Comment by ephbit | Hacker News Reader

Comment by ephbit | Hacker News Reader

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> I think about 2 ton battery is closer to truth for a truck that has about 400-500 mile range.

A Tesla Model 3 has a ~ 500 kg battery and weighs ~ 2 tons. With the drag coefficient of a truck being significantly greater and its gross weight amounting to ~ 15-20 times that of the Model 3, I'd say your 2 tons are way off. Sure, the truck will go slower than the Model 3, but still.

Also electric motors do weigh a few kg's as well, so I'd guess the less heavy drive train of the electric vehicle isn't going to save all that much weight.

> Charging at the starting point while loading, at (mandatory) breaks and at the destination should be enough; a BEV truck done right shouldn't require extra waiting time.

I think this needs to be compared to the procedure with a diesel truck. The diesel truck needs a few minutes at the gas station to refill and get some Ad Blue or what. Then it's all flexible to go anywhere for a few hundred miles. Compared to that, your BEV truck is going to need careful planning ahead of charging and any small deviation from the plan is going to be a lot more of a hassle than for the diesel truck.

Point being. Nope, there aren't many H2 gas stations either as of now. But once there are, the refuelling of a FCEV will be much more similar to that of a diesel/gasoline vehicle than that of a BEV, aka more convenient/resilient.

"A Tesla Model 3 has a ~ 500 kg battery and weighs ~ 2 tons. With the drag coefficient of a truck being significantly greater and its gross weight amounting to ~ 15-20 times that of the Model 3, I'd say your 2 tons are way off. Sure, the truck will go slower than the Model 3, but still."

A truck should consume about 5x than what a Model 3 LR does. 5x 445 kg (actual M3LR battery weight) is 2225 kg.

> A truck should consume about 5x than what a Model 3 LR does.

Do you have any hints as to why "5x" should be roughly correct?

Model 3 Cd: ~ 0.23 [1]

Tesla Semi Cd: ~ 0.36 [2]

A shipping container measures w x h: 2.438 x 2.591 m The total height of the truck will be > 3 m so we're talking about 2.438 m x 3 m projected area. That's ~ 7.3 m2

A Tesla S apparently has 0.562 m2 drag area so let's assume 0.6 m2 for the Model 3. [1]

This amounts to a factor of Semi to Model 3 of:

7.3 m2 * 0.36 / (0.6 m2 * 0.23) = 19

So at the same speed the aerodynamic drag of a truck will be almost 20 times that of a Model 3. Yes, a truck typically drives slower and speed goes into calculation of engine power at a power of 3. But the truck would have to go slower than the Model 3 by a factor of 19^(1/3) ~ 2.7 to have roughly the same drag.

If the Model 3 drives at 150 kph and the Semi at 100 kph, the Semi still has more than 5 times the aerodynamic drag.

And you'll have to add friction to that and losses for accelerating the greater mass. (I doubt regenerative braking will scale well with increased vehicle mass)

[1] https://en.wikipedia.org/wiki/Automobile_drag_coefficient

[2] https://insideevs.com/news/345710/tesla-semi-details-on-truc...

Your drag area figure for Tesla Semi is absurdly high.

Note that drag area is cross-sectional area times drag coefficient. If your numbers are otherwise correct, Semi's "drag area" should be 0.36 * 7.3 m^2 = 2.62800 m^2.

2.62800 m^2 (Semi) / 0.562 m^2 (Model S) is approximately 4.68. So I think 5x energy consumption is completely feasible.

> I doubt regenerative braking will scale well with increased vehicle mass

Why would that be an issue? 500 kWh magnitude battery can absorb about 7x power compared to a Model 3 LR AWD battery. Regenerative braking is probably only ever issue when the battery is somewhere above 95% full.

> Your drag area figure for Tesla Semi is absurdly high.

Don't think so. But I made a different error. See further down.

Drag equation [1]:

> FD = 1/2 * rho * u² * cD * A

> The reference area A is typically defined as the area of the orthographic projection of the object on a plane perpendicular to the direction of motion.

So A in the case of a truck carrying a standard container cannot be smaller than the section of the container. And because the container cannot hover millimetres above the ground but must rather be carried at a height of at least half a metre you'll have A > greater than the cross section of the container.

Which is what I calculated above.

I did make an error though by multiplying the drag area of the Model S by the drag coefficient, since the 0.562 m² already takes the coefficient into account.

So you're right, the factor Semi/Model S is ~ 4.68 based on the numbers I assumed.

It does look more feasible indeed based on this number.

Yet I'm still sceptic a battery 5 times larger will suffice because of higher friction and because I doubt regenerative braking will recover the same proportional amount of energy for the Semi as for the Model S.

Let's see. Decelerating the 20,000 kg Semi going at 100 kph at mild 0.10 g requires a force of 0.1 * 9.81 m/s² * 20,000 kg = 19,620 N.

At a velocity of 100 kph that equals (not taking drag and other friction into account) an initial (lossless) braking power of 545 kW that could be regained by regenerative braking. Okay, could be feasible as well, if charging can be ramped up to this rate within the fraction of a second.

If you brake at 0.5 g though, you'd have to suddenly feed in the ball park of 2 MW into the battery. Not sure that's possible.

[1] https://en.wikipedia.org/wiki/Drag_equation

> If you brake at 0.5 g though, you'd have to suddenly feed in the ball park of 2 MW into the battery. Not sure that's possible.

MCS [0] charging standard goes up to 3.75 MW. I don't think Semi can charge at that power, but 2 MW — why not.

Of course, the catch is that the higher the battery state of charge (SoC), the lower the charging current can be. 2 MW might not be possible, say, somewhere above 50-80% SoC.

[0]: https://en.wikipedia.org/wiki/Megawatt_Charging_System

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