LOL TCH — If the ICE is producing power, its torque is upward in my diagrams. Since we must have torque balance at each shaft of the planetary-gear set, the torque of MG1 must therefore be downward. If MG1's direction of rotation is also negative, the work it's doing is positive ('-' x '-' = '+'). If MG1 is doing positive work, it's acting as a motor. This means that MG1 is doing positive work on the ICE. There's really no alternative possibility. But this is the definition of "heretical" mode operation, I believe. [The power needed by MG1 to enable it to do this work comes from MG2 and/or the NiMH battery.]

Stan

 

You guys are kidding,,,,right?

 

Stan,

I see I am not the only one confused here! I believe the issue is more nomenclature than anything else. You refer to the MG1 direction as though you are describing the spin direction, when you are referring to a torque x rotation vector. For the sake of clarity, please use the term rotational direction in reference to MG1 such that direction of spin, i.e., RPM, of MG1 when the ICE is not turning (0 ICE RPM) is positive when the car is moving forward, and negative when the car is moving backwards, and not use rotational direction to refer to rotation x torque. Perhaps you could use a term like "negative work product" rather than "spinning backwards" for these quantities.

When the ICE is running, and the car is moving forward, MG1 rotation can be positive or negative, and in either case MG1 can be consuming current (motor) or producing current (generator), depending on conditions such whether the car is accelerating/decelerating, climbing/decending, speed of vehicle and ICE, etc., correct?

-- Alan

 

alan_in_tempe — I was trying to be as clear as I could be. Sorry if I was confusing. In my (and Toyota's) nomograms, the speed (rpm) of the MGs is shown on the vertical axis. Zero is in the middle, positive is upward, and negative is downward. The directions of the torques acting on each shaft are shown by arrows, positive upward and negative downward. Now (MG power) = (MG torque) x (MG rpm). If MG power is positive, work is being done by the MG, and so it is acting as a motor. If MG power is negative, work is being done on the MG, and so it is acting as a generator. [And, if no torque acts on it (i.e., if it's spinning freely), then no work is being done.] There can be no argument about this I'm afraid.

In the nomograms, when the ICE is providing power its torque vectors Te, Tes, and Ter are upward-pointing. Since the ICE's rpm is always positive (or zero) because it's not allowed to spin backwards, these upward torque vectors represent positive work being done by the ICE (Pe = Te x Ne > 0). See Figs. A-C. During fuel-cut engine braking, the ICE's rpm is still positive of course, but now work is being done on the ICE, and so now the ICE torque vectors are negative (i.e., downward-pointing). See Figs. F-J.

If the ICE is not spinning, then as the nomogram shows, MG1's rpm (Ns) must be negative whenever the car is moving forward (Figs. D and E). So, here I disagree with you. You seem to want me to change the sign of MG1's shaft rotation! Why? What I'm doing is straightforward, consistent, and standard. If I were to label MG1's rotation direction negative-upward, then I'd need a different rotation axis for MG1 compared with the ICE and MG2. It wouldn't be sensible in my view. I use the terms "spin", "rpm", and "shaft speed" interchangeably; I consider the direction of rotation to be positive if the rpm > 0, and negative if rpm < 0. Perhaps what's causing the confusion is the fact that we have these two quantities being displayed somewhat differently here:

• torque is being displayed as a vector [either up (positive) or down (negative)]
• rpm is being displayed as a scalar along the vertical axis [either above 0 (positive) or below 0 (negative)]

The reason for using vectors for torques is to make apparent the torque balance at each shaft, as mandated by Newton's Third Law — the vector sum of the torques must be zero at each shaft. A scalar is all we need for shaft speed and direction. Finally, work (or power) can be either positive or negative, depending on whether the device in question is doing work (positive) or having work done on it (negative).

I thus cannot agree with your statement: "When the ICE is running, and the car is moving forward, MG1 rotation can be positive or negative, and in either case MG1 can be consuming current (motor) or producing current (generator) ..." (emphasis mine). I claim that, with the ICE "on" and the car moving forward, MG1 is acting as a motor only when its rpm is negative (i.e., Ns below zero, torque vector Tm1 negative-pointing — Fig. C), and as a generator only when its rpm is positive (i.e., Ns above zero, torque vector Tm1 negative-pointing — Fig. B). [MG1's current has nothing to do with this argument.]

I hope that this helps to clarify the point I'm trying to make.

Stan

 

No argument here, except that I think you often refer to the torque [I]arrows[/] when you write about rotational direction.
No problem, and I apologize for suggesting the opposite convention as Toyota; I just want to clearly separate out the rotation direction from the torque/power/work issues, which is where I believe the confusion is.I believe the confusion is mixing discussion of rotation with discussion of power.This is true for the ICE, which can only run in one direction, but absolutely not true for MG1 and MG2, which can run forward or backwards (positive or negative RPMs) and can act as either a motor or a generator whether running forwards or backwards. If you do not agree with this statement, you can skip the rest until we have sorted this out. This is a critical point to my argument, and a fundamental reality about these kinds of motor/generators.I will assume you agree MG1 can rotate forwards or backwards when the ICE is running, even if the car is moving forward (depending on the ratio of the ICE RPM to the wheel speed).

Current has everything to do with whether a motor is acting as motor as generator, but I was not clear on this point. The direction of rotation determines the open circuit (no current) voltage polarity, and when the current is in the same direction, it is consuming electrical power and producing mechanical power (motor), and when the current is in the opposite direction, it is consuming mechanical power and producing electrical power (generator), which I was over-simplifying by the terms consuming current (motor) or producing current (generator), and did not mean to imply by "consuming" or "producing" the direction of the current, as that is relative to the direction of rotation.

Consider this very limited example: car is moving forward at a speed such that when the ICE is near idle RPM, then MG1 is moving in the forward direction. Now assume this is a steep decline such that the wheels are powering the PSD (ICE in fuel cut). 72% of the wheel induced power is going towards speeding up the ICE, and 28% is going to slow down MG1. When you are inputting mechanical power to a motor (in either direction), it can generate electrical power. You seem to be claiming this is impossible. I admit that this is a very restrictive example, and may only last for a brief moment until the ICE spins up fast enough to bring the MG1 speed negative (in the range you have no problem understanding MG1 acting as a generator). The only trick here is to have the drive voltage stay more negative (less positive counts!) than the motor speed, which is a trick the inverter can easily handle.

-- Alan

 

alan_in_tempe — I'm afraid that I am unable to find the supposedly misleading terminology that you say I used in earlier posts in this thread. If you point me to where these examples occurred, I could perhaps understand your complaint.

I myself am unable to measure either the MG voltages or the MG currents. I doubt that you have the ability to monitor them either. But, I can measure or compute the shaft speeds using the formulas I gave, and if pressed, I could do a reasonable job of estimating the torques too. We don't however, need the torques' magnitudes, but only their signs (i.e., directions), in order to deduce the direction of energy flow (i.e., whether motor or generator action is occurring). It's thus much more sensible in my view to base our arguments on these mechanical quantities, as I have done.

Let me elaborate a bit on the nomograms, since they seem still to be a source of confusion. The sloping line in the nomogram merely serves to show the linear relationship between the shaft speeds of the sun-gear (Ns — MG1 acts here with torque Tm1, and the ICE with torque Tes), the planetary carrier (Ne — the ICE acts here with torque Te), and the ring-gear (Nr — MG2 acts here with torque Tm2, as does the ICE with torque Ter, and the car itself with torque Tr). The shaft speeds Ns, Ne, and Nr are scalars; i.e., signed real numbers — they can be negative, zero, or positive, representing respectively backwards rotation, stationarity, or forwards rotation. The value of each shaft speed is given by its ordinate, and I have labelled all of them on the vertical axis of each Figure. The torques Tes and Tm1 at the sun-gear; Te at the planetary carrier; and Ter, Tm2, and Tr at the ring-gear are also scalar quantities; they are represented by signed real numbers corresponding to the respective shaft torques — negative if the torque is in the direction of negative shaft rotation, zero, or positive if the torque is in the direction of positive shaft rotation. As such , there's no need to represent them vectorially. But, how could one display these torques on the nomogram as scalars? Not easily. That's why they are universally represented on the nomogram as vectors, but vectors that can point upwards (positive torques) or downwards (negative torques) only. The direction of the arrow indicates the sign of the corresponding torque, and the length of the arrow is proportional to its magnitude. This mode of representation has benefits: Newton's Third Law requires that torque balance hold at each shaft, and this is easily seen visually in the vector representation.

So, I disagree with you when you say (I'm paraphrasing you here) that a scalar can be used to represent the ICE's shaft speed, since the ICE can only rotate in one direction, but that a scalar cannot represent MG1's and MG2's shaft speeds because they can be either positive or negative. A scalar is all we need for all three shaft speeds! Yes, indeed, MG1 and MG2 can act as either motors or generators (when terminated electrically in the appropriate way by the inverter electronics), and in isolation, they could do so in either rotational direction. However, when coupled to the planetary-gear set, you cannot switch from motor to generator behavior or vice versa without reversing the sign of the corresponding torque (Tm1 or Tm2), and we do not have the freedom to do so in our nomograms because of the torque-balance requirement.

For example, consider MG1. The torques Tes and Tm1 are the only torques acting on the sun-gear, and must always be in balance, i.e. equal and opposite. Now, the direction of Tes is always the same as the directions of Te and Ter because of the constant 28:72% torque split from Te to these other shafts. If the ICE is providing motive power to the car, these vectors are all upward (i.e., positive torques). This forces Tm1 to be downward in this condition. There's no choice in this matter! A negative Tm1 means that the motor/generator action of MG1 is determined solely by whether its shaft speed Ns is negative/positive (because then the product of shaft speed and torque — representing mechanical power — is respectively positive/negative). My Figures C and B show these two possible cases — "heretical" and "normal" modes of operation respectively. You cannot reverse the sign of the torque of MG1 (which would force it to switch between motor/generator) without simultaneously causing all the ICE's torque vectors Tes, Te, and Ter to reverse direction! But then the car is doing work on the ICE, and this is then the case of engine braking not of powered driving. Indeed, this is the case you consider in the example you give in your last paragraph. That means that you're considering the case shown in my Figures I and J. And, yes, I agree with you that in this case MG1 is acting as a motor (but not in the cases of Figures G, and H1, where it's behaving as a generator).

Does this help clarify what I'm saying, and why I claim that it must be true? I hope so! It's actually a highly nontrivial point. Other technically-knowledgeable readers of this thread (and there appear to be many of them!) might care to express their opinions on this matter.

Stan

 

I'm short on time now, but you deserve as much reply as I can give. As best I recall, the power/torque/rotation direction terminology confusion was in the other related thread. It is possible I am recalling verbiage from a different poster, however. I will search and get back on this.
I am not measuring, just reasoning with the known physics involved.You are entirely correct here, and I apologize. I am not sure why, but I was reading into your words unsigned scaler when that is not what you said.I think this may be the core of the issue. I believe the the torques on the sun gear are Tm1 and Tp (the torque of the planetary gears acting on the sun gear). Tp consists of two inputs, Tes (the planetary carrier) and Tr. However, I need more time to think this through and determine if this really is the source of our disagreement.

My thoughts are that there are three inputs to the PSD: MG1, the ICE, and the combination of the wheels and MG2, say MG2W. Power contributions from all three must be zero (looking at the PSD as a passive black box). Since MG2W and ICE can both add mechanical power simultaneously (downhill, ICE w/fuel), in which case MG1 is absorbing that power, generating current. This can occur at road speeds and engine speeds such that MG1 is turning backwards or forwards. If your nomograms say otherwise, then I suggest the error is in your nomograms, but I need more time to review your equations and nomograms. For clarification, I am not saying the HSD uses this mode, but that it can (just like Toyota can do things not explicitly spelled out in their relevant patents). That is, I am only arguing with your "impossible" comments. I believe this is the same, or very similar mode where MG1 starts up the ICE and acts as a generator simultaneously.

-- Alan

 

I don't think it is quite that simple. 72% of the wheel induced power is not necessarily being felt by the ICE carriers at the ring/carrier mesh. Power includes a torque component and is not simply RPM, and MG2, acting as a generator, is drawing off a significant amount of that power. Once MG2 generates electriciy from this power, it can be used by the ECU to apply opposing electrical force, which can be converted into an opposing torque by MG1 accross the sun gear/carrier mesh.

Think about what has to happen if you are coasting down a mountain at high speed, say 80 MPH or so and accelerating becasue the grade is getting steeper. Also assume that you have been doing it for long enough that the battery is fully topped up. The ICE cannot stop without overspeeding MG1 in the negative direction. In this case, a positive going torque (which is actually negative to the situatiion, a sort of decreasing Heretical Mode with Fuel Cut), is needed on MG1 to force the ICE to spin faster. How else can the ECU protect MG1 (and the ICE in other cases) except to use the power available from MG2 to directly control the sun gear, carrier (ICE) mesh torque direction and magnitude, which provides it the ability to control both RPMs, and therefore indirectly the torque magnitude and direction across the ring/carrier (ICE) mesh as well? It would make no sense at all to run MG2 open and use the battery for this.

An interesting question here is what happens in this situation when you tap the brakes lightly to take a downhill curve? The battery is topped up, so what is done with the regen energy? Is regen even used at this point? The obvious answer is to apply it in the same manner to MG1, but is it the real case?

 

To try to inject some life into this stalled thread, let me provide some additional information that I have computed regarding fuel-cut engine-braking in 'B,' to illustrate the utility of my equations. In this braking mode, all the energy being transferred from the car is dissipated only in the ICE. The MGs don't dissipate any power, at least in principle, but they do shuttle energy around. However, not all the braking power flows through the MGs — part of it flows from the wheels to the ICE directly through the planetary-gear set. The following is what my formulas say about this (see my posts # 27 and 34).

Assuming, as I did before, that no battery charging is taking place (it's "full" say), and that the MGs are 100% efficient in their energy conversion (this isn't seriously wrong — they are apparently ~95% efficient at each conversion between mechanical and electrical energy, so that their back-to-back conversion efficiency is ~90%), then I can calculate the energy flow in say the fuel-cut coasting scenario shown in my Figure I (in the post #34 cited above). We have:

Transmission in 'B';
Car travelling at 100 km/h (~62 miles per hour) (this is the Road Speed RS);
ICE spinning at 3000 rpm (this is Ne);
Ring-gear spinning at 2822 rpm (this is Nr);
Sun-gear (i.e., MG1) spinning at +3463 rpm (this is Ns);

and I compute that, of the total braking power flowing from the wheels to the ICE, fully 67.9% flows mechanically directly through the planetary-gear set from the ring-gear to the planetary-carrier (i.e., the ICE), while only 32.1% flows electrically from the ring-gear through MG2 to the sun-gear (i.e., MG1). [I won't bore you with the calculations!] The ICE creates all the drag, but its effect is felt by the wheels partially directly through the planetary-gear set and partially indirectly through the electrical path of the MGs.

Stan

 

No argument from me. That is exactly what I said. The only essential difference between "D" and "B" is the use of MG1 by the ECU to force the ratio across the ICE/axle boundry down so that the ICE is forced to spin faster. Doing this means that MG2 must generate more power (via increased current), and that MG1 is using more power to accomplish the controlling change accros the sun gear PSD boundry.

A small by-product is that a 10% coupling loss in the electrical patch will also mean a slightly higher total electrical loss that is ratiometric to the torque changes across the PSD, not a significant factor.