F1 2011 Fuel

Roger Snead · Mar 18, 2012

Hello to all.

Here is the scenario.....
Fuel sim is enabled in Career (xbox) and I am doing a 40% race w/e. Then say, I set up with10 laps fuel for Qually.
Come Race time I am unable to regulate how much fuel is put in. How much fuel is put in car for a 60 lap race track? Is it 24 laps worth of fuel, (40% of 60) based on fuel mapping set on Standard mix the whole 24 laps?
And then it's up to me to use fuel mapping strategy...i.e. sometimes on rich, sometimes on standard, sometimes on lean?

And if Fuel Sim is OFF for the race w/e, am I always loaded with a full load of fuel in Qually and Race? i.e. 60 laps of fuel, even while only doing a 40% Race? (24 laps) Doing this, I am not able to change Fuel mapping in RACE, correct?

Do I have it right?

TheHemps · Mar 19, 2012

You can't choose your fuel mix when the fuel sim is off.

If you are doing 20%/50% races, the fuel use is relative. So in a 50% race, it will burn up in half the time as a 100% race.
You can't change your fuel load before a race, it automatically outs in enough laps of fuel for a 100% race, but as I said, it is still relative to the race distance.

Hope this helps.

Roger Snead · Mar 19, 2012

Thanks, Also.....KERS and DRS does not use any extra fuel does it? Seems that by using them you would burn LESS fuel, correct? You would get from point A to B faster w/o using any more fuel......correct? Many thanks.

TheHemps · Mar 19, 2012

KERS usage increases tyre wear, if you use full KERS every lap you will lose a lap on the tyres. Doesn't affect fuel though.

DRS should actually save fuel, but I'm not sure of the physics of it in game.

David O'Reilly · Mar 19, 2012

My take on DRS is that by reducing drag it will increase speed with no change in engine power.
However to be really pedantic the car achieving a higher top speed will use a few more revs so actually burn a tiny bit more fuel.
However #2: As less fuel is being used to generate downforce and push the car into the road that saving may more than compensate for the extra revs.
Thrust must equal drag plus lift. Thrust is constant, lift (or its opposite downforce) is reduced so speed is increased until either an equivalent amount of drag is found at a higher speed or the rev limit is reached.
Result probably eqal or less fuel use if all other things constant.
Damn I love these kind of questions!

Roger Snead · Mar 19, 2012

I don't see how RPM's would change by reducing drag which in turn increases the speed of the vehicle(DRS)
The wheels would be turning faster when DRS is enabled, but would that increase RPM's?

David O'Reilly · Mar 21, 2012

Zebrawire said:
I don't see how RPM's would change by reducing drag which in turn increases the speed of the vehicle(DRS)
The wheels would be turning faster when DRS is enabled, but would that increase RPM's?

Yes, there is a direct link from wheel speed to engine speed called a gearbox.
If wheel speed increases engine speed increases and vice versa , there is zero slip once clutch fully engaged.
BTW This is how engine braking works, wheels trying to increase engine speed, engine fighting it off throttle.
One more example, consider why 7th needs to be longer for DRS? (more revs!)

Paul Lock · Mar 21, 2012

Zebrawire said:
I don't see how RPM's would change by reducing drag which in turn increases the speed of the vehicle(DRS)
The wheels would be turning faster when DRS is enabled, but would that increase RPM's?

Here is an analogy. Maybe not a very good one, but you are gonna get it anyway!

If you were cycling a push-bike, that had fixed gearing (meaning there is no free-wheel) and consider going downhill as the 'free' boost that DRS gives an F1 car.
Although you are getting a boost which doesnt actually require you to be pushing the pedals down and using up energy to achieve it - your legs will still be going around at a faster speed, possibly until the point they cant go any faster and there is smoke coming out of your groin and then your legs fall off (leg limiter). Obviously none of that has really required you to exert energy (use fuel) into the pedals, but the fact the pedals have been controlling your legs means that your legs (the engine) is still the limiting factor.

Caesar Westland · Mar 22, 2012

Paul Lock said:
Here is an analogy. Maybe not a very good one, but you are gonna get it anyway!

If you were cycling a push-bike, that had fixed gearing (meaning there is no free-wheel) and consider going downhill as the 'free' boost that DRS gives an F1 car.
Although you are getting a boost which doesnt actually require you to be pushing the pedals down and using up energy to achieve it - your legs will still be going around at a faster speed, possibly until the point they cant go any faster and there is smoke coming out of your groin and then your legs fall off (leg limiter). Obviously none of that has really required you to exert energy (use fuel) into the pedals, but the fact the pedals have been controlling your legs means that your legs (the engine) is still the limiting factor.

The analogy would be beter if you say you remove the head wind for the biker. Resulting in less drag... so with same energy this person can produce more RPM.

Paul Lock · Mar 22, 2012

Caesar Westland said:
The analogy would be beter if you say you remove the head wind for the biker. Resulting in less drag... so with same energy this person can produce more RPM.

I will keep that in mind next time I use it

David O'Reilly · Mar 22, 2012

Well in fact the effect of going downhill would be almost identical as in Pauls analogy (or was it a metaphor?) as
Thrust must equal drag plus lift.
Reducing drag (less headwind) would have the same impact as reducing lift (negative lift IE downhill)
With a given thrust speed increases until again drag plus lift equals thrust albeit at higher RPM.
The total energy output would be higher due to friction and pumping losses.

BTW we seeem to have lost the OP

Roger Snead · Mar 22, 2012

You have'nt lost me ..... lol ( OP) Just tak'n it all in.

BTW, What is the final verdict, from a virtual engineering viewpoint? Will I burn more fuel my using DRS and/or KERS?

ENDR · Mar 22, 2012

DRS ON = less drag, so less energy needed to reach a certain speed, but at the same time
DRS ON = less drag, so you'll gain more speed and reach higher revs, which means more fuel being used.
Soo... Wha?

Roger Snead · Mar 22, 2012

ENDR said:
DRS ON = less drag, so less energy needed to reach a certain speed, but at the same time
DRS ON = less drag, so you'll gain more speed and reach higher revs, which means more fuel being used.
Soo... Wha?

Yes, but these higher rev's are generated via gearbox not by fuel consumption....right?

ENDR · Mar 22, 2012

So you're saying, that while using DRS the cars don't burn any fuel?

Roger Snead · Mar 23, 2012

ENDR said:
So you're saying, that while using DRS the cars don't burn any fuel?

No,,,,,,of course they are burning fuel. You said " DRS ON = less drag, so you'll gain more speed and reach higher revs, which means MORE fuel being used"
.I said they would not be burning MORE fuel.

David O'Reilly · Mar 23, 2012

Zebrawire said:
You have'nt lost me ..... lol ( OP) Just tak'n it all in.
BTW, What is the final verdict, from a virtual engineering viewpoint? Will I burn more fuel my using DRS and/or KERS?

In my virtual view:
KERS-No fuel used. Its kinetic energy harvested under brakes and re used via electric hence the name KINETIC Energy Recovery System.
DRS Its a given that you will be using mix 3 and bouncing off the limiter with DRS laps so fuel use is high as is speed. But if you used a constant lap time as the control it is a fuel saving device due to Drag reduction. IE a fuel saving compared to any given lap time achieved without its use.
This is to say that if you cruised around on mix 1 and opened DRS all the time you could then it would save fuel.

ENDR · Mar 23, 2012

David O'Reilly said:
KERS-No fuel used. Its kinetic energy harvested under brakes and re used via electric hence the name KINETIC Energy Recovery System.

But it's not like they're running on KERS ONLY, it's just a small boost of power that helps the engine.