Since air density is proportional to the square of the elevation this can lead to significant efficiency gains. Believe it or not, partly as a result of this, the SR-71 had it’s best mpg at peak speeds.
Since having lower air density also means you need a higher lift coefficient (angle of attack) to produce the required lift, and then you have more lift-induced drag (which goes with the square of the lift coefficient). I think air density more or less washes out when it comes to its impact on range. That being said, you cover the full vehicle range at a higher velocity at higher altitudes, so it certainly seems like there would be significant benefit from a travel-time perspective.
All that being said, there are significant high voltage insulation challenges at higher altitudes, which is something we are working on.
The issue is a certain aircraft has an optimum cruise altitude. If you try to fly fast at low altitude, it'll be horrendously inefficient. If you try to fly higher, you'll often be beyond the maximum lift coefficient so you'll be less efficient or you'll stall.
To first order, efficiency is independent of cruise velocity.
The range for an electric aircraft (this is basic physics) is: Range = (battery specific energy) * efficiency * (L/D) * (mass_battery/mass_total)/gravity.
Altitude and air density and velocity do not directly figure into the calculation as you pick your cruise altitude to maximize your (L/D). And maximum L/D depends somewhat loosely on Reynolds number (which, granted, does depend on speed) and especially Mach Number. If you can keep totally subsonic flow (i.e. usually up to about Mach 0.5), your maximum (L/D) doesn't directly depend on speed.
Sailplanes increase their speed (at optimal glide ratio) by putting on ballast. You can achieve the same effect by cruising at higher altitudes.*
Just using the drag polar approach and neglecting second-order effects (assume negligible dependence on Re, sufficiently subsonic so negligible impact of M, and linear lift coefficient region aka no stall), we get the following (I'm skipping a lot of intermediary steps):
Cd = Cd0 + k*Cl^2 -> Cd0 is the parasitic drag coefficient -> k is the lift-induced drag coefficient -> Cd is the overall drag coefficient
Range is maximized when Cd0 = k*Cl^2 (parasitic drag = lift-induced drag) -> Cl is a function of speed: since the required lift is constant, more speed = less Cl required = less lift-induced drag -> maximum L/D is achieved at this range-optimal speed
This speed can be calculated exactly from the total weight (W), air density (rho), lifting area (S), and drag coefficients:
range-optimal speed = sqrt((2*W/(rho*S))*sqrt(k/Cd0))
As long as you always operate at this range-optimal speed (aka speed for maximum L/D) which is a function of air density (and therefore altitude), the equation for range reduces significantly:
R = endurance*velocity, where endurance = battery energy / drag power, and we know the equation for drag power...
Simplifies to:
R = E*eta/(2*sqrt(Cd0*k)*W) -> R is range -> E is battery energy -> eta is total system efficiency
Dimensionally, this equation is of course the same as yours, with an energy being divided by a force to get a distance. The key point I am trying to make is that if you just look at that equation with no context, speed and air density are not present anywhere. But what is hidden in the assumptions is that you are assuming that you are operating at the maximum L/D speed given the air density at any particular altitude. Going back to my other comment, range at the range-optimal speed does not depend on air density or velocity directly, but lower air density at higher altitudes will result in a higher range-optimal speed, and hence less travel time for a given range.
This is not really true for any other transportation method. Cars and buses and boats and even trains have an efficient vs speed trade off especially at higher speeds.
And there is an efficiency advantage of speed in that you can get by with just a cramped seat because your trip time is short, a few hours. A similar trip in a conventional train, cruise ship, zeppelin, or sailboat may require bringing along basically a small apartment (or “sleeper car”) which is much heavier and can destroy the efficiency advantage you might have otherwise had. And the same vehicle can be used many more times for the same route if its speed is much greater, which (combined with the lower vehicle weight per person) reduces the effective embodied emissions of the vehicle per passenger mile significantly.
What are these challenges? How does having a near vacuum cause trouble with ~1kV potentials?
I can't go into much detail, but we are working on addressing this in a couple different ways in our insulation system design.
Like I said in the other comment, if the plane is operating at the range-optimal speed, I think the air density does not impact the range capability (it cancels out) but it does increase the range-optimal speed, allowing for faster travel.
A car for example doesn’t need to produce lift, but it still displaces air which causes drag. The same is true of an aircrafts fuselage, which is generally not used to generate lift but still increases total drag. https://en.wikipedia.org/wiki/Parasitic_drag
Also, an aircraft is generally designed so that at cruse speed and altitude the wing incidence angle provides appropriate lift. https://en.wikipedia.org/wiki/Angle_of_incidence_(aerodynami.... Which means at optimal curse distance the cabin would almost perfectly level independent of optimal cruse speed or altitude.
If you go through the analysis, the air density drops out of the range equation if you assume are operating at the range-optimal speed (which is higher at lower air densities).
Not quite true -- range-optimal speed is where the sum of those terms is minimal. With some assumptions, this is where the derivative is 0, dDrag/dv = 0, and since derivative is linear, this means: the range-optimal speed is where the (infinitesimal) increase of parasitic drag (with speed) is equal to the decrease of lift-induced drag (in other words, opposite derivatives).
D = A*v^2 + B/v^2 (D is total drag, first term is parasitic drag, second term is lift-induced drag)
dD/dv = 0 where v = (B/A)^(1/4)
Plug in v = (B/A)^(1/4)
D = sqrt(AB) + sqrt(AB), aka dD/dv = 0 exactly when parasitic drag is equal to the lift-induced drag