5 ms·

>cables that can carry megawatts of energy are probably pretty heavy Required cable thickness (and thus weight) is proportional to current, not voltage. Conduc

>cables that can carry megawatts of energy are probably pretty heavy Required cable thickness (and thus weight) is proportional to current, not voltage. Conductors to carry megawatts of electricity could be fairly light as long as the voltage is extremely high and the amperage relatively low.

deleted 4y ago
[deleted]
jvanderbot 4y ago
And magnetic force is a function of current, and magnetic force is what drives wheels. Technically you are correct, but thick cables are actually needed to do lots of translation from electricity to movement.
- willmadden 4y ago
  I don't think there's a problem with cable weight with electric motors. Production EV's today have ridiculous amounts of torque and don't require prohibitively heavy cables. The issue with acceleration using wheels is gravity. You are limited to 1g, the downforce on the wheels, which you use for traction to push the vehicle forward. Spoilers increase the downforce on the vehicle beyond 1g, but the energy required to do that is proportional to the drag on the spoiler. Short of a rocket engine, the way to accelerate beyond 1g with a wheeled vehicle is to have a downforce fan. That can be powered by combustion or by electricity. I don't know which would offer the highest power to weight ratio.
  - deleted 4y ago
    [deleted]
  - lazide 4y ago
    I’m not sure having a ‘reverse Helicopter’ on the back of your vehicle really qualifies as ‘ground vehicle’ at that point. Especially if it’s providing enough force to literally drive upside down on a ceiling somewhere.
    - stefs 4y ago
      it's as ground as can be, short of a tunneling machine maybe.
    - sacrosancty 4y ago
    - et2o 4y ago
      This was the basis of a very famous F1 car in the 1970s. It was ultimately banned.
    - benj111 4y ago
      What's the down force for an F1 car? Are they not ground vehicles?
  - solidr53 4y ago
    BEV used to have around 80kg of copper [1], nowadays, manufacturers are moving towards 800V architecture attempting to cut the weight in half. I also feel like 800V cars also maintain their maximum power better on lower SoC than 400V systems. [1] https://www.copper.org/publications/pub_list/pdf/A6192_ElectricVehicles-Infographic.pdf https://www.copper.org/publications/pub_list/pdf/A6192_Elect...
  - mannykannot 4y ago
    > Short of a rocket engine, the way to accelerate beyond 1g with a wheeled vehicle is to have a downforce fan. This electric vehicle is accelerating to 100kph (~28 m/s) in less than 1.5 seconds, which is an average acceleration of about 2g, if I am not mistaken. Looking around, I have seen 4.0 given as a representative friction coefficient for drag racing tires, which would imply a corresponding peak acceleration of 4g without aerodynamic assistance (again, if I am not mistaken.)
  - usrusr 4y ago
    If we are already what-if-ing with wired power delivery, just skip the pantograph and look at what wire-guided missiles do: they have their data lines on a spool, unrolling as they go. For power you certainly wasn't thicker wires, and while you're at it, you'll certainly power the unrolling of the spool instead of just dragging it to spin up. Then if the spool unrolling is for some reason faster than the wheels, the wire will serve as reaction mass, pushing the vehicle. Now we have what's essentially a copper rocket that isn't constrained by downforce and tire grip at all.
- kroeckx 4y ago
  For the torque, you can use a low current using more windings, which has the same effect of having a higher current with less windings. The acceleration you can reach is mostly limited by the peak current and the force constant, but the maximum speed you can reach at that acceleration is mostly limited by the voltage you have available because of the back-EMF. A simplified formula is that the voltage you need is R * I + v * BEMF. The higher the voltage is, the better the insulation between the windings needs to be. You need to find a balance between the different properties.
  - happyopossum 4y ago
    More windings = more weight - you're gonna pay for it somewhere.
    - sudosysgen 4y ago
      Sure but the wiring is thinner, thereby somewhat equalising.
omegalulw 4y ago
This seems...wrong? Let's say you can control voltage, cross section area A. Now, you want a constant power across a load P = V * I. Your net resistance R is C + D * L / A where C and D are constants and L is your cable length. I is proportional go V * A / (C * A + E) where E is also a constant. So your load power is proportional V^2* C * A / (C * A + E) which is proportional to V^2 * A / (A + F) where F is also a constant. With a large "enough" A, this is effectively V^2. With a small A, this is V^2 * g where g is A / F. So the smaller the area you have the more power you are wasting (roughly equal to V^2 * (1-g) which is heat in the wires). So the smaller area you have, the less efficient your power delivery. And juicing up your source power is a lot more expensive than juicing up your source voltage.

deleted 4y ago

[deleted]

jvanderbot 4y ago

And magnetic force is a function of current, and magnetic force is what drives wheels. Technically you are correct, but thick cables are actually needed to do lots of translation from electricity to movement.

willmadden 4y ago
I don't think there's a problem with cable weight with electric motors. Production EV's today have ridiculous amounts of torque and don't require prohibitively heavy cables. The issue with acceleration using wheels is gravity. You are limited to 1g, the downforce on the wheels, which you use for traction to push the vehicle forward. Spoilers increase the downforce on the vehicle beyond 1g, but the energy required to do that is proportional to the drag on the spoiler. Short of a rocket engine, the way to accelerate beyond 1g with a wheeled vehicle is to have a downforce fan. That can be powered by combustion or by electricity. I don't know which would offer the highest power to weight ratio.
- deleted 4y ago
  [deleted]
- lazide 4y ago
  I’m not sure having a ‘reverse Helicopter’ on the back of your vehicle really qualifies as ‘ground vehicle’ at that point. Especially if it’s providing enough force to literally drive upside down on a ceiling somewhere.
  - stefs 4y ago
    it's as ground as can be, short of a tunneling machine maybe.
  - sacrosancty 4y ago
  - et2o 4y ago
    This was the basis of a very famous F1 car in the 1970s. It was ultimately banned.
  - benj111 4y ago
    What's the down force for an F1 car? Are they not ground vehicles?
- solidr53 4y ago
  BEV used to have around 80kg of copper [1], nowadays, manufacturers are moving towards 800V architecture attempting to cut the weight in half. I also feel like 800V cars also maintain their maximum power better on lower SoC than 400V systems. [1] https://www.copper.org/publications/pub_list/pdf/A6192_ElectricVehicles-Infographic.pdf https://www.copper.org/publications/pub_list/pdf/A6192_Elect...
- mannykannot 4y ago
  > Short of a rocket engine, the way to accelerate beyond 1g with a wheeled vehicle is to have a downforce fan. This electric vehicle is accelerating to 100kph (~28 m/s) in less than 1.5 seconds, which is an average acceleration of about 2g, if I am not mistaken. Looking around, I have seen 4.0 given as a representative friction coefficient for drag racing tires, which would imply a corresponding peak acceleration of 4g without aerodynamic assistance (again, if I am not mistaken.)
- usrusr 4y ago
  If we are already what-if-ing with wired power delivery, just skip the pantograph and look at what wire-guided missiles do: they have their data lines on a spool, unrolling as they go. For power you certainly wasn't thicker wires, and while you're at it, you'll certainly power the unrolling of the spool instead of just dragging it to spin up. Then if the spool unrolling is for some reason faster than the wheels, the wire will serve as reaction mass, pushing the vehicle. Now we have what's essentially a copper rocket that isn't constrained by downforce and tire grip at all.
kroeckx 4y ago
For the torque, you can use a low current using more windings, which has the same effect of having a higher current with less windings. The acceleration you can reach is mostly limited by the peak current and the force constant, but the maximum speed you can reach at that acceleration is mostly limited by the voltage you have available because of the back-EMF. A simplified formula is that the voltage you need is R * I + v * BEMF. The higher the voltage is, the better the insulation between the windings needs to be. You need to find a balance between the different properties.
- happyopossum 4y ago
  More windings = more weight - you're gonna pay for it somewhere.
  - sudosysgen 4y ago
    Sure but the wiring is thinner, thereby somewhat equalising.

omegalulw 4y ago

This seems...wrong? Let's say you can control voltage, cross section area A. Now, you want a constant power across a load P = V * I. Your net resistance R is C + D * L / A where C and D are constants and L is your cable length. I is proportional go V * A / (C * A + E) where E is also a constant. So your load power is proportional V^2* C * A / (C * A + E) which is proportional to V^2 * A / (A + F) where F is also a constant. With a large "enough" A, this is effectively V^2. With a small A, this is V^2 * g where g is A / F. So the smaller the area you have the more power you are wasting (roughly equal to V^2 * (1-g) which is heat in the wires). So the smaller area you have, the less efficient your power delivery. And juicing up your source power is a lot more expensive than juicing up your source voltage.