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Plaid power train (out of investor thread)

hacer

Active Member
Apr 13, 2016
1,066
4,449
Clarksville, MD
This is my "dumbed down" explanation post that @CorneliusXX asked for regarding my post Tesla, TSLA & the Investment World: the 2019 Investors' Roundtable.
My original post was trying to correct the mistake that @Fact Checking made about the plaid drive system having 3 different fixed gear ratios. If two of the motors independently drive the rear motors but have different fixed gear ratios, that would mean that one of them would always have less available torque than the other one except at very high speeds. That's useful only if you always drive the same direction on a circular track. Of course I was way behind on the thread and there were many others who had already challenged the idea, but I hadn't read them yet. My post did a little more in speculating why it wouldn't be necessary for even a change of gear ratio to have better high speed power output. This post is going to try to explain (and justify) that statement a bit better. It will be dumbed down but still somewhat technical.

The "problem" (to the extent there is one) for high power at high speed in the Tesla is mainly caused by the motor back-EMF (EMF stands for electromotive-force, and it is the voltage generated by the spinning motor, and it opposes the driving voltage). To make this simpler to understand I'll be using a DC motor to describe in general how it works (because the DC motor is much easier to understand and explains the issue).

To start with the problem will be greatly simplified: I will use the 1st law of thermodynamics (energy in = energy out if there is no stored energy) but in a steady-state per-unit-time metric. That is: power in = power out which ignores inefficiency (heat generation) and energy storage. Of course real motors have these other effects but they aren't important for understanding the issue. Power is the time-rate-of-change (the derivative with respect to time) of energy.
For a motor the input power is electrical. Electrical power is the product of voltage (V) and current (I). The output power is mechanical and is the product of torque (T) and angular speed (W). Thus the motor analysis begins with V*I = T*W. Note that if you connect a DC motor to a battery and have no load on the shaft (it is free to spin with no torque), the motor will spin at high speed, but T=0 so it must be that V*I = 0. Since it is hooked to the battery, we know that V is the battery voltage so (I) must be zero. Sweeping several things under the rug here, the current is zero because the motor back-EMF is exactly matching the battery voltage. Similarly, if we fix the shaft so the motor can't turn (even though it's hooked to the battery) again the output power is zero because W=0. Torque will be large and the input current will be really large yet the input power must still be zero. In this case, the motor Voltage is zero (even though it is connected to the battery). All of the battery voltage will drop across the resistances I'm neglecting in my idealized model here.

In reality, there is some resistance in the motor winding, battery and wires and the motor current is limited to the battery voltage minus the back-EMF divided by the total resistance.
I = (V - BEMF)/R​
R is the sum of all resistance in the battery, the motor windings, electronics, cables, etc. But it is small, so the power-in=power-out model is still reasonable to understand the motor mechanical power but some resistance is necessary to understand how current is limited in the motor. In the real world if you connect a battery directly to a DC motor with a fixed shaft, chances are some smoke will appear and something may melt or catch on fire.

You can see that for the DC motor, the back-EMF is directly proportional to motor speed. The proportional constant is called "the motor velocity constant" or "back-EMF constant". The motor torque is also proportional to the current (I) through the "motor torque constant". You can read more about both of these here: Motor constants - Wikipedia.

Now let's look at the model 3 motor torque-power curves:
Tesla-Model-3-SOC-Dyno-Results.jpg

We can see that at low speeds the torque is largest and constant, even independent of battery state of charge (SOC); These are the curves just below 325 lb-ft on the left axis. They are flat and independent of speed because the motor controller does not apply full battery voltage to the motor at low speeds; instead it uses electronic magic to very efficiently lower the applied voltage in order to produce the desired current (and thus torque). For the dyno curves, the desired torque was "give it all you got". For various reasons, not the least of which is fuses within the battery there is a maximum current limit that Tesla's controller will not let you exceed. For performance S vehicles, this is somewhere near 1,500 Amps. This dyno run was done on a model 3 which has a smaller battery (fewer parallel cells) so it probably also has smaller peak current allowed but I haven't checked. Note that model 3 LR battery still has the same (unloaded) voltage as the S's 100kWh battery.

Observe that the power curves (blueish) are going up linearly as the speed increases within the constant torque region as expected because T is constant while W is linearly increasing. At some speed, the Torque begins to decline, and the speed at which this decline happens depends on the SOC. The highest SOC (and thus highest available voltage) gives the highest speed before this decline begins. Power output is still increasing because speed is going up faster than torque is going down. The motor back-EMF throughout all speeds (and SOCs) is always less than the battery voltage because the torque never goes to zero. But back-EMF gets larger and larger so that eventually less current than the battery would be capable of (and thus less power) can be delivered to the motor at high speeds. When this happens, the motor controller is no longer trying to reduce the applied voltage to the motor, instead it is providing the full voltage available because even that will not exceed the current limit.

The gear ratio (and wheel diameter) controls the scaling between the motor RPM and the vehicle speed. Gearing allows a trade-off between speed and torque because the gears also obey the 1st law of thermodynamics. So you can have higher vehicle speed for a given motor RPM, but that also means less torque at the wheels. This is what Porche did by adding a gear box - they can shift the maximum power point of motor-battery system to a higher vehicle speed with the higher gear. To be sure, this could also be done electronically by adding a voltage boost capability to the electronics but that might be less efficient than what a transmission could achieve.

Now let's consider the Plaid drive train where you have two motors instead of one for the rear wheels. I'm going to ignore the front motor which sometimes will use some of the available battery current because this is dumbed down. With the one motor Raven, the transaxle splits the torque, giving half to each wheel. With two motors on the Plaid, each motor directly provides that wheel's torque. At first glance you might think that you could double the torque but at low speeds you don't gain anything because the battery current is the limiting factor. If one motor was capable of taking all the current that the battery had left to give then you must instead send only half of the current to each motor. The total torque on the ground is the same as the Raven in that case. However, you have cut each motor's current in half! Transforming my equation above:
BEMF = V - IR​
You can see that for half the current, we can tolerate more back-EMF (and thus more speed) with the same battery voltage. If the dyno curves were re-run with a half torque target, you would see that the torque curves have half their value, but they would remain flat up to much higher speeds. This is because even though the back-EMF is just as high as before, you are only trying to get half the current through the motor. You want only half the current because you are forced to split the battery's limited current between two motors.
You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.
 

CorneliusXX

Active Member
Jun 19, 2015
2,104
16,588
London
This is my "dumbed down" explanation post that @CorneliusXX asked for regarding my post Tesla, TSLA & the Investment World: the 2019 Investors' Roundtable.
My original post was trying to correct the mistake that @Fact Checking made about the plaid drive system having 3 different fixed gear ratios. If two of the motors independently drive the rear motors but have different fixed gear ratios, that would mean that one of them would always have less available torque than the other one except at very high speeds. That's useful only if you always drive the same direction on a circular track. Of course I was way behind on the thread and there were many others who had already challenged the idea, but I hadn't read them yet. My post did a little more in speculating why it wouldn't be necessary for even a change of gear ratio to have better high speed power output. This post is going to try to explain (and justify) that statement a bit better. It will be dumbed down but still somewhat technical.

The "problem" (to the extent there is one) for high power at high speed in the Tesla is mainly caused by the motor back-EMF (EMF stands for electromotive-force, and it is the voltage generated by the spinning motor, and it opposes the driving voltage). To make this simpler to understand I'll be using a DC motor to describe in general how it works (because the DC motor is much easier to understand and explains the issue).

To start with the problem will be greatly simplified: I will use the 1st law of thermodynamics (energy in = energy out if there is no stored energy) but in a steady-state per-unit-time metric. That is: power in = power out which ignores inefficiency (heat generation) and energy storage. Of course real motors have these other effects but they aren't important for understanding the issue. Power is the time-rate-of-change (the derivative with respect to time) of energy.
For a motor the input power is electrical. Electrical power is the product of voltage (V) and current (I). The output power is mechanical and is the product of torque (T) and angular speed (W). Thus the motor analysis begins with V*I = T*W. Note that if you connect a DC motor to a battery and have no load on the shaft (it is free to spin with no torque), the motor will spin at high speed, but T=0 so it must be that V*I = 0. Since it is hooked to the battery, we know that V is the battery voltage so (I) must be zero. Sweeping several things under the rug here, the current is zero because the motor back-EMF is exactly matching the battery voltage. Similarly, if we fix the shaft so the motor can't turn (even though it's hooked to the battery) again the output power is zero because W=0. Torque will be large and the input current will be really large yet the input power must still be zero. In this case, the motor Voltage is zero (even though it is connected to the battery). All of the battery voltage will drop across the resistances I'm neglecting in my idealized model here.

In reality, there is some resistance in the motor winding, battery and wires and the motor current is limited to the battery voltage minus the back-EMF divided by the total resistance.
I = (V - BEMF)/R​
R is the sum of all resistance in the battery, the motor windings, electronics, cables, etc. But it is small, so the power-in=power-out model is still reasonable to understand the motor mechanical power but some resistance is necessary to understand how current is limited in the motor. In the real world if you connect a battery directly to a DC motor with a fixed shaft, chances are some smoke will appear and something may melt or catch on fire.

You can see that for the DC motor, the back-EMF is directly proportional to motor speed. The proportional constant is called "the motor velocity constant" or "back-EMF constant". The motor torque is also proportional to the current (I) through the "motor torque constant". You can read more about both of these here: Motor constants - Wikipedia.

Now let's look at the model 3 motor torque-power curves:
Tesla-Model-3-SOC-Dyno-Results.jpg

We can see that at low speeds the torque is largest and constant, even independent of battery state of charge (SOC); These are the curves just below 325 lb-ft on the left axis. They are flat and independent of speed because the motor controller does not apply full battery voltage to the motor at low speeds; instead it uses electronic magic to very efficiently lower the applied voltage in order to produce the desired current (and thus torque). For the dyno curves, the desired torque was "give it all you got". For various reasons, not the least of which is fuses within the battery there is a maximum current limit that Tesla's controller will not let you exceed. For performance S vehicles, this is somewhere near 1,500 Amps. This dyno run was done on a model 3 which has a smaller battery (fewer parallel cells) so it probably also has smaller peak current allowed but I haven't checked. Note that model 3 LR battery still has the same (unloaded) voltage as the S's 100kWh battery.

Observe that the power curves (blueish) are going up linearly as the speed increases within the constant torque region as expected because T is constant while W is linearly increasing. At some speed, the Torque begins to decline, and the speed at which this decline happens depends on the SOC. The highest SOC (and thus highest available voltage) gives the highest speed before this decline begins. Power output is still increasing because speed is going up faster than torque is going down. The motor back-EMF throughout all speeds (and SOCs) is always less than the battery voltage because the torque never goes to zero. But back-EMF gets larger and larger so that eventually less current than the battery would be capable of (and thus less power) can be delivered to the motor at high speeds. When this happens, the motor controller is no longer trying to reduce the applied voltage to the motor, instead it is providing the full voltage available because even that will not exceed the current limit.

The gear ratio (and wheel diameter) controls the scaling between the motor RPM and the vehicle speed. Gearing allows a trade-off between speed and torque because the gears also obey the 1st law of thermodynamics. So you can have higher vehicle speed for a given motor RPM, but that also means less torque at the wheels. This is what Porche did by adding a gear box - they can shift the maximum power point of motor-battery system to a higher vehicle speed with the higher gear. To be sure, this could also be done electronically by adding a voltage boost capability to the electronics but that might be less efficient than what a transmission could achieve.

Now let's consider the Plaid drive train where you have two motors instead of one for the rear wheels. I'm going to ignore the front motor which sometimes will use some of the available battery current because this is dumbed down. With the one motor Raven, the transaxle splits the torque, giving half to each wheel. With two motors on the Plaid, each motor directly provides that wheel's torque. At first glance you might think that you could double the torque but at low speeds you don't gain anything because the battery current is the limiting factor. If one motor was capable of taking all the current that the battery had left to give then you must instead send only half of the current to each motor. The total torque on the ground is the same as the Raven in that case. However, you have cut each motor's current in half! Transforming my equation above:
BEMF = V - IR​
You can see that for half the current, we can tolerate more back-EMF (and thus more speed) with the same battery voltage. If the dyno curves were re-run with a half torque target, you would see that the torque curves have half their value, but they would remain flat up to much higher speeds. This is because even though the back-EMF is just as high as before, you are only trying to get half the current through the motor. You want only half the current because you are forced to split the battery's limited current between two motors.
You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.
Thanks for the time and effort on this response. It's very much appreciated and I get what you are saying about the gear ratios not needing to change with the inclusion of the additional motor to increase performance.

Out of interest do you see a way that additional performance on top of what you explained could be achieved with separate gear ratios on the rear wheels and varying the current split between each motor?
 

Fact Checking

Well-Known Member
Aug 3, 2018
7,517
120,116
Vienna
You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.

Thanks a lot for this excellent description!

Am I correct that the peak of the power curve will shift from the current ~70 kmh to about ~140 kmh - which is already pretty close to the ~160 kmh average speed of the Nürburgring?

Tesla might still change the gearing ratios as well, to further optimize this - as they'd have plenty of torque at lower speeds.

At what gearing ratios could Tesla expect trade-offs at lower speeds? I.e. why aren't the gearing ratios much higher than the current 7:1 they are using, while torque is still maxed out at lower speeds? There must be some low speed consideration that favors lower gearing ratios, right?
 

hacer

Active Member
Apr 13, 2016
1,066
4,449
Clarksville, MD
Thanks for the time and effort on this response. It's very much appreciated and I get what you are saying about the gear ratios not needing to change with the inclusion of the additional motor to increase performance.

Out of interest do you see a way that additional performance on top of what you explained could be achieved with separate gear ratios on the rear wheels and varying the current split between each motor?
No doubt the front and rear gear ratios would continue to be different because these cars are mostly optimized for everyday driving performance, which I think they should continue to be. Even if there is a little performance gain to be had on a high-speed track by making the rear ratio higher, it's still probably not the best overall choice for a car that will mostly be driven on regular roads. You do want to vary the current split between left and right on the rear only when you are doing regen or acceleration while turning - that is where the wheels are turning at different speeds and you actually want different torques. I am sure the plaid system already does this since it is a no-brainer.
 

mongo

Well-Known Member
May 3, 2017
12,953
38,373
Michigan
With two motors on the Plaid, each motor directly provides that wheel's torque. At first glance you might think that you could double the torque but at low speeds you don't gain anything because the battery current is the limiting factor.

Not quite. The drive electronics function as a DC-DC converter (buck in this case). As such power out = power in / efficency.
Current from pack does not equal current to motor unless you are at 100% duty cycle and the motor is at full pack voltage.

The max torque is SW limited to what the drive unit can handle (and beyond that, by available friction). With two rear motors, each can only output half the torque due to the tires, but they could indeed put out double the torque of one motor at low speeds. Concider that at 0 RPM power =0, so current can be arbitrarily high without putting a load on the pack (current recirculates through the motor windings).


If the dyno curves were re-run with a half torque target, you would see that the torque curves have half their value, but they would remain flat up to much higher speeds. This is because even though the back-EMF is just as high as before, you are only trying to get half the current through the motor. You want only half the current because you are forced to split the battery's limited current between two motors.

Good point on the back-emf curve vs the halved value of desired current and motor power limit.
The small flat power section may indicate that the pack power limit is being reached. If so, that level will set the point at which the dual motor system tapers off, if it occurs before the back-EMF effects.
 
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Reactions: VT_EE

VT_EE

Active Member
Apr 22, 2017
2,025
2,419
Maryland
This is my "dumbed down" explanation post that @CorneliusXX asked for regarding my post Tesla, TSLA & the Investment World: the 2019 Investors' Roundtable.
My original post was trying to correct the mistake that @Fact Checking made about the plaid drive system having 3 different fixed gear ratios. If two of the motors independently drive the rear motors but have different fixed gear ratios, that would mean that one of them would always have less available torque than the other one except at very high speeds. That's useful only if you always drive the same direction on a circular track. Of course I was way behind on the thread and there were many others who had already challenged the idea, but I hadn't read them yet. My post did a little more in speculating why it wouldn't be necessary for even a change of gear ratio to have better high speed power output. This post is going to try to explain (and justify) that statement a bit better. It will be dumbed down but still somewhat technical.

The "problem" (to the extent there is one) for high power at high speed in the Tesla is mainly caused by the motor back-EMF (EMF stands for electromotive-force, and it is the voltage generated by the spinning motor, and it opposes the driving voltage). To make this simpler to understand I'll be using a DC motor to describe in general how it works (because the DC motor is much easier to understand and explains the issue).

To start with the problem will be greatly simplified: I will use the 1st law of thermodynamics (energy in = energy out if there is no stored energy) but in a steady-state per-unit-time metric. That is: power in = power out which ignores inefficiency (heat generation) and energy storage. Of course real motors have these other effects but they aren't important for understanding the issue. Power is the time-rate-of-change (the derivative with respect to time) of energy.
For a motor the input power is electrical. Electrical power is the product of voltage (V) and current (I). The output power is mechanical and is the product of torque (T) and angular speed (W). Thus the motor analysis begins with V*I = T*W. Note that if you connect a DC motor to a battery and have no load on the shaft (it is free to spin with no torque), the motor will spin at high speed, but T=0 so it must be that V*I = 0. Since it is hooked to the battery, we know that V is the battery voltage so (I) must be zero. Sweeping several things under the rug here, the current is zero because the motor back-EMF is exactly matching the battery voltage. Similarly, if we fix the shaft so the motor can't turn (even though it's hooked to the battery) again the output power is zero because W=0. Torque will be large and the input current will be really large yet the input power must still be zero. In this case, the motor Voltage is zero (even though it is connected to the battery). All of the battery voltage will drop across the resistances I'm neglecting in my idealized model here.

In reality, there is some resistance in the motor winding, battery and wires and the motor current is limited to the battery voltage minus the back-EMF divided by the total resistance.
I = (V - BEMF)/R​
R is the sum of all resistance in the battery, the motor windings, electronics, cables, etc. But it is small, so the power-in=power-out model is still reasonable to understand the motor mechanical power but some resistance is necessary to understand how current is limited in the motor. In the real world if you connect a battery directly to a DC motor with a fixed shaft, chances are some smoke will appear and something may melt or catch on fire.

You can see that for the DC motor, the back-EMF is directly proportional to motor speed. The proportional constant is called "the motor velocity constant" or "back-EMF constant". The motor torque is also proportional to the current (I) through the "motor torque constant". You can read more about both of these here: Motor constants - Wikipedia.

Now let's look at the model 3 motor torque-power curves:
Tesla-Model-3-SOC-Dyno-Results.jpg

We can see that at low speeds the torque is largest and constant, even independent of battery state of charge (SOC); These are the curves just below 325 lb-ft on the left axis. They are flat and independent of speed because the motor controller does not apply full battery voltage to the motor at low speeds; instead it uses electronic magic to very efficiently lower the applied voltage in order to produce the desired current (and thus torque). For the dyno curves, the desired torque was "give it all you got". For various reasons, not the least of which is fuses within the battery there is a maximum current limit that Tesla's controller will not let you exceed. For performance S vehicles, this is somewhere near 1,500 Amps. This dyno run was done on a model 3 which has a smaller battery (fewer parallel cells) so it probably also has smaller peak current allowed but I haven't checked. Note that model 3 LR battery still has the same (unloaded) voltage as the S's 100kWh battery.

Observe that the power curves (blueish) are going up linearly as the speed increases within the constant torque region as expected because T is constant while W is linearly increasing. At some speed, the Torque begins to decline, and the speed at which this decline happens depends on the SOC. The highest SOC (and thus highest available voltage) gives the highest speed before this decline begins. Power output is still increasing because speed is going up faster than torque is going down. The motor back-EMF throughout all speeds (and SOCs) is always less than the battery voltage because the torque never goes to zero. But back-EMF gets larger and larger so that eventually less current than the battery would be capable of (and thus less power) can be delivered to the motor at high speeds. When this happens, the motor controller is no longer trying to reduce the applied voltage to the motor, instead it is providing the full voltage available because even that will not exceed the current limit.

The gear ratio (and wheel diameter) controls the scaling between the motor RPM and the vehicle speed. Gearing allows a trade-off between speed and torque because the gears also obey the 1st law of thermodynamics. So you can have higher vehicle speed for a given motor RPM, but that also means less torque at the wheels. This is what Porche did by adding a gear box - they can shift the maximum power point of motor-battery system to a higher vehicle speed with the higher gear. To be sure, this could also be done electronically by adding a voltage boost capability to the electronics but that might be less efficient than what a transmission could achieve.

Now let's consider the Plaid drive train where you have two motors instead of one for the rear wheels. I'm going to ignore the front motor which sometimes will use some of the available battery current because this is dumbed down. With the one motor Raven, the transaxle splits the torque, giving half to each wheel. With two motors on the Plaid, each motor directly provides that wheel's torque. At first glance you might think that you could double the torque but at low speeds you don't gain anything because the battery current is the limiting factor. If one motor was capable of taking all the current that the battery had left to give then you must instead send only half of the current to each motor. The total torque on the ground is the same as the Raven in that case. However, you have cut each motor's current in half! Transforming my equation above:
BEMF = V - IR​
You can see that for half the current, we can tolerate more back-EMF (and thus more speed) with the same battery voltage. If the dyno curves were re-run with a half torque target, you would see that the torque curves have half their value, but they would remain flat up to much higher speeds. This is because even though the back-EMF is just as high as before, you are only trying to get half the current through the motor. You want only half the current because you are forced to split the battery's limited current between two motors.
You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.
This has got to be the best written technical post I’ve seen in over two years on TMC. Thank you for taking the time to write it.
 

hacer

Active Member
Apr 13, 2016
1,066
4,449
Clarksville, MD
Thanks a lot for this excellent description!

Am I correct that the peak of the power curve will shift from the current ~70 km/h to about ~140 km/h - which is already pretty close to the ~160 kmh average speed of the Nürburgring?

Tesla might still change the gearing ratios as well, to further optimize this - as they'd have plenty of torque at lower speeds.

At what gearing ratios could Tesla expect trade-offs at lower speeds? I.e. why aren't the gearing ratios much higher than the current 7:1 they are using, while torque is still maxed out at lower speeds? There must be some low speed consideration that favors lower gearing ratios, right?
If we use the same front/rear power split as the Raven (366 kW rear, 177 kW front) that would be 183 kW peak power in each of the two rear motors for the plaid system. If the Raven peak current capability is 1500 A, with a battery internal resistance of 0.0461 Ohms (Model X 100D Battery Internal DC Resistance) that would give a battery terminal voltage of 331V for full SOC, and 535 A would be taken by the front motor. That leaves 482.5 A for each of the rear motors. Take these next calculations with a big grain of salt.

The LR-3 battery has 46 parallel cells, but they are 2170 cells vs 18650 in the Raven. Each cell should be about 1.46 times less resistance. But the Raven has 86 parallel cells so the LR-3 battery resistance I estimate at 0.059 Ohms. From that I guestimate LR-3 battery peak current is 1400 * 0.046/0.059 = 1090 A (it was not a performance model). If someone has a measured value for this it would help a lot. If that is the peak current during the dyno test above, the motor torque constant is then 320 lb-ft/1090 A = 0.293 lb-ft/A. From this we can compute the back emf constant Kb = 0.293 lb-ft/A * 1.36 N-m/lb-ft * 2 * pi / 60 = 0.0418 V/RPM.

So the question is at what RPM would the peak torque no longer be possible on the Plaid:

BEMF = 0.0418 * SPEED (in RPM) = 331 - 482.5 * R where R is the resistances excluding the battery. Going back to the LR-3 it loses steam at 90% SOC at:
BEMF = 0.0418 * 5200 = 399 - 1090 * 0.059 - 1090 * R which gives R = 0.107
Pluggin this in for the Plaid we get
SPEED = (331 - 482.5 * 0.107)/0.0418 = 6683 RPM. The power output will peak just a little higher and lasts about 800 RPM more so at about 100 km/hr the power would start declining for the same gear ratio. The power would also tail off more slowly.

So short answer is NO, this doesn't put the peak out nearly as far as you were hoping.
 

hacer

Active Member
Apr 13, 2016
1,066
4,449
Clarksville, MD
Not quite. The drive electronics function as a DC-DC converter (buck in this case). As such power out = power in / efficency.
Current from pack does not equal current to motor unless you are at 100% duty cycle and the motor is at full pack voltage.

The max torque is SW limited to what the drive unit can handle (and beyond that, by available friction). With two rear motors, each can only output half the torque due to the tires, but they could indeed put out double the torque of one motor at low speeds. Concider that at 0 RPM power =0, so current can be arbitrarily high without putting a load on the pack (current recirculates through the motor windings). ...
You are right the buck converter can raise the current far above the battery current at low Voltage. (I was suckered in by my dumbed down analysis and missed this). This is actually great because then you can raise the fixed mechanical gear ratio and still have the same low-speed performance. Doing that would allow the peak power curve to be pushed as far out as @Fact Checking wanted!
 

VT_EE

Active Member
Apr 22, 2017
2,025
2,419
Maryland
So @hacer , what are your thoughts on the effect of a larger battery in the Plaid to increase the maximum available current along with different gearing than the Raven? The cooling system is obviously different so it's not a stretch to assume the battery is not the same as the Raven.
 

hacer

Active Member
Apr 13, 2016
1,066
4,449
Clarksville, MD
So @hacer , what are your thoughts on the effect of a larger battery in the Plaid to increase the maximum available current along with different gearing than the Raven? The cooling system is obviously different so it's not a stretch to assume the battery is not the same as the Raven.
I'm not totally sure what you are asking. My opinion is that the battery in the Plaid model S at Nurbergring has a normal battery from a P100D. The new roadster will definitely have a larger battery.
 
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mongo

Well-Known Member
May 3, 2017
12,953
38,373
Michigan
So @hacer , what are your thoughts on the effect of a larger battery in the Plaid to increase the maximum available current along with different gearing than the Raven? The cooling system is obviously different so it's not a stretch to assume the battery is not the same as the Raven.

Low speed torque is limited by the drive unit current limit not the battery power. With two rear motors, the max torque for each is cut in half. With the same drive electronics, you could double the output ratio and keep the same wheel torque. Slower motor -> less back-EMF -> more available torque.
 
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Ulmo

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You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.
Oh! That would help fix one of the big problems I always had with my Tesla Model S: it was a dog above 60MPH, and couldn't sustain high speeds for more than a lick. Two Plaid motors in back would improve the acceleration power above 60MPH. Also, in a later message, you stated it would allow a higher gear ratio and therefore this acceleration improvement could be continued into higher speed.

Model 3 already made continuous high speed better. Would continuous high speed also be possible with two Plaid motors in back as you described, the same way continuous high speed is available with Model 3 today?

Speaking of which, why hasn't the Lucid Air come out? It was supposed to have 4 motors or something.
 
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Olle

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Jul 17, 2013
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This is my "dumbed down" explanation post that @CorneliusXX asked for regarding my post Tesla, TSLA & the Investment World: the 2019 Investors' Roundtable.
My original post was trying to correct the mistake that @Fact Checking made about the plaid drive system having 3 different fixed gear ratios. If two of the motors independently drive the rear motors but have different fixed gear ratios, that would mean that one of them would always have less available torque than the other one except at very high speeds. That's useful only if you always drive the same direction on a circular track. Of course I was way behind on the thread and there were many others who had already challenged the idea, but I hadn't read them yet. My post did a little more in speculating why it wouldn't be necessary for even a change of gear ratio to have better high speed power output. This post is going to try to explain (and justify) that statement a bit better. It will be dumbed down but still somewhat technical.

The "problem" (to the extent there is one) for high power at high speed in the Tesla is mainly caused by the motor back-EMF (EMF stands for electromotive-force, and it is the voltage generated by the spinning motor, and it opposes the driving voltage). To make this simpler to understand I'll be using a DC motor to describe in general how it works (because the DC motor is much easier to understand and explains the issue).

To start with the problem will be greatly simplified: I will use the 1st law of thermodynamics (energy in = energy out if there is no stored energy) but in a steady-state per-unit-time metric. That is: power in = power out which ignores inefficiency (heat generation) and energy storage. Of course real motors have these other effects but they aren't important for understanding the issue. Power is the time-rate-of-change (the derivative with respect to time) of energy.
For a motor the input power is electrical. Electrical power is the product of voltage (V) and current (I). The output power is mechanical and is the product of torque (T) and angular speed (W). Thus the motor analysis begins with V*I = T*W. Note that if you connect a DC motor to a battery and have no load on the shaft (it is free to spin with no torque), the motor will spin at high speed, but T=0 so it must be that V*I = 0. Since it is hooked to the battery, we know that V is the battery voltage so (I) must be zero. Sweeping several things under the rug here, the current is zero because the motor back-EMF is exactly matching the battery voltage. Similarly, if we fix the shaft so the motor can't turn (even though it's hooked to the battery) again the output power is zero because W=0. Torque will be large and the input current will be really large yet the input power must still be zero. In this case, the motor Voltage is zero (even though it is connected to the battery). All of the battery voltage will drop across the resistances I'm neglecting in my idealized model here.

In reality, there is some resistance in the motor winding, battery and wires and the motor current is limited to the battery voltage minus the back-EMF divided by the total resistance.
I = (V - BEMF)/R​
R is the sum of all resistance in the battery, the motor windings, electronics, cables, etc. But it is small, so the power-in=power-out model is still reasonable to understand the motor mechanical power but some resistance is necessary to understand how current is limited in the motor. In the real world if you connect a battery directly to a DC motor with a fixed shaft, chances are some smoke will appear and something may melt or catch on fire.

You can see that for the DC motor, the back-EMF is directly proportional to motor speed. The proportional constant is called "the motor velocity constant" or "back-EMF constant". The motor torque is also proportional to the current (I) through the "motor torque constant". You can read more about both of these here: Motor constants - Wikipedia.

Now let's look at the model 3 motor torque-power curves:
Tesla-Model-3-SOC-Dyno-Results.jpg

We can see that at low speeds the torque is largest and constant, even independent of battery state of charge (SOC); These are the curves just below 325 lb-ft on the left axis. They are flat and independent of speed because the motor controller does not apply full battery voltage to the motor at low speeds; instead it uses electronic magic to very efficiently lower the applied voltage in order to produce the desired current (and thus torque). For the dyno curves, the desired torque was "give it all you got". For various reasons, not the least of which is fuses within the battery there is a maximum current limit that Tesla's controller will not let you exceed. For performance S vehicles, this is somewhere near 1,500 Amps. This dyno run was done on a model 3 which has a smaller battery (fewer parallel cells) so it probably also has smaller peak current allowed but I haven't checked. Note that model 3 LR battery still has the same (unloaded) voltage as the S's 100kWh battery.

Observe that the power curves (blueish) are going up linearly as the speed increases within the constant torque region as expected because T is constant while W is linearly increasing. At some speed, the Torque begins to decline, and the speed at which this decline happens depends on the SOC. The highest SOC (and thus highest available voltage) gives the highest speed before this decline begins. Power output is still increasing because speed is going up faster than torque is going down. The motor back-EMF throughout all speeds (and SOCs) is always less than the battery voltage because the torque never goes to zero. But back-EMF gets larger and larger so that eventually less current than the battery would be capable of (and thus less power) can be delivered to the motor at high speeds. When this happens, the motor controller is no longer trying to reduce the applied voltage to the motor, instead it is providing the full voltage available because even that will not exceed the current limit.

The gear ratio (and wheel diameter) controls the scaling between the motor RPM and the vehicle speed. Gearing allows a trade-off between speed and torque because the gears also obey the 1st law of thermodynamics. So you can have higher vehicle speed for a given motor RPM, but that also means less torque at the wheels. This is what Porche did by adding a gear box - they can shift the maximum power point of motor-battery system to a higher vehicle speed with the higher gear. To be sure, this could also be done electronically by adding a voltage boost capability to the electronics but that might be less efficient than what a transmission could achieve.

Now let's consider the Plaid drive train where you have two motors instead of one for the rear wheels. I'm going to ignore the front motor which sometimes will use some of the available battery current because this is dumbed down. With the one motor Raven, the transaxle splits the torque, giving half to each wheel. With two motors on the Plaid, each motor directly provides that wheel's torque. At first glance you might think that you could double the torque but at low speeds you don't gain anything because the battery current is the limiting factor. If one motor was capable of taking all the current that the battery had left to give then you must instead send only half of the current to each motor. The total torque on the ground is the same as the Raven in that case. However, you have cut each motor's current in half! Transforming my equation above:
BEMF = V - IR​
You can see that for half the current, we can tolerate more back-EMF (and thus more speed) with the same battery voltage. If the dyno curves were re-run with a half torque target, you would see that the torque curves have half their value, but they would remain flat up to much higher speeds. This is because even though the back-EMF is just as high as before, you are only trying to get half the current through the motor. You want only half the current because you are forced to split the battery's limited current between two motors.
You should immediately see that the power curve will now ramp up for even higher speeds, and fall off more slowly for the very same gear ratio. The Plaid system, with identical batteries, motors, and gear ratios will have about the same 0-60 MPH time, but will have much better 60-120 MPH time. This is what I was saying in my original post.
The characteristics that you describe seem to match a DC motor, especially the part about the back EMF limiting motor speed. DC motors are also generators, where while you increase motor a speed, the generated voltage will eventually match the battery voltage and the motor speed will thus level off.
In a variable inverter controlled AC motor like in a Tesla, this is not the case. The only way to make a Tesla motor counter balance the battery and therefore act as a generator is to shift the phases, which is what happens when you take your foot of the accelerator and regenerate.
I may be going out on a limb here now but I think that the leveling off of power as you reach higher rpm is in the Mountainpass diagram above is due not to back emf, but rather to the inverter's inability to supply full tension through a whole revolution while also shifting phases at high speed.
I agree with your post that both rear motors will have the same gear ratio. Since today's Model S is about 10:1, with two motors we will probably see the somewhere near 5:1 with sustained wheel torque for the reason the @mongo explained. That probably lets Tesla delete the intermediate gear so there will only be the pinion and and the ring gear, meaning higher efficiency and lower production cost. Four cogged wheels total for the rear drivetrain. Current Teslas have 4 wheels in the reduction gear and another 4 in the differential for a total of 8, so the Plaid powertrain will be simpler, probably more reliable and cheaper to make.
 
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mongo

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May 3, 2017
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The characteristics that you describe seem to match a DC motor, especially the part about the back EMF limiting motor speed. DC motors are also generators, where while you increase motor a speed, the generated voltage will eventually match the battery voltage and the motor speed will thus level off.
In a variable inverter controlled AC motor like in a Tesla, this is not the case. The only way to make a Tesla motor counter balance the battery and therefore act as a generator is to shift the phases, which is what happens when you take your foot of the accelerator and regenerate.
I may be going out on a limb here now but I think that the leveling off of power as you reach higher rpm is in the Mountainpass diagram above is due not to back emf, but rather to the inverter's inability to supply full tension through a whole revolution while also shifting phases at high speed.
I agree with your post that both rear motors will have the same gear ratio. Since today's Model S is about 10:1, with two motors we will probably see the somewhere near 5:1 with sustained wheel torque for the reason the @mongo explained. That probably lets Tesla delete the intermediate gear so there will only be the pinion and and the ring gear, meaning higher efficiency and lower production cost. Four cogged wheels total for the rear drivetrain. Current Teslas have 4 wheels in the reduction gear and another 4 in the differential for a total of 8, so the Plaid powertrain will be simpler, probably more reliable and cheaper to make.

In an AC induction motor, the rotor acts as a magnet due to the induced current. There is an extra parameter known as slip that is the relation between the stator field rotational speed and the rotor rotational speed. The slower the rotor vs the stator, the more torque is developed. If the rotor is spinning faster than the stator, you get regen.

The drive electronics do have limit in terms of controlability due to their switching frequency being divided by the motor RPM and number of poles, thus reducing resolution at high speeds. This could reduce peak stator current if the time per commutation step is too short and windiing inductance too high.

Interesting point regarding eliminating a gear reduction stage. They may keep the same number to achieve the required offset from the motor to the drive shaft.
 

Olle

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Jul 17, 2013
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In an AC induction motor, the rotor acts as a magnet due to the induced current. There is an extra parameter known as slip that is the relation between the stator field rotational speed and the rotor rotational speed. The slower the rotor vs the stator, the more torque is developed. If the rotor is spinning faster than the stator, you get regen.

The drive electronics do have limit in terms of controlability due to their switching frequency being divided by the motor RPM and number of poles, thus reducing resolution at high speeds. This could reduce peak stator current if the time per commutation step is too short and windiing inductance too high.

Interesting point regarding eliminating a gear reduction stage. They may keep the same number to achieve the required offset from the motor to the drive shaft.
 

mongo

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May 3, 2017
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Michigan
@mongo: Unless they make both gears larger for distance. Got to watch the ground clearance though

Given the entire housing is changing anyway, they could do it. Although, it would be less impact to manufacturing to leave the motor gear the same and change the intermediate and final gears. Final is a new part anyway due to no diff.
 

cduzz

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Jun 6, 2019
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429
boston ma
I suspect the "plaid" power-train is made up of a battery sled that's 80, 120, 200 or 400kwh pack using 2170 cells.

The same architecture (battery pack / motors) used across the S/X/Roadster/Truck/Van/Semi.

Not all pack configurations fit in all platforms (IE the 400kwh configuration is 2 200kwh packs, and only fits in the Semi.)

The motor architecture will be such that the rear assembly can house 1 or 2 motors and there will be several motor/gear sets for optimizing different torque ranges at different wheel speeds. Different platforms will have different reduction gear ratios for the motor sets; the Semi will optionally have 4 rear motors where each pair has a different ratio.

Truck (cab forward like a 60s dodge van truckScreen Shot 2019-09-16 at 9.56.49 AM.png)
  • 80, 120, 200kwh battery (for towing capacity or off-road range)
  • optional portal axles plus air suspension, air suspension, or coil suspension
  • optional 3 motor, 2 motor, or 1 motor (2r+1f, 1f/1r, 1r)
  • Different length beds / size cabs


Minivan / Van
  • 80 or 120kwh battery
  • 1r or 1r/1f motors
S/X:
  • 120 or 200kwh battery
  • 1/1 or 2/1 "plaid" or "maximum plaid" modes
Roadster
  • 120 or 200kwh battery
  • 2/1 (maximum plaid) only
Semi / box truck
  • 120, 200 or 400kwh
  • 4/1 or 2/1 or 2, depending on intended use
  • Optional box truck or Semi rear architecture
Econobox / Hot Hatch
  • half length 80kwh*
  • 1f or 1r/1f
 

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Olle

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Jul 17, 2013
783
402
Orlando, FL
Quote. Given the entire housing is changing anyway, they could do it. Although, it would be less impact to manufacturing to leave the motor gear the same and change the intermediate and final gears. Final is a new part anyway due to no diff. quote

True. Perhaps be the current motor gear is already the perfect size to mesh with a five times bigger final gear? Another aspect is width. By eliminating intermediate shaft the could make the gearbox a lot narrower in one shared case for both motors. Both motors need to sit behind the axle as not to infringe on battery space if they use today’s model s battery. With motors next to each other width becomes a premium.
 

mongo

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May 3, 2017
12,953
38,373
Michigan
Quote. Given the entire housing is changing anyway, they could do it. Although, it would be less impact to manufacturing to leave the motor gear the same and change the intermediate and final gears. Final is a new part anyway due to no diff. quote

True. Perhaps be the current motor gear is already the perfect size to mesh with a five times bigger final gear? Another aspect is width. By eliminating intermediate shaft the could make the gearbox a lot narrower in one shared case for both motors. Both motors need to sit behind the axle as not to infringe on battery space if they use today’s model s battery. With motors next to each other width becomes a premium.

On current units, the drive electronics sit on the opposite side of the output gear as the motor. So they do have a major width problem. Removing the gear would gain them one half of its width. I suppose they will move the electronics to the same side of the gear as the controlled motor. We do know they already figured it out for the Roadster. Fun puzzle.
 

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