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Neutral Gear in a HSD Toyota (Again)

Discussion in 'Toyota Hybrids and EVs' started by RGeB, Oct 26, 2023.

  1. RGeB

    RGeB Member

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    Yes, I know, neutral has been discussed many times in PriusChat. But where I checked, the ‘explanations’ were a bit glib, and the threads diverged into such vital issues as “Will my Prius go flat in a carwash?” or “Can I use neutral when gliding to hyper-mile?” I am interested in how it works. Where does the ICE torque go?

    We know that (in modern THDs) ICE is connected by a “one-way clutch/damper” to the planetary carrier (C). This is not a “clutch” that can disengage ICE from the drive-train. Instead it stops ICE from being rotated backwards, and it takes some of the shock out of ICE stop/start. The rotor of MG1 is splined to the planetary sun (S). So MG1 can start and control the speed of ICE or serve as a generator using power from ICE. The rotor of MG2 is connected by conventional gears to the planetary ring (R) and also to the differential/wheels. So MG2 can function as a motor to power the wheels, or as a generator taking power from the wheels during braking.

    TIS for most models does not seem to deal with neutral (N), except to say that “MG1 and MG2 are basically shut down”. I take this to mean that the motor stator wiring is made open circuit. So the rotors will spin freely. This will prevent MG1 from starting ICE. But ICE continues to run, though not to charge, if it is on when N is selected.

    The owners’ manuals warn about possible dangers of N, like a flat HB and towing damage (all good, but no explanation of where torque goes if ICE is on).

    The oft-cited Wikipedia entry on HSD seems (to me) misleading on this point, as stationary planet gears are neither necessary nor sufficient for N in HSD.

    In theory with stationary state, when ICE runs (without mechanical separation from the drive train) 72% of ICE torque will go to R. If there is torque to R, there is torque to the wheels (even if not enough to overcome the counter-torque from friction between the tyre and road). So when ICE runs in N, the car should be easier to push forward, and harder to push backward. This should be more noticeable at higher ICE rpm (= more ICE torque); but ICE will not rev in N or in Park (P) while charging HB; so you cannot perform this test.

    TIS for some models indicates that motor shut-down is temporarily over-ridden if N is used while driving and braking, or if ABS operates. So (electronic) control seems quite sophisticated, but (mechanical) torque is unexplained. I tried an experiment:

    Park your HSD Toyota on a level surface and chock the rear wheels very securely, front and back. Use a trolley jack to lift both front tyres (or tires if you are in the USA) a small distance clear of the ground. Things are complicated by use of the parking brake*, so turn it off.

    While the car is in Park (P), READY or not, with ICE off, try to turn a front tyre forwards then back. The wheel rotates, albeit stiffly in both directions, and the other front wheel rotates through the same angle in the opposite direction! So I conclude that in P this is an effect through the front differential, without rotation of planetary R.

    Shift to neutral (N). You can only do this from READY, and while in N you cannot turn the car OFF, only to accessory or ignition. Again try to turn a front tyre forwards then back. It will move, a bit less stiffly than in P, in both directions. Now you can hear a noise which could be the planetary gear set or MG1 or MG2 or all. Also, rotation of either front wheel, forwards or backwards, has no effect on the other front wheel. So I conclude that in N the effect involves rotation of planetary R.

    Now turn on the aircon on and wait for the HB SOC to drop enough to start ICE (or speed things up by depressing the accelerator pedal to start ICE). While ICE is running, repeat the above tests.

    In P, the effect is as before.

    In N, forward rotation of the wheel is much easier than backward rotation of the wheel. When one front wheel is rotated forward, there is a lesser rotation of the other front wheel forward. When a wheel is rotated backward, there is no rotation of the other wheel. So I conclude that:
    1. This is not an effect between wheels through the differential; it involves the planetary gear set.
    2. There is a small amount of torque from ICE to the wheels (just not enough to rotate them against friction, even with tyres off the ground)!!
    Don’t get me wrong, N in HSD gives an effect almost like N in a conventional transmission. ICE delivers power only into the path of low resistance (freely spinning / open circuit motors). The car pushes fairly easily in either direction, and it rolls fairly freely backwards down a slight incline. But HSD differs in that ICE is not physically disengaged from the drive train, so when ICE runs it delivers some (slight at low ICE rpm) torque to the wheels.

    Consistent with this, torque PIDs, though open to interpretation, can be read to show that MG2 (and MGR) give most ‘wheel torque’ at low car speed, whereas ICE dominates at high car speed.

    I did the tests in a gen5 rav4 hybrid (AXA-H54), but I don’t think that matters much. Unlike some early prii, the AXA-H52/4 will not need to start ICE to keep a safe MG1 rpm at any realistic kph (the motors and gearing differ), but unless the Prius invokes quite a high ICE rpm, there will not be a substantial increase in torque from ICE.

    * My car is set to engage the parking brake in P, but this can be over-ridden manually using the parking brake switch. Sometimes with parking brake switch on, the front wheels are locked; sometimes not. When a raised front tyre is rotated in P, the parking brake usually turns off (so the pawl to planetary R is disengaged). In N, the parking pawl does not engage, even though the parking brake light sometimes stays on. I do not understand all this, so I have advised to conduct the above tests with the parking brake off.
     
  2. ChapmanF

    ChapmanF Senior Member

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    Right; a simple answer to the "where does the engine torque go" is (a) in unloaded conditions there isn't much, net after what the engine needs to keep twirling itself, and (b) of that small net torque, 78/108 of it goes to the ring gear (and thence to the wheels) and 30/108 of it goes to MG1, because that's what always happens to the torque.

    If you think in terms of power rather than torque, as power is torque ✕ rpm, if the car is stationary, then of the small net power from the engine, the wheels are getting none (78/108 ✕ 0) of it, and what MG1 is doing (net engine torque ✕ 30/108 ✕ MG1 rpm) accounts for all of it. MG1 is open-circuited in N, so none of this is coming out as electrical power; it's just making MG1's bearings and the sloshing oil warm.

    The delivery of a slight torque to the wheels in N is not super distinctive to this transmission. In a manual with the clutch disengaged, the transmission input shaft is still usually supported in a pilot bearing in the crankshaft. None of the synchronizers in the transmission are in mesh, but the countershaft is turning, and so all of the gears on the mainshaft are turning, not all at the same speed but still all in the same direction, and manual gearbox oil is usually pretty thick stuff, so they'll be putting a bit of torque on the mainshaft. You can work out a similar story for the torque converter and planetaries in an old automatic. The brake bands will be released in N so the rings that would be grounded in D are allowed to spin, but that doesn't mean absolutely no torque gets through.
     
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  3. RGeB

    RGeB Member

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    From what I understand, it is not always but rather when:
    (i) all friction is ignored,
    (ii) the system is in steady state (no acceleration etc),
    (iii) there is only one input (so torque split to two ideal outputs).

    We generally make these assumptions because the theory and the calculations are otherwise too difficult (for me at least).
     
  4. ChapmanF

    ChapmanF Senior Member

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    (i) is a simplification in the model, yes. Not a very big one; when the car is stationary in N there is nothing going to friction in the final drive output (as none of that is moving), and the two components moving are the engine (whose friction is accounted for in the net output you're seeing) and MG1 (whose friction is accounted for in the torque being applied to it). Friction between gear teeth isn't being modeled.

    That aside, the 108:78:30 torque split holds at constant or changing velocities, and whatever you choose to call inputs or outputs. A good schematic picture has to be drawn, of course, showing just where the three imagined strain gauges measuring these three torques would go, and what is on which side of each of those locations.
     
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  5. RGeB

    RGeB Member

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    No, the simple Torque / HSD model applies only during steady state (so the first post was correct to use it as an approximation).

    It does not always apply. During acceleration (or deceleration) it is necessary to consider inertia. See Miller 2006.

    The question I still wonder about is the scenario where the driver of an early Prius selects neutral at high speed (steady state or not).
    Then the wheels and planetary R have substantial rpm, and ICE may need to run at fairly high speed to protect MG1.
    Eg, from the EAHart animation, at 70 mph ICE would need to be at >1500 rpm to keep MG1 <5000 rpm.
    The Toyota hybrid ICE variants have fairly flat torque curves, and at >1500 rpm should be making enough torque to notice.
    So will substantial power then flow from ICE to wheels?
     
  6. ChapmanF

    ChapmanF Senior Member

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    Thanks for that link, a good paper I had not seen before.

    The key in reconciling different descriptions, as I alluded in post #4, is to draw carefully where one imagines the strain gauges being placed to measure these torques, and what will be on which side of each.

    For example, where Miller writes mg.png on page 4, the second term is the torque required to change the rpm of the engine crankshaft itself. Whether you "see" that torque depends on where your gauge is (and why the engine rpm is changing). If the engine speed is increasing because you've opened the throttle, the torque required for that acceleration is substantial but will not be seen on a strain gauge placed on the input shaft between the crankshaft and planet carrier. That gauge is measuring net torque out of the engine, and that doesn't include what accounted for the acceleration of the engine's own internal bits. The fixed ratios of torque distribution by the PSD will apply to the torques you do see.

    If the crankshaft speed is accelerating because you are engine braking, then you do see that component of torque on your input shaft strain gauge, and properly so; you also see 30/108 of it between the sun gear and MG1, and so on, still as expected.

    Those "lumped" inertial terms are further expanded in the appendix of the paper. Distribute the multiplication, separate the terms, and decide between which terms your imagined strain gauges will go. The terms you can't practically eliminate are, of course, the inertia of the PSD carrier and gears themselves, which are indeed non-zero but probably not the droids you're looking for; the PSD fits in your hand.

    The torque curves link is dead for me, but I assume it's similar to one of the familiar graphs that have been attached here before:

    [​IMG]

    The curve that's shown is a 1-D path through a 2-D operating region. (The beauty of the hybrid drivetrain is that the ECU mostly gets to pick the path that goes through the most fuel-efficient points, and operate the engine only on that path for the most part.)

    But imagine an engine mounted on test stand. Take a nice wireless strain gauge and mount it between the end of the crankshaft and the, well, free air behind it. Now start the engine. (How? Well, wrap a rope around the flywheel and pull, I guess.)

    Can you manipulate the throttle to run this test-stand engine at 1500 rpm? Sure you can.

    The vertical line rpm=1500 intersects the plotted curve around 10 kW, which would be a torque around 64 Nm. But is your strain gauge reading 64 Nm? No, it isn't. The engine simply isn't operating on that curve in these conditions. It is operating at a point that is on the vertical rpm=1500 line, but far below that curve. It is producing enough torque internally to keep itself spinning at 1500 rpm, and delivering practically zero net torque or power to anything outside itself.
     
    #6 ChapmanF, Oct 29, 2023
    Last edited: Oct 29, 2023
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  7. RGeB

    RGeB Member

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    The link is live for me, but in case it is problematic here are some of the figs:
    Toyota.PNG Toyota2.PNG
    They are for the 2.5l engine (Camry rav4 etc hybrids) and agree roughly with those from the US EPA:
    Toyota3.PNG
    No doubt smaller Toyota engines will make less torque, but they must be way above zero at 1500 rpm. ChapmanF, I don't understand much of your recent post ("probably not the droids you're looking for; the PSD fits in your hand"?) but I am glad that you (and Fred_H?) are happy with your creative accounting of relocated torques. Personally, I would back Miller.

    Getting back on topic, when I get home next week, I will try the same experiment on a non-HSD automatic. I no longer have a manual transmission to test, but maybe someone else will try it.
     

    Attached Files:

  8. ChapmanF

    ChapmanF Senior Member

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    I am backing Miller. Even using his algebra.

    As you surely understand, an engine at any RPM is producing no more net torque than it is being opposed by. Mr. Newton gives us that.

    In neutral, there is nothing opposing MG1's rotation other than the friction of its bearings and the medium in which it spins. That will amount to a certain torque when MG1 is rotating at a certain speed. If you were to apply to it any torque higher than that, you would accelerate MG1; its speed would not remain constant.

    That torque, as reflected through the PSD (multiply it by 108/30) is an upper bound on the torque that is opposing the engine. Perhaps by raising the car and letting the wheels spin freely, you could reduce the torque opposing the engine even below that bound. But in any case, the engine is not at that point producing any more net torque than that upper bound; if it were, you would see the engine's and MG1's RPM increasing together.
     
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  9. RGeB

    RGeB Member

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    Apologies to ChapmanF (and Fred_H?)

    1. I tested a 2006 Honda Civic 1.8l VTi auto, and it is also easier to rotate a front wheel forward with ICE on. I am not set up to measure forces, but the difference is tangible, and this car is definitely not a HSD.

    2. Also I think I now understand some of the advice in part of post #6. To paraphrase: The car computers can alter ICE torque (at any wheel or engine rpm) by altering fuel supply. There will be a no-load torque if ICE idles, and there may be additional torque to run ICE-driven accessories (eg to control MG1 speed). There is a maximum (WOT) torque available at any rpm (depending on the engine design), but when load is low the actual torque at that rpm may be much lower.

    The US EPA has a neat set-up, using an engine dynamometer to hold set rpm levels while logging engine parameters at various throttle levels (loads) to create a torque-rpm matrix in which other parameters (like BTE) can be mapped. Our hypothetical HSD Toyota can also give various torques at any rpm while gliding in N.

    We do not know the Toyota ICE torques under all relevant circumstances. Even the torque needed to control speed of free-spinning MG1 seems to be unreported. Torque PIDs remain open to interpretation (because we are not given required information about how and where they are measured or estimated). So it is hard to calculate how much power would flow from ICE to the wheels under the scenario where a driver of an early Prius engaged N at a speed sufficient for ICE to run in order to control rpm of MG1. Hopefully, it would be small.

    3. But I am still with Miller on the need to consider factors like inertia when apportioning ICE torque between HSD sinks outside of steady state. Seems that we now agree on that.
     
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  10. ChapmanF

    ChapmanF Senior Member

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    We are both "with Miller", though our interpretations of what Miller is saying have not been shown to match.
     
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  11. RGeB

    RGeB Member

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    This is drifting off the original topic of the thread (What happens to ICE torque in N? ... which I think has been largely answered thanks to Chapman F) into the esoteric world of torque distribution from a planetary gearbox. But someone may be interested in these esoterics. So, here is my understanding:

    Steady-state analysis is not adequate for a system undergoing acceleration (messy things like inertia, shaft rigidity and load-proportional losses all get in the way). To understand the car, it is necessary to consider these along the entire drive train.


    The equations are complex (at least to a non-mathematician, -engineer or -mechanic like me). We do not know the value for every parameter in them under most conditions. But the consequence is simple enough:

    The proportion of engine torque to planetary R (and eventually to the wheels) or to planetary S (and thence to the rotor of MG1) will vary in the same car under different conditions. Even under ‘steady state’ conditions, the proportions probably vary with load. Torque from ICE will always be split, but not always exactly 78:30 (~72%:28%). These proportions of ICE torque will never reach the wheels and MG1.

    But most drivers under most conditions will not notice or care about the difference. The car sensors and computers take care of it all (and MG2 and MGR) without fuss in response to conditions including driver demand (even if the algorithms are secret). There will be a much bigger impact on affected factors of potential interest (like fuel efficiency) from terrain and the nature of commands sent through the pedals.
     
  12. RGeB

    RGeB Member

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    This is drifting off the original topic of the thread (What happens to ICE torque in N? ... which I think has been largely answered thanks to Chapman F) into the esoteric world of torque distribution from a planetary gearbox. But someone may be interested in these esoterics. So, here is my understanding:

    Steady-state analysis is not adequate for a system undergoing acceleration (messy things like inertia, shaft rigidity and load-proportional losses all get in the way). To understand the car, it is necessary to consider these along the entire drive train.


    The equations are complex (at least to a non-mathematician, -engineer or -mechanic like me). We do not know the value for every parameter in them under most conditions. But the consequence is simple enough:

    The proportion of engine torque to planetary R (and eventually to the wheels) or to planetary S (and thence to the rotor of MG1) will vary in the same car under different conditions. Even under ‘steady state’ conditions, the proportions probably vary with load. Torque from ICE will always be split, but not always exactly 78:30 (~72%:28%). These proportions of ICE torque will never reach the wheels and MG1.

    But most drivers under most conditions will not notice or care about the difference. The car sensors and computers take care of it all (and MG2 and MGR) without fuss in response to conditions including driver demand (even if the algorithms are secret). There will be a much bigger impact on affected factors of potential interest (like fuel efficiency) from terrain and the nature of commands sent through the pedals.
     
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  13. ChapmanF

    ChapmanF Senior Member

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    The key thing that a mathematician or engineer would add to that is quantification: by how much, at absolute maximum, under any conditions, could the ratio measured at the PSD differ from exactly 108:78:30 for any reason (and for how long, before metal bits go flying)?

    Are you thinking it could ever be as much as 5%? 0.5%? 0.05%? 0.005%? Even that much?

    Again, part of the way we are talking at cross purposes here comes from vagueness about what torques to measure; this goes back to the invitation in post #4 to indicate exactly, in a diagram, where three notional strain gauges would be placed, and in post #6 to expand the expressions in Miller to see which terms would lie on the + and which on the − side of each measurement location.

    Nothing wrong with that as far as it goes, but it's important to remember that different questions can have different answers, and a question about "the car" or "the entire drive train" is not the same question as one about "the PSD", a small component found inside the transmission. And the different questions can matter to different people.

    In the tetrahedron of state, the three "messy things" you've mentioned correspond to the I edge (where energy can be stored, as in an electrical inductor, by changing the momentum of a moving mass), the C edge (where energy can be stored, as in an electrical capacitor, but by twisting up a not-perfectly-rigid shaft), and the R edge (where, as in an electrical resistor, energy goes to heat and is lost to the system). tets.png

    There are definite limits on the I and C cases. An inertia term is only nonzero while something is being accelerated, and you can only accelerate something so long before it's going too fast. The limits are even stricter for the C case: you may measure an instantaneous torque split of not exactly 108:78:30 while some energy is being stored in the twist of a shaft or deflection of a gear tooth, but there is a very very very brief limit on how long that can go on before the shaft looks like this:

    [​IMG]

    Energy storage on the I and C edges is (ideally, and very nearly in practice) reversible: taking the C case for example, for every instant you might measure a torque split slightly one side of 108:78:30 because a shaft is twisting up or a tooth is deflecting, you will measure it on the other side as the shaft untwists or the tooth springs back and returns the energy. If we talk about that as a reason for saying the PSD torque split "isn't always exactly" 108:78:30, we are talking about an effect that definitely matters to engineers tasked with questions like "how thick do I need to make that shaft or those teeth?", but is a vanishing part of the answer when that's not the question.

    The R losses are, of course, always losses, which the engineers have done their best to keep to a handful of %. The amount by which they might be "load-proportional" cannot, of course, exceed some fraction of that. Whether all of them matter, or only the losses within the PSD itself, depends again on the question being asked. The question "what torque split happens at the PSD?" is more specific than one about "the drivetrain" or "the car".
     
    #13 ChapmanF, Nov 6, 2023
    Last edited: Nov 6, 2023
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  14. RGeB

    RGeB Member

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    This theoretical discussion has not much predictive value. We can imagine sensors wherever we wish, but it would be good to know where Toyota imagines or places the sensors used for their reported PID values. (Ockham would place them at the primary output shaft of each device, but Toyota does not always do this {MG2 regen torque ≈ 3 * MG2 torque}, and seems to regard the locations as secret.)

    THS is of course more than the planetary gearset (sometimes called PSD). THS has 4 inputs (MG1, MG2, Wheels, damped ICE) at times, and the calculated ratios from real-world dymanic PID values are all over the place compared to predictions from the model of a steady-state planetary gearset with a single input. I have shown an example below (logged at about a 0.16 sec PID refresh rate).
    Log.PNG
    In an earlier post I complained about the OBD refresh rate, and blamed OBDLink (unfairly). The primary limitation arises in the Toyota gateway module, which limits OBD PID rate (probably as a safety measure to prevent overloading of a CAN bus that is busy with ADAS). Even so, I think 0.16 sec is fast enough for the purpose here.
     
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  15. ChapmanF

    ChapmanF Senior Member

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    That's the crux of the confusion here. Toyota simply has not placed any sensors in the locations where you would need sensors to be placed, if you needed to be convinced of how a planetary gearset splits up torque. They don't need any sensors placed there, because they're already convinced. The PIDs you can read with a scan tool are reporting other things.

    To see the effect of how the PSD splits torque, you would measure three specific torques:

    1. The torque by which the engine and the planet carrier oppose each other (as if you cut the input shaft and inserted a strain gauge there)
    2. The torque by which the sun gear and MG1 oppose each other (as if you cut the stubby shaft connecting the two and inserted a strain gauge there)
    3. The torque by which the PSD ring gear and the rest of the downstream drivetrain oppose each other (as if you could take that ring gear / final drive gear / MG2 spline assembly and stretch it, cut it right between the ring gear part and the other two connections, and insert a strain gauge there)

    It is those three torques (whether actual sensors are placed there or not) that observe the fixed 108:30:78 (respectively) ratio of the PSD, and that is what allows one to understand the role of that component in the context of the complete transmission.

    It is as if you are looking at an electronic circuit and you spot a pair of resistors in series with a tap between. You know that is a voltage divider, and even if the circuit was not built with easy test points at the ends and junction of those resistors, you know what voltage relationship you would see if you did measure at those points, because you already know what a voltage divider does. It might be that "we can imagine sensors wherever we wish", but it is the skill to imagine them in the right places, and know what they then would tell us, that allows us to then go on and understand the role that voltage divider plays in the larger overall circuit.

    As far as the Toyota PIDs you can read, its "MG1 torque" is not torque (2) above, and its "MG2 torque" is not torque (3) above. Those PIDs are the torques to be applied to the MG1 and MG2 rotors, respectively, by the stator electrical currents.

    The torque (2) above would be the same as the electrically-applied "MG1 torque" (if any), but for the other terms like (R) frictional losses in MG1's rotation and any (I) inertial torque while its angular velocity is changing. This is where those terms are properly accounted for (and they are worth accounting for, because MG1 being otherwise 'free', when the electrically-applied torque is zero, the whole story is these other terms).

    The torque (3) above equals the sum (with care to get the signs right) of the electrically-applied torque at MG2 (if any) and the torque and losses through the final drive train to move the car. (A fastidious accountant can throw in more terms, like inertia of MG2's rotor, but that's less important here as MG2 isn't 'free'—the whole drive train and the car (if not on a lift) move whenever it does, and the MG2 rotor inertia is a fairly small part of that picture).

    The torque (1) above would directly equal the (net) engine torque, if there were any PID directly giving that, but again, that isn't a quantity that any sensor exists in the car to measure directly. The ECM does compute an "actual engine torque" PID, but given that there is no such sensor, the ECM is clearly using other inputs it actually has along with a model of engine behavior, and while impressed they can do that at all, I don't have high expectations of its accuracy, or its ability to separate what is internal from what is net.

    In short, it isn't surprising when observed values that are not torques (1), (2), and (3) above do not show the relationship imposed by the 108:30:78 PSD; they are measurements taken elsewhere in the system, and that's to be expected.

    That doesn't mean people shouldn't understand how the PSD itself works and the way it splits torque. That's an important part of understanding the drivetrain as a whole.
     
    #15 ChapmanF, Nov 8, 2023
    Last edited: Nov 8, 2023
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  16. RGeB

    RGeB Member

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    ChapmanF, you are free to discuss PSD as if it were THS (the topic of this thread) and to assume that there is no difference between Toyota "request torque"and "torque" PIDs and to use sloppy accounting when the system is not in steady state. I do not. But I still appreciate your ideas.
     
  17. ChapmanF

    ChapmanF Senior Member

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    Equally, I suppose you are free to write posts suggesting that that is what I am doing when it is not. It takes all kinds to make a world.

    But other readers deserve a chance to see exactly how you do it. For example

    Clever phrasing, but of course what I said in #15 was that to "understand how the PSD itself works and the way it splits torque" is "an important part of understanding the drivetrain [THS, if you prefer] as a whole".

    The last paragraph of #15 literally said that. Discussing the PSD "as if it were THS" is not anything that has happened in this thread.

    I could offer the analogy that when you find, in the midst of a larger electronic circuit, a pair of resistors forming a voltage divider, your knowledge of what a voltage divider does is what allows you to understand (1) the function of that part of the circuit, and thereby (2) its role in the function of the circuit as a whole. But that would just be repeating post #15.

    I have made no such assumption, nor spoken of the "requested engine torque" PID at all. I took care in #15 to refer to the "actual engine torque" PID by its full name to avoid inviting any such confusion. Indeed, there are two PIDs:

    pids.png

    Since you have mentioned the "requested engine torque" PID, which I did not, it's there to show the torque the ECM is being asked to generate from the engine, by command received from the power management control ECU. The "actual engine torque" PID (which I did mention) is the ECM's idea of what torque it is currently getting the engine to produce (and ideally something close to the request). However, as mentioned in #15, there is no torque sensor anywhere to supply this value directly. The ECM is constrained to estimate it from other data.

    Before leaving the "requested engine torque" PID, an alert reader will notice that either its label (torque) or its unit (kW) is wrong in the manual. It should either be requested power in kW, or requested torque in Nm. This post shows how to clarify which interpretation is right.

    Indeed you do not; you haven't yet taken up any of the invitations earlier in the thread to demonstrate understanding of the accounting at all. Your insistent repetition of "when the system is not in steady state" is coming in place of any visible effort on your part to quantify, or just place an upper bound on, the difference you think that will make.

    Referring back to the tetrahedron of state in #13, in the conditions you're calling "steady state" (presumably constant RPM and load), only the R terms (losses) are significant. If you want to model the gear train so finely that you have C terms for the elastic deformation of each gear tooth as it goes by, you can, and will then have such C terms even in this "steady state" (and you can plug in some plausible values and get an idea how big they are, and decide if it was worth your time to model things that finely).

    Let the torque be non-constant, and you can include C terms that represent the twisting of shafts that aren't perfectly stiff; let the RPMs be non-constant, and you can include I terms representing inertia of various components. Again, you can find plausible values for the torsional stiffness of shafts and the moments of inertia for various components, and plug those values in to get some bounds on how large those effects can even be, and for what purposes it would or would not be important to include them in your model. To do that, of course, you must keep careful track of which shafts and which inertial moments are on which sides of the points at which you are asking about torque. "Sloppy accounting" indeed is what will result from not doing so.

    You have made interesting and valuable contributions in other threads on PriusChat, and I would prefer to be able to engage with you in respectful discussion. Your penchant for mischaracterizing my posts rather than responding to them, however, complicates that.
     
    #17 ChapmanF, Nov 14, 2023
    Last edited: Nov 14, 2023
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