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California to Virginia in my S60! Advice/tips? Specifically owners near I-70 & I-64

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One note: the ideal is to arrive at a Supercharger with 1 mile of range left. As you get closer to a Supercharger, you can increase your speed to use up your buffer. I like to think of my buffer as a %, not absolute miles. If you're 20 miles from the SC and have 40 miles of range left, you can go as fast as you'd like; but the same 20 mile buffer would make me nervous if the SC is 120 miles away.

The rule of thumb to minimize travel time is to drive as fast (in MPH) as you can charge (in miles-of-range/hour) at the next stop. Because Superchargers charge faster than you can drive, at low state of charge, you should drive as fast as you like once you're sure you have enough juice in the cells.
 
This is a strategy that works in short, rolling hills. On long mountain grades, it will use more energy. On long mountain grades, a constant and identical speed up and down hills will use the least energy for a given total travel time.

While a little OT, can you elaborate? The idea of going slower up and fast down is to take the edge off the high power draw going up thus limiting ohmic losses and then on the downhill section make up the speed loss to get the same average speed on the trip. Why would that be different on longer hills compared to small ones?
 
While a little OT, can you elaborate? The idea of going slower up and fast down is to take the edge off the high power draw going up thus limiting ohmic losses and then on the downhill section make up the speed loss to get the same average speed on the trip. Why would that be different on longer hills compared to small ones?

OK, here is a short set of facts without a lot of explanation:

Ohmic losses are small relative to aerodynamic losses at normal speeds and elevations. For short steep hills, the losses to regen down a hill and power up a hill (10% in and 10% out of the battery) can be larger than the aerodynamic losses coasting through the bottom of the hill and letting the speed increase. On long downhills, you will have more energy in the battery at the bottom of the hill than if you go a more moderated speed and add energy to the battery than to let the speed increase to match the aerodynamic force with the gravitational force times the downward slope of the hill. Because aerodynamic drag is a square law force (cube law power needed), it is a non-conservative force, and you minimize total energy used by traveling the minimum constant speed to cover the distance in the time allotted.

I have a LOT of mountain, hypermile driving experience and can tell you that these facts correlate strongly with those experiences.
 
OK, here is a short set of facts without a lot of explanation:

Ohmic losses are small relative to aerodynamic losses at normal speeds and elevations. For short steep hills, the losses to regen down a hill and power up a hill (10% in and 10% out of the battery) can be larger than the aerodynamic losses coasting through the bottom of the hill and letting the speed increase. On long downhills, you will have more energy in the battery at the bottom of the hill than if you go a more moderated speed and add energy to the battery than to let the speed increase to match the aerodynamic force with the gravitational force times the downward slope of the hill. Because aerodynamic drag is a square law force (cube law power needed), it is a non-conservative force, and you minimize total energy used by traveling the minimum constant speed to cover the distance in the time allotted.

I have a LOT of mountain, hypermile driving experience and can tell you that these facts correlate strongly with those experiences.

Actually that makes sense. A long downhill section is definitely better driven slower and getting some regen than going faster and losing more in the wind drag.
 
Actually that makes sense. A long downhill section is definitely better driven slower and getting some regen than going faster and losing more in the wind drag.

Since I was the one giving this advice, let me first start by saying I agree completely with Cottonwood.

With that being said, if you want to average 60mph, then driving 55mph uphill and 65mph downhill will result in a significant increase in range. But not as much range as driving uphill at 50mph and downhill at 50mph.

That's how the advice was given. It is ALWAYS more efficient to allow gravity to take you downhill at the slowest speed bearable while storing as much energy as possible.

But . . . I have allowed gravity to work it's magic on me between Silverthorne and Denver. I think gravity will accelerate you to over 90mph . . . but I can't be totally sure:wink:
 
A twist on the uphill/downhill discussion...for true passes, traffic often slows down on the uphill by a decent amount, but traffic won't speed up on the downhill by that same amount. So, a modified 'cottonwood theorem' of slowing down on the uphill [with the flow of traffic-ish, so it won't feel like you're slowing down out of the ordinary] but resuming your flatland speed-ish on the downhill will provide a balance between drive time and power consumption.

Another tip I heard for maximizing energy use is to run the hvac on recirculation as much as possible. Especially in extreme cold temps, the system is doing less work because its heating previously heated cabin air instead of constantly heating fresh, new, frigid outside air. Same goes for cooling, though apparently to a lesser degree. FWIW, I dont have any personal data to confirm/deny this method's validity, but it sounds good. :tongue:
 
This discussion is somewhat academic. It's always better to drive more slowly. If I drive 45 mph uphill and 45 mph downhill, capturing regen power, then it's better than 55mph. At some hill length/speed increase it's more energy efficient to coast and increase speed than it is to regen because there are regen losses. At some increase in speed it's more efficient to regen because drag is more of a penalty than regen losses.

Define what you're optimizing and then the math can be done. For example, I would pose the question: "For 10 mph speed increments from 40-70mph, what increase in speed on downhill sections would be equivalent to the energy losses from regen to keep my speed constant?"
 
... drag is more of a penalty than regen losses.

I see this mentioned a lot here. Regen is not 100% efficient. In fact only 15-20% of the kinetic energy can be recaptured. But the mistake that many people do is conclude since it is not very efficient it should be avoided. That's is a false logic. Any alternative to regen (except going slower in the first place) is giving you zero energy back, while regen gives at least something back. So it is irrelevant what the efficiency is on regen, it is always better. Using regen over friction brakes is better. Using regen over high speed downhill in neutral is better.
 
I see this mentioned a lot here. Regen is not 100% efficient. In fact only 15-20% of the kinetic energy can be recaptured. But the mistake that many people do is conclude since it is not very efficient it should be avoided. That's is a false logic. Any alternative to regen (except going slower in the first place) is giving you zero energy back, while regen gives at least something back. So it is irrelevant what the efficiency is on regen, it is always better. Using regen over friction brakes is better. Using regen over high speed downhill in neutral is better.

I think we're overall agreeing here. But I have issues with your 15-20% number for recapture. Tesla has indicated that the recapture efficiency, ignoring losses due to drag and rolling resistance at speed is 80%. So, if I'm traveling at 70 mph, and reduce my speed to 50 mph, I will recapture 80% of the kinetic energy difference less the energy losses to drag and rolling resistance over the period of time I engage regen. The full discussion is from 7 years ago but the math is still valid.

The Magic of Tesla Roadster Regenerative Braking | Blog | Tesla Motors

We have to look at Wh/mi. Is it more efficient, cresting a hill at 55mph, to coast down the hill and pick up speed (going faster at no energy cost) or to regen for a constant 55mph with regen (going the same speed for "negative" energy cost). That optimization problem depends on the length and slope of the hill vs the regen efficiency.
 
My physics isn't what it used to be, so I'm not going to provide an answer, but let me try to frame the problem with bounds that make it answerable.

Suppose there is a 10 km stretch of road. The first 5 km descends at a uniform 6% grade, and the second 5 km ascends at a uniform 6% grade. (6% is the maximum grade on the U.S. Interstates at speeds above 60 mph.) What is the most energy efficient way to drive this stretch on a windless day while achieving an average speed of 100 kph?