3 ms·

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 tal

by ephbit 4y ago

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 https://en.wikipedia.org/wiki/Automobile_drag_coefficient [2] https://insideevs.com/news/345710/tesla-semi-details-on-truck-aerodynamics-and-drag-coefficient/ https://insideevs.com/news/345710/tesla-semi-details-on-truc...

vardump 4y ago
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.
- ephbit 4y ago
  > 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 https://en.wikipedia.org/wiki/Drag_equation
  - vardump 4y ago
    > 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 https://en.wikipedia.org/wiki/Megawatt_Charging_System

by ephbit 4y ago

vardump 4y ago

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.

ephbit 4y ago
> 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 https://en.wikipedia.org/wiki/Drag_equation
- vardump 4y ago
  > 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 https://en.wikipedia.org/wiki/Megawatt_Charging_System