Rough Estimate of Energy Used to Climb + Implications for M3 Efficiency

[ForeverFree]

53 miles Sherman Oaks to Mt. Waterman (+6000 ft) = 22 kWh

53 miles Mt. Waterman to Sherman Oaks (-6000 ft) = 2 kWh

Implies 53 miles flat would be about 12 kWh

Implies roughly 10 kWh to gain 6000 ft = Roughly 1.67 kWh per 1000 ft

Overall energy use was 227 Wh/mi ... or roughly 4.5 mi/Wh

Implies roughly 1.67 x 4.5 or about 7 extra range miles to climb 1000 ft ... or about 7 recouped when descending 1000 ft

Because regen isn’t perfectly efficient, actual numbers may be more like 8 to climb and 6 upon descent​

These jibe reasonably well with calculations I did on Model S years ago along Sherwin Grade near Mammoth



Putting those numbers to use today:

We used 375 range miles to go 297 miles Sherman Oaks to Mammoth Lakes (7000 ft gain)

Back out 55 miles for the gain (at 8 kWh/1000 ft), and you’ve got 320 RM used to go 297 actual miles.

Used 140 RM to go 100 miles Lone Pine —> Mammoth Lakes.

Back out 30 - 35 RM for this 4000 ft climb, and you’ve got 105 - 110 RM to cover 100 actual miles.

This implies a flatland equivalent use of 255 Wh/mi at our autopilot speeds of 72- 74 mph. Compares with about 300 in our Model S. Impressive.

Five-sixths the energy usage —> sixth-fifths the efficiency. 20% more range per kWh.

​

Where this really becomes impressive is that ...

The 3 seems able to pull as many kWh Supercharging as the S.

That means that it adds miles 20% faster.

And, for us, moving from an 85D (265 Mile range after three years) to a 3LR (at least 310), one can add more miles at this faster rate before succumbing to the charging taper.

​

The practical difference:

No Mojave charging stop necessary to complete the uphill, headwind 200-mile run to Lone Pine (arrived with 78 RM).

Much shorter stop in Lone Pine. No walk to McDonalds, eat, wait around, and then walk back. Stay at Supercharger, eat sandwich, and return to the road. Arrive in Mammoth with 70 miles still in tank.

​

Very impressed with Model 3 as road trip car.

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[swaltner]

Regen isn't 100% efficient - if your descent is steep enough, regeneration will kick in. That would shift it so your average usage on flat ground would be closer to 22 kWh than 2 kWh. While I don't know the end adjustment that would be required, that shifts the rest of your info slightly.

To calculate the raw energy required to raise a Model 3 up 1000', you can slightly modify the calculations I ran at Thought Experiment - Full Recharge Going Down Imaginary Steep Hill and then corrected at Thought Experiment - Full Recharge Going Down Imaginary Steep Hill for a Model S.

For a Model 3, the empty weight I found was 3,549 to 3,838 lbs. Take the larger value (long range battery), add a few pounds for a driver and stuff and set the weight of the vehicle at 4,078 pounds. That converts to 1,850 kg. Also 1,000' converts to 305 meters. That yields:

305 m * 1850 kg * 9.8 m/s^2 = 5,530,000 joules
5,530,000 joules / 3,600,000 joules/kWh = 1.47 kWh

So, there's an added 1.47 kWh in potential energy for a Model 3 that is 1,000' higher than another one. Add in a little bit of inefficiency for the electric drivetrain and you're probably somewhere right around 1.6 kWh. It's somewhere between the perfect no-friction physics calculation of 1.47 kWh and the 1.67 kWh value you came up with from the assumption that the difference between up and down was only attributable to rise and fall. All in all very close to what you came up with.

 

[ForeverFree]

Regen isn't 100% efficient - if your descent is steep enough, regeneration will kick in. That would shift it so your average usage on flat ground would be closer to 22 kWh than 2 kWh. While I don't know the end adjustment that would be required, that shifts the rest of your info slightly.

To calculate the raw energy required to raise a Model 3 up 1000', you can slightly modify the calculations I ran at Thought Experiment - Full Recharge Going Down Imaginary Steep Hill and then corrected at Thought Experiment - Full Recharge Going Down Imaginary Steep Hill for a Model S.

For a Model 3, the empty weight I found was 3,549 to 3,838 lbs. Take the larger value (long range battery), add a few pounds for a driver and stuff and set the weight of the vehicle at 4,078 pounds. That converts to 1,850 kg. Also 1,000' converts to 305 meters. That yields:

305 m * 1850 kg * 9.8 m/s^2 = 5,530,000 joules
5,530,000 joules / 3,600,000 joules/kWh = 1.47 kWh

So, there's an added 1.47 kWh in potential energy for a Model 3 that is 1,000' higher than another one. Add in a little bit of inefficiency for the electric drivetrain and you're probably somewhere right around 1.6 kWh. It's somewhere between the perfect no-friction physics calculation of 1.47 kWh and the 1.67 kWh value you came up with from the assumption that the difference between up and down was only attributable to rise and fall. All in all very close to what you came up with.

Very interesting that the physics potential energy calculation jibes with what I observed. (And, my observations were probably not good beyond 1 significant digit, because trip meter usage is rounded to the nearest kWh.)

Given the inefficiencies, probably something on the order of a 1.7 kWh/mi “tax” going up and 1.3 or 1.4 recouped on the way down? What do you think?