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by mliben 1943 days ago
We are definitely saying the same thing in very different ways.

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.

1 comments

Oh, yes, precisely. That’s one thing about air travel that people don’t really understand. They think fast = inefficient, but as long as you’re supersonic, speed is roughly independent of efficiency. You can have you cake and eat it, too!

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.