this post was submitted on 15 Aug 2026
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[โ€“] ChickenLadyLovesLife@lemmy.world 55 points 6 days ago* (last edited 6 days ago) (37 children)

This seemed intuitively wrong to me (like, way too low a cost), but: 25,000 pounds moving 100 mph is equal to 11,331,007 J of kinetic energy. Since 3.6 million J equals 1 kWh and 1 kWh on average costs $0.17, that means you could accelerate 25,000 pounds to reasonable bare minimum flying speed for about fifty cents (not considering efficiency of the machinery). My mind still can't process this, but math is math.

On the other hand, looking at it from a potential energy perspective it's a bit more expensive. 25,000 pounds at a cruising altitude of 10,000 ft. (still quite low from an airliner perspective) is about 339 million J, 94 kWh or about $16 -- the cost of lunch at MacDonald's.

Since a plane requires the most thrust at takeoff, you could use ground-based catapults to get the plane to takeoff speed (or faster even) and then you could carry smaller batteries and propelling machinery. For extra fun, you could have landing planes snag a wire and use their momentum to accelerate a plane taking off.

To save even more weight, since you're going airport-to-airport you could leave off the landing gear and just have the planes come down on a bouncy trampoline-like surface. If you think that's batshit crazy, the British actually experimented with this idea for their aircraft carriers in the 1950s.

Edit: to make these numbers more realistic I'm going to assume something like a 737, which can weigh something like 150,000 pounds fully loaded (this includes fuel but you'd need batteries instead for an electric plane). Getting this to a 150 mph takeoff speed would take about 100 million J (getting it then to a cruising speed of 500 mph would be another 233 million J, but that's pretty minor compared to the other costs). Climbing this plane to 30,000 ft would take 6.1 billion J. Resisting a drag force of 5000 pounds (about what a 737 experiences at cruising speed at 30,000 ft) for 500 miles (the distance from Cleveland to New York City) would need 17.6 billion J. Assuming landing is free (fuck TANSTAAFL) that means a typical trip needs 23.8 billion J or 6618 kWh or $1125. Assuming a real-world efficiency of 25% means the actual cost would be $4500 (which is in the ballpark of what jet fuel costs). Assuming 200 passengers, that's $22.50 per person. Not exactly "$5 of electricity" but surprisingly small.

Feel free to check my math, my brain hurts.

[โ€“] Natanael@infosec.pub 2 points 6 days ago

It will be too hard to time landings and take offs at small ports, but if you replace this with a massive flywheel you can have gearing to both spin it up on landings and draw from it to launch.

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