Fuel usage at speeds and rpm's
 

I was actually looking to put this in an offtopic-section, but couldn't find it.

I have noticed that my car is more economic at 80km/h then on 120km/h. For some this might sound logical because you are making les rpm, but on the other hand you go faster and cover more miles per hour. So is the usage of fuel in ratio with the miles you make in a certain gear? To make this at least partly understandable for myself, I'm leaving rolling-resistance and air resistance behind for now, even though I am aware that this plays a roll.

Very simple comparison:
When doing 120 km/h I do about 3200 rpm in 5th gear.
That means 1 hour of 3200 rotation per minute transports me 120 kilometer.
The ratio between rpm and kilometer would be 3200/120= 26.66 rpm/km/h

When i do 80 km/h I do about 2400 rpm in 5th gear.
When doing 80 km/h, the same calculation would give me a ratio of: 30 rpm/km/h.

I was curious if the ratio would get better at lower speeds as well, so I tried to see how many rpm I have to do to drive really slow.
900 rpm gives me about 35 km/h. Which would calculate into: 25 rpm/km/h.

Being the curious cat I am. I throw this calculation in Excel to see where the sweetspot would be. But I didn't have the time to fully test the speed I drive when doing a certain amount of rpm (so, to be continued)

From the quick calculation, I would guess that the higher ratio means a better relation between speed and rpm.


However, I wanted to take it a bit further. So I dove into the average fuel usage for petrol cars.

I have a 1.8 engine, and the usage is roughly 12 kilometer for 1 liter petrol. As far as Google could tell me, the average fuel/air mixture is 1:12.9. 1 part petrol and 12.9 parts air. That should mean 7% fuel with 93% air.
I guess that 1 full rpm of the engine fires all 4 cylinders in the engine, which would mean that 1800cc is equal to 1.8ml of air/fuel. Using the mixture as described above, this would mean 0.13ml fuel with 1.67ml air for 1 full rotation (counting all 4 cylinders).
Lets say that I'm doing 2000 rpm, than I would use 0.13 (fuel) * 2000 = 259.2ml fuel per minute.

Lets say that I'm doing 60 km/h with 2000 rpm, then I would need (259.2 * 60 minutes =) 15552 ml to cover those 60 kilometers within that hour.
This is 15.552 liters.
Looking at that fuel usage, it would mean that I get 3.9 kilometer per liter fuel. Which doesn't match the real-life fuel usage.

So where am I messing this up?

Not a specialist on this, just thinking out loud. So I might be overlooking something very obvious.

My next thing is going to be to look up the gear ratio of my transmission box and see if I can use this to estimate the speed at a certain rpm.

As it says under your name..you really must get out more :wine:, think most cars reach an optimum efficiency at a certain speed. Not wanting to get in to calculations, the old coal burning steam ships had some kind of querky usage like 10t coal to do 10 knots and 15t coal to do 12 knots, it didn't relate to the fuel used.

But it has to be calculatable...

I'm looking into transmission ratios now, because I was hoping to make a graph with theoratical speeds at certain rpm's in certain gears.
The information I could find says the 1.8 engine has a gear ratio of:
1st gear: 3.58:1
2nd gear: 2.02:1
3rd gear: 1.35:1
4th gear: 1.03 :1
5th gear: 0.81:1

If I take the first gear, with its 3.58:1 ratio, this would mean that 3.58 rpm of the engine should give 1 rpm on the axle. So 500 rpm of the engine would give me 139.7 rpm on the axle.
The wheels are 225/40/R18 and have a circumference of 200.18 cm (but I'm going to use 200 cm to make it easier).
This would mean that at 500 rpm the wheels would spin 139.7 times, covering 139.7*200=27958 cm per minute. This is 279.6 meter per minute (??) and eventually calculates into 16.7 km/h (??)

Seems a bit high to make 16 km/h at 500 rpm?
I tried to turn it around, and calculate what would happen if it means that 3.58:1 means that the axle spins 3.58 times when the engine spins 1 time. But this eventually calculated into doing 215 km/h at 500 rpm.

The problem with the calculation above is that this gives me a speed of 370 km/h at 2500 rpm in 5th gear. Seems a tad high, even for the mighty 75.

Are you not missing the final drive ratio from your calculations?

Possibly, but what would that be? :D

I believe it's 3.9:1

Just had a long Google and something really cool: http://roverclub-bg.com/index.php_fi...eneraldata.pdf
Amazing what info it gives :O

According to this PDF, the final drive ratio would be 4.41:1.
Now just to figure out where to fit this into the calculations. Back to Google.

Well, I fitted in the Final Drive ratio at the end of the RPM calculation, and it matches pretty good. I drive roughly 80 km/h when making 2400 rpm. When I look in my calculations, I see the list says 84 km/h with 2500 rpm. That sounds about right.

There is just 1 downside, when I look at my first post, I used the relation between the rpm and the speed I was going. For example:

But it seems this myth is busted by the calculations because they show a very steady and linear relation between speed and rpm. Meaning the relation between both is always the same, regardless of speed or RPM (unless you change gear). So the quote above is probably miscalculated, or it is down to air- and rollresistance which is not taken into account in the list below.

So the only thing left now, is to figure out how to calculate the fuel usage based on rpm. And then I should be able to make a theoratical list of fuel usage at certain speeds in certain gears.

How are you going to be able to calculate that without torque request, injection timings and throttle opening?

I don't know, I have yet to learn how these affect the fuel consumption. :D

It will all be theoratical anyway, there is still roll resistance, air resistance, angle of driving (gravity) etc. that plays a part.