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High Altitude Operation

How would the Tesla Model S perform at high altitudes such as Pike's Peak which is over 14,000 ft? Would oxygen, temperature, or pressure affect how it runs?

From what I have read naturally aspirated gas engines loose 3% of HP for every 1,000 ft. So at 14,000 ft that is a loss of around 42%.

A gas engine with 416 HP like the Tesla would instead have 241 HP.

A gas engine would have to have 744 HP sea level to have 416 HP atop Pike's Peak.

The Tesla will perform even better at 14,000' than at sea level: less aero drag. Motor output should be the same at any altitude.

Lithium ion batteries do not interact with the atmosphere in any significant way. I think they and the motor would suffer no loss of performance at high altitudes. That being said, I wonder if operating at low pressure might have an effect on the longevity of the battery cells. If the cells are 'air tight', they would try to expand when the ambient pressure dropped. This would put mechanical stress on the casing of the cell. My best guess is that the battery might eventually get damaged if the car was regularly (say every day) driven up to high altitudes and back.

It might not have to be driven to high altitudes and back. For example, the highest incorporated city in the US is Leadville, Colorado, it is at 10,000 ft. It is relatively flat up there, so the car would be driven and stored at those altitudes. Plasma televisions do not work up there either as they are internally pressurized. Maybe next year we'll see some testing at Pike's Peak, many car manufacturers all around the world test there. I'd love to try it, the cars I have driven there starve except for the turbocharged cars I've seen.

It may be OK going up there: On Panasonic's web site, docs say the battery has a pressure release valve. Not sure if you can come down, without squishing the batteries.

http://www.panasonic.com/industrial/includes/pdf/Panasonic_LiIon_Overvie...

Given that the 18650 is a small, round pressure vessel, I don't think a 5ish PSI imbalance at 14,000 feet would be a bother at all. SteveZ is likely on to something as the aero drag will lead to better efficiency.

In contrast, a flat, pressurized panel would in fact have some issues with a pressure imbalance due to it's large unsupported surface area.

I will be taking my Model S to Tahoe in spring and look forward to the extra power over Echo Summit.

The Colorado contingent is planning a Model S road rally to the summit of Mt Evans next year after the snow melts. For those not familiar, the road to the top of Mt Evans is the highest paved road on the continent, at 14,265'. We'll let you know how it goes.

http://www.teslamotors.com/forum/forums/model-s-mt-evans-road-rally

http://www.mountevans.com/

Better yet, take your S on a road trip and join us...

I wonder if the thinner air affects cooling; liquid cooling also must have air flow to dump into via a radiator.

Normally, thinner air does not pull heat away from electronics as well, so anything that requires cooling will suffer. However, since the batteries are liquid cooled/heated, the system will compensate as long as it's not overstressed.

As I said, the heat taken from the electronics liquidly must end up in the air via a radiator. Thin air slows that last stage for both ICE and BEV.

The air is usually colder up high so the radiator would be more effective and compensate for thinner air. So the net effect while driving up could be a wash?

Drop the tire pressure by about 14 pounds, beef up the battery pack and brake system to withstand vacuum, change to radiant cooling pannels, and harden the display screens. Should be able to drive on Mars. Perhaps not long, but a bit.

They said you could have it delivered anywhere that made you happy.

Seriously though, only electric cars and electronic robotic vehicles have ever been driven off-world. Pikes Peak should not present an issue.

I don't believe that small pressure change has any impact on drivetrain itself (drivetrain probably still works well in conditions where humans would die due lack of oxygen). Only tiny impact I can see is the cooling, but even that would not be a big deal.

Lithium ion batteries can't interact with air directly, it would destroy the battery (lithium is alkali metal, it goes boom with contact to water, so moisture inside the battery is a bad bad thing).

Batteries are liquid-cooled, so pressure for batteries remains pretty constant (water pressures are in completely different scale than air). Electric motors do not depend on air in any way. Wires don't care about tiny pressure changes.

So the only effect that low air pressure has is in in heat transfer from radiators, but since temperature ranges where that works are between -30C and something like +50C I don't think it does matter.

Conclusion: you don't see any impact on performance between sea level and top of the Mount Everest.

Only thing I've got is this: would if affect tire pressure?

Vawikus

Yes, it does. When we normaly talk about pressure, we are usualy referring to "gague pressure". A tire that you measure at 35 PSI at sea level is at 35 PSI gague, and at 50 PSI absolute over vacuum because air pressure at sea level is about 15 PSI absolute.

But it's pretty much unsaid that we usualy refer to gague. Divers, pilots, and mountineers need to be able to convert between pressure and gague pressure, but most people dont consider this much day to day.

Roughy, 14,000 feet is about 10 PSI absolute. If you fill your tire up at 35 PSI at sealevel, and measure again you'll find it reads at 40 PSI on Pike's peak. Come back down to sea level and the tire gague reads 35 again.

So yes, the tires will be a bit overinflated, but by about 5 pounds. If you bleed them at altitude, you'll need to refill when you get to the flats or you'll be 5 PSI low.

Look for a full report next spring or early summer from us Mt. Evans Road Rally participants. But Stevez has already experienced 14,000+ feet driving up Trail Ridge Road in Rocky Mountain National Park.

RE: air pressure in tires, yes they will increase in pressure slightly at higher altitude, but I don't see them being a concern with performance at 10-14k feet above sea level, mainly because there are no straight roads. Everything is pretty much twisty, windy roads with a bunch of sharp switchbacks; so you are not going too fast at these elevations. If you do go too fast you will not be around long, as we don't have many, if any, guardrails along these roads. It's our way of thinning the herd.

Besides, I ride by bicycle to the top of Mt. Evans 3-4 times per year, and never notice a difference in handling as I descend. Amazingly, I can descend on a bike faster than folks do in their cars, primarily since I can get through the switchback much faster.

If interested in the Mt. Evans Road Rally, keep an eye on the Colorado forum group threads. Once the highway opens for the season, we will set a date for the event.

Where I live there are lots of mountain roads. I was wondering: how much range will I loose if I drive over a mountain road with a 1000(3280 ft)meter ascent and 1000 meter descent compared to driving the same distance flat?

I've read that the Roadster has a battery to wheele efficiency of 88%. Assuming that the model S has the same efficiency and that the same efficiency applies to regenerative braking, I made a simple calculation:

The work bringing the 2108 kg vehicle 1000 m up is:
W=mgh= 2108 kg x 9,81 m/s^2 x 1000 m = 20676 kJ = 5,7443 kWh.

That means you must pull 5,7443/0,88 = 6,5276 kWh from the battery driving up.
Going down you regain 5,7443 x 0,88 = 5,0550 kWh.

If this is right you loose 1,4726 kWh. 200 Wh/km(332 Wh/mile) means you only loose 7,4 km(4,6 miles).

Does this make sense? Are my assumptions reasonable?

@Knut - can't say for sure how efficient the regen is, but real world results elsewhere in the forums talking about going over the Altamont Pass (which admittedly is only 300m) suggest that your calculations are probably fairly close.

Regen is way lossier than that, and moreover is limited to kW 60 top on the MS. You're far off. The 1000' rise costs you about 7 mi., of which you'd regain/regen about 3 on the way down.

@KnutNorway | DECEMBER 2, 2012: I was wondering: how much range will I loose if I drive over a mountain road with a 1000(3280 ft)meter ascent and 1000 meter descent compared to driving the same distance flat?

Check out the Green Race website at http://www.jurassictest.ch/GR/ where you can enter your actual route and it will tell you the range of your vehicle.

@Brian H, I don't thing regen is lossier than that, you just need to remember that there are other losses where this needs to be added. You lose far more than 6,5276 total going up and going down you gain that 5,7443 back (minus losses) until you actually go to regen which is actually rather strong deceleration, just to keep the speed uses 15-20kW, so you need to slow down at 15-20kW before you even go to regen. After that you start to gain that .88, not before.

Timo is correct. It's not that the regen is lossier, it's that the rest of the system uses power also, through air drag, rolling resistance, etc.

Brian/Timo: what I tried to do was to remove the 15-20 kW from the equation by comparing my mountain trip with driving the same distance flat. You are right, I would not actually have 5,0550 kWh more on the battery after a 1000 m decline. But if my decline is 20 km long I would expect to have 5,0550 kWh more on the battery than if I had driven 20 km flat. Are you with me?

This is of course very theoretical. In practice I think it would be difficult to avoid using the brake pedal on steep down hillls with narrow 180deg bends. Also I think the efficiency driving up and down would be lower than when driving flat. The internal resistance in the battery eats a bigger part of the cake when the power output/input is larger.

Alex K: thanks!

I'd say you end up with 5,7443 kWh more if you do it that gradually. Potential energy has to go somewhere and only thing where it can go is the car. In that case you just press the accelerator less than normal which should give you pretty much exact potential energy difference in battery.

I think this is the only case where losses don't take much anything away from the gain/loss unless the drivetrain itself has more losses with less/more load.


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