6 ms·

I agree about the article, I just want to mention that improving motor efficiency would also help with some of the other losses you mention. Going from 80% to 9

I agree about the article, I just want to mention that improving motor efficiency would also help with some of the other losses you mention. Going from 80% to 90% means that you need half as much cooling, batteries may be more efficient when you draw less power from them, etc. That might it worth it even if the motor efficiency gain alone doesn’t look like much on paper.

bArray 7y ago
> Going from 80% to 90% means that you need half as much > cooling At the very most this will be < 5% performance increase because they'll need to match the best of BLDC (~90%) and 100% is simply impossible as there are losses that cannot be engineered out (thanks Physics). Also that 80% -> 90% isn't all heat, I imagine the amount of cooling required to stay roughly the same. > batteries may be more efficient when you draw less power > from them, etc. From memory, a switch mode power supply is one of the most efficient at about 90% - but you really have to design it well to get that kind of efficiency [1]. The batteries were a soft point for cars. But there's quite a bit of efficiency to get from phase control algorithms which would be in the motor control circuit. [1] https://en.wikipedia.org/wiki/Switched-mode_power_supply https://en.wikipedia.org/wiki/Switched-mode_power_supply
- codeflo 7y ago
  What do you mean, it’s not all heat — of course it is, that’s kind of the definition of efficiency.
hwillis 7y ago
> I just want to mention that improving motor efficiency would also help with some of the other losses you mention. It's not that simple. Increases in torque are strongly associated with increases in current- torque is directly proportional to total magnetic flux, so more torque in a smaller package generally means more current in your wires. The alternative is to add more turns of thinner wire, but thinner wire has lower packing efficiency. Losses from current rise as RI^2 in simple wires, slightly faster in transistors, and some very complex factor in batteries that I can't remember but is between I^3 and I^4. You can make a 100% efficient motor, but if it quadruples the current draw then it will be almost useless for vehicles.
- raxxorrax 7y ago
  But the missing torque necessitates a transmission as the article mentions that will also have a limited degree of efficiency.
  - hwillis 7y ago
    That single reduction at the motor is not very important. Remember that there are many more gears in the car besides just the drive reduction! There's one gear mesh (spot where two gears meet) in that reduction, then five per differential, of which there are two in 4wd vehicles. Even more importantly there are also eight CV joints[1] which have higher losses than gears, plus dozens of bearings and elastic losses in the wheels. The losses of the motor reduction are small compared to everything else. Also, a single stage spur or helical reduction has ~99% efficiency[2]. It's not at all like a full transmission[3], which has a ton of parts that are always spinning and churning oil, plus sliding friction. The reduction on an electric motor uses a more efficient grease and does not churn oil. Fun fact- greases (mixes of soap and oil, normally silicone oils) are non-newtonian fluids. They're shear-thinning, like ketchup, and have much lower friction under pressure while still sticking in place, unlike oil. Amazingly sophisticated for something that has existed for centuries! Anyway, what I'm saying in the above post is that even if the efficiency is much higher, even counting the gearing, it does not necessarily lead to higher efficiency elsewhere in the car. If this motor gives 20% improvement on an 85% efficient motor+gearset, that's only a 3.4% decrease (85/88) in power required. Say that 3.4% of power would have been operating in a regime that had 10% higher losses(which would be insane)- that's only .34% in cascading savings. 3.74% total. That decrease will be very, very easily overwhelmed if the motor requires higher current or is otherwise less ideal for the drivetrain. Resistive losses alone mean that if the current is 1.85% higher, it will be a net loss. [1]: https://en.wikipedia.org/wiki/Constant-velocity_joint https://en.wikipedia.org/wiki/Constant-velocity_joint [2]: https://khkgears.net/new/gear_knowledge/abcs_of_gears-b/gear_types_and_characteristics.html https://khkgears.net/new/gear_knowledge/abcs_of_gears-b/gear... [3]: https://youtu.be/vOo3TLgL0kM?t=779 https://youtu.be/vOo3TLgL0kM?t=779
  - petre 7y ago
    Or another motor. Or a motor optimized for high torque (lots of pole pairs) and another one for high speed (two pole pairs as you can hardly do 360° with a single pair). The issue is usually operating at high speeds, as electrical motors do develop high torque. Even a bus needs to operate with flux weakening. The motor in the article seems to combine a classical cylindrical design with a planar design (poles places on a disc).

bArray 7y ago

> Going from 80% to 90% means that you need half as much > cooling At the very most this will be < 5% performance increase because they'll need to match the best of BLDC (~90%) and 100% is simply impossible as there are losses that cannot be engineered out (thanks Physics). Also that 80% -> 90% isn't all heat, I imagine the amount of cooling required to stay roughly the same. > batteries may be more efficient when you draw less power > from them, etc. From memory, a switch mode power supply is one of the most efficient at about 90% - but you really have to design it well to get that kind of efficiency [1]. The batteries were a soft point for cars. But there's quite a bit of efficiency to get from phase control algorithms which would be in the motor control circuit. [1] https://en.wikipedia.org/wiki/Switched-mode_power_supply https://en.wikipedia.org/wiki/Switched-mode_power_supply

codeflo 7y ago
What do you mean, it’s not all heat — of course it is, that’s kind of the definition of efficiency.

hwillis 7y ago

> I just want to mention that improving motor efficiency would also help with some of the other losses you mention. It's not that simple. Increases in torque are strongly associated with increases in current- torque is directly proportional to total magnetic flux, so more torque in a smaller package generally means more current in your wires. The alternative is to add more turns of thinner wire, but thinner wire has lower packing efficiency. Losses from current rise as RI^2 in simple wires, slightly faster in transistors, and some very complex factor in batteries that I can't remember but is between I^3 and I^4. You can make a 100% efficient motor, but if it quadruples the current draw then it will be almost useless for vehicles.

raxxorrax 7y ago
But the missing torque necessitates a transmission as the article mentions that will also have a limited degree of efficiency.
- hwillis 7y ago
  That single reduction at the motor is not very important. Remember that there are many more gears in the car besides just the drive reduction! There's one gear mesh (spot where two gears meet) in that reduction, then five per differential, of which there are two in 4wd vehicles. Even more importantly there are also eight CV joints[1] which have higher losses than gears, plus dozens of bearings and elastic losses in the wheels. The losses of the motor reduction are small compared to everything else. Also, a single stage spur or helical reduction has ~99% efficiency[2]. It's not at all like a full transmission[3], which has a ton of parts that are always spinning and churning oil, plus sliding friction. The reduction on an electric motor uses a more efficient grease and does not churn oil. Fun fact- greases (mixes of soap and oil, normally silicone oils) are non-newtonian fluids. They're shear-thinning, like ketchup, and have much lower friction under pressure while still sticking in place, unlike oil. Amazingly sophisticated for something that has existed for centuries! Anyway, what I'm saying in the above post is that even if the efficiency is much higher, even counting the gearing, it does not necessarily lead to higher efficiency elsewhere in the car. If this motor gives 20% improvement on an 85% efficient motor+gearset, that's only a 3.4% decrease (85/88) in power required. Say that 3.4% of power would have been operating in a regime that had 10% higher losses(which would be insane)- that's only .34% in cascading savings. 3.74% total. That decrease will be very, very easily overwhelmed if the motor requires higher current or is otherwise less ideal for the drivetrain. Resistive losses alone mean that if the current is 1.85% higher, it will be a net loss. [1]: https://en.wikipedia.org/wiki/Constant-velocity_joint https://en.wikipedia.org/wiki/Constant-velocity_joint [2]: https://khkgears.net/new/gear_knowledge/abcs_of_gears-b/gear_types_and_characteristics.html https://khkgears.net/new/gear_knowledge/abcs_of_gears-b/gear... [3]: https://youtu.be/vOo3TLgL0kM?t=779 https://youtu.be/vOo3TLgL0kM?t=779
- petre 7y ago
  Or another motor. Or a motor optimized for high torque (lots of pole pairs) and another one for high speed (two pole pairs as you can hardly do 360° with a single pair). The issue is usually operating at high speeds, as electrical motors do develop high torque. Even a bus needs to operate with flux weakening. The motor in the article seems to combine a classical cylindrical design with a planar design (poles places on a disc).