. I brush hog from about 1800-2400 depending on conditions. I never work it hard below 1800 RPM or so to avoid lugging. I get more acres mowed per fillup running 1800-2400 than screaming along at 2650. It has a governor so down the highway I just give it full throttle and let 'er rip. I don't go far on the highway and I'm more interested in lowering the closure rate of fast drivers thsn saving a couple oz of diesel.

I want to apologize for the civility of my post. I would have been sarcastic, superior in tone and condescending to all you ignorant rabble but I got here too late to take sides.

Pat

 

[foreman Etexas]

I had to run around 2500 rpms cause I was using all 46horse's

 

[DarkBlack]

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As to the other comments, aerodynamic drag increases as the square of speed. In other words, double the speed, quadruple the aerodynamic drag. However, remember that power is work divided by time. We are doing 4 times the work to move an object the same distance twice as fast, but doing it in half of the time. 4 divided by 0.5 is 8. This is where we find the "8 times as much power to go twice as fast" principle.

Your #'s would make a good approximation for pure aerodynamics,maybe for approximating a windmill, but the origional post was for comparing power for a car @30 mph and 60mph. Wind resistance is only a part of the load, and it's proportion varies with the speed, At 30mph there are many significant, much more linear losses than wind drag,