B Mode, Battery Charge and Charge Rate

You're getting the overall application about right, but you're still missing the idea that these ratios can only be understood as fixed and static when one of the 3 elements is somehow held still. As soon as the third one moves, the ratios change.

 

The one dependable anchor point in the shifty sands is the division of torque. If you see 1.0 at the planet carrier shaft, it's 0.28 at the sun gear shaft and 0.72 at the ring gear, always and whatever else is going on.

Speeds vary, and power varies (because it is torque times speed, and the speeds vary). The speeds of all three components will always lie on some straight line drawn on the nomograph.

 

The nomograph shows relations of RPM. Other discussions that have established the 0.28 and 0.72 torque split have proceeded straight from the geometries and tooth counts of the gears. But of course, reality being somewhat consistent, you can reveal them in the nomograph if you're interested.

Because the torque split is what's constant, it will be the same under all conditions. Usefully, that includes the conditions where MG1 isn't moving, or where MG2 isn't moving, so you can look at the controlling equations in the top right of that nomograph, and put zeros in convenient places to make derivations easy.

Say you take the first equation shown. If you pick a moment when MG2 RPM is zero, at that moment MG1 RPM is 3.6 times ICE RPM. At that moment this is a simple one-input one-output gear system where the effect on torque is just the inverse of what happens to RPM. So what's 1 / 3.6? Hmm ... about 0.28.

You can do the same with the last equation, and pick a moment when MG1 isn't turning, which leaves you with ICE = 2.6*MG2/3.6, or MG2 RPM = 3.6/2.6 times ICE RPM. Again inverting to look at torque, what's 2.6 / 3.6? Hmm ... about 0.72.

Finally, whatever torque the engine produces only has those two places to go ... if it is going 28% to the sun gear shaft and 72% to the ring, well, those two had better add up to 100%. Hmm ... yup, looks like they do.

 

'bout as close as you're likely to get by eyeballing a graph, right? If you pick MG1=0 and ICE=3500 then "5000" for MG2 is pretty close to the actual 4846, and ".7" for the ratio is pretty close to the actual .72....

 

Yes, the planetary gear set distributes torque in fixed proportion: whatever torque is on the input shaft from engine flywheel to planet carrier, 28% of that torque appears on the stubby splines between the sun gear and MG1.

The other 72% appears—it pays to speak carefully here—at the ring gear, through its internal teeth. That's the force with which the planet teeth are pushing it around, times its effective radius.

You'll notice I didn't say MG2, even though I did say the ring gear and MG2 is directly connected to it (or through a reduction gear in some models). That's because there's also something else directly connected to the ring gear: the output to the final drive and wheels.

Of the 72% of engine torque that reaches the ring gear, you can't say offhand how much MG2 is seeing and how much the final drive is seeing: the torque at the ring gear is the difference of those two, but they can be anywhere from nearly the same to very different.

A couple of limiting examples: (1) you're driving electrically at 30 mph, engine stopped. MG2 is supplying all the torque needed to move the car. The torque seen by all the components of the PSD is diddly. Strictly, the engine output shaft is seeing diddly, the sun gear splines are seeing 0.28✕diddly, with 0.72✕diddly at the ring gear teeth. MG1 is spinning around at −4500 RPM or so, but essentially freely, at 0.28✕diddly torque, accounting for a negligible amount of power (−4500✕0.28✕diddly).

(2) you're cruising down the highway at 50 MPH using 27 HP from the engine (71 foot-pounds of torque at 2000 RPM). MG1 is barely turning at all and accounting for none of that power (0.28✕71 foot-pounds✕0 RPM, no power), and the ring gear is mechanically getting 0.72✕71 foot-pounds✕2854 RPM, all 27 HP from the engine), and all of that is going to the wheels to move the car. MG2 is spinning along for the ride at the same 2854 RPM, but diddly torque. In this condition the electrical power path is insignificant (MG1 20 foot-pounds✕diddly RPM = diddly = MG2 2854 RPM✕diddly foot-pounds).

It's kind of a funny fact that direct mechanical connections, as seen between the ring gear, MG2, and the final drive) constrain the RPMs of the connected components to be the same, but don't enforce any particular apportionment of torque. If you're riding a tandem bicycle, both your pedal cranks go around at the same speed, whether you are both putting in the same effort, or you are busting your butt and your partner is kind of letting their feet go around.

By contrast, a gear train like the PSD constrains the torques at each component into fixed proportions, but their speeds are allowed to vary.

Each arrangement fixes the apportionment of one of the factors of power, while allowing the other factor to vary, so in either case it is variable just how the power is split between components.

Messages crossed while I was writing this one. As already mentioned above, torque and power are not the same: power is the product (torque ✕ RPM). Taking units into account, if power is in HP and torque in foot-pounds, power = torque ✕RPM / 5252. If you're in doubt of that here, there are better resources around than PriusChat to clear it up for you.

 

If you were analyzing a resistive electrical circuit you would want to use an identity for power in terms of the electrical values you could measure; an identity in terms of RPM and foot-pounds would be of little use to you there. You would also want to get the identity right: P = I²R, the squared matters (otherwise you're just restating Ohm's law and talking about volts, not power).

Likewise you wouldn't bring an electrical identity for power to an analysis of a gear train.

Luckily, a Prius lets you have both kinds of fun. A gear train with motors! Just use each identity where it makes sense.

At this point, yes; we're slipping into rehashing what's already been covered.

Because the gear train apportions torque in an unchanging breakdown, a graph of that would be really boring. The nomograph shows the relationships of RPM, which can vary. Post #50 already showed how you can unearth the .28, .72 torque split from the governing equations used to make the graph.

They hold "at equilibrium", meaning the "equilibrium" where the forces between gear teeth are equal and opposite and the gears are moving as those forces dictate ... meaning, at all times when you aren't hearing fragging sounds from inside the transmission.

 

Your mind quite clearly is scrambling torque and power here. That is contributing to your misunderstandings.

Those 28% and 72% numbers are for torque only. You seriously misunderstand them when you try to apply those same figures to power.

And they are not "just the limits of the range of possibility". Apart from small friction losses, they are fixed constants of the machine.

It needs to apply torque to its shaft, so that the other two shafts, and thus the drive wheels, can also have torque.

 

The word salad scramble continues ...

Torque is rotational force.

In the post I was responding to, you took a torque split figure and tried to apply it to power. That just doesn't work, and appeared to be contributing to your confusion.

The engine control units tell them where to go, and adjust various conditions to make it happen as intended.

(assuming you meant to include current) Now try doing the same with torque and rotational speed. And with force and linear speed. Those also represent power, the same as voltage and current.

 

... keep going ...