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chuq

[New Car] Tesla Cybertruck

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I'm surprised this thread didn't already exist 🙂

Obviously they are not in the wild, so can't get precise driving/charging model info.  However ABRP includes the Tesla Roadster which is in the same situation so I'm assuming estimates can be made.

We don't have a lot of the data which is required for the analytical model as mentioned in the blog post.  What is the workaround? 

Models (with EPA estimated range, in case it helps):

  • Single motor - 250 mi / 400 km EPA est
  • Dual motor - 300 mi / 480 km EPA est
  • Tri motor - 500 mi / 800 km EPA est.

We don't have info on the batteries, other than the Tri motor is a double stacked version of the single motor.

We probably can't even estimate the mass due to the unusual materials used.  Motor1 says "our educated guess is that the Cybertruck is some 1,000-1,500 pounds (450-680 kilograms) heavier [than the F150] but this is just an assumption as we don't have exact and official details."  F150 is 4400 to 5300 lb.

Front area is probably the easiest to calculate: 78.8 inches wide x 75.0 inches high = 41.04 sq ft / 3.813 sq m (not allowing for the gap under the car - subtract about 14% if this area is to be excluded)

Charging curve - assume the same as the Model 3?

Anyone else have any insights?

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This is great stuff! I'm about to disappear for a week on Thanksgiving (Road tripping to Big Bend National Park in my Bolt, should be an adventure!), when I get back I'll implement (unless @Bo (ABRP) gets there first).  

For charging curve, I suspect we can multiply the charge curve of the Model 3 by the capacity ratio, and then cap performance at ~250kW since that's the max capability of V3 Supercharging.  Will toy with that a little bit too when I get back.

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Thanks @Jason (ABRP)!  I've shared links to this thread on reddit /r/teslamotors as well as the Now You Know discord channel.  Any idea which of the unknown parameters are the most important to find?

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If you can find/calculate the frontal area then that would be the only other missing component for the drag equation.  I don't know what system abrp uses to calculate aero losses over time but if it uses the drag equation then that would be the only other thing you would need.

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Also, the drag coefficient has got to be way lower than .48, because the ram 1500 has a drag coefficient of .36 with no aerodynamic back cover like the cybertruck.

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7 hours ago, Arial said:

Also, the drag coefficient has got to be way lower than .48, because the ram 1500 has a drag coefficient of .36 with no aerodynamic back cover like the cybertruck.

I found another source ( https://youtu.be/7741YdnAcR4 ) that gave a figure of between 0.39 and 0.4, so that is closer than what you said!

 

 

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5 hours ago, chuq said:

I found another source ( https://youtu.be/7741YdnAcR4 ) that gave a figure of between 0.39 and 0.4, so that is closer than what you said!

 

 

Ran a simulation overnight on a crude model of the truck, got a Cd of .363 unknown.png

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