So what happened

Bloders

Member
Location
Ruabon
Err they are designed to make full power at low rpms.
Tractor engine not so much(y)
so if the tractor engine cannot produce much power at low rpm, then it will have to run faster to power the machine.

At the end of the day, Power = force x velocity. So to shift a machine at 60km/hr will use a specific amount of power. Whatever engine speed this power is generated at, the requirement is the same.
The efficency of the engine will vary with speed somewhat but if its powering along at 60km/hr at 1000rpm, the fuel consumption will not be a significantly (depending on the definition of significantly but im 25%) different than at 1500rpm.

Fendt, nor their drivers cannot change physics im afraid.
 

Pint2

Member
Screenshot_20190730-210525_Gallery.jpg
Screenshot_20190730-210832_Gallery.jpg
 

Speedstar

Member
Location
Scottish Borders
so if the tractor engine cannot produce much power at low rpm, then it will have to run faster to power the machine.

At the end of the day, Power = force x velocity. So to shift a machine at 60km/hr will use a specific amount of power. Whatever engine speed this power is generated at, the requirement is the same.
The efficency of the engine will vary with speed somewhat but if its powering along at 60km/hr at 1000rpm, the fuel consumption will not be a significantly (depending on the definition of significantly but im 25%) different than at 1500rpm.

Fendt, nor their drivers cannot change physics im afraid.
you are all most thinking out side the box there , tractor was only doing 1000rmps at 65kph
 
Frankzy's graph is interesting. I presume ad blue use is metered in at low engine RPM because combustion temperature is lower or something- seems to be gently falling as engine speed increases?

The fuel consumption is interesting, too. At 1000rpm the engine is burning 25kg/hour, but only putting out 130 kW.

I always thought that your best absolute fuel consumption would be where peak torque occurs- the speed where fuel is giving you the maximum bang for your buck in terms of Nm vs grams of fuel consumed- normally some way above idle and some way below peak engine power. If this is the case, then it appear to be around 1500rpm in the case of the 939 in this example, where the engine, according to PTO power, must be producing around 250 kW and using around double the fuel (50kg/hour). But, compare this to the point where the lowest specific fuel consumption occurs which seems to be a sniff under 1300 rpm, interesting and bang goes my old-school theory.:unsure:

Fendt drivers will be able to confirm at what revs a 900 series tractor will 'settle' at once the transmission is running at top speed under load (I presume this is somewhere around the 1300 and below mark) and, how steady the revs will be when putting 24 tonnes all told.

Time for some old-fashioned calculatoring. :wacky:

The 939 in this example is burning 210 grammes of fuel per kWh and running at a power output of around 190 kW. (Interestingly this should mean fuel use at that engine speed and power output of around 40kg/hour).

90 miles is around 145 KM. Assuming the journey was at 65kph the entire time, so the journey took about 2 hours 14 minutes or 2.23 hours.

Need to work out total energy used in this journey- relatively simple, just times that 190 kW engine power (assuming rock-steady engine speed at all times) by duration of the journey.

2.23 hours x 190kW is 423.7 kWh.

Take the total kWh of energy used, and times that by the absolutely best specific fuel consumption on that graph (210 grams/kWh) gives 88.977 kg of fuel used.

Using the standard EC figures for off-road diesel, we can see best-case, the off road diesel will have a density of 845 kg/m3 or 0.845L/kg. Take the weight of fuel consumed (88.977kg) and times it by that density to get total fuel consumption in litres: 88.977 x 0.845= 76.185565 Litres.

Need to convert that to (UK) gallons, divide total litres by 4.54609= 16.538512216 gallons.

Divide miles (90) by gallons to give best-case fuel consumption: 90/ 16.53812216 = 5.4418437901 miles per gallon, representing a best case scenario and with a transmission/drivetrain that gives zero loss from engine to road and with no change in engine speed whatsoever to get the vehicle moving initially (or any use of the brakes at any time).
 

Servac

Member
Location
Wales
Interestingly I have found some figures regarding the fuel consumption of trucks and I have done some calculations- highly rough mind- that suggest you need about 0.2 kWh to move a tonne one kilometre so my figures I reckon, aren't half bad.o_O
Yea, but a truck isn't half as aerodynamic as a Fendt 939
 
Last edited:

Kiwi Pete

Member
Livestock Farmer
Frankzy's graph is interesting. I presume ad blue use is metered in at low engine RPM because combustion temperature is lower or something- seems to be gently falling as engine speed increases?

The fuel consumption is interesting, too. At 1000rpm the engine is burning 25kg/hour, but only putting out 130 kW.

I always thought that your best absolute fuel consumption would be where peak torque occurs- the speed where fuel is giving you the maximum bang for your buck in terms of Nm vs grams of fuel consumed- normally some way above idle and some way below peak engine power. If this is the case, then it appear to be around 1500rpm in the case of the 939 in this example, where the engine, according to PTO power, must be producing around 250 kW and using around double the fuel (50kg/hour). But, compare this to the point where the lowest specific fuel consumption occurs which seems to be a sniff under 1300 rpm, interesting and bang goes my old-school theory.:unsure:

Fendt drivers will be able to confirm at what revs a 900 series tractor will 'settle' at once the transmission is running at top speed under load (I presume this is somewhere around the 1300 and below mark) and, how steady the revs will be when putting 24 tonnes all told.

Time for some old-fashioned calculatoring. :wacky:

The 939 in this example is burning 210 grammes of fuel per kWh and running at a power output of around 190 kW. (Interestingly this should mean fuel use at that engine speed and power output of around 40kg/hour).

90 miles is around 145 KM. Assuming the journey was at 65kph the entire time, so the journey took about 2 hours 14 minutes or 2.23 hours.

Need to work out total energy used in this journey- relatively simple, just times that 190 kW engine power (assuming rock-steady engine speed at all times) by duration of the journey.

2.23 hours x 190kW is 423.7 kWh.

Take the total kWh of energy used, and times that by the absolutely best specific fuel consumption on that graph (210 grams/kWh) gives 88.977 kg of fuel used.

Using the standard EC figures for off-road diesel, we can see best-case, the off road diesel will have a density of 845 kg/m3 or 0.845L/kg. Take the weight of fuel consumed (88.977kg) and times it by that density to get total fuel consumption in litres: 88.977 x 0.845= 76.185565 Litres.

Need to convert that to (UK) gallons, divide total litres by 4.54609= 16.538512216 gallons.

Divide miles (90) by gallons to give best-case fuel consumption: 90/ 16.53812216 = 5.4418437901 miles per gallon, representing a best case scenario and with a transmission/drivetrain that gives zero loss from engine to road and with no change in engine speed whatsoever to get the vehicle moving initially (or any use of the brakes at any time).
Bloody hell!

You'll never be a farmer with a calculator like that!!
 

Kiwi Pete

Member
Livestock Farmer
Interestingly I have found some figures regarding the fuel consumption of trucks and I have done some calculations- highly rough mind- that suggest you need about 0.2 kWh to move a tonne one kilometre so my figures I reckon, aren't half bad.o_O
I got as far as working our trucks avg fuel burn out in MPG, it's not flat open going, mind; but it's around that 5.5 MPG mark that you mentioned in your post above.

How does that translate to your findings? Pretty close?

The 530's have the emission control stuff on them (or did) but the 460's and old 400's don't. Consumption seems largely similar, you just get there faster with more HP.
 

Mouser

Member
Location
near Belfast
Frankzy's graph is interesting. I presume ad blue use is metered in at low engine RPM because combustion temperature is lower or something- seems to be gently falling as engine speed increases?

The fuel consumption is interesting, too. At 1000rpm the engine is burning 25kg/hour, but only putting out 130 kW.

I always thought that your best absolute fuel consumption would be where peak torque occurs- the speed where fuel is giving you the maximum bang for your buck in terms of Nm vs grams of fuel consumed- normally some way above idle and some way below peak engine power. If this is the case, then it appear to be around 1500rpm in the case of the 939 in this example, where the engine, according to PTO power, must be producing around 250 kW and using around double the fuel (50kg/hour). But, compare this to the point where the lowest specific fuel consumption occurs which seems to be a sniff under 1300 rpm, interesting and bang goes my old-school theory.:unsure:

Fendt drivers will be able to confirm at what revs a 900 series tractor will 'settle' at once the transmission is running at top speed under load (I presume this is somewhere around the 1300 and below mark) and, how steady the revs will be when putting 24 tonnes all told.

Time for some old-fashioned calculatoring. :wacky:

The 939 in this example is burning 210 grammes of fuel per kWh and running at a power output of around 190 kW. (Interestingly this should mean fuel use at that engine speed and power output of around 40kg/hour).

90 miles is around 145 KM. Assuming the journey was at 65kph the entire time, so the journey took about 2 hours 14 minutes or 2.23 hours.

Need to work out total energy used in this journey- relatively simple, just times that 190 kW engine power (assuming rock-steady engine speed at all times) by duration of the journey.

2.23 hours x 190kW is 423.7 kWh.

Take the total kWh of energy used, and times that by the absolutely best specific fuel consumption on that graph (210 grams/kWh) gives 88.977 kg of fuel used.

Using the standard EC figures for off-road diesel, we can see best-case, the off road diesel will have a density of 845 kg/m3 or 0.845L/kg. Take the weight of fuel consumed (88.977kg) and times it by that density to get total fuel consumption in litres: 88.977 x 0.845= 76.185565 Litres.

Need to convert that to (UK) gallons, divide total litres by 4.54609= 16.538512216 gallons.

Divide miles (90) by gallons to give best-case fuel consumption: 90/ 16.53812216 = 5.4418437901 miles per gallon, representing a best case scenario and with a transmission/drivetrain that gives zero loss from engine to road and with no change in engine speed whatsoever to get the vehicle moving initially (or any use of the brakes at any time).
Nice try but Somethings not adding up if you only get 5mpg assuming 100% efficiency and 0 friction. I'm guessing it's the 40kg/hour figure which seems way too high. (y)
 
I got as far as working our trucks avg fuel burn out in MPG, it's not flat open going, mind; but it's around that 5.5 MPG mark that you mentioned in your post above.

How does that translate to your findings? Pretty close?

The 530's have the emission control stuff on them (or did) but the 460's and old 400's don't. Consumption seems largely similar, you just get there faster with more HP.

If someone posts the graphs of a truck engine doing what Frankzy has done I would be able to work it out.

40 kg of fuel and hour is only 33.8 litres per hour so not far away from what a lot of us would say a tractor will drink when being properly worked.

EDIT

I have made a mistake with the diesel weight vs litres calculation, I should have seen that but didn't (along with the numerous typos) late last night when I was doing all this. 40kg divided by 0.845 would be 47.33727811 litres per hour.
 
Last edited:
Nice try but Somethings not adding up if you only get 5mpg assuming 100% efficiency and 0 friction. I'm guessing it's the 40kg/hour figure which seems way too high. (y)

The results from dyno testing I guess would simulate maximum load, or near enough, if the load is less then the engine won't need a much fuel to attain a given rpm- it's like when you car is in neutral and the lights- the slightest twitch of throttle sends the engine toward the red line no sweat.
 

Drillman

Member
Mixed Farmer
Quite right. The lorry is easily twice as aerodynamic as the Fendt with its completely exposed huge wheels and drag inducing styling.
I do wonder if someone will maybe share with us all what the amazing 29mpg fendt does now to the gallon after the flux capacitor was modified?

After all if economy has dropped from 29mpg to 5mpg I wouldn’t be very happy would you?

There’s certainly no post engine modification (total rebuild/replacement) figures that have been put on here that I’m aware of!
 

bumkin

Member
Mixed Farmer
Location
pembrokeshire
diesel is lighter than water isnt it, so 40kg will be 48 litres not 33.8?
what the hell are you lot going on about this is a pointless thread some drivers can get better fuel consumption than others because they are constant they dont change gear all the time and rev the bol--cks of the machine they just keep grinding on if one machine is more fuel efficient than another depends on weight and different transmissions tyre pressure as long as you have a clean burn there isnt a lot you can do,i knew a chap who could use more fuel per hour than any one else when he left the yard it was as if he was in a drag race he would have changed gear four times only to have to stop at the gate sixty yards away then do it all again when he got on the road
 

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