Why does the engine wind out when in B mode?

Discussion in 'Gen 2 Prius Technical Discussion' started by jerlands, May 15, 2019.

  1. ChapmanF

    ChapmanF Senior Member

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    Any transmission has a gozinta and a gozouta. Here the gozinta is the input shaft from the engine (leading to the PSD planet carrier). The gozouta is the ring gear (followed by the final drive reduction and the differential, which we can set aside for now).

    Keep in mind that the torque presented at the gozouta is the (signed) sum of two torques that act on the ring gear simultaneously, one from the planet gears and one from MG2. Sometimes both torques have the same sign, and add; sometimes the signs differ.

    The first of those is always, without exception, 0.72 times the torque at the input shaft. Also without exception, the sun gear/MG1 torque is 0.28 of that at the input shaft.

    Everything else the tranny does comes down to what games you can play with that one fixed relationship, when you don't get to change anything but the timing of electric pulses you allow between MG1 and MG2. (Turns out that allows plenty of interesting games.)

    As for the PSD torque relationships, nothing stops you calling any of the three torques "1" and scaling the other two so the proportions are the same. Calling the engine shaft torque "1" and the other two 0.28 and 0.72 is just the way we do it 'round here.
     
  2. jerlands

    jerlands Member

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    gozinta are three things... and those three things all have varying degrees of impact in relation to the gozouta (ring gear.) MG1 + MG2 + ICE = some result via PSD
    Isn't the ring gear the sum of 3 forces... MG1, MG2 and ICE?

    The gear ratio between the ICE and MG1 is 2.6 (one revolution of ICE and you get 2.6 revolutions of MG1)... that's where you're getting this torque conversion. .72/.28= 2.6 That becomes irrelevant in my mind because the range of torque possible from MG1 is very large! Also ICE has a range (varying input) that is also rather large.

     
  3. jerlands

    jerlands Member

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  4. jerlands

    jerlands Member

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    It really all comes down to what the ring gear is doing doesn't it and the forces acting on the ring gear that cause it to go with X amount of force forward or X amount of force backward, the rest of the system is power management. And MG2 is solely responsible for the ring gear moving any direction. Right?

    Actually your power ratios may be coming from the H.P of MG2 vs H.P. of MG1 and since MG2 is much larger (more windings) it has more effect in producing electricity and more effect when using electricity. But this doesn't really matter because the overall effect is how the ECU balances the forces. In the PSD, mechanically connected is the ring Gear to MG2 and the other side is MG1 coupled with ICE. The sun gear's spline (MG1) fits into the socket in the planetary which is engine driven. The engine cannot drive MG2 because there is no mechanical fixed connection between the planetary and the ring. the planetary just spins within the ring which acts as a raceway for it.
     
  5. fuzzy1

    fuzzy1 Senior Member

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    I suspect it is more likely the other way around. The tooth counts set the ratios, then the MGs were sized accordingly.

    ----------

    Also, when providing full propulsion power, MG2 takes in the power from MG1 plus the power available from the battery.
     
  6. Kevin_Denver

    Kevin_Denver Active Member

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    Just some additional information to add - In B mode my scanguage will show 5000rpm + or - 50rpm or so with 9999MPG instantaneous fuel economy displayed (no fuel being burnt) when on long downhills with a full battery. If the battery is not full, in B mode it normally only revs to about 3k rpm. This is normal in modern fuel injected cars - under high rpm, no throttle, the engine will stop injecting fuel and just pump air, using the car's momentum to keep the engine spinning.
     
  7. jerlands

    jerlands Member

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    Yes and the gearing (other than the ICE and MG1) is from the Ring Gear to the Differential?
    and I understand it takes 4.13 turn of the ring gear to make 1 turn of the differential gear?

    Which is why the ICE needs to turn faster than MG2 in order for MG1 to produce electricity.
     
  8. jerlands

    jerlands Member

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    Those are interesting numbers I'll put to use when I test this new battery (that just arrived :) I'm also doin' the rear struts but get waylaid by the weather...
     
  9. Skibob

    Skibob Senior Member

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    wrong thread
     
  10. jerlands

    jerlands Member

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    This thread was started to learn about how dynamic braking works while the other was started to try and investigate possible run away causes. The components for both issues are the same just different conditions that might bring about an effect leading to whatever I was curious about :) But it's true.. in both threads I'm trying to learn something about the drive train...
     
  11. Skibob

    Skibob Senior Member

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    No, the thing I posted was in the wrong thread. Was supposed to be in the AC thread. :ROFLMAO:
     
  12. ChapmanF

    ChapmanF Senior Member

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    MG1 and MG2 are actually inside-the-box details of this transmission ... just like clutches and bands, or push belts, or whatever other inside-the-box details you find in another transmission. The gozinta is still a single mechanical rotation, as is the gozouta.

    I'm leaving the battery aside for clarity. There are moments when some current from the battery will slosh in (when either MG1 or MG2 is consuming more power than the other MG is producing), or slosh out to the battery (when either MG1 or MG2 is producing more power than the other MG is consuming), but under conditions of balance the chief electrical path is simply between the MGs with very little to or from the battery, and (except for a PiP or Prime) the car shoots for those balanced conditions and achieves them for large stretches of the time. That's why you really can learn the operation of the transmission without too much attention to the battery to begin with.

    In balanced condition, the electric path between the MGs is all an inside-the-box detail; it is neither a gozinta nor a gozouta. It happens to flow on some short orange cables out of the tranny case, through the inverter box on top, and back into the tranny ... but the inverter is also inside-the-(conceptual)-box.

    One true thing doesn't become irrelevant because you put another true thing next to it. In fact, once you put the two of them together, you're incredibly close to seeing what's going on. Yes, the engine has a wide range of torque (rotating in a single direction, at speeds around 1,000 to 5,000 rpm). Yes, MG1 has a wide range of torque (and can rotate in either direction, around −10,000 to 10,000 rpm or so). Yes, at any moment, the torque of either one can be in the same or the opposite direction as its rotation at that same moment (the MG can be turning electric power into rotary power, or rotary power to electric power; the engine can be turning fuel into rotary power, or rotary power into noise and heat).

    And still yes, at every single moment, the algebraic relationship between the torques at those three points in the mechanism holds, exactly as determined by their tooth counts. Irrelevant? Without that fact, you'd have too little information to solve for what's going on.

    Not right at all. Torque is applied to the ring gear from two "inside-the-box" sources, MG2 and its own teeth in mesh with the planet gears, as well as from outside-the-box (the gozouta, to the final drive, differential, and wheels).

    Look more closely. MG1 is no more and no less "coupled" with the engine than MG2 is. The sun gear and the input shaft from the engine are concentric, but they turn independently. There is no way for any of those three pieces to "drive" any other one of the three, until you also constrain the third one. Therein is the whole game.

    Ok, start there (which is a reasonable operating mode, like when the engine's at idle and the car is stopped). The engine is spinning the planet carrier, the ring gear is stationary, and so as the planet carrier turns, its planet gears have to spin, as they roll along the inner teeth of the nonmoving ring gear.

    That means the sun gear/MG1, right now, is doing what, and in which direction?

    Now change the story a little. Keep the engine doing the same thing it's doing, and now start applying some MG1 torque opposed to its current motion (which happens naturally, if the inverter switches start letting MG1 generate some juice). What happens out at the ring gear?

    True enough, but that's part of the "slosh tank" function of the battery. It can't keep that up all day, and it's possible (and easier, when learning) to understand what the tranny is doing under balanced conditions, when the power in the gozinta is just equal to the power out the gozouta, just possibly in a different torque/speed combination.

    It doesn't need to. Either MG1 or MG2 can produce electricity any time it is spinning, in either direction. The only time MG1 is ever not spinning is when the engine and ring gear happen to be spinning at an rpm ratio of 1:1.39. At any other time, MG1 is spinning in one direction or the other, and will produce electricity if the inverter starts gating its windings through with the proper phasing.

    Have you ever checked out the nomographs and explanations on pages TH-11 through TH-25 of the New Car Features manual (more info)? 2004 edition (in the later Gen 2 years, it just has updates). Useful information there.
     
    #72 ChapmanF, May 17, 2019
    Last edited: May 17, 2019
  13. jerlands

    jerlands Member

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    Well I got the replacement battery in the car and all I have to do is put in the interlock and mount the 12v cable back on then I can see if it'll work :) and reset the alarms then put all the interior back on.. I could be giving it a test drive tomorrow afternoon :)
     
  14. fuzzy1

    fuzzy1 Senior Member

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    I am not aware of any such constraint. As Chapman points out, MG1 ought to be able to generate electricity any time it is spinning, and the engine is providing some torque.
     
  15. jerlands

    jerlands Member

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    My understanding is that for MG1 to turn sufficiently to produce enough energy to charge the battery and supply power to MG2 is that it needs to be physically spinning faster than MG2 (which makes sense seeing it's size :)
     
  16. fuzzy1

    fuzzy1 Senior Member

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    That would be a significantly narrower claim than the version I responded to earlier.
     
  17. jerlands

    jerlands Member

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    I'm still trying to get it straight in my head because both MG2 and IGE work on the magnitude of MG1 (how much voltage it will generate.)
     
  18. jerlands

    jerlands Member

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    somehow an answer popped in my head, vanished and I think I have it now :) what causes MG1 to rotate is the pinion gears. They have to be turning. If the rpm of the ring gear is the same rpm as the ice then the pinion gears don't turn and hence neither does MG1?

    EDIT: that's what I was thinking anyway but MG1 will spin at the same speed as ICE and MG2 but not make use of the 2.6 gear ratio
     
    #78 jerlands, May 18, 2019
    Last edited: May 18, 2019
  19. fuzzy1

    fuzzy1 Senior Member

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    Their RPMs, torques, and powers are linked by simple algebra. Mere voltages can be arbitrary. In actual design, the voltages have been linked by other engineering choices not relevant to this line of discussion, but this doesn't factor in to understanding how the PSD works.

    Because the gears on each shaft have different tooth counts, the system cannot hold any one shaft still by spinning the other two at the same speed. Their speeds will be different. Put another way, if any two shafts are spinning at the same speed, the third shaft must also be spinning at some speed, not be stationary.
     
  20. ChapmanF

    ChapmanF Senior Member

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    But where is it that you're getting these "understandings"?

    It's common for newcomers on PriusChat to come in with a certain amount of misinformation picked up one place or another, but if there's an especially prolific supply somewhere that you've stumbled into, maybe we should be looking there, to have a better heads-up about what future new members might show up burdened with.

    The edit is right about the speeds. If the engine and ring are doing the same rpm, that will be MG1's rpm also. At that operating point, all three have the same rpm.

    The gear ratio hasn't changed, nor has the division of torque (the one thing that is unchanging at all operating points). The torque applied to the sun gear by the pinions is 0.28 of the engine shaft torque, and that applied to the ring gear by the pinions is 0.72 of the engine shaft torque.

    Power is torque ✕ rpm, so at this particular operating point where all three have the same rpm, the algebra is easy. The rpms all cancel, and MG1 accounts for 0.28 of the engine shaft power, with the other 0.72 being delivered to the teeth of the ring gear.

    In a balanced condition (not charging or using the battery), the 28% going via MG1 is just electrically shuttled over to MG2, where it rejoins the 72% at the ring gear, and the power leaving the gozouta to the final drive is 100% of what was on the gozinta. (Idealized picture, disregarding mechanical and electrical losses, usual disclaimer.)

    This same picture could be seen when the car is decelerating using engine braking; nothing changes but the signs of the torques and of the electrical power flow between the MGs.

    At any other operating point, where the components have different rpms, the torque breakdown stays the same (it never changes), but the rpms won't cancel, you'll need them in the formula to be able to solve for the power taking each path.

    Check.

    In contrast, if you let two shafts spin at different speeds, then you can find the right different speeds so that the remaining shaft will be stationary.

    This happens sitting at stoplights with the engine running; the engine will be at idle (about 1,000 rpm), MG1 will be doing about 3,570, and the gozouta is stationary. The engine is producing a small amount of power, none of which is going to the drive wheels (0 rpm, power is torque ✕ rpm, 0 ✕ anything is 0), so it's all going to MG1. Sure enough: engine power ✕ 0.28 (torque to MG1) ✕ (3,570 rpm / 1,000 rpm) ≈ engine power.

    That picture can be a pure idle, where the battery needs no charging, the throttle is at idle position, and no electric power is being accepted from MG1. In this case, MG1 is doing nothing but stirring some oil around, and the power from the idling engine is no more than what's needed to do that.

    It can also be charging the battery, in which case the engine might still be close to idle rpm so those numbers don't change, but the throttle opens further and the engine is producing more power through increased torque (you can hear the exhaust note deepen). MG1 is still getting 100% of that power, which is now more than needed just to stir the oil around, and the inverter is accepting the rest and stuffing it into the battery.

    Another combination you could have fleetingly under the right driving conditions would be the engine doing, say, 2,160 rpm, and the gozouta at about 3,000. At that moment, MG1 would happen to be stationary and the electrical path carrying zero power; you'd be cruising down the road with essentially all of the engine power delivered mechanically through the PSD. Sure enough: engine power ✕ 0.72 (torque to ring) ✕ (3,000 rpm / 2,160 rpm) ≈ engine power.

    That's a good cruising situation if you hit the right conditions, because it avoids the electrical losses on the electrical power path.
     
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