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.
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:
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:
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:
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.
