Gen 3 PSD Question

Discussion in 'Gen 3 Prius Technical Discussion' started by Mark Monroe, Feb 25, 2013.

  1. Mark Monroe

    Mark Monroe Junior Member

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    I have been reading everything I can find on on the PSD. I have watched the videos and whatnot. I feel like I have a pretty good understanding of the PSD and the general powertrain of the Prius.

    I am a bit lost one on thing though.. <smile> I am going to try to explain my question and hopefully it will make sense.

    I think I understand how the ICE and MG1 and MG2 are tied together in the PSD. But.. When the ICE is not running - basically stopped and at zero RPM - what keeps the ICE from spinning "backwards" when MG1 and MG2 are applying lots of power in the PSD.

    For example, if I am understanding correctly, at a dead stop with the ICE turned off - as soon as you start off MG1 and MG2 are going to apply power to the PSD. That power would be applied on both the output shaft running to the tires and to the ICE. There would need to be something in place to keep the ICE from spinning backwards in order for full power to be applied to the output shaft running to the tires.

    Thoughts/Ideas

    MarkM
     
  2. fuzzy1

    fuzzy1 Senior Member

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    When the ICE is stopped, MG2 is the only thing applying power to the wheels. MG1 is spun backwards at whatever varying RPM is needed to keep the ICE stopped. MG1 is not actually helping propel the vehicle.
     
  3. cablesm

    cablesm New Member

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    This is correct -- I'd like to add that the only torque the engine might feel is from friction in the planet gears, and that should be negligible.
     
  4. telmo744

    telmo744 HSD fanatic

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    MG1 has to apply some torque also. The system acts like a differential.
    Some with a scangauge can confirm that with Amps/power to MG1?

    edit: some info here

    Two MG1 Questions | PriusChat
     
  5. fuzzy1

    fuzzy1 Senior Member

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    Just barely, only enough to overcome the very small friction of the gears, bearings, windage, and to drive its own rotational inertia when RPM must change.

    Without something else mechanically locking or applying torque to the ICE shaft, MG1 cannot contribute to the car's propulsion.
     
  6. vincent1449p

    vincent1449p Active Member

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    Hi telmo744,

    The NCF says MG1 does not generate any torque. Do you have any official sources of your info?

    Vincent
     

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  7. kbeck

    kbeck Active Member

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    Mark,

    The right way to think about the Prius's HSD is to think about Torques.

    With the exception of the ICE, any motor/generator can be spinning one direction or the other. For example, MG1 can be spinning clockwise with torque in the clockwise direction (energy into the HSD) or spinning clockwise with torque in the counterclockwise direction (energy out of the HSD). Same for MG2. For the ICE, it'll either be stopped or, if it's spinning, it can have torque in the same direction as the ICE spin (energy out of the engine) or it can have torque in the opposite direction as the ICE spin (energy into the engine, or engine braking).

    Now, Energy (as in Joules) is Torque * (change of angle of shaft). (For those of you listening, this is exactly like Energy = Force*distance. In fact, a force of one Newton applied to some object over a distance of one meter expends exactly one Joule of energy (which is why a the units of a Joule is a N-m). If you do it in one second, then you've done a Joule/s, and that's using 1 Watt of power, power being the rate of use of energy.

    Deep breath. Those of us who muck with this stuff measure angles in radians, one complete turn of a shaft being 2*pi radians. (It just so happens that the numbers all come out straight if one plays with radians rather than, say, degrees. Of one insists upon using degrees, one ends up with terms like 2*pi/360 sprinkled all over the place to make the numbers come out right.) So, if one applies a torque of 1 Newton-meter (that is, a force of one Newton at a distance of one meter from the shaft we're turning), and that shaft turns one radian (57.3 degrees), then one Joule of energy has been expended. (Note for the unwary: The units are still Newton-meters; an angle has no units). If you do that in one second, then you've been expending energy at one Joule/second, and there's your Watt. Energy = Torque * Theta, Theta being how far the shaft/whatever has been turned.

    Now, consider ye electric motor/generator. Put a clamp on the shaft and run current through the stator: That translates, 1:1, to putting a torque on the shaft. The more current, the more torque. To a first order approximation, if the shaft isn't turning, energy isn't being transferred anywhere. If it is turning, then energy is going into the shaft. On the other hand, if you turn the shaft externally, that shifts the magnets in there, and current comes out of the motor, only now we're calling it a generator. And, unlike a motor, where the torque is in the same direction as the rotation, the shaft resists turning, so the torque is opposite the direction of rotation.

    Hokay. So, let's take an example: Suppose we're motoring along in a straight line at 40 mph, the ICE is running, and everything is ideal (i.e., we're not going to lose power in resistors and the like). Let's suppose the motor is doing 100 Joules a second (100W), and that same 100W is going out the wheels; air resistance and rolling resistance are going to absorb that all of that 100W of energy, so we don't speed up or slow down.

    So, let's suppose the ICE puts 100W of energy into the HSD. We then let MG1 take 50W out of the HSD (the MG1 rotates one way, but the torque is in the other direction). 50W then goes along the shaft of the HSD and straight into MG2 and the wheels, and, for this example, MG2 puts 50W of energy into the shaft (torque points the same way as the MG2 shaft direction). Total power to the wheels = 50W on the shaft + 50W from MG2 = 100W, and away we go.

    But there's two, count 'em, two! degrees of freedom in here that the engineers from Toyota can exploit.

    First: If we keep the power split (50W through MG1 to MG2, 50W to the shaft) steady, this says nothing about the RPM of the ICE. The ICE can be spinning really fast with a little torque for 100W; or it can be spinning really slow with tons of torque to get to that 100W that keeps the car moving steady. The fixed gear ratios in the HSD then dictate how fast MG1 and MG2 rotate, but we can take more or less current out of (MG1) or into (MG2) to keep that 50-50 power split steady.

    Second: We don't have to keep the power split steady. We can hold MG1 still (torque, but no rotation means no energy transfer) and make all the energy go through the shaft and turn the wheels (MG2 may rotate, but if no current goes through the windings, there's no torque, so no energy is transferred); We can allow MG1 to rotate and pull current out of it, thus creating negative torque (it's a generator in this guise) and pull 20W out of the ICE and over to MG2, and 80W through the shaft; or put 20W in the shaft and 80W through MG1/MG2. Or any combination in between, that's why it's a degree of freedom.

    So, what sets the RPM and the gearing, here?

    First: For a given power output, the ICE has a preferred RPM at which it is most efficient. That sets one of the rotation speeds; if the car is moving at 40 mph, as in our example, that sets the rotation speed of the shaft in the middle of the HSD and, since that's the same as the shaft for MG2, MG2's speed.. The speed of MG1 is perforce set. Now, if the engine torque is greater than the torque presented by the shaft and MG1, the engine will speed up - but we can keep that from happening by increasing the load from MG1, and putting that additional energy back into MG2. If the engine torque is less than the torque from MG1 and the shaft, the engine will slow down, but we can fix that by decreasing the torque on MG1, thereby holding the engine steady.

    Result: a continuously variable transmission gear ratio. Yeah, no gears, per se, but that's the point of the HSD.

    Things that throw monkey wrenches into the works:
    1. One loses energy in the power flow into MG1 over to MG2, to tune of 10%-20% or so, or so I've heard. But keeping the engine at its high efficiency point is what it's all about. (Note: Engines can be made to run at a variety of RPMs with good torque, but then they're not efficient. But if you're after efficiency, a single RPM, or relatively small range of RPMs, is your friend.)
    2. There's limits on the RPMs of the two MGs, and they, too, have places of peak efficiency.
    All of which is why Toyota has probably got patents up, down, and sideways, and why it took such a long time for this thing to be developed.

    Now: Given the gear ratios, it's possible that MG1 can, for a particular set of conditions where the ICE is actually running, be rotating clockwise or counter clockwise. But think about energy transfer: If the torque put on the shaft by the electrics is in the opposite direction from the shaft rotation, energy goes into MG1. And it happens naturally; if the electronics are sucking 10A out of MG1, it doesn't matter which direction it's rotating, it going to resist turning.
    For some reason the idea of MG1 sometimes running backwards, but absorbing energy anyway, gives some people the fits. MG2 seems, well, "normal" because torque applied to it goes straight into the wheels, and it always rotates with the wheels. Of course, we can have negative torque with MG2 - we call that regenerative braking, so that's all right.

    So, like I said, it's not quite that useful to think about the direction of rotation of MG1 and/or MG2, or even the ICE: Instead, think about which way the torques are being applied with which way the various pieces are turning. That'll tell you the energy flow directions, and that's the fun.

    KBeck
     
  8. telmo744

    telmo744 HSD fanatic

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    I haven't been online, so bit late on this.

    It's true, as Fuzzy1 just pointed. MG1 does not apply torque in this condition (only a bit to overcome movement by gear mesh, as it is free)
     
  9. Dan4500

    Dan4500 Junior Member

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    Thanks for this great post. Explains the power distribution and partitioning in an understandable way. Reads like the answer to the question asked of Toyota engineers in the hiring interview that they have to answer.
     
  10. vincent1449p

    vincent1449p Active Member

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    Your post came before fuzz1, so it is natural to ask you first.

    Hi fuzzy1,

    I have SGII and Techstream monitoring MG1 torque on both my Gen1 and C, and it always show 0 torque when ICE is stopped and powered only by MG2. This agrees with all the official documents that I can find online. I must have missed something that you have. Would you share where you got this info?

    Here is another document that says MG1 rotates backwards and just idles; it does not generate electricity.

    http://www.autoshop101.com/forms/Hybrid01.pdf

    Not generate electricity means no power from MG1. We know Power = Torque * Angular Velocity. Since RPM is not zero, Torque must be zero to give zero Power.

    Vincent
     
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  11. telmo744

    telmo744 HSD fanatic

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    I've given a thought about this, and if MG1 can rotate freely and its inertia momentum low enough so that ICE does not move, it makes sense. ICE then acts as a "ground", because its static friction is higher.
     
  12. vincent1449p

    vincent1449p Active Member

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    I don't think it has anything to do with inertia. As long as MG1 does not apply any torque, there is no way ICE can move. The PSD need to have 2 of its components to apply torque to have an effect on the 3rd component. In this case, ICE is the 3rd component. If ICE need to be start/stop, MG1 is the one that control it by applying torque either positive or negative depending on the speed and torque of MG2. So, even a slightest torque from MG1 can upset the balance of ICE. That is why MG1 must only idle and rotates in the opposite direction of MG2.

    Vincent
     
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  13. fuzzy1

    fuzzy1 Senior Member

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    To keep the ICE stationary, MG1 must spin synchronously with MG2, though multiplied by a scale factor determined by the PSD gears. When MG2's RPM changes, so must MG1.

    MG1 does have some angular momentum, so some torque must be applied to change its speed. If MG1 is idled, that torque must come from MG2 and through the ICE, risking having the ICE turn a bit if its stationary friction is not sufficient. Instead, it is better to actively drive MG1 at the desired speed to ensure the ICE remains stationary.

    That is why MG1 should supply at least a little torque, to at least spin itself up and down and a overcome a bit of windage and bearing friction. But none of that torque helps propel the car.

    I can easily imagine that this torque may be too small to show on your gauges, or may not be included in the amount reported by the ECUs. It doesn't really count as torque delivered to the PSD.
     
  14. vincent1449p

    vincent1449p Active Member

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    The ICE is not kept stationary by friction. It is stationary because there is no "pivot" for it to turn. MG1 has to act as the "pivot" by exerting some torque. No matter how much torque is exerted by MG2, ICE cannot turn if MG1 doesn't exert any torque.

    So you 're only imagining this "non-existence" torque and no reference to substantiate your theory?

    Vincent
     
  15. fuzzy1

    fuzzy1 Senior Member

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    To keep the ICE stationary, MG1 must change its RPM in concert with MG2 RPM and vehicle speed.

    MG1 doesn't have zero mass, so therefore it does have rotational momentum, or inertia. Therefore basic physics requires some torque in order to change its RPM. (As a reference, I will point to the theories of motion commonly attributed to Isaac Newton.)

    Since the ICE is not kept stationary by friction, where do you think this torque comes from? Pixie dust? Magic fairies? Some other imagined theory? Or active drive applied to its stator windings?
     
  16. vincent1449p

    vincent1449p Active Member

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    The torque come from MG2 and not generate by MG1 itself, there is a difference.

    Vincent
     
  17. vincent1449p

    vincent1449p Active Member

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    There is no magic or imagined theory. It is not easy to understand how can two mass; MG1 & MG2 rotating at high speed and the ICE kept still. Pls read the attached Gen2 NCF which I think is by far has the most details in PSD operation. It not only tell you the direction of RPM (+ve for forward and -ve for reverse), it also indicates torque whether it is Drive or being Driven.

    Vincent
     

    Attached Files:

  18. fuzzy1

    fuzzy1 Senior Member

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    That is your problem, not mine. The basic engineering mechanics is quite easy.
    The relevant portion I see is this, and it lacks sufficient detail to resolve our dispute or say how MG1 gets the torque to spin itself up and down as the vehicle speed changes:

    "Driving With MG2
    When the vehicle is started off, the vehicle operates only by MG2. At this time, the engine remains stopped, and MG1 is spinning in the opposite direction without generating electricity."

    If you see a better passage, please highlight it.

    A hint, though not definitive, is provided elsewhere:

    "Start The Engine
    ...
    During this operation, to prevent the reactive force of the sun gear of MG1 from rotating the ring gear of MG2 and driving the drive wheels, an electrical current is applied to MG2 in order to apply a brake. This function is called 'reactive control'."

    If Toyota can actively drive MG2 to a desired RPM (zero) when it is not actually participating in propulsion, in order to keep something else (the car) from undesired movement, they can and should also do the analogous action in another mode with the other MG (MG1) to prevent undesired motion (of the ICE). In the absence of enough friction in the ICE to do this parasitically, this must be done actively. No extra hardware is needed, it is already in place for other modes. So this action is 'just software', really firmware.
     
  19. vincent1449p

    vincent1449p Active Member

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    Let's look at the Nomograph:

    Vehicle starting off.jpg

    When the vehicle is stopped, all components at 0 rpm and 0 torque. When vehicle starting off, only MG2 is applying the torque, look at the black arrow. Both the ICE and MG1 does not apply any torque (no arrow). ICE still at 0 rpm but MG1 is idling (no arrow) and spinning in the opposite direction (-ve rpm) of MG2.

    Start the engine.jpg

    Now, to start ICE, MG1 has to apply torque (black arrow) and rotate in the same direction as MG2. When 2 components are applying torque (Drive), the 3rd component (ICE in this case) becomes the driven (shaded arrow).

    Generating electricity.jpg

    After ICE has started, it is applying torque (black arrow). Now, both ICE and MG2 are driving, the 3rd component (MG1 in this case) becomes the driven.

    I said it is not easy to understand because you need to understand which component is driving and which component is being driven. When only 1 component is driving, there is no "pivot". You need 2 components to create a "pivot" in order to have an effect on the 3rd component.

    Vincent
     
  20. fuzzy1

    fuzzy1 Senior Member

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    I wonder how they get the mass of MG1 from stationary to rotating, or changing rotation speed, without some sort of torque from somewhere? Pixie dust? Magic fairies? Some other violation of basic physics? Or is the earlier statement about not enough static friction in the ICE incorrect?

    Or could this just be a simplified description that lacks full detail, e.g. MG1 is still actively driven to provide its own needed torque for speed changes, but not enough (as I stated earlier) to supply any torque to the PSD?
     
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