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At a very high level, a Class 8 truck is meant to transport as heavy a load as possible as far as possible at as low cost as possible. On US highways the maximum gross vehicle weight for a Class 8 truck is 80,000 lbs. (36,280 kg). A typical diesel powered class 8 tractor will have a weight of 17,000 pounds (7,711 kg) and the empty trailer "tare weight" might be 15,000 pounds (6,803 kg). Thus, a diesel semi-tractor trailer may be able to haul 48,000 pounds (21,772 kg).

There are many significant considerations with respect to the viability of the Tesla Truck:

In order to evaluate the practicality of the Tesla Truck, I'll start with the energy required per highway mile at 65 m.p.h. on a smooth and level highway in good condition and in good weather. We'll assume that the semi tractor trailer is loaded to its maximum weight of 80,000 pounds. As I've detailed in other posts, for such unaccelerated motion, the sum of the external forces acting on the vehicle must be zero. The forces resisting the forward motion are aerodynamic drag and tire rolling resistance. The sum of these is the total that the electric motor must provide to the pavement through the drive train and the tires.

Starting with the drag, to a reasonable degree of accuracy, the drag force is ~D=1/2\,C_{D}\,A\rho\,{v}^{2}~ where D is the drag, C

For rolling resistance, we'll look at four driven wheels, two steering wheels, and four free rolling tires on the tractor and eight free rolling tires on the trailer. I'll give Tesla the benefit of the lowest rolling resistance tires that I've been able to find. Then, to first order, rolling resistance is dependent only on the coefficient of rolling resistance of the tires and the normal force (weight) on those tires.

While the actual rolling resistance will depend on how much weight is on which tires (because the rolling resistance will vary) and I won't know how much weight can be in the trailer until I've determined battery weight. I'll assume tires with "state of the art" low coefficient of rolling resistance (C

Thus, the approximate force that the motor must apply to the pavement through the drive train and tires is 3,954 Nt. And, since force times distance is work (and energy), the motor must do 3,954 joules of work (that is, supply 3.954 joules of energy) to move the truck one meter at 65 m.p.h. And, since there are 1609.3 meters in a mile, the motor must do 6.36317*10

In order to travel 500 miles on a level road in good conditions with no starts and stops at 65 m.p.h., Tesla will need a battery system that can supply 2.079*500 or 1,039 kilowatt hours, or 1.039 megawatt hours. If we use a number of 140 watt hours/kilogram for specific energy of a Li ion battery, such a battery pack would weigh 7,421 kg, or 16,360 pounds.

This is as ideal as it can possibly be. The truck will climb hills and, though some of the potential energy paid for in kilowatt hours can be recovered coming downhill and even energy normally wasted by braking as the truck rolls downhill will be partially recovered by a regenerative system. Nevertheless, there will be waste associated with climbing and descending. Similar considerations apply to stops and starts for traffic, stoplights, stop signs, meals, etc.

Therefore, we need a "fudge factor" for the various starts and stops, accelerations, etc. While I've seen numbers such as 90% bandied about for how much of the kinetic and potential energy in the Tesla truck can be recaptured by regenerative braking, my experience in the Lexus CT 200h makes that number seem very high. I've calculated that, in my CT 200h, about 39% of the potential energy from a hill descent went into the batteries. But I'll be generous and speculate that Tesla is much more efficient at 75%.

Suppose that the truck does the equivalent of stopping 100 times in 500 miles. An 80,000 pound vehicle travelling at 65 m.p.h. has a kinetic energy of 1.537*10^7 joules. Losing 25% of this number 100 times wastes 3.830*10^8 joules, or 106.4 kWh. Adding this to the 1,039 kWh we find that the battery pack must supply 1,145 kWh or 1.145 mWh. This will require a battery pack weighing 8179 kg or 18,031 pounds. Call it 18,000.

We now need to determine how much the battery pack weight will reduce the payload that can be hauled. To compare apples to apples, we'll figure that a very fuel efficient diesel powered semi tractor gets about 7 m.p.g. and thus will use about 71 gallons of diesel fuel weighing about 490 pounds (numbers for the weight of a gallon of diesel fuel are all over the place, but I think this represents a good average). The engine and transmission might weigh something like 3,500 pounds. The total of the materials not needed in the Tesla truck (diesel fuel, engine, transmission) is about 4,000 pounds. Thus, the available payload for the Tesla is some 14,000 pounds less than that of the diesel powered semi. And this understates the issue since we've removed the internal combustion engine and transmission, but the electric motors weigh something!

A typical heavy hauling semi tractor trailer can legally haul somewhere around 44,000 to 48,000 pounds of payload, so the 14,000 pound reduction represents a 29% to 32% reduction in payload. I imagine that many loads are not at the maximum allowable to have the total vehicle weight not exceed 80,000 but, as best I can find, most intermediate and long haul loads do exceed the approximately 30,000 pounds available in the Tesla truck.

As far as weight is concerned, the Tesla truck with a sufficiently sized battery to achieve a 500 mile range is at a significant disadvantage. I'm sure there are applications where this disadvantage would not be relevant, but the average over the road trucker would be severely disadvantaged. For the 500 mile range version, I believe that significant advances in battery technology will be necessary.

Next time: cost.

There are many significant considerations with respect to the viability of the Tesla Truck:

- Cost (both initial purchase and lifetime, including maintenance)
- Range
- Time to recharge
- The effect of the weight of the battery on the load that can be transported

And, of course, there is interplay between these items. For example, to achieve longer range, larger battery packs are needed. This will add significant cost and charging time and, due to the added weight of the larger pack, will reduce the payload that can be carried.

Image Credit: Tesla |

Starting with the drag, to a reasonable degree of accuracy, the drag force is ~D=1/2\,C_{D}\,A\rho\,{v}^{2}~ where D is the drag, C

_{D}is the vehicle's drag coefficient, A is the frontal area presented to the relative wind, ~\rho\,~ is the density of air, and ~v~ is the vehicle's speed. From a youtube video presentation, Musk states that the drag coefficient is 0.36, which is very low for such a vehicle. I haven't found the frontal area of the vehicle, but various sources seem to indicate that a reasonable estimate is about 10.75 m

^{2}. We'll use 1.2 kg/m

^{2}for air density and convert 65 m.p.h. to 29.06 m/s. Plugging, this yields a drag force of 1,961 Nt (440.8 pounds).

For rolling resistance, we'll look at four driven wheels, two steering wheels, and four free rolling tires on the tractor and eight free rolling tires on the trailer. I'll give Tesla the benefit of the lowest rolling resistance tires that I've been able to find. Then, to first order, rolling resistance is dependent only on the coefficient of rolling resistance of the tires and the normal force (weight) on those tires.

While the actual rolling resistance will depend on how much weight is on which tires (because the rolling resistance will vary) and I won't know how much weight can be in the trailer until I've determined battery weight. I'll assume tires with "state of the art" low coefficient of rolling resistance (C

_{rr}of an average of 0.0056. And I'll assume that each of the tires carries an equal load so that the rolling resistance, R, is determined by ~R=C_{rr}w~, where ~C_{rr}~ is the coefficient of rolling resistance and w is the vehicle weight. Converting 80,000 pounds to 355,858 Nt and plugging and chugging, we find that the approximate rolling resistance is 1,993 Nt.

Thus, the approximate force that the motor must apply to the pavement through the drive train and tires is 3,954 Nt. And, since force times distance is work (and energy), the motor must do 3,954 joules of work (that is, supply 3.954 joules of energy) to move the truck one meter at 65 m.p.h. And, since there are 1609.3 meters in a mile, the motor must do 6.36317*10

^{6}joules of work. The battery system must supply enough energy to do this work, and must supply more, given that the motor/drive train combination is not 100% efficient. If we assume 85% overall efficiency, the battery system must supply 7.486*10

^{6}joules/mile. This is 2.079 kilowatt hours. Note that Tesla says "less than 2 kWh/mile." I'm sticking with my number but it shows that my estimates can't be too far off.

In order to travel 500 miles on a level road in good conditions with no starts and stops at 65 m.p.h., Tesla will need a battery system that can supply 2.079*500 or 1,039 kilowatt hours, or 1.039 megawatt hours. If we use a number of 140 watt hours/kilogram for specific energy of a Li ion battery, such a battery pack would weigh 7,421 kg, or 16,360 pounds.

This is as ideal as it can possibly be. The truck will climb hills and, though some of the potential energy paid for in kilowatt hours can be recovered coming downhill and even energy normally wasted by braking as the truck rolls downhill will be partially recovered by a regenerative system. Nevertheless, there will be waste associated with climbing and descending. Similar considerations apply to stops and starts for traffic, stoplights, stop signs, meals, etc.

Therefore, we need a "fudge factor" for the various starts and stops, accelerations, etc. While I've seen numbers such as 90% bandied about for how much of the kinetic and potential energy in the Tesla truck can be recaptured by regenerative braking, my experience in the Lexus CT 200h makes that number seem very high. I've calculated that, in my CT 200h, about 39% of the potential energy from a hill descent went into the batteries. But I'll be generous and speculate that Tesla is much more efficient at 75%.

Suppose that the truck does the equivalent of stopping 100 times in 500 miles. An 80,000 pound vehicle travelling at 65 m.p.h. has a kinetic energy of 1.537*10^7 joules. Losing 25% of this number 100 times wastes 3.830*10^8 joules, or 106.4 kWh. Adding this to the 1,039 kWh we find that the battery pack must supply 1,145 kWh or 1.145 mWh. This will require a battery pack weighing 8179 kg or 18,031 pounds. Call it 18,000.

We now need to determine how much the battery pack weight will reduce the payload that can be hauled. To compare apples to apples, we'll figure that a very fuel efficient diesel powered semi tractor gets about 7 m.p.g. and thus will use about 71 gallons of diesel fuel weighing about 490 pounds (numbers for the weight of a gallon of diesel fuel are all over the place, but I think this represents a good average). The engine and transmission might weigh something like 3,500 pounds. The total of the materials not needed in the Tesla truck (diesel fuel, engine, transmission) is about 4,000 pounds. Thus, the available payload for the Tesla is some 14,000 pounds less than that of the diesel powered semi. And this understates the issue since we've removed the internal combustion engine and transmission, but the electric motors weigh something!

A typical heavy hauling semi tractor trailer can legally haul somewhere around 44,000 to 48,000 pounds of payload, so the 14,000 pound reduction represents a 29% to 32% reduction in payload. I imagine that many loads are not at the maximum allowable to have the total vehicle weight not exceed 80,000 but, as best I can find, most intermediate and long haul loads do exceed the approximately 30,000 pounds available in the Tesla truck.

As far as weight is concerned, the Tesla truck with a sufficiently sized battery to achieve a 500 mile range is at a significant disadvantage. I'm sure there are applications where this disadvantage would not be relevant, but the average over the road trucker would be severely disadvantaged. For the 500 mile range version, I believe that significant advances in battery technology will be necessary.

Next time: cost.

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