2014-2020 Formula One 1.6l V6 turbo engine formula

[bill shoe]

I don't agree with the power and torque curve. The power will definitely reduce from 10,500 rpm. The friction will cause losses at higher rpms and the combustion efficiency is also likely to suffer.

I agree friction will always be worse at higher rpm. I don't agree that net power will always peak at 10,500 rpm.

Manufacturer A decides to design their engine with the goal of maximizing peak power at 10,500 rpm. This means certain levels of boost and compression ratio. The boost, compression ratio, and ignition timing are carefully selected to work together optimally at 10,500 rpm.

Manufacturer B decides to design their engine with the goal of maximizing area under the power curve from 10,500 to 15,000 rpm. This might mean peak power at 12,750 rpm. At this rpm, compared to Manufacturer A at 10,500 rpm, the boost will be lower and therefore the optimum compression ratio would be a bit higher. The boost, compression ratio, and ignition timing will work together optimally at 12,750 rpm. OK, what happens in this same engine at 10,500 rpm? Boost will be higher so therefore the compression ratio will be too high. Detonation will then be a problem so therefore ignition timing will have to be retarded a bit. This retarded ignition might more than cancel out the lower friction at 10,500 rpm. Therefore this engine makes more net power at 12,750 rpm than 10,500 rpm. And it has more area under the power curve to boot.

This is simply one example of how a 2014 engine might not peak at 10,500 rpm.

[WhiteBlue]

I don't agree with the power and torque curve. The power will definitely reduce from 10,500 rpm. The friction will cause losses at higher rpms and the combustion efficiency is also likely to suffer.

I agree friction will always be worse at higher rpm. I don't agree that net power will always peak at 10,500 rpm.

Manufacturer A decides to design their engine with the goal of maximizing peak power at 10,500 rpm. This means certain levels of boost and compression ratio. The boost, compression ratio, and ignition timing are carefully selected to work together optimally at 10,500 rpm.

Manufacturer B decides to design their engine with the goal of maximizing area under the power curve from 10,500 to 15,000 rpm. This might mean peak power at 12,750 rpm. At this rpm, compared to Manufacturer A at 10,500 rpm, the boost will be lower and therefore the optimum compression ratio would be a bit higher. The boost, compression ratio, and ignition timing will work together optimally at 12,750 rpm. OK, what happens in this same engine at 10,500 rpm? Boost will be higher so therefore the compression ratio will be too high. Detonation will then be a problem so therefore ignition timing will have to be retarded a bit. This retarded ignition might more than cancel out the lower friction at 10,500 rpm. Therefore this engine makes more net power at 12,750 rpm than 10,500 rpm. And it has more area under the power curve to boot.

This is simply one example of how a 2014 engine might not peak at 10,500 rpm.

You keep forgetting the fuel limit. Above 10,500 fuel remains constant. Effectively you can add excess air by higher boosting but you will not get more power unless you increase the combustion efficiency or reduce the friction losses. Both is not very likely. In fact both effects will work against getting more power above 10,500. So the conclusion will be a falling power curve and not a conatant power curve.

[bill shoe]

You keep forgetting the fuel limit. Above 10,500 fuel remains constant.

I'm remembering the fuel limit. This is how I see it-- Constant fuel above 10,500 means roughly constant mass airflow above 10,500.

A fixed amount of mass airflow at 10,500 rpm (with a given volumetric efficiency) implies a certain amount of boost.

The same amount of mass airflow at 12,750 rpm (with same volumetric efficiency) necessarily implies lower boost.

If the engine is going faster than 10,500 rpm then the boost vs rpm will drop off in a similar manner to the torque vs rpm curve that Ringo posted earlier. The lower boost at 12,750 results in a power-maximizing compression ratio that is high. The higher boost at 10,500 results in a power-maximizing compression ratio that is (slightly) lower.

However, when the "12,750-peak" engine is operated at 10,500, it will already have the higher compression ratio, so it may require retarded ignition at 10,500 rpm to avoid detonation.

[ringo]

The friction is unpredictable. It can be lower at higher speeds in some cases.

I agree friction will always be worse at higher rpm. I don't agree that net power will always peak at 10,500 rpm.

This is indicated power curve not brake. However i don't see why any team would refuse to maximum power as soon as they can.

Manufacturer B decides to design their engine with the goal of maximizing area under the power curve from 10,500 to 15,000 rpm. This might mean peak power at 12,750 rpm. At this rpm, compared to Manufacturer A at 10,500 rpm, the boost will be lower and therefore the optimum compression ratio would be a bit higher. The boost, compression ratio, and ignition timing will work together optimally at 12,750 rpm.

When you say optimum compression raito you mean boost pressure? Most teams will put compr ratio as high as it's sensible to do.

That power curve posted is for the highest power that can be made at any rpm.
It wont work any more optimally at 12,500 than at 10500. We may need the friction characteristic to determine that.
Nonetheless i think it is better to make maximum power for as wide a rev range as possible.

edit: i see what you a getting at. And it led me to look a turbo compressor map.
The boost does look like the torque curve that i posted. But the thing that stands out is the drop in pressure ratio after 10,500 if the fuel mass flow is held constant.

The implication of dropping the boost with the same mass flow of air, is a drop through of turbine efficiencies.


for example super imposing my pressure ratio with air mass flow graph with this garret turbo show what happens over 10,500rpm. The compressor operates with a dropping efficiency. It goes from 78 to 68 percent.

It affects the amount of energy the HERS can collect. Also affects the charge temperature.

It's a lot of implications and this is why i think millions may be spent on create some kind of alien turbo that will be perfectly matched to constant power after 10,500. You can imagine how that map would look.
Anyway this is all assuming that the volumetric efficiency stays the same from 10,500 to 15000. It wont, as it could go over 1 since it's a turbo engine.

[Pierce89]

I don't agree with the power and torque curve. The power will definitely reduce from 10,500 rpm. The friction will cause losses at higher rpms and the combustion efficiency is also likely to suffer.

I agree friction will always be worse at higher rpm. I don't agree that net power will always peak at 10,500 rpm.

Manufacturer A decides to design their engine with the goal of maximizing peak power at 10,500 rpm. This means certain levels of boost and compression ratio. The boost, compression ratio, and ignition timing are carefully selected to work together optimally at 10,500 rpm.

Manufacturer B decides to design their engine with the goal of maximizing area under the power curve from 10,500 to 15,000 rpm. This might mean peak power at 12,750 rpm. At this rpm, compared to Manufacturer A at 10,500 rpm, the boost will be lower and therefore the optimum compression ratio would be a bit higher. The boost, compression ratio, and ignition timing will work together optimally at 12,750 rpm. OK, what happens in this same engine at 10,500 rpm? Boost will be higher so therefore the compression ratio will be too high. Detonation will then be a problem so therefore ignition timing will have to be retarded a bit. This retarded ignition might more than cancel out the lower friction at 10,500 rpm. Therefore this engine makes more net power at 12,750 rpm than 10,500 rpm. And it has more area under the power curve to boot.

This is simply one example of how a 2014 engine might not peak at 10,500 rpm.

You keep forgetting the fuel limit. Above 10,500 fuel remains constant. Effectively you can add excess air by higher boosting but you will not get more power unless you increase the combustion efficiency or reduce the friction losses. Both is not very likely. In fact both effects will work against getting more power above 10,500. So the conclusion will be a falling power curve and not a conatant power curve.

you're making the assumption that the engine rules will allow them to use the full fuel alotment at 10.500 rpm. Just because the alotmnet quits rising at 10,500 doesn't mean the engine has enough boost to use that alotment.It might need 13,500 before it uses all the fuel. The teams pushed for the 15,ooo and I figure they had a reason for it. The engineers in the TWG must've thought they could use the RPM if they asked for it.

[ringo]

what speed will these engines idle at?

the curent ones idle at 4000 right? or that was in the v10 days.

[xpensive]

Fact is you have 38 cc per second to deal with, end of story. What Cosworth can do with that is doubtful, but BMW or toyota?

[ringo]

BMW need to return. They must be licking their chops at this formula.

[WhiteBlue]

I agree friction will always be worse at higher rpm. I don't agree that net power will always peak at 10,500 rpm.

Manufacturer A decides to design their engine with the goal of maximizing peak power at 10,500 rpm. This means certain levels of boost and compression ratio. The boost, compression ratio, and ignition timing are carefully selected to work together optimally at 10,500 rpm.

Manufacturer B decides to design their engine with the goal of maximizing area under the power curve from 10,500 to 15,000 rpm. This might mean peak power at 12,750 rpm. At this rpm, compared to Manufacturer A at 10,500 rpm, the boost will be lower and therefore the optimum compression ratio would be a bit higher. The boost, compression ratio, and ignition timing will work together optimally at 12,750 rpm. OK, what happens in this same engine at 10,500 rpm? Boost will be higher so therefore the compression ratio will be too high. Detonation will then be a problem so therefore ignition timing will have to be retarded a bit. This retarded ignition might more than cancel out the lower friction at 10,500 rpm. Therefore this engine makes more net power at 12,750 rpm than 10,500 rpm. And it has more area under the power curve to boot.

This is simply one example of how a 2014 engine might not peak at 10,500 rpm.

You keep forgetting the fuel limit. Above 10,500 fuel remains constant. Effectively you can add excess air by higher boosting but you will not get more power unless you increase the combustion efficiency or reduce the friction losses. Both is not very likely. In fact both effects will work against getting more power above 10,500. So the conclusion will be a falling power curve and not a constant power curve.

you're making the assumption that the engine rules will allow them to use the full fuel allotment at 10.500 rpm. Just because the allotment quits rising at 10,500 doesn't mean the engine has enough boost to use that allotment.It might need 13,500 before it uses all the fuel. The teams pushed for the 15,ooo and I figure they had a reason for it. The engineers in the TWG must've thought they could use the RPM if they asked for it.

I'm not making an assumption! It is in the rules. You are kidding yourself if you think that the engines will not run on the highest allowed fuel rate at any given rpm. Of course they will. And it follows logically that the power will fall beyond 10,500 rpm. The teams only asked for it to appease the noise lovers and traditionalists who think that racing engines must rev high. If you have max power at 10,500 then 12,000 is plenty enough to have head room for your gearing.

[WhiteBlue]

BMW need to return. They must be licking their chops at this formula.

They wanted an efficient formula for 2011. When they realised that they would not get it and the credit crunch hit they exited. The formula is still dominated by the money makers and the chassis constructors. Every appeal for automotive players is only there by accident and when the FiA fights for it. So its not a big surprise that the bigger and better car makers are not attracted.

[rjsa]

I'm not making an assumption! It is in the rules. You are kidding yourself if you think that the engines will not run on the highest allowed fuel rate at any given rpm. Of course they will. And it follows logically that the power will fall beyond 10,500 rpm. The teams only asked for it to appease the noise lovers and traditionalists who think that racing engines must rev high. If you have max power at 10,500 then 12,000 is plenty enough to have head room for your gearing

You don't shift at max power, you shift at the point that gives you the maximum area under the power curve. And where eaxctly that falls dependes on a lot of things that we just don't know yet. But I'm pretty sure that it's quite highier than 10500.

[rjsa]

BMW need to return. They must be licking their chops at this formula.

They wanted an efficient formula for 2011. When they realised that they would not get it and the credit crunch hit they exited. The formula is still dominated by the money makers and the chassis constructors. Every appeal for automotive players is only there by accident and when the FiA fights for it. So its not a big surprise that the bigger and better car makers are not attracted.

BMW quit after taking a beating. Just like Toyota and Honda used the USA crisis as an excuse. Period.

[FW17]

With the 100l/hr rule for the engines, the so called power circuits such as Monza and Silverstone will have lesser power available than in places like Monaco.

For all we know engine power may be maximum available to the drivers at Monaco

[machin]

And it follows logically that the power will fall beyond 10,500 rpm. The teams only asked for it to appease the noise lovers and traditionalists who think that racing engines must rev high. If you have max power at 10,500 then 12,000 is plenty enough to have head room for your gearing.

The question is; will the efficiency drop above 10500rpm faster than the fuel quantity allowance drops below 10,500rpm? If the efficiency drop-off is low then the teams will gear the cars to use the engine above 10500rpm, if the efficiency drop off is higher than the fuel allowance drops off below 10500rpm then they'll gear the cars to use the revs probably at some point around the 10500rpm level.

My personal view is that the efficiency drop-off above 10500rpm will be lower than the fuel allowance drop off below 10500rpm, so they will use the engines above 10500rpm because that will give them the widest power band.

[rjsa]

It the power decay is at least as steep as the power building from the fuel flow (which I doubt) something bellow 15k rpm could be the shifting point.

So I'm still going with full revs for the engines. Until we get a more complex model or reality to prove me wrong.

Because who ever designed the rules have more tools than us and they wouldn't set a 15k limit if it was not to be reached.

And it follows logically that the power will fall beyond 10,500 rpm. The teams only asked for it to appease the noise lovers and traditionalists who think that racing engines must rev high. If you have max power at 10,500 then 12,000 is plenty enough to have head room for your gearing.

The question is; will the efficiency drop above 10500rpm faster than the fuel quantity allowance drops below 10,500rpm? If the efficiency drop-off is low then the teams will gear the cars to use the engine above 10500rpm, if the efficiency drop off is higher than the fuel allowance drops off below 10500rpm then they'll gear the cars to use the revs probably at some point around the 10500rpm level.

My personal view is that the efficiency drop-off above 10500rpm will be lower than the fuel allowance drop off below 10500rpm, so they will use the engines above 10500rpm because that will give them the widest power band.

I think I said this before. The area under the power curve.
Machin, can you produce graphs showing how much energy is transfered to the ground by integrating the power curve? One shifting at peak power and another maximizing it?