Why does the engine wind out when in B mode?

False.

In normal cruise conditions, a lot of power takes a direct mechanical (not electrical) path from the ICE to MG2's shaft (via the planetary gears) and on to the wheels, never getting converted to and from electric power. As Chapman already pointed out, the near ideal condition is when the ICE and MG2 (and wheel) speed ratios are just right such that MG1 becomes still. This means no power flows to MG1 to take the electrical path to MG2, thus minimizing electrical conversion inefficiencies through the generators/inverters/motors. Some electric power is still needed to hold MG1 still, as there is still torque on its shaft, but this is less than the normal power flow conversion losses.

But this is a fairly narrow condition. Most of the time, while some power takes that most efficient mechanical path, some other power must take the ICE -> MG1 -> inverter -> MG2 path to the wheels. Not quite as efficient, but it does create the 'continuously variable' part of the 'transmission' that is needed to reasonably efficiently match ICE RPM to wheel RPM.

A mechanical lock on the MG1 shaft could make that special MG1-RPM=0 condition even more efficient, as no electrical power at all would be needed to hold it still. But then it would be more like an original Honda IMA system, which needs different engine controls when ICE and wheel speeds and power aren't perfectly lined up for best ICE efficiency. The Honda system is more efficient under certain conditions, but the Toyota system works better in a broad variety of less ideal conditions.

 

Both MG2 and ICE spin algebraically 'forward' in these modes or 'gears'. Actual physical directions are packaging issues that I don't care about. I'm an electrical engineer, not a mechanical or industrial engineer.

In R gear, MG2 spins 'reverse', while ICE still spins 'forward', if it spins at all.

 

Be careful not to confuse power and torque. You did it once already in #89, but that's like confusing watts and volts.

Stated correctly in terms of torque, the 72% of torque transmitted from the engine to the ring gear is "theoretical" in the exact sense that gravity is "theoretical"—the sense that means verified, true, and unchanging.

Now, if you try to claim the same breakdown for power, that is only true in one specific case, when all three PSD components are spinning the same speed so the rpms fall out of the formula. At any other time, while the torque split never changes from 1:.28:.72, the power split will differ from that ratio, as the rpm ratios change. It can vary all the way from MG1 stationary and all the power going mechanically to the ring gear, or the ring gear stationary and all the power going to MG1.

I think I see part of what's going wrong here. jerlands actually has the dimensions right for HP, but not the unit. It's back to that confusing part of physics (already covered in #88) where both torque and work end up having the same dimensions (a force · a distance) even though they are not the same concept, and they have different units.

When measuring work, the (distance) refers to how far the object of the force has been made to move.

When measuring torque, the (distance) refers to the length of the moment arm: how far is the force applied from the center of rotation.

You can do work by putting a torque on something, but the thing has to turn. If you're just applying a torque and the thing's not turning, work hasn't been done (just the same as applying a voltage but no current flowing).

Once the thing does turn, you can turn it through some angular distance θ. If you turn it by one radian, you have done an amount of work equal (numerically) to the torque. If you turn it one full revolution (2π radians), then you've done work equal (numerically) to 2π times the torque.

So to get from torque to work, you have to multiply by an angular distance covered, in radians. But that doesn't show up when you look only at the units' dimensions, because the radian is a dimensionless unit. It has a numerator and denominator that are both distances, so the dimensions cancel.

So you might be putting T = 10 newton·meters of torque on something, and turn it by θ = half a revolution (π radians). So the work you've done is T·θ = 10 newton·meters · π radians, but the radian being dimensionless, that comes out 10π newton·meters, or about 31.4 joules, the SI unit of work. A newton·meter is a joule when it means an amount of work ... but not when it means a torque.

So when jerlands says "H.P. is torque over time", that's the confusion. HP is work over time, and that is not the same as torque over time, but work units and torque units have matching dimensions, so they are easy to mix up if you aren't careful to keep them straight.

Ok, that one's a little weirder, but the dimensionless radian may be to blame here too.

First the obvious thinko out of the way, "TIME" should have been "TIME⁻¹". There is time in the dimensions of any speed unit, but it's in the denominator.

Linear speed, like speed down a road, has dimensions distance/time, not just time⁻¹ by itself.

But everything we're talking about inside the tranny here is angular speed: not distance/time, but radians/time (or rpm, with a conversion factor; one radian/second = 60/2π rpm). And the radian is dimensionless, remember, so this ends up just looking like time⁻¹ by itself. You have to remember the angular distance is a necessary part of the deal.

Good example to clarify the point. Let T be the torque, 33,000 ft·lbs, and let ω be the angular speed, 1 RPM (or 2π/60 radians/sec).

The power, T·ω, is 33,000 ft·lbs · 2π/60 radians/sec = 3456 ft·lb·radian/sec = 3456 ft·lb/sec because radian is dimensionless = 6.28 HP (550 ft·lb/sec = 1 HP).

Again, where are you getting your material? MG2 spins opposite the engine like my sister is the queen of England.

It happens that the engine spins in the same direction that the wheels do in forward motion. Sure, there's no reason an engineer couldn't build it the other way, but it's pretty common in transverse-engine FWD cars, and definitely true in the Prius.

So about the MGs. If the car is not moving, neither is MG2. The engine drives the planet carrier (also in the same direction the wheels would spin going forward). The pinions have to rotate opposite that direction (because their outward-facing teeth have to roll along the stationary ring gear). That means their inward-facing teeth are moving in the same direction the planet carrier is turning, only faster, and the sun gear (and MG1) is being driven, in the same direction as the engine, at 3571 rpm if the engine is idling at 1000. Let's say the battery is charged, so nothing but idling is happening. MG1 is stirring up some oil at 3571 rpm, and that happens to take all the power the engine produces at idle (engine power = engine torque · 1000 rpm = MG1 torque · 3571 rpm, when MG1 torque is .28 · engine torque).

Now say you want to start driving forward, as gently as possible, without goosing the engine, just letting off the brake so motion is possible. This whole time, the pinion teeth have been exerting a torque (72% of the engine output torque) on the ring teeth trying to make it turn, but it hasn't been turning, so they were delivering no power (zero rpm times 72% of anything is still zero). But now the ring gear can turn, and there's a torque on it, so it will. It will turn in the same rotation direction as the engine, and now the wheels too.

As the ring gear rpm increases, the MG1 rpm has to decrease. Nothing else can possibly happen. The pinions aren't spinning as fast, because the outward facing teeth can move forward now. As long as the ring gear rpm is below that of the planet carrier, they still spin some, and so the MG1 rpm will still be faster than the engine, but it will be coming down from 3571, as the ring gear rpm goes up from zero.

MG1 is still getting 28% of the engine torque, but MG1 rpm is now less than 3571, so (assuming the engine's still doing 1000), MG1 is getting less than 100% of the engine power. That makes sense; the only work it was doing was stirring oil, and now it's doing less of that.

The ring gear teeth are still getting 72% of the engine torque, but the ring rpm is now greater than zero, so now there's actual power going this way. Not very much power yet: 72% percent of engine torque times an rpm that's still close to zero. It's the same amount of power that now isn't being used to stir oil by MG1, and the engine output power is 100% accounted for.

Now, you won't get anywhere very fast this way, so you step on the go pedal a little. The engine speeds up, producing more torque and more power.

Now a funny thing happens. The new extra torque out of the engine is split, as usual, 28% to MG1 and 72% to the ring gear. The trouble is, MG1 just isn't very hard to spin. It's a rotor in some bearings and it stirs up some oil, but that's it. It takes very little torque to just make MG1 spin faster.

And the identity that MG1's torque is 28% of the engine's is true both ways. If the engine crankshaft isn't working against some substantial torque, giving more gas is only going to send the engine revs sky high, just like goosing the gas pedal in a regular car in neutral. And the only torque the engine sees to work against is 1/0.28 of the torque to spin MG1, and that just ain't much torque.

For anything more to happen here, we need to make MG1 harder to spin. Having it do a tiny bit of oil stirring just isn't enough.

So, the HV ECU can start gating the switches on MG1's stator windings, and let it generate some electrical power. (It has been generating voltage this whole time, but no power, because the switches were not letting any current flow). Now there is electrical power coming out of it, and there's no free lunch; it is now a lot harder to turn, because the rotor permanent magnets are now doing work in the stator magnetic fields.

The torque the engine gets to work against is still only 1 / 0.28 of the torque to drive MG1, but the torque to drive MG1 just went up a bunch, so the torque at the engine shaft did also. The engine is no longer free-revving, but delivering power into a significant load. As always, its torque is splitting 28% to MG1 and 72% to the ring; its power is splitting according to those fractions times the rpms of those components.

Meanwhile, we've got this electrical power here. To make MG1 hard to turn, we had to let it produce electrical power. But now what do we do with that?

Well, we send it over to MG2, phased so that it adds to the torque already arriving through the PSD. So some of the engine power went mechanically through the PSD, and some went electrically around it, but all of it ends up at the gozouta, just as it should in any self-respecting transmission.

Is that a trick question? In EV D mode, presumably the engine is off, spinning in no direction, so you could say anything about the "MG1 spin in relation to ICE" and not be wrong.

If you mean in relation to the normal direction of the engine , MG1 will be counter-rotating then, as it does any time the ring gear is spinning in the D direction faster than the engine is (which certainly holds when engine rpm is zero).

You can find that information (and more) in the nomographs on pages TH-11 through TH-25 of the NCF, as already mentioned in #72.

Edit:

That might have been fuzzy1; I don't think it was me. (Not to say it would be a bad thing to do.)

 

Could you be more specific on the details causing "have to take the drive out of B"? If the battery was never fully charged, and there was never any noticeable increases in engine RPM, let alone exceeding redline, what were those current dumps that made you take it out of B?

 

If I remember right, this whole thread began with the OP thinking that the sound of the engine braking in B mode meant having to take the drive out of B to protect the engine from overspeeding.

That led into whether the car ever lets the engine overspeed in B mode (it doesn't), and to the OP claiming that, well, what it's doing in this case is all different because it's a (new invented term) "current dump to MG1", and that led to six pages of trying to help the OP understand the transmission, because heaven knows that's easier than looking at a tach.

 

In a battery with reduced capacity, engine braking will happen more readily. Replacing the battery with one having greater capacity will reduce how often engine braking will happen under typical circumstances.

Whenever engine braking does happen, it is under control of the HV ECU, does not exceed a safe speed for the engine, and does not indicate any need to shift out of B.

That is the range. In my Gen 3, on a long descent, the car will noticeably begin trading battery charge for engine braking somewhere around 77 or 78 % SoC. You can hear the engine beginning to wind up. Once the SoC has reached 80%, my Gen 3 has fully traded charging for engine braking, and if the descent is steep enough to keep the engine at the maximum safe RPM that the ECU allows, it will stay there for the rest of the descent.

 

The best thing about the page you found there is it has not one but two posts on it by hobbit, who is one of the long-time members of this community with a deep understanding of how the stuff works, and in those posts, true to his usual form, his explanation of how the MGs are controlled is spot-on.

The rest of that thread makes a neat story, as you can see maxsolar coming in with a few confused notions about how the stuff works, and getting them straighter as the thread goes on, showing an ability to learn when explanations are given.

The part you quoted here comes from maxsolar's last post in the thread, and there's nothing wrong with it, though there's also nothing about it that supports your interpretation.

I greened the part you were probably seeing as support for "have to take the drive out of B". But if you look again, he's saying that B mode is less than ideal "from an energy management perspective" (because you would rather recover and store that energy, instead of using the engine to dissipate it). He is not saying that there is any need to exit B mode to protect the engine. The car keeps its rpm safe at all times.

He goes on to confirm that while you'd rather avoid B mode when you have the battery capacity to cover the whole descent, there are times when you just don't, and those are the times B mode is made for, and it will dissipate as much energy for you as it safely can, so you only need to use the friction brakes to take care of the rest.

That was the part where just getting a tach (or an OBD reader able to show a tach) could have saved you a bunch of anxiety right from the start. We can (and do) make a lot of judgments based on what sounds "good" or "bad" intuitively, but sometimes a good instrument helps adjust those intuitive judgments to match what's objectively going on.

 

I see what you are referring to in the graphs uploaded by that poster. During a descent in D, then switching to B, about one second after selecting B, there is a transient current about one second wide, of triangular shape with about an 82 amp battery discharge at its peak. That integrates to about 41 amp seconds, or (at nominal battery voltage) about 2.3 watt hours. For scale, that's a little less than a twentieth of one of the 50 Wh little cars you can earn on the MFD for regeneration.

It's an interesting little blip, and the B mode description in NCF doesn't mention it. The graph does not show any reason for concern that this "dump" leads to overspeeding the engine at all. The same graph shows the engine speed rising smoothly to something under 3616 rpm and being held there rock steady as long as B mode is in effect, dropping smoothly again after B mode is exited. During the transition out of B mode, the engine speed decrease is also accompanied by a current blip, not as tall but of longer duration, and in the other direction. It probably integrates to about the same twentieth of a little car icon.

Why the transition is handled that way is an interesting question. Since switching from D to B means telling the car "I would like less of this regen energy to go toward the battery and more of it toward the engine, please", the transition could be accomplished by simply taking some of the power flowing from MG2 to the battery, and diverting it to MG1, in the co-rotating direction (before the transition, it is counter-rotating and the engine is stopped). That would bring MG1 and the engine eventually to the target engine-braking speed. And all of that is exactly what's happening once that brief transient is over.

Likewise, the transition from B back to D could be accomplished just by saying "ok, not sending part of this power from MG2 over to MG1 anymore, just sending it to the battery." That would stop the electrical part of the power flow toward the engine immediately, and the mechanical part would also taper off as the engine naturally slowed to a stop.

But Toyota seems to involve a tiny amount of battery interaction in both of these transitions. One explanation could be that the D mode and the B mode both work by using (different) algorithms to pick target rpms for the engine and MG1, and when you switch from one mode to the other, the new target rpms just get fed to a different layer of the firmware whose job is to hit the target rpms, and that layer says "schezbzflat! I'm way off my target!" and gets there the fastest way it knows, by borrowing or lending with the battery.

Whatever the reason, it only lasts a second and is a small amount of energy; any time after that transitional second, what you are hearing is not any special or alarming "current dump", but just engine braking doing what it does.