2014-2020 Formula One 1.6l V6 turbo engine formula

[Cold Fussion]

which are very very important

Not really, knowing a torque figure tells you very little about the performance of the car.

[Brian Coat]

Thinks: "Please God don't let another round of Torque/Power/Gearing posts start ... "

[trinidefender]

Thinks: "Please God don't let another round of Torque/Power/Gearing posts start ... "

This repetitive conversation happens so often because there are constantly new members who come and inevitably ask the same question. Ergo they get the same answer from our members. As unfortunate as it is the only way that this can be stopped is if we were to introduce a technical section that has information such as this that can be linked whenever somebody comes with these age old questions.

[Cold Fussion]

The problem is that the general public has no concept of power, torque and energy and to make matters worse, the colloquial terms for 'torque' and 'power' have little to do with the technical terms. Consequentially, media outlets, automotive reviews, commentators are constantly misrepresenting the meaning of all these terms and non technical find their way here and we have these silly conversations endlessly. I don't think a technical section would change anything because if they were really interested they could just read wikipedia which suitably explains it.

At least he hasn't said torque is energy like some other members have tried..

[R_Redding]

The problem is that the general public has no concept of power, torque and energy ~

I think that inertia and mass should be added to your list aswell.

Inertia of a rotating mass is very counter intuitive.

These videos,mainly by Walter Lewin show how so...

https://www.youtube.com/watch?v=cB8GNQuyMPc
https://www.youtube.com/watch?v=CHQOctEvtTY
https://www.youtube.com/watch?v=NeXIV-wMVUk

Rob

[hemichromis]

In the case of a theoretical power level of 900hp at max fuel allowance available at 10500rpm it would be 610nm at 12500rpm with the same fuel allowance and hp it would be 512nm. I have assumed that as the engine is turbo charged and limited to 15000rpm the power level would be the same between 10500 and the limiter, i could be wrong but i expect the manufacturers are aiming for this.
we can also see that thanks to the turbo and ers the cars have plenty of torque below their peak as well.

[Tommy Cookers]

fundamentally, what determines the ICE torque and power/rpm curves is the fuel rate limit falling with rpm below 10500
(since compressor rpm is modulated to charge the cylinders optimally at all rpm - going beyond N/A engine's mapping method)
fundamentally, what determines the (best) PU torque and power/rpm curves is the mu-k power limit not falling with rpm below 10500

ie the constant (120 kW crankshaft-referred) mu-k power limit means mu-k torque increases as rpm go below 10500
(until the mu-k power limit falls as rpm go below c5700 rpm, mu-k torque being limited to 200 Nm crankshaft-referred at all rpm)

so yes, below 10500 rpm the PU has typically a 'better-shaped' torque/rpm curve than the ICE alone
(and correspondingly, below 10500 rpm the PU also has typically a 'better-shaped' power/rpm curve than the ICE alone)

however track time spent below 10500 rpm is spent there by choice, and in various ways at controlled partial power and torque
there is no particular benefit from the torque curve shape, it probably saves a few gear shifts

above 10500 rpm the torque from mu-k action must fall with rpm due to its 120 kW power limit
the PU power and torque/rpm curves will be rather similar to those of the ICE alone

yes, still 'better-shaped' than in the N/A F1 engine
those had closer gear ratios, optimised for each event
these cars don't have that, and they don't need it

[stevesingo]

If energy released and therefore torque of the ICE is in proportion to fuel burnt, should we be looking at the fuel flow limits to indicate ICE output? More importantly fuel flow per revolution will indicate torque as this removes the time component of power calculation - Imp power =Tq*rpm/5252

The regulations state Fuel Mass up to 10500rpm= 0.009*rpm+5.5

example 0.009*10500+5.5=100kg

If we calculate the fuel flow limit across the usable rpm range we get

7k___68.5kg/hr
7.5__73
8____77.5
8.5__82
9____86.5
9.5__91
10___95.5
10.5__100
11___100
11.5_100
12___100

If we calculate how much fuel per cycle -

Fuel per Cycle (g)= Fuel Flow ((Kg/hr)/60/(rpm/2))*1000

7k___0.3262g
7.5__0.3244
8____0.3229
8.5__0.3203
9____0.3216 -edited typo
9.5__0.3193
10___0.3183
10.5_0.3175
11___0.3030
11.5_0.2899
12___0.2778

As we can see as rpm rises we get less fuel per cycle.

Assuming that the ICE uses all fuel available at a similar AFR and that friction rises with rpm, the ICE should make more torque lower in the rpm range than at 10500rpm, after which torque will drop significantly.


Thoughts?

[FW17]

How is it there is an increase in rpm without increase in fuel flow?

[Facts Only]

How is it there is an increase in rpm without increase in fuel flow?

Fuel flow limit doesnt increase above 10,500rpm (I believe)

[hemichromis]

If energy released and therefore torque of the ICE is in proportion to fuel burnt, should we be looking at the fuel flow limits to indicate ICE output? More importantly fuel flow per revolution will indicate torque as this removes the time component of power calculation - Imp power =Tq*rpm/5252

The regulations state Fuel Mass up to 10500rpm= 0.009*rpm+5.5

example 0.009*10500+5.5=100kg

If we calculate the fuel flow limit across the usable rpm range we get

7k___68.5kg/hr
7.5__73
8____77.5
8.5__82
9____86.5
9.5__91
10___95.5
10.5__100
11___100
11.5_100
12___100

If we calculate how much fuel per cycle -

Fuel per Cycle (g)= Fuel Flow ((Kg/hr)/60/(rpm/2))*1000

7k___0.3262g
7.5__0.3244
8____0.3229
8.5__0.3203
9____0.3193
9.5__0.3193
10___0.3183
10.5_0.3175
11___0.3030
11.5_0.2899
12___0.2778

As we can see as rpm rises we get less fuel per cycle.

Assuming that the ICE uses all fuel available at a similar AFR and that friction rises with rpm, the ICE should make more torque lower in the rpm range than at 10500rpm, after which torque will drop significantly.


Thoughts?

Sound correct to me.

[bergie88]

How is it there is an increase in rpm without increase in fuel flow?

Fuel flow limit doesnt increase above 10,500rpm (I believe)

Exactly, this is defined in the rules by a formula which increases with rpm till 10.500rpm, after which it is limited to a certain value. This is also shown by the calculation of stevesingo. For this reason the ICE is designed to reach its peak efficiency around 10.500 rpm, because the most fuel per cycle times maximum efficiency results in the highest peak power.

[stevesingo]

How is it there is an increase in rpm without increase in fuel flow?

Fuel flow limit doesnt increase above 10,500rpm (I believe)

Exactly, this is defined in the rules by a formula which increases with rpm till 10.500rpm, after which it is limited to a certain value. This is also shown by the calculation of stevesingo. For this reason the ICE is designed to reach its peak efficiency around 10.500 rpm, because the most fuel per cycle times maximum efficiency results in the highest peak power.

10500rpm is the point of highest flow, but not the point of highest fuel per cycle. The way the formula works (I don't know why the FiA did this) means the lower the rpm the higher the fuel flow per cycle at a rate of about 1% increase per 1000rpm below 10500rpm.

[Tommy Cookers]

WITHDRAWN (with apologies to the following poster)

[stevesingo]

Moot point now Tommy has had some coffee and withdrawn his comment.