2025/2026 Hybrid Powerunit speculation

[wuzak]

The standard ramp rate is 100kW/s, which means the cars will go from 300km/h to 216km/h in 2.5s.

That equates to a deceleration rate of about 0.95G.

But, if the drivers stay in the same gear, the rpm will be reduced.

If it is 12,000rpm @ 300km/h (equates to a top speed of 375km/h @ 15,000rpm), the engine speed will be reduced to 8,653rpm.

At that engine speed, by the fuel flow rules and esimated 48% TE (gives 400kW at maximum fuel flow), the ICE will produce only 334kW, which would mean 84kW left to propel the car.

Speed would be down to 178km/h, the deceleration would be around 1.38G.

The energy recovered during this would be approximately 250kW * 2.5s / 2 = 312.5kJ.


Of course, slowing the car this much would mean that the braking time is short and the time that the maximum braking recovery can occur will be very short.

So it would be balancing using the ICE to drive the generator and using brakes.

[mzso]

If you really want to make the racing unsafe, you could simply specify cable-operated drum brakes. Much cheaper.

Braking earlier and from less speed in most cases, would be safer. Why do you want to put this strawman everywhere?

that would require about 1500 kW rear braking capacity and about 3000 kW front braking capacity
ie about 12 times the 2026 capacity and about 30 times the present capacity

If you want to exactly match F1 levels. I don't see braking at 150m or 200m instead of 100 from top speed necessarily a bad thing.
On the other hand. I don't think anyone honestly knows how far F1 style technology could take these. Special insulation could allow them to heat up a lot more than normal generators/motors, and the high speed airflow could cool them down.

since the banning of Renault's mass damper it's illegal to have concentrated masses of stuff dotted around the car
apart from unavoidable items such as the wheels - these must be tethered to the car

Okay. Why do you bring it up?

re so-called 'axial-flux technology' - this only means electrical machines that are disc-shaped not drum-shaped
we already have disc-shaped things called brakes so perhaps we should devise brakes that also generate electricity

Well, that would be the axial-flux (M)GU, a brake that produces electricity via braking. They can design axial flux motors with a much better power density, that's why it's trending.

[wuzak]

We want braking, not breaking.

[michl420]

I think we underestimate the influence of the smaller and slippery cars.
It would be worth considering to save some energy at the beginning/middle of the straight and not lost it with the air resistance of the higher speed.

[diffuser]

I think we underestimate the influence of the smaller and slippery cars.
It would be worth considering to save some energy at the beginning/middle of the straight and not lost it with the air resistance of the higher speed.

If you're trying to get to the end of straight in the shortest amount of time, getting to your top speed as fast as you can, is always the quickest. The longer you stay at the top speed, the more beneficial. In most straights that will make you impossible to over take, even if the car behind has a 10KPH higher top speed. Obviously it depends how much faster the lead car with the slower top speed can out accelerate the car with the higher top speed. The higher the top speed the earlier you need to brake, cause it takes longer to stop.

[karana]

The standard ramp rate is 100kW/s, which means the cars will go from 300km/h to 216km/h in 2.5s.

That equates to a deceleration rate of about 0.95G.

But, if the drivers stay in the same gear, the rpm will be reduced.

If it is 12,000rpm @ 300km/h (equates to a top speed of 375km/h @ 15,000rpm), the engine speed will be reduced to 8,653rpm.

At that engine speed, by the fuel flow rules and esimated 48% TE (gives 400kW at maximum fuel flow), the ICE will produce only 334kW, which would mean 84kW left to propel the car.

Speed would be down to 178km/h, the deceleration would be around 1.38G.

The energy recovered during this would be approximately 250kW * 2.5s / 2 = 312.5kJ.


Of course, slowing the car this much would mean that the braking time is short and the time that the maximum braking recovery can occur will be very short.

So it would be balancing using the ICE to drive the generator and using brakes.

I don't think your calculations are correct. If the terminal velocity of a car is 300km/h at 400kW, it will be 216km/h at 150kW. But even if the output is reduced immediately to 150 kW, the velocity won't reach 216km/h in 2.5 seconds.

The drag force at 300km/h will be F_D = P/v = 400kW/300km/h ~ 4802N, the resulting force slowing down the car would be F_D - 150kW/300km/h ~ 3000N.

The resulting deceleration would be 3000N/800kg ~ 4m/s^2.

Even if we assume constant deceleration, in 2.5 seconds the velocity would be only reduced to ~290km/h. And this assumes instant reduction of power, which isn't possible.

[wuzak]

The standard ramp rate is 100kW/s, which means the cars will go from 300km/h to 216km/h in 2.5s.

That equates to a deceleration rate of about 0.95G.

But, if the drivers stay in the same gear, the rpm will be reduced.

If it is 12,000rpm @ 300km/h (equates to a top speed of 375km/h @ 15,000rpm), the engine speed will be reduced to 8,653rpm.

At that engine speed, by the fuel flow rules and esimated 48% TE (gives 400kW at maximum fuel flow), the ICE will produce only 334kW, which would mean 84kW left to propel the car.

Speed would be down to 178km/h, the deceleration would be around 1.38G.

The energy recovered during this would be approximately 250kW * 2.5s / 2 = 312.5kJ.


Of course, slowing the car this much would mean that the braking time is short and the time that the maximum braking recovery can occur will be very short.

So it would be balancing using the ICE to drive the generator and using brakes.

I don't think your calculations are correct. If the terminal velocity of a car is 300km/h at 400kW, it will be 216km/h at 150kW. But even if the output is reduced immediately to 150 kW, the velocity won't reach 216km/h in 2.5 seconds.

The drag force at 300km/h will be F_D = P/v = 400kW/300km/h ~ 4802N, the resulting force slowing down the car would be F_D - 150kW/300km/h ~ 3000N.

The resulting deceleration would be 3000N/800kg ~ 4m/s^2.

Even if we assume constant deceleration, in 2.5 seconds the velocity would be only reduced to ~290km/h. And this assumes instant reduction of power, which isn't possible.

Yes, my calculations were rather simplistic and wrong. Apart from the terminal velocity.

[Mattchu]

These last few pages are the main reason I still come to this forum.
Some of the other "topics" are just hot garbage, whereas in these threads everyone is mostly civil, sometimes a bit cheeky, some great knowledge/technical input, at the end of the day pretty much all seem respectful of one another.

[mzso]

We want braking, not breaking.

Fixed.

[wuzak]

The drag force at 300km/h will be F_D = P/v = 400kW/300km/h ~ 4802N, the resulting force slowing down the car would be F_D - 150kW/300km/h ~ 3000N.

The resulting deceleration would be 3000N/800kg ~ 4m/s^2.

Even if we assume constant deceleration, in 2.5 seconds the velocity would be only reduced to ~290km/h. And this assumes instant reduction of power, which isn't possible.

Using your drag number of 3,000N, the deceleration would be 3,000N/800kg = 3.75m/s^2.

a = (V1-V2)/t => V2 = V1 - a*t

V1=300km/h = 83.333m/s
a = 3.75m/s^2
t = 2.5s

V2 = 73.96m/s = 266km/h.

But I'm not sure that the drag slowing the car would be that.
If you average the drag slowing the car you get 1,500N and a resultant speed of 283km/h.

Calculating using kinetic energy.
KE1 = 1/2 * m *V1^2
m = 800kg

KE1 = 2777.78 kJ.

Recovery starts at 0 and ends at 250kW and takes 2.5s (at 100kW/s ramp rate).

The energy recovered is 250kW * 2.5s /2 = 312.5kJ.

The kinetic energy at the end of the ramp is KE2 = 2777.78 kJ - 312.5kJ = 2465.28kJ

Speed at that kinetic energy is:

V2 = ((2 * KE2)/m)^(1/2) = 78.5m/s = 282.6kph.

[Tommy Cookers]

... Special insulation could allow them to heat up a lot more than normal generators/motors, and the high speed airflow could cool them down.
... the axial-flux (M)GU, a brake that produces electricity via braking. They can design axial flux motors with a much better power density, that's why it's trending.

heat is the greatest enemy of magnets (and they can self-heat)

F1 shows us that the (M)GU as a brake has about 1% of the power density of a friction brake

axial-flux MGs will contribute a degradation in the acceleration of the car (even worse than radial-flux MGs will)
even the 2026 car's inertia is disproportionately increased by the high rpm rotation of the MG
using bigger or more MGs would make this worse

btw I thank those posters explaining to us the rules on how the 2026 cars will be operated

[mzso]

heat is the greatest enemy of magnets (and they can self-heat)

Heat is an issue for electromagnets. That is why is proposed better insulation. And a disk shape would help with cooling as well.

F1 shows us that the (M)GU as a brake has about 1% of the power density of a friction brake

axial-flux MGs will degrade the acceleration of the car (even worse than radial-flux MGs will)
people must understand that car's inertia is disproportionately increased by these rotating bits

I think your overstating the effects of the disk layout. As well as being a fair bit heavier the cylindrical motors can't have a lot smaller diameter because they need to produce torque.

Comparing to current F1 MGU's is not a reasonable comparison from a design goal aspect. Also, does the K have a minimum weight as well as the ICE? Not that AFAIK anyone published weight values for the K.

I mean brake disc would be well out of range, but I think with a bleeding-edge design it would be much closer than you'd expect.

[michl420]

I think we underestimate the influence of the smaller and slippery cars.
It would be worth considering to save some energy at the beginning/middle of the straight and not lost it with the air resistance of the higher speed.

If you're trying to get to the end of straight in the shortest amount of time, getting to your top speed as fast as you can, is always the quickest. The longer you stay at the top speed, the more beneficial. In most straights that will make you impossible to over take, even if the car behind has a 10KPH higher top speed. Obviously it depends how much faster the lead car with the slower top speed can out accelerate the car with the higher top speed. The higher the top speed the earlier you need to brake, cause it takes longer to stop.

You mean like in the red bull/vettel days and I am aware of this.
What I mean is, it is better, energy wise, to drive 1km with 300 kph instead the first 500 m with 350kph and the last 500m with 250kph.(worst case scenario)

[DenBommer]

heat is the greatest enemy of magnets (and they can self-heat)

Heat is an issue for electromagnets. That is why is proposed better insulation. And a disk shape would help with cooling as well.

F1 shows us that the (M)GU as a brake has about 1% of the power density of a friction brake

axial-flux MGs will degrade the acceleration of the car (even worse than radial-flux MGs will)
people must understand that car's inertia is disproportionately increased by these rotating bits

I think your overstating the effects of the disk layout. As well as being a fair bit heavier the cylindrical motors can't have a lot smaller diameter because they need to produce torque.

Comparing to current F1 MGU's is not a reasonable comparison from a design goal aspect. Also, does the K have a minimum weight as well as the ICE? Not that AFAIK anyone published weight values for the K.

I mean brake disc would be well out of range, but I think with a bleeding-edge design it would be much closer than you'd expect.

So you also believe in axial-flux motors/generators?

[wuzak]

Also, does the K have a minimum weight as well as the ICE? Not that AFAIK anyone published weight values for the K.

The minimum mass of the MGU-K is 16kg and the MGU-K Mechanical Transmission is 4kg. The MGU-K Mechanical Transmission mass can be applied to the MGU-K, the ICE, or half each.

C5.18.7 Depending on where the MGU-K Mechanical Transmission is located, the total mass of the MGU-K “PU Mass group” elements as referred to in Appendix C4 must be no less than the values defined below:

a. If all of the speed ratio of the MGU-K Mechanical Transmission is located in the RV-PU-ERS the total mass of the MGU-K must be no less than 20.0kg.
b. If all of the speed ratio of the MGU-K Mechanical Transmission is located in the RV-PU-ICE the total mass of the MGU-K must be no less than 16.0kg.
c. If part of the speed ratio of the MGU-K Mechanical Transmission is located in both the RV-PU-ICE and the RV-PU-ERS the total mass of the MGU-K must be no less than 18.0kg.

The total mass of item 29 (MGU-K torque sensor) and item 31 (K torque sensor shaft and mechanical connection between MGU-K and ICE) must be allocated to either the MGU-K or the ICE to assess compliance with this article and article C5.5.1.


C5.5.1 Depending on where the MGU-K Mechanical Transmission (as defined in item 27 of Appendix C4) is located, the overall mass of the ICE “PU Mass group” elements as referred to in Appendix C3 must be no less than the values defined below:
a. If all of the speed ratio of the MGU-K Mechanical Transmission is located in the RV-PU-ERS the total mass of the ICE must be no less than 130.0kg.
b. If all of the speed ratio of the MGU-K Mechanical Transmission is located in the RV-PU-ICE the total mass of the ICE must be no less than 134.0kg.
c. If part of the speed ratio of the MGU-K Mechanical Transmission is located in both the RV-PU-ICE and the RV-PU-ERS the total mass of the ICE must be no less than 132.0kg.


RV-PU-ERS = Reference Volume Power Unit ERS
RV-PU-ICE = Reference Volume Power Unit ICE