Honda Power Unit Hardware & Software

[Webber2011]

Ok guys, so me being a bit of a dumbo with regards to some of the discussion about how all this energy is transferred between the different elements I have a few questions

They have most likely already been answered but not in a way I really understand.
( Or they maybe have but I've missed it in among a billion or so posts ? )

PZ, your last post raises a question I've been wondering about.
When a driver decides to "lift and coast" at high speed I imagine at times they are still using part throttle, as to use none at all would cause massive braking simply because of the drag.

So how much energy do they recover with a simple "lift and coast" as opposed to a hard braking event at the end of the same high speed straight ?

Is there some sort of equation like if you lift and coast at 300 metres, you gain the same as braking hard at 100 metres ?
Or is that too track dependent ?

I know this will differ from team to team, but is there some sort of rule of thumb ?

Thank you
Simo'

[PlatinumZealot]

Ok. I admit it is not fresh in mind the vehicle dynamics to calculate the energy they can skim off from harvesting at end of straight and lift and coast.
But the gist of it that drag froce is proportional to velocity squared. And power is force times velocity, so power is proportional to velocity cubed.
Ok what this means.. Is that i can sacrifice some amount of power at end of straight and my velocity barely drops and also my lap time is negligably affected. This power is "invested" into the ERS to use at a more power sensitive part of the track.

I lift and coast when i drive to save fuel. But as you say the formula 1 guys do not fully come off the throttle before the true lift and coast. I presume this is where the MGUK adds additional electrical load and leeches some power from the ICE. The same for the MGUH to the turbine. It will not cause a significant speed drop at end of straight I think.. I would estimate you need 100hp to go 15kph faster in F1 but most of that has to do with accerating from low speed to high speed. At end of straight when acceleration has tapered off, the rate of change of horsepower with speed is much smaller.

Example... I chose to give up 5kph from 330 kph end of straight for three seconds. That roughly 35 bhp (26kW) engine power sent to mguk***. Energy wise It is 78kJ. Sounds small but it can be extended and done on all the straights. It would add up.
Laptime would only slow by 0.004 seconds on that particular end of straight. But it would be a bigger laptime gain if i use that energy at another critical part of the track to accelerate.

Say, this is only for the harvesting done before lift and coast. Lift and coast itself is a direct fuel saving measure i think.

EDIT
*** a conundrum.. I have i no idea what proportion of battery power is used down the straights! I assumed 800 hp from the ice alone. However you need battery power for initial acceleration! So i cannot confirm if ice alone is deployed at END of straigts typically, but common sense says you can only charge or discharge the battery at any one time. So therefore, very likely harvesting at end of straight means the ice alone is driving the flywheel and the battery is charging only.

[gruntguru]

A few stated assumptions: 50s duration, flywheel of 4Kg (the minimum in the rules), 100mm diameter (seems sensible enough for the turbo sizing alone), 40 pulses per second, 2000rpm rev change (of the MGU-H) during a pulse. Energy per pulse depends on the mass distribution on the MGU-H (mass in the centre has less inertia than the edge), but lets say that it's something like 180J because the majority of the mass is on the outer ring of the MGU-H, due to some clever designer putting weight out there.
In this case, we have 50s per lap * 40 events per second * 180J per event = only an additional unmetered 360KJ per lap. That's only 3s of full deployment and hardly worth the trouble. You could go for a larger RPM change, at cost of the ICE seeing more and more variable levels of manifold pressure.

You have assumed a 2000 rpm delta but not stated the initial speed. If the initial speed is 120,000 the energy delta is 26 kJ (assuming 50 mm radius of gyration). If the initial speed is zero, the energy delta is 219 J which I assume is the scenario you used.

Although this looks more optimistic for your theory, I still don't buy it. The rules don't allow energy (beyond 4 MJ/lap) to be transferred from the ES to the 'K and this is nothing but a loophole which the FIA would close in a heartbeat once discovered. The unusual energy flow would be obvious in data reviewed by officials.

[PlatinumZealot]

A few stated assumptions: 50s duration, flywheel of 4Kg (the minimum in the rules), 100mm diameter (seems sensible enough for the turbo sizing alone), 40 pulses per second, 2000rpm rev change (of the MGU-H) during a pulse. Energy per pulse depends on the mass distribution on the MGU-H (mass in the centre has less inertia than the edge), but lets say that it's something like 180J because the majority of the mass is on the outer ring of the MGU-H, due to some clever designer putting weight out there.
In this case, we have 50s per lap * 40 events per second * 180J per event = only an additional unmetered 360KJ per lap. That's only 3s of full deployment and hardly worth the trouble. You could go for a larger RPM change, at cost of the ICE seeing more and more variable levels of manifold pressure.

You have assumed a 2000 rpm delta but not stated the initial speed. If the initial speed is 120,000 the energy delta is 26 kJ (assuming 50 mm radius of gyration). If the initial speed is zero, the energy delta is 219 J which I assume is the scenario you used.

Although this looks more optimistic for your theory, I still don't buy it. The rules don't allow energy (beyond 4 MJ/lap) to be transferred from the ES to the 'K and this is nothing but a loophole which the FIA would close in a heartbeat once discovered. The unusual energy flow would be obvious in data reviewed by officials.

Guru, take a look at the Honda graph in the last image. The data is there and it is obvious! Look at the rapid osciallting part of the mguh plot in yellow. That is the alternating generator/motor action.

https://i.imgur.com/k0cys2U.jpg

[gruntguru]

Can't see it.
The MGUH current oscillates a lot during upshifting (not surprising) and there are downspikes during extended braking (electric turbine spooling not required during downshift throttle blips).

Is there something else you are looking at?

[wuzak]

A few stated assumptions: 50s duration, flywheel of 4Kg (the minimum in the rules), 100mm diameter (seems sensible enough for the turbo sizing alone), 40 pulses per second, 2000rpm rev change (of the MGU-H) during a pulse. Energy per pulse depends on the mass distribution on the MGU-H (mass in the centre has less inertia than the edge), but lets say that it's something like 180J because the majority of the mass is on the outer ring of the MGU-H, due to some clever designer putting weight out there.
In this case, we have 50s per lap * 40 events per second * 180J per event = only an additional unmetered 360KJ per lap. That's only 3s of full deployment and hardly worth the trouble. You could go for a larger RPM change, at cost of the ICE seeing more and more variable levels of manifold pressure.

You have assumed a 2000 rpm delta but not stated the initial speed. If the initial speed is 120,000 the energy delta is 26 kJ (assuming 50 mm radius of gyration). If the initial speed is zero, the energy delta is 219 J which I assume is the scenario you used.

Although this looks more optimistic for your theory, I still don't buy it. The rules don't allow energy (beyond 4 MJ/lap) to be transferred from the ES to the 'K and this is nothing but a loophole which the FIA would close in a heartbeat once discovered. The unusual energy flow would be obvious in data reviewed by officials.

Guru, take a look at the Honda graph in the last image. The data is there and it is obvious! Look at the rapid osciallting part of the mguh plot in yellow. That is the alternating generator/motor action.

https://i.imgur.com/k0cys2U.jpg

Is it obvious?

Can you cut the relevant bit out and show the obvious bit?

[gruntguru]

Interesting though that apart from braking the K is almost always motoring at maximum. I didn't think they had that much energy available.???
Also the H has an almost flat line when generating. It looks as if it has an electric limit lower than the surplus turbine power available.

[wuzak]

The energy flow diagram shows the extra energy flowing from the MGUK to the MGUH and back to the ES.

But if they are dragging the MGUK to get that extra 1MJ to give to the MGUH, what is the MGUH doing?

[hurril]

Thank you very much, Craigy, for a brilliant post!

I agree with you all the way through and this is exactly as I understand what is suggested too. I can see what it is that some of you are calling flywheel(:ing) but there's no such thing required here. The spinning mass is enough but I'm not going to argue wording. Red herring, etc

The problem, as I see it, is that while the logic is sound, there's a skyhook in it. You say that braking the turbine only lowers the engine output and raises the fuel consumption. Well... what does the MGU-k do to provide the means of transferring energy in an "un-taxed" way through the MGU-h to the ES? It deletes 160kW worth of power that has been bought with fuel. Thus raising the specific fuel consumption.

This is not to say that this strategy is not viable nor that it isn't being used. Just that the value of it is probably a tad overstated because it is so sexy. (My inner nerd is drooling at this.)

My understanding is also that it is not that easy to even reach the allowed 2Mj per lap in "normal" MGU-k harvesting. There just isn't enough area under the time-under-breaking x 160kW graph.

(Again: what an excellent post!)

[henry]

I think we need to differentiate between qualifying energy flows and race energy flows.

In qualifying fuel consumption isn’t really an issue so I imagine the following based on the rules

Energy positive
Start the lap with somewhat less than 4 mJ in the ES , some will have been used getting from the last corner to the start line end with 0

Recover 2 mJ from the K during braking and motoring against the K in sections not at WOT

Energy negative
Drive the beginning of straights in electric supercharger mode, ES drives K and H

Drive the next part of straights in self-sustain-plus , ES and H driving the K

Drive the final part of the straight in self-sustain, H driving the K

Putting some numbers to this

Start the lap with 3.5 mJ in the ES and add 2 making 5.5 to spend on a qualifying lap

In electric supercharge spend ES at 180 kw , assume 90 kw to drive the compressor with 30 from blow down and 120 to the K

In SS+ spend ES at 60 kw, assume the H provides 60

Spend some on lag control, I don’t know this so I’ll ignore it, just as I’ll ignore H to ES in part throttle, I would expect some but can’t begin to quantify.

Because of 4 mJ rule we have 1.5 mJ to spend driving the H, which gives 25 seconds (1.5 mJ at 60 kw)
In this mode the K consumes 120 kw for 25 seconds, 3 mj.
Leaving 17 seconds in SS+ for a total of 42 seconds at WOT.
The rest of the WOT would be in SS

If braking provides 1mJ to the ES from the K we would need to motor for 8 seconds against the K to provide the other 1 mJ.

If the “flywheel” exists any contribution it makes would extend the time in Electric suppercharger mode and reduce SS+ and reduce lap time.

It is also possible that at the end of the straight they switch to ICE only mode and divert the H to the ES allowing a little more time in supercharge.

In Race mode the numbers are very different. Consumption matters and so burning extra fuel to keep the ES charged becomes an issue to be managed.

Most of the time they start and end the lap with the same charge, 2 mJ.

If they get 1 mJ from braking they need to trade off power at the end of straights to increase power at the beginning,

If they switch to ICE only they get, say, 60 kw into the ES and if they go to ICE minus, I.e. dragging the K they get 180 kw, topping up the ES from the K in only 5 seconds.

If they do burn fuel above this the “flywheel” mode might be useful to build up the ES ready for some use of the overtake button

I’m afraid this gets too complex for me. With 4 power modes, 4 recovery options, track configuration, weather, tyres, aero setup, etc optimising all of the parameters for a race looks remarkably hard. Qualifying seems so much easier.

[dren]

You have assumed a 2000 rpm delta but not stated the initial speed. If the initial speed is 120,000 the energy delta is 26 kJ (assuming 50 mm radius of gyration). If the initial speed is zero, the energy delta is 219 J which I assume is the scenario you used.

Although this looks more optimistic for your theory, I still don't buy it. The rules don't allow energy (beyond 4 MJ/lap) to be transferred from the ES to the 'K and this is nothing but a loophole which the FIA would close in a heartbeat once discovered. The unusual energy flow would be obvious in data reviewed by officials.

Guru, take a look at the Honda graph in the last image. The data is there and it is obvious! Look at the rapid osciallting part of the mguh plot in yellow. That is the alternating generator/motor action.

https://i.imgur.com/k0cys2U.jpg

Is it obvious?

Can you cut the relevant bit out and show the obvious bit?

The flow diagram states it. Also, look at the orange in the MGUH plot vs the blue. You can see the periods where it looks like the H is oscillating back and forth between motoring and generating. There are two white leader lines showing the difference between the H in orange and blue. These periods line up with where the K is generating more.

[dren]

This might help.

It says in Extra Harvest mode the Deployment = MGUH->MGUK energy + 2MJ from MGUK generation + 1MJ Extra Harvest (MGUK->MGUH->ES). Sadly, it doesn't state how much the MGUH is capable of generating.

It says the MGUK generation is from braking and from partial (assuming this means partial throttle).

You can see where the MGUH oscillates in the MGUH trend in orange.

[etusch]

https://youtu.be/M8VdsWn-Q9Y

[Craigy]

... I still don't buy it. The rules don't allow energy (beyond 4 MJ/lap) to be transferred from the ES to the 'K and this is nothing but a loophole which the FIA would close in a heartbeat once discovered. The unusual energy flow would be obvious in data reviewed by officials.

The ES doesn't need to transfer >4MJ from the ES to the K directly in order for this to be viable. The direct route from ES to K is max 4MJ per regulation and I am not suggesting it needs to be breached.

So long as the ES SoC doesn't go below a nominal 0 or above a maximum 4MJ,
The ES can transfer any amount of energy you like to the H, at any rate.
The H can transfer any amount of energy you like to the K, at any rate (though the K can only "consume" energy at 120kW).

The suggestion is that this ES to H transfer, then H to K transfer is interleaved, and the switch is done many times per second, and that it's also a reversible process (that is to say the K can charge up the ES using the H as a go-between).

The "unusual energy flow" is legal according to my reading of the rules. What rule would it breach, according to your reading of the rules?

[dren]

It's just a way to send to the ES more energy from the K than the direct allowed 2MJ/lap. It's basically electric fuel accumulation as TC stated earlier.

It's a novel idea, but you're forced to carry more fuel in order to accumulate it electrically. I wonder if Mercedes does this since it's fuel load is a bit lower than the rest.