How to increase mpg with physics?
 

Everybody knows that a well maintained engine runs better and more efficient than a neglected one.

Diesel Bombers in their high output engines know this better than anyone else.
But what can be done to increase mileage?

How you use your vehicle, normal cruising speed and tire size are the most important influences in milrage.
A simple furmula will help you to get the maximum out of your fuel:

Ideal cruising speed= RPM x tire diameter/ gear ratio x 336

You need: numbers as an example from my truck:
- rpm at highest torque - 2000rpm
- tire diameter - 28,2"
- highest tranny gear - 0.79
- axle gear ratio - 3.73

ideal cruising speed= 2000 x 28.2/(0.79 x 3.73 x 336)
= 56,96mph or 91km/h

As you can see that's exactly the limit because above 55mph wind resistance does increase fuel consumption.
You also can see that tire size does influence fuel consumption.
Bigger isn't always better, or you have to regear.

With this simple formula I increased my mpg from 26mpg up to 29mpg.
Knowing that 57mph is my personal limit to gain maximum mileage.

What's your max mpg?:humm:

With 30" tires, ideal cruising speed for my truck is 61.5mph or 99km/h. :humm:
Best fuel economy to date so far on a trip was 22.0mpg. Averaged 20.8mpg in the mountains in July :rocking: lol that was from the overhead though....

Would this apply to all vehicles (diesels) given air flow/drag around the vehicle being that they are different body styles? I havent done any hard figure tests but based on the computer read out on the overhead console ( I know not exact mpg's from there) but mine does get alot better around 53mph. Right now hand calculating every tank i am getting a whole wopping 15.9mpg :jump:

Yes, it works for all cars.
Remember, if you go faster than 55mph you also have to work against the wind resistance.

The ideal crusing speed just reflects the highest torque of your engine with the lowest needed power to keep your truck rolling.

You can aid in fuel savings by the common rules like:
- lowering weight.
- deceasing rolling resistance by proper tire inflation or appropriate tires for roads
- using a bed cover or lowered tail gate.
- dismounting unused roof bars and roof baskets.

I always have a good smile when listening to buddies doing a tune up to save fuel and driving around with ultra aggressive 35" offroad tires and with a load full of stuff in their back.

How do you "seat of the pants" find the rpm's at highest torque?

I found mine in the factory service manual and in the user manual.
You also can use your Dyno print out. Just have a look at your torque peak and go to the rpm.

last not least- ask your dealer about the rpm at top torque.

My top torque starts, right behind when the turbo starts to kick in. But I'm sure thats not a good method to find out...

How about proper tire pressure?:w2:

I'd like to know where you got your formula because it is misleading. What is the exact defenition of "ideal cruising speed"?

Your formula implies that your vehicle gets its best mileage at 56.96 mph. This is not true. It will get higher mpg at lower and lower speeds.

So, try it- it's a compromise between fuel consumption, highest gear and torque.

This formula is taken from the Bosch automotive technical handbook, german issue no.:22

you will also find it in John Lawlor's Auto Math Handbook in combination with Larry Sheppard.

You will find further hints in "On the Physics of drag Racing written by Geoffery T. Fox.

I also recommend Isaac Newtons and his Newton's Laws.

Have a look at the Hp, Tq and BSFC charts of a Diesel and find out that the power band provides the lowest BSFC at Tq peak.

Yes, you can decrease speed down to 50mph- and gain a bit due to the reason that wind resistance starts to develop here. Which is not mentioned in the formulas above. but nobody goes 50mph on a highway- do you?

Ideal crusing speed is a compromise between speed and fuel consumption- nothing else.

You can save a lot-with the truck in the garage...

I know I am always amazed to see 21 mpg from my truck weighing 9600 lbs, I believe its the 36 inch tires that do it for me

This formula correlates rpm and speed... nothing else. It was govornment mandated that peak torque should coincide with 55 mph. That is why vehicles with standard rear-end ratios and standard tires all have there peak at 55 mph. It has nothing to do with with wind resistance. There is no factor for wind resistance. This simple formula will NOT get the maximum out of your fuel. If (ideal cruising speed) is a compromise between speed and fuel consumption why did you claim best mileage? A compromise with speed indicates an arbitrary value being placed on your windshield time. When in traffic, it places a value on everyone elses time. According to this, Your ideal cruising speed would increase with each additional passenger. You have made incorrect assumptions about the defenition of this formula. A more practical use of this formula would be to plug in your cruising speed (mine is ten over by the way) and extrapolate what tire size or gear ratio will optomize that mileage. If you want to find out what your "ideal cruising speed" is (according to your formula), Drive down the road until you reach 2000 rpm, then look at the speedometer. This is the ONLY thing this formula tells you.

Ok, simple prove me the opposite and I will be happy.
Including the tire size and "knowing your speed is over 60 shows simply an oversized tire not suiting your gears. I think even Ford engineers know that.
I guess you know about speed and correlated wind resistance. Power to overcome wind resistance is a squared function not linear.

Please perform a coast down test and tell me your Brake Specific fuel consumption. I will be happy to hear about your results.

You didn't read carefully. As you state in your first sentence and absolutely right it correlates rpm and speed and neglects wind resistance because speed limit prohibit to drive faster than 70 miles anyway.

So your usable range is between 0 and 70 miles.

But to have or to get these rpm numbers you have to know your torque peak which in easy words explains best engine effi. I don't talk about peak hp-it's peak torque.
And yes, Diesel engine manufactures have a relative flat torque cuve compared with gassers. That makes gearing in a diesel so important.

So what about germany, we don't have your 55 miles speed limit.
We have either 62mph or no speed limit, aren't those cars sold to the US?
They don't have a US specific gearing...

Please let me know is you are interested in those calculations, I will pull them out and copy them for you. Shouldn't be a problem to share knowledge.
Those formulas aren't black magic:wizard:

Can you back pedal any faster? Are we now speaking irrelevent jibberish? I have proven your initial statement false. This formula does not give you the speed at which you get the "maximum out of your fuel". Slower will increase mielage by an inverse square relationship. If you graphed mielage vs. speed, You would see only a miniscule insignificant inflection point at max torque rpm.

it gives you the speed the engine is running at it's most efficient

I believe thats where this is going.

I'll agree to that as long as no more is read into it.

http://up.picr.de/4154962.jpg

Otherwise you can keep it simple.

6% fuel saving by decreasing 10% weight ( barely possible)
3% fuel saving by decreasing 10% air resistance ( not possible at all, fixed)
2% fuel saving by decreasing 10% rolling resistance ( possible with using tires with street thread and a huge amount of silica)

I used the formula because no one here modifies his truck to gain higher mpg.
It's a static value. Don't try to tell me you are able to work with dynamics.

You still don't get the combination of torque and speed and related fuel consumption.

Thinking about it... you would want to be driving at a stable rpm (one above peak torque) if you cruise above peak torque any hills or headwinds will cause the rpms to lower and cause the motor back into peak torque giving you the power to maintain speed, cruising below peak torque will be harder to regain rpm under the same situation leading to more speed variances..

and then we could go into volumetric efficiency of a diesel and..... i'll just stop right there :s:


think about it... cruising faster up a grade and letting the rpms fall some while adding fuel to maintain speed is a lot easier than starting at speed and trying to increase. eg. downshifting a gear vs lugging the current one.

I neglected rolling resistance, inclination, acceleration and braking resistance for that reason.
As you surely recognized(because you mentioned it:tu:) it's all in the formula above.
But it's simply not necessary to get a number where to start. Just keep the number of the highest speed as the static limit.
There is no increase above this number.

anyways.. it's way to many variables to calculate the true efficiencies..

what is really needed is to measure the amount of POWER required to maintain a certain speed, the smaller the better..

if we could manage to get a gauge to show current power being used...........
Thats peaked my interest, going to look into designing this.

Scan Gauge II should show that if you have OBDII...

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Or you could perform a coast down test to gain those numbers to evaluate the needed hp at 50mph- but that needs a good hour for the calculations.

so...

power = acceleration * (mass * velocity)


weight and speed (mass velocity) would be known.. but if maintaining a set speed acceleration would be 0..... and the end result of that formula to be zero, I would imagine acceleration would have to do with friction of tires,air?

it is easier to use negative acceleration as kown as braking or coast to get those numbers in a time frame. It usually done from 60mph coasting to 50mph and taking the time as exactly as possible.
I will copy the complete process, if you need it. It will be tommorrow in your messages.