4 ms·

So how far is the trip down, 10 miles? The potential energy of moving 65 tons 10 miles down (think of pulling a rope with a 65 ton weight) is a freaking lot of

So how far is the trip down, 10 miles? The potential energy of moving 65 tons 10 miles down (think of pulling a rope with a 65 ton weight) is a freaking lot of energy.

sp332 9y ago
It carries 45 tons + 65 tons downhill and then has to carry 45 tons back up. That's the hard part.
- brianwawok 9y ago
  With 0 friction and 100% effeciency, the break even would be 45 tons of load? So this has 20 extra tons to make up for the various losses..
  - hexane360 9y ago
    Not 20 extra tons. 65 Extra tons.
    - brianwawok 9y ago
      Huh? The weight is 45 tons going up, and 45 + 65 = 110 going down, right? In a 100% efficient no friction wonder world, the load needed would be 45 tons going up, and 45 + 45 = 90 tons going down. This would balance out to 0 battery usage. Since our going down weight is 110 tons and not 90 tons, we have 110 - 90 = 20 "extra tons" that can go to things like friction and imperfect conversion of energy forms.
      - sp332 9y ago
        With no friction etc you would just store all the energy on the way down and use it exactly the same (reversed) on the way up. 45 tons down is all you need to get 45 tons up. No extra. 90 tons down would get you 90 tons up.
      - Roodgorf 9y ago
        I think you just described 0 friction with 50% efficiency. With 100% efficiency the weight of the truck alone would be enough to recoup the energy on the way down.
        brianwawok 9y ago
        Opps you are right. So we need like 40% effeciency to make this work?
        hexane360 9y ago
        Yep. efficiency * 110 tons * gh = 45 tons * gh efficiency = 45 tons / 100 tons = 40.9%
ChuckMcM 9y ago
That it the beauty of doing as an energy/energy analysis. Energy at the top is total mass x 9.8 x height, energy to get the unit to the top is unloaded mass x 9.8 x height. The ratio will tell you how much "excess" energy you get, the efficiency of converting that from mechanical energy to stored charge and back to mechanical energy will tell you how much energy you will have 'left'. And as long as you don't run out of battery capacity to hold energy you're capturing on the way down, the math just works.
ams6110 9y ago
The linear/road distance down doesn't matter. It's elevation that provides the potential energy. Pretty sure they are not descending 10 miles in elevation.

So how far is the trip down, 10 miles? The potential energy of moving 65 tons 10 miles down (think of pulling a rope with a 65 ton weight) is a freaking lot of energy.

sp332 9y ago

It carries 45 tons + 65 tons downhill and then has to carry 45 tons back up. That's the hard part.

brianwawok 9y ago
With 0 friction and 100% effeciency, the break even would be 45 tons of load? So this has 20 extra tons to make up for the various losses..
- hexane360 9y ago
  Not 20 extra tons. 65 Extra tons.
  - brianwawok 9y ago
    Huh? The weight is 45 tons going up, and 45 + 65 = 110 going down, right? In a 100% efficient no friction wonder world, the load needed would be 45 tons going up, and 45 + 45 = 90 tons going down. This would balance out to 0 battery usage. Since our going down weight is 110 tons and not 90 tons, we have 110 - 90 = 20 "extra tons" that can go to things like friction and imperfect conversion of energy forms.
    - sp332 9y ago
      With no friction etc you would just store all the energy on the way down and use it exactly the same (reversed) on the way up. 45 tons down is all you need to get 45 tons up. No extra. 90 tons down would get you 90 tons up.
    - Roodgorf 9y ago
      I think you just described 0 friction with 50% efficiency. With 100% efficiency the weight of the truck alone would be enough to recoup the energy on the way down.
      - brianwawok 9y ago
        Opps you are right. So we need like 40% effeciency to make this work?
        hexane360 9y ago
        Yep. efficiency * 110 tons * gh = 45 tons * gh efficiency = 45 tons / 100 tons = 40.9%

ChuckMcM 9y ago

That it the beauty of doing as an energy/energy analysis. Energy at the top is total mass x 9.8 x height, energy to get the unit to the top is unloaded mass x 9.8 x height. The ratio will tell you how much "excess" energy you get, the efficiency of converting that from mechanical energy to stored charge and back to mechanical energy will tell you how much energy you will have 'left'. And as long as you don't run out of battery capacity to hold energy you're capturing on the way down, the math just works.

ams6110 9y ago

The linear/road distance down doesn't matter. It's elevation that provides the potential energy. Pretty sure they are not descending 10 miles in elevation.