4 ms·

Read the article. The idea is about batteries and motors on each car. Apparently this attacks the inefficiencies in joining/splitting up cars at their destina

Read the article. The idea is about batteries and motors on each car. Apparently this attacks the inefficiencies in joining/splitting up cars at their destinations.

skulk 5y ago
Right, but will this leave any room in the car for actual cargo? I'd like to see actual numbers (kg of battery / mile) here, my gut feeling is that lithium ion batteries are not good enough for this application, and they'll just end up using mostly diesel. From the article: > Parallel Systems cars can carry 128,000 pounds, or 2.8 times more than a semitruck. They also recharge in an hour: Without being plugged in or pulled off the track, the wheels can be recharged to drive another 500 miles. Is this saying that a single charge can move a full load of cargo 500 miles?
- NikolaeVarius 5y ago
  I dont think it needs to pull all of it by itself. I can imagine a system where in the areas the Train needs to stop/reload/whatever being surrounded by electrical infrastructure that at all moments is shoving electricity into the battery as fast as possible. Then during departure, for the first few miles, electrical lines follow the track continuing to recharge the train while it gets up to speed. Afterwards, the battery is only really needed to overcome air resistance and friction, since so much kinetic energy has been given to the train for "free" This is all just a random thought really, no idea if this is anything like what they are going for.
  - skulk 5y ago
    EDIT: should have used rolling resistance, not friction, so the final figure here is at least an order of magnitude too big, maybe two. To me, the numbers just don't add up. Let's assume - 0.2 as the coefficient of kinetic friction between the wheels and the rail [1] - the car holds 100 tons [2] - batteries can store 250 Wh/kg [3] How much battery is needed to combat friction for one mile? (normal force times friction coefficient times distance, then scaled to battery capacity) > (100 tons * 10 m/s^2 * 0.2 * 1 mile) / (250 Wh/kg) > = 324 kg So, 324 kg of lithium ion battery is needed per mile? Am I doing something wrong here because this looks completely untenable. [1]: https://the-contact-patch.com/book/rail/r1717-friction-between-wheel-and-rail https://the-contact-patch.com/book/rail/r1717-friction-betwe... [2]: https://www.csx.com/index.cfm/customers/resources/equipment/railroad-equipment/ https://www.csx.com/index.cfm/customers/resources/equipment/... [3]: https://www.cei.washington.edu/education/science-of-solar/battery-technology/ https://www.cei.washington.edu/education/science-of-solar/ba...
    - flyingfences 5y ago
      I don't see how the limit of friction is relevant here. Trains don't drag their wheels along the rail; the whole point of a wheel is that it rolls. A useful calculation would be based on rolling resistance.
      - skulk 5y ago
        totally forgot about that part. Rolling resistance seems to be at least an order of magnitude lower than friction so that does bring the numbers into a reasonable range.
        credit_guy 5y ago
        wikipedia indeed says the rolling resistance for rail cars is indeed 0.002, so you are off by a factor of a round 100. 3.24 kg/mile is not that bad. Actually even 10kg/mile sounds very acceptable to me, so they may end up being able to use cheaper batteries like NiMH.
        sudosysgen 5y ago
        Rolling resistance, for trains, is just a small part of the equation. Air resistance and changes in altitude are also quite significant.
- ZetaZero 5y ago
  The article, with pictures, has a better description. This isn't about replacing diesel train engines with battery engines. Instead, it gives each individual car a battery and motors to propel that one car.

Read the article. The idea is about batteries and motors on each car. Apparently this attacks the inefficiencies in joining/splitting up cars at their destinations.

skulk 5y ago

Right, but will this leave any room in the car for actual cargo? I'd like to see actual numbers (kg of battery / mile) here, my gut feeling is that lithium ion batteries are not good enough for this application, and they'll just end up using mostly diesel. From the article: > Parallel Systems cars can carry 128,000 pounds, or 2.8 times more than a semitruck. They also recharge in an hour: Without being plugged in or pulled off the track, the wheels can be recharged to drive another 500 miles. Is this saying that a single charge can move a full load of cargo 500 miles?

NikolaeVarius 5y ago
I dont think it needs to pull all of it by itself. I can imagine a system where in the areas the Train needs to stop/reload/whatever being surrounded by electrical infrastructure that at all moments is shoving electricity into the battery as fast as possible. Then during departure, for the first few miles, electrical lines follow the track continuing to recharge the train while it gets up to speed. Afterwards, the battery is only really needed to overcome air resistance and friction, since so much kinetic energy has been given to the train for "free" This is all just a random thought really, no idea if this is anything like what they are going for.
- skulk 5y ago
  EDIT: should have used rolling resistance, not friction, so the final figure here is at least an order of magnitude too big, maybe two. To me, the numbers just don't add up. Let's assume - 0.2 as the coefficient of kinetic friction between the wheels and the rail [1] - the car holds 100 tons [2] - batteries can store 250 Wh/kg [3] How much battery is needed to combat friction for one mile? (normal force times friction coefficient times distance, then scaled to battery capacity) > (100 tons * 10 m/s^2 * 0.2 * 1 mile) / (250 Wh/kg) > = 324 kg So, 324 kg of lithium ion battery is needed per mile? Am I doing something wrong here because this looks completely untenable. [1]: https://the-contact-patch.com/book/rail/r1717-friction-between-wheel-and-rail https://the-contact-patch.com/book/rail/r1717-friction-betwe... [2]: https://www.csx.com/index.cfm/customers/resources/equipment/railroad-equipment/ https://www.csx.com/index.cfm/customers/resources/equipment/... [3]: https://www.cei.washington.edu/education/science-of-solar/battery-technology/ https://www.cei.washington.edu/education/science-of-solar/ba...
  - flyingfences 5y ago
    I don't see how the limit of friction is relevant here. Trains don't drag their wheels along the rail; the whole point of a wheel is that it rolls. A useful calculation would be based on rolling resistance.
    - skulk 5y ago
      totally forgot about that part. Rolling resistance seems to be at least an order of magnitude lower than friction so that does bring the numbers into a reasonable range.
      - credit_guy 5y ago
        wikipedia indeed says the rolling resistance for rail cars is indeed 0.002, so you are off by a factor of a round 100. 3.24 kg/mile is not that bad. Actually even 10kg/mile sounds very acceptable to me, so they may end up being able to use cheaper batteries like NiMH.
        sudosysgen 5y ago
        Rolling resistance, for trains, is just a small part of the equation. Air resistance and changes in altitude are also quite significant.
ZetaZero 5y ago
The article, with pictures, has a better description. This isn't about replacing diesel train engines with battery engines. Instead, it gives each individual car a battery and motors to propel that one car.