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I did my first 100+ mile round trip today, with a hefty (20 mph steady with gusts higher) wind out of the south.

First leg of the trip was heading south at 70 mph, I was seeing 29 to 31 KW at a steady 70 mph on level ground, about 40 miles of the trip was south at 70 mph and then headed east at 55 for 10 miles, and then another 10 miles at 40 mph. Total for the first leg of the trip was 61.1 miles with an average of 2.9 miles per kWh (20.6 kWh consumed)

Second leg of the trip was the reverse of the southbound leg, started out going west and then turned North with a nice tail wind, but winds had died down a bit, closer to 15 mph with gusts higher. On this leg in the exact same areas where I was seeing 29 to 31 KW at a steady 70 mph I was now seeing 17 to 20 KW. Trip North was 60.4 miles with average consumption Northbound of 4.0 miles per kWh (15.1 kWh consumed).

Average them both together and you get 121.6 miles and 35.8 kWh consumed for 3.4 miles per kWh average consumption for the round trip.

I know from reading here and my own experience that a head wind hurts more than a tail wind helps, so other trips to the same location in calm weather should use less energy.

Later,

Keith

PS: each leg of the trip took about an hour, and was 60ish miles... so average speed 60ish mph.
 

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a head wind hurts more than a tail wind helps.
Without having any facts to prove it, I think that's true for the Bolt.

Because: headwind causes additional aerodynamic drag. Which at same speed puts more load (kw's required) on Bolt's power package. Which in turn results in more of the battery's energy getting dissipated as battery discharging loss, inverter loss, motor loss.

So that the battery-to-road wheel energy loss is not linear with speed. For example at constant 30 mph battery to wheel loss might hypothetically be 8%, whereas at 60 mph it could be 10% and at 90 mph it could be 13%.

Would be nice to have test data indicating battery to wheel losses associated with Bolt's speed and load.
 

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Not just the Bolt. The physics at work apply to birds, bicycles, cars, etc.

https://physics.stackexchange.com/questions/137144/the-effect-of-windspeed-on-a-car
Side note: I believe it's worse for objects in the air. For the sake of simplicity, let's assume a steady and direct 30 MPH wind exactly along the path of travel.

A Bolt travelling at 60 MPH for 60 miles only spends an hour in the headwind and will be in the tailwind for an hour on the return. But the Bolt will be moving at 90 MPH through the wind in one direction and at 30 MPH on the return. As aero drag is non-linear, the harm of going upwind will outweigh the benefit of going downwind.

However, a plane or bird will take two hours to fly upwind, and 40 mins to fly back. The upwind portion adds a full hour to their travel, but the downwind only saves 20 mins. So the round trip would take 2:40 instead of 2:00.

Even if the wind was perfectly perpendicular to the road, you'd end up running against a ~67 MPH wind in both directions.. THANKS PYTHAGORAS!!!

There is no beneficial wind. Unless you can always fly west to east in or near the jetstream :) Or you're sailing...
 

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The headwind effect is actually worse for anything that flies. Even a wind coming from behind, but off to the side has a negative effect for air "vehicles" because energy must be expended to correct the course (IE, you get blown sideways and have to correct). So for airplanes, it's not uncommon to have a "headwind" in both directions, even if the wind direction doesn't change direction! The wind direction needs to be within 45deg of your course to actually aid forward progress
 

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Aerodynamic drag varies with the square of the air speed, so a relatively small difference in air speed can make a big difference in drag. For example, 70 mph versus 60 mph is a 17% difference in air speed, but it's a 36% difference in aerodynamic drag. Of course, there are other factors that contribute to overall efficiency (including rolling resistance and drivetrain losses), but a big chunk of the total is aerodynamic losses. With a CV of 0.32, the Bolt is definitely not among the slipperiest cars out there.
 

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Enzeder, LooseBolt, GlenandhisBolt1. All absolutely correct.

When I set aside my curiosity about power package @ load loss. And look again at the equation for aerodynamic drag* and blow some Bolt numbers through it. It’s clear that aerodynamic drag increases at an increasing rate as speed (or wind speed) increases. Just like everyone has said.

So if you’re driving along at 60 mph going directly into a 10 mph headwind: aero force** increases by perhaps 36% (as stated earlier) as opposed to no wind. Then you turn around and go directly downwind (tail wind) at the same 60 mph, all other things equal: aero force decreases by about 30% as opposed to no wind. So net-net you lose. Versus having no wind at all and going the same total distance at 60 mph.

I’d still like to see though, whether Bolt’s battery to wheel energy efficiency is affected by motor load. And by how much at various speeds & loads.
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*- the general “drag equation” is ½*Cd*A*ρ*v^2
v – speed in metres per second
Cd – drag coefficient. 0.308 is latest published for Bolt EV. (0.32 was prelim design target).
A – frontal area of Bolt in square metres. (I’m currently using 2.4)
ρ – density (weight) of air in kg. per cubic metre.

** energy required to overcome aerodynamic force is only one of the major things determining Bolt’s energy usage. Others being: rolling resistance, temperature, mass (weight), battery-to-wheel losses, elevation change (hills), altitude, etc.
 

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I have been bitten by range problems due to head winds in both Tesla's and Bolt's - it really really sucks to lose range to head winds - it's an EV killer and one of the details IMHO that ultimately prevents EV adoption until we have bigger batteries with range that is more than sufficient to absorb losses due to environmental factors...

there is just too much planning required to drive distance with an EV...
 
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