this post was submitted on 15 Aug 2026
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[–] ChickenLadyLovesLife@lemmy.world 26 points 5 hours ago* (last edited 1 hour ago) (4 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.

[–] aesthelete@lemmy.world 5 points 59 minutes ago (1 children)

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.

Don't you dare talk about catapulting using electric technologies in America though. Steam only! 🇺🇸🗽🦅🏈

[–] ChickenLadyLovesLife@lemmy.world 1 points 38 minutes ago

I didn't say which type of catapult. I don't need ICE showing up at my door.

[–] AlteredEgo@lemmy.ml 2 points 56 minutes ago (1 children)

Since a plane requires the most thrust at takeoff, you could use ground-based

Actually dragging (or wheeling in) a wire or having a wire car supplying the plane with electricity during takeoff would work too.

I think the only realistic use case is going to be short trips. For long range offsetting the carbon for jet fuel just makes more sense.

But really small personal vehicles could be interesting. There is the Pivotal BlackFly which can VTOL and uses less electricity than a big electric car - and it needs to roads. So for commuting this could actually work to save on infrastructure.

[–] ChickenLadyLovesLife@lemmy.world 2 points 35 minutes ago

Actually dragging (or wheeling in) a wire or having a wire car supplying the plane with electricity during takeoff would work too.

After doing more of the math, I realized that the energetic cost of takeoff is quite a small fraction of the overall cost. So the only real benefit of the catapult would be to reduce the size and weight of the propulsive machinery on the plane. So externally providing just the electricity wouldn't be much of a benefit.

[–] Abyssian@lemmy.world 2 points 2 hours ago

For extra fun, you could have landing planes snag a wire and use their momentum to accelerate a plane taking off.

What about Flintstones style breaks, where everyone's legs stick out under the plane and they need to use them to stop?

Sure, it wouldn't be effective, but one or two of these new flights being on the news and global emissions would be down even farther than with your plan.

[–] Neocorporation@lemmy.world 7 points 4 hours ago (1 children)

I fucking hate everyone and would love to subject you fucks to 4g of pain taking off with a stupid catapult system. Nice math.

[–] Abyssian@lemmy.world 1 points 2 hours ago

All those screaming kids would get a quick education on how relatively nice everything was before the plane was launched.

"Billy, why are you crying? Do I have to take you on another plane ride? Oh, you're gonna cry harder now? That's it, I'm getting the vomit bags. I got this nice new child design one for you that wraps around your head and ties closed at your neck. Won't that be a treat?"

[–] captain_aggravated@sh.itjust.works 11 points 5 hours ago (4 children)

Every time I hear of an all electric aircraft of any size, I always wonder what they're going to do about landing weight.

Every modern transport category jet has a higher takeoff weight than landing weight, because of the simple unavoidable fact that landings are rougher than takeoffs. Taking off, the load gradually comes off of the landing gear, on landing it's suddenly applied. Jets burn tons, literally tons, of fuel enroute, so they're considerably lighter on approach. It's why aircraft have dump valves to jettison fuel overboard in case of forced landing early in the flight.

Batteries don't get lighter as they're discharged, so...?

[–] dgriffith@aussie.zone 2 points 46 minutes ago

Batteries don’t get lighter as they’re discharged, so…?

JETTISON ZE PACKS!! PREPARE FOR LANDING!!!

[–] teuto@lemmy.teuto.icu 3 points 2 hours ago (1 children)

The same way we make bigger aircraft with higher landing weights: beefier landing gear and structural reinforcement. Airliners don't have a lower max landing weight than takeoff because they have to, they do it because it's cheaper and more efficient. Add some more structural weight and you lose some payload, but if the efficiency gains from fuel cost savings make it worthwhile, then manufacturers will make them.

[–] captain_aggravated@sh.itjust.works 1 points 2 hours ago (2 children)

You'll hit a point where the payload is so poor it isn't worth operating.

[–] scratchee@feddit.uk 2 points 1 hour ago

That’s true regardless, batteries will never power intercontinental wide-bodies (short some major new developments in battery design). This will probably reduce the maximum aircraft size where batteries remain viable, but they are obviously very viable for short hops in small jets, the point where they cannot compete is somewhere but it’s not “never”, even with this limitation.

That all said, maybe someone will explore just dropping batteries along the way? It sounds ridiculous, but it also sounds like something we have the tech to solve…

[–] teuto@lemmy.teuto.icu 1 points 1 hour ago

It's a bigger deal on bigger aircraft, but on mid-size narrow bodies it's not as much of a problem. A 737, depending on the model, only has a difference of about 20-30k lbs between MTOW and MLGW and no fuel dump capability. An overweight landing isn't really a big deal in one, just a quick maintenance inspection. Closing the gap so MTOW and MLGW are equal is doable. It's hard to overstate how huge of an expense fuel is to an airline, if they have to lose some passengers and cargo they would absolutely do it if it got rid of the fuel expense.

Of course all that is contingent on having batteries with enough energy density to get somewhere close to current MTOWs while having something of a useful range.

[–] yes_this_time@lemmy.world 7 points 5 hours ago* (last edited 5 hours ago) (2 children)

You can rethink the engineering with electric motors.

The planes you are describing are designed assuming they will be lighter on landing because of fuel. So why design them for take off weight?

Electric motors are condusive of blown wing design for example, and would have a unique landing profile. (The plane can land at much slower velocity)

Edit: yeah they've moved the engine shroud which allows for lower speeds. Given the same runway you would be able to trim vertical speed.

You can rethink the engineering with electric motors.

One of the other limitations is that you're stuck with propellers if you're using electric motors, so your top speed is going to be significantly slower than jets. Unless you do the Tu-95 thing with contra-rotating props whose tips exceed the speed of sound, and then you have monstrous noise problems.

[–] captain_aggravated@sh.itjust.works 5 points 4 hours ago (1 children)

That ain't gonna happen on a civilian airliner.

Blown wings are basically powered lift. You're planning on bringing a civilian passenger plane down final approach at a speed it can't glide at if the power plant fails?

I could see that for a carrier based aircraft where STOL is a factor but no you're not doing that in airline operations.

[–] yes_this_time@lemmy.world 1 points 3 hours ago (1 children)

Why couldn't it actually be safer since you could have distributed power centers, across multiple motors?

You could also have hybrid approaches - there is space between not being able to glide and smashing your landing.

I'm just getting at it being a different system so some old assumptions can be reexamined.

Bigger challenge than landing is energy density.

Regardless it's a very interesting space.

Multiple redundant motors are heavy.

[–] jaschen306@sh.itjust.works 3 points 5 hours ago (1 children)

Fun fact, batteries DO become lighter when they discharge. But obviously not like fuel. But it's still a fun fact.

[–] ammonium@lemmy.world 1 points 4 hours ago

You mean because of e=mc²? That's true but basically unmeasurable. Air batteries do get mesurable heavier.

[–] melsaskca@lemmy.ca 51 points 10 hours ago (2 children)

Is there no simple anymore? Plane A went this far on 5 dollars electricity. Plane B went the same distance on X dollars worth of jet fuel. I want to know the distance travelled and I want to know what "X" is.

[–] betanumerus@lemmy.ca 11 points 7 hours ago* (last edited 7 hours ago) (1 children)

$5 for 30 minutes in the air. pick any speed you want. it doesn't matter. avgas and jet fuel don't compete with $5.

[–] mirshafie@europe.pub 14 points 7 hours ago (2 children)

I pick 30 minutes at Mach 7, here's your $5 Canadian and please get out of my way, I do intend to board now.

[–] UnderpantsWeevil@lemmy.world 5 points 6 hours ago (3 children)

I pick 30 minutes at Mach 7

How long do you want to spend at Mach 7 relative to stopping and starting? Because that could be a lot of G-force.

[–] fartographer@lemmy.world 1 points 2 hours ago

I want people to find my face at the starting runway, and my skull in some field 7 miles away from where the plane landed.

[–] supersquirrel@lemmy.ca 7 points 6 hours ago (1 children)

Meh just accelerate with F-force and then switch to G-force when you are up to speed and vice versa when slowing down, problem solved!

[–] UnderpantsWeevil@lemmy.world 5 points 6 hours ago

Using N-force, which is the little can of Nox under my seat

[–] betanumerus@lemmy.ca 4 points 7 hours ago

i definitely won't be in your way. bring back pictures.

[–] xthexder@l.sw0.com 16 points 8 hours ago* (last edited 8 hours ago) (2 children)

Well, the data we do have is that the flight was just under 30 minutes and very likely under 100 miles.

From the Wikipedia on airline fuel efficiency:

The worst-performing flights are short trips of from 500 to 1500 kilometers because the fuel used for takeoff is relatively large compared to the amount expended in the cruise segment, and because less fuel-efficient regional jets are typically used on shorter flights.

In the example values table, the most efficient plane for a 560km trip burns 0.92 kg of fuel per km, so doing some rough math and assuming the electric plane travelled 100 miles, that would be roughly 148kg of fuel, or 50 gallons (190L).

At current jet fuel prices ( $3.76/gallon ) that's about $188 US in jet fuel as a rough estimate. It's unclear if the test flight went up to full altitude or if the plane was at full weight, so a fair comparison might have used even half as much jet fuel.

Edit: From some of the other comments, it seems like they might have only considered flight time as cruising time, not takeoff and landing, so my numbers will be quite far off if that's the case. My gut feeling is that this is probably the case, because this seems like too big a difference otherwise.

[–] UnderpantsWeevil@lemmy.world 7 points 6 hours ago* (last edited 6 hours ago)

$5 of electricity to lift an airplane 10,000 feet definitely seems low.

If you want to be really smug, you could say a gas powered plane requires $0 of fuel to glide for 30 minutes

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[–] RememberTheApollo_@lemmy.world 16 points 10 hours ago* (last edited 10 hours ago) (3 children)

I did my best to find any technical data about the flight. Couldn’t find any actual numbers. FWIW it’s not intended to be a standalone method of powering the aircraft for commercial use; they plan on making it a hybrid, which makes far more sense as far as range and payload are concerned. Best guess a 25000 lb aircraft like this will probably cruise around 120-150Kt at a nice, slow, efficient airspeed for a test like this. So maybe a 40-50 mile flight because “air time” probably started as soon as they lifted off.

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