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Comment by ChuckMcM | Hacker News Reader

That had me pause as well. I'm still a bit dubious. Given potential energy is mass x grav x height, since the vehicle mass cancels top to bottom. You have load mass x grav * height of "extra" energy on the way down, which if you convert at 50% efficiency to charge would be a quantity of energy E which on the way up, if you have a 50% efficient power train would take the mass of the truck to the top of the hill. That suggests that the truck can carry 4x it's mass down hill. But in the phys.org article [1] it says it weighs 45 tons and can carry 65 tons. That is 44% additional mass. And because its linear 44% "additional" potential energy at the top. Which has to be converted twice (mechanical -> electrical, electrical -> mechanical).

I'm not expecting them to meet their goal of producing excess energy.

[1] weighing in at 45 tons when empty plus 65 tons of loading capacity -- https://phys.org/news/2017-09-e-dumper-world-largest-electri...

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I think your efficiency numbers are too low.

Large electric motors, 90-95% efficient.

Battery round trip efficiency, 80-85%.

High voltage inverters, likely above 98%.

Roughly seems like regen round trip efficiency would be at least 63.5%.

The weight ratio is (45 tons)/(65tons + 45 tons) = .409

Unfactored are rolling friction loses.

40 kWh regen per trip on 700 kWh total is very likely to have 98%+ charge/discharge efficiency. If it's doing 20 trips per day and a "day" is 8 hours (I suspect it's actually more), then that's what- a ten minute trip? That would be charging at .34 C. A standard slow charge is between .2-.8 C.

   > The weight ratio is (45 tons)/(65tons + 45 tons) = .409

Sproing! Discussing it here I realize that I kept thinking about it as 'can the energy of bringing 65 tons down equal the energy of 45 tons up' but you are absolutely right that really its 110 tons down, and 45 tons up. That certainly helps.

The efficiency of an electric powertrain is much higher than 50%. Tesla claims that the total roundtrip efficiency for regen -> battery -> driving for the Tesla Roadster is around 64%: https://www.tesla.com/blog/magic-tesla-roadster-regenerative...

That would be an excellent comparison. So at least one data point of existing practice.

Although as mentioned previously it needs 69.2% and 100% loading to break even. Could be a challenge.

Compared to fuel wasted on pre-heating and efficiency below 50% for a diesel engine with huge co2 emmissions, this doesn't sound too bad though O:)

It is absolutely a win, no question there. Just being able to recapture any of the energy going down hill and applying it to the return trip is better than the existing practice! I just enjoy the physics of the question, the sort of "How would we compute just what it would take to achieve this?" kind of thing.

Right. This seems like an ideal application for an EV. I'm surprised no one has done it before.

They only recently got around to adding electric wheels to airplanes[1], which may save up to 1.1 billion dollars annually since using jet engines to move a few mph is like, zero efficient.

[1]: http://gizmodo.com/these-little-electric-wheels-will-save-ai...

Added benefit: spin up the wheels prior to landing to reduce wear on the tires when they first touch the ground.

... and then make them regenerative, capturing the braking energy into supercapacitors.

Airplane wheels don't really do any substantial braking. Thrust reversers and aerodynamic friction do the work there.

Wow, that is a great idea. I mean, propelling a vehicle at 5 mph using jet engines optimized for 600 mph has to be one of the most energy inefficient means ever invented.

The Roadster's drivetrain is relatively inefficient for an EV (notably, it uses an AC induction motor rather than a permanent magnet motor) so I wouldn't be surprised if this truck could reach the breakeven point. But I don't have hard numbers to back that up!

Actually the power to weight ratio of all of Tesla's motors far exceeds that of permanent magnet motors. I am sure that the truck would use similar motors, they are both more cost effective and much higher torque (good for trucks).

Is power to weight important for a big truck? Seems like it would be less important there.

Note that Tesla is switching to a permanent magnet motor for the Model 3, which is an interesting development. I'm not sure why, but I hear the magnets have become a lot cheaper than they once were.

Is there any more info on this? My impression was that neodymium permanent magnets synchronous motors beat everything else on power-to-weight ratio, especially regular induction motors (well, I guess Tesla's motor isn't "regular", but I still wonder how they achieved that).

First of all, its 144% additional mass. But the calculation can be done very simply. If the regeneration efficiency would be 50%, then it needs to weight twice down what it weights going up to be a zero sum game. The regneration efficiency should be higher than 50% (Tesla has over 60%) and it is much more than twice the mass going down than going up. So there should be a healthy surplus of energy generated by going down.

Yup, read the additional comments. Net is:

   E(loaded) -> %eff capture -> E(bottom) -> %eff drive -> E(unloaded)

And as long as you can multiply the two efficiency fractions and stay above 41% you win.

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.

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

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..

Not 20 extra tons. 65 Extra tons.

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.

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.

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.

Opps you are right. So we need like 40% effeciency to make this work?

Yep.

efficiency * 110 tons * gh = 45 tons * gh

efficiency = 45 tons / 100 tons = 40.9%

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.

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.

You should probably edit this comment too- 45 tons up and 110 tons down means the efficiency just has to be >41% (45/110).

I just tried to, and just missed the edit window.

Is 50% a reasonable number? It seems low to me, given that electric motors are ~90% efficient, and battery charge efficiencies seem to be around 80%. To get that to 50% overall, the gears and powertrain would have to be < 70% efficient.

Also, the trash has to be compacted. I doubt this will be a net positive. Just a nice side effect.

It's hauling mineral deposits, not trash, so no compacting.

Right you are. There was another thread that was talking about this for garbage trucks. Whoops.

Hard to say, perhaps the most efficient conversion of potential energy to electricity is hydroelectric which can hit 90%, but if you put it into batteries and take it out again that drops some additional points and what ever you lose from heating the batteries. You would want to look at friction losses in the wheels, and coupling losses between the wheels and the brakes. And on the way up, you're going through the power train.

Based on the mass ratio the magic number is 69.2% to hit break even assuming you always load to 65 tons. If you load to 55 tons (15% margin of error) the magic number would be 82% efficiency.

EDIT: See another comment that the ratio is 110 tons to 65 tons, not 65 tons to 45 tons. That improves things. Total system regen efficiency has to be above 41%. That number crosses 60% when the load is down to 30 tons.

In a typical car with gears, you tend to lose about 15% of your power through the drive train on a 2wd vehicle. It approaches 30% with a 4wd vehicle. Since this is based on a diesel vehicle, I imagine it won't be too far off. Maybe worse because of the huge size of everything, but I'm not a mechanical engineer.

Though it is not - I believe - the case of this specific Komatsu model, electric driven dumper trucks are in production, getting rid of the transmission, examples, JFYI:

http://www.komatsuamerica.com/equipment/trucks/electric

http://www.cat.com/en_ID/products/new/equipment/off-highway-...

Basically this kind of trucks have a huge diesel engine and electric generator and four electric motors, one for each wheel.

Power is lost in the generation and powering of the electric motors of course, but there is no transmission overhead (though there is a reduction gearbox on each motor anyway).

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