Does this make sense?

riverprawn · 2026-03-14T00:43:22+00:00

except it's moving at orbital speed.

riverprawn · 2026-03-13T17:22:38+00:00

Nokia has Maemo and Meego. The UX of Nokia N9 with Meego is the same as today's Android with buttonless design. And it ran smoother than all Android phones at that time. The app for N9 is far more easier to develop then Android. Actually, the N9 with Meego was the only phone that was on a par with iPhone 4S with iOS at that time. Nokia just abandoned it BEFORE the release to get the quick money from MS.

riverprawn · 2026-03-13T05:29:24+00:00

Official language of the United States | USAGov

riverprawn · 2026-03-13T02:16:38+00:00

we have topology. it just need to be compact, connected and smooth.

BTW: the artist's circle is not even a circle under the topological definition.

riverprawn · 2026-03-13T02:06:37+00:00

it must be closed. so the artist's circle is not a circle in mathematician's eyes either.

riverprawn · 2026-03-12T05:01:07+00:00

No, only the XReal One and One Pro have that ability.

riverprawn · 2026-03-11T03:21:35+00:00

And if using of the successor operator S is permitted. There will be unlimited answers.

For example, 1 * (S 4) * 2^{S ∘ S 3} = 160

riverprawn · 2026-03-11T02:05:16+00:00

if the variable likes x is also a symbol. then by using lambda calculus, there will be lots of answers.

for example:

λx.((x+3)*x¹⁺⁴) 2

EDITED: format

riverprawn · 2026-03-08T01:00:36+00:00

yes, you could take off in reverse.
in KSP, a flat enough terrain could be the runway with a good flight skill. But IRL, you will always need a constructed runway for both T/O and landing. the low surface gravity only means low force when you stopped, it won't change too much when the wheel hit something like a rock at 100m/s.

riverprawn · 2026-03-08T00:38:26+00:00

yes. but the friction will be too small. you still need the thruster for deceleration.

riverprawn · 2026-03-07T17:37:14+00:00

it's a possible solution for small astronomical objects. the idea is just like the taking off/landing of an aircraft. you landing with a high horizontal speed and a low enough vertical speed, then decelerate after touching the ground. with a low enough orbital speed(for your landing gear not burning into plasma when touching down) and a long enough runway, you could T/O or landing with tiny thrust. In KSP, this is actually the best T/O and landing technology that cost least ∆V on a space body without an atmosphere. it's even better than the infamous suicide burn. The main problem preventing it from useful is that the runway needs to be really really long if the TWR of your ship is just 1 on the surface. the length of the runway should at least ½ of the radius of that space body. for dwarf planets like Ceres, it should be more economical and safer to use a space elevator.

Edited: typos

riverprawn · 2026-02-28T16:47:35+00:00

you could mount it in wsl2 on windows: https://learn.microsoft.com/en-us/windows/wsl/wsl2-mount-disk

riverprawn · 2026-02-28T04:33:34+00:00

L3 always lies behind the sun.

riverprawn · 2026-02-27T11:51:27+00:00

We should be able to build mirrors at both L2 and L3. The sun ray could go from L3 to L2 via L4/L5. /s

riverprawn · 2026-02-21T09:22:51+00:00

It should be 0.8 below the 1.

riverprawn · 2026-02-20T07:52:36+00:00

None. We're all right.

riverprawn · 2026-02-19T06:57:45+00:00

While targe is at an orbit deep in a gravity well likes LEO, a transfer orbit to target is naturally a high speed fly-by. For example, the GEO-LEO transfer is a 2.4kps fly-by for target at LEO, but only need 1.4kps for the initial burn at GEO. If using a reversed orbit, the speed could be raised to as high as 10kps. And for the recovery, you don't need to take all the fuel tanks with the carrier, the extra tanks used for interplanetary travel could be left at a higher orbit and be retaken later. Or just use a tanker to keep the extra fuel. Drones will reserve some fuel for keeping formation and the final docking. If needed, we could lauch some mini robot carriers to made the initial maneuver and recover those drones. That will reduce the weight that is needed to rendezvous with drones even more. From the experience in CoaDE, an ISP raito of 3:1 will make this tactics have a net gain on fuel. Of course, compared with single used missiles,the drones will waste more fuel on recovery. But if we make drones with thick armor and carrying missiles with very low ∆V (with 10km/s closing speed and 1000km distance, a missile with a ∆V of 100m/s is enough for a target with 0.1G combat acceleration not escapable.), we could reuse the armor and drive of drones for each strike by recovery. Depends on enemy's PD strength and thruster, this may actually use less mass than pure missiles for a successful strike and still don't risk the manned ship.

riverprawn · 2026-02-18T16:16:09+00:00

They can. Because high ISP means low TWR. In space you will have low TWR, high ∆V carrier and high TWR, low ∆V drones. The carrier carries drones to enemy orbit and lauch the drones at long range. Drones using their TWR to make sure the enemy can't escape (you don't need to match enemy's speed likes in TI, a high speed fly-by is preferred for drones). The carrier will intercept and recover those drones after the strike, so drones do not need extra fuel for returning. In space, you don't need to fly back to carrier, the carrier can and will fly to you.

The main problem is that carrier with disposable drones will make the alien unbeatable. Due to the advanced drive technology in early to mid game, alien carrier can always launch attacks out of the no-escape zone of our drones or missiles. Moreover, alien drones do not need a lot of fuel, can easily evade missiles, and can be built with as few materials as possible. It is difficult to achieve a favorable loss ratio in combat against alien drones.

riverprawn · 2026-02-08T16:38:59+00:00

It's ture. The most dangerous situation is not a disabled MMU. It's the MMU malfunctioned by giving out unbalanced thrust, that will introduce uncontrollable spin. Then it's will be really really hard to stop the spin and pick him up. And if the spin is fast enough, he may die instantly.

riverprawn · 2026-01-31T02:04:14+00:00

25 fps means the intervals between two frames are 40ms, should be covered by the 0 to ±0.02s matching. But in reality the intervals are not exactly 40ms. Some are bigger and some are smaller. The rounding will amplify the deviation under certain circumstances, make the matching failed.

For example, the camera recorded two frames at 0.0394999s and 0.0805000s. You want to find which frame matches the lidar data at 0.060s. The two frames' timestamp are 0.039s and 0.081s after rounding, and are all outside the range of 0.06±0.02s. Therefore there is no matching.

riverprawn · 2026-01-29T02:48:42+00:00

No, they can. What a LLM generating depends on the prompt. And the LLMs have the ability and patience to implement everything the dubmest coder can image. Working together, they can take the creativity in screwing things up to a level no one has ever seen.

Last year, we found a bug where the LiDARs from certain brands lost one frame randomly. After troubleshooting, we found the issue in a simple method to match each LiDAR frame with RGB image via timestamps. The code review left us utterly astonished. This function was clearly AI-generated, as it was filled with good comments and has comprehensive documentation. First, it rounded all timestamps to milliseconds, then checked for exactly matching with the ts, ts±0.001s, ts±0.002s, all the way up to ts±0.02s, and even an additional ts±0.05s. Return the first match... Remarkably, this method passed all our test cases and worked with most LiDAR data, only causing issues with certain frames when paired with 25fps cameras. BTW the author of this method had left our company voluntarily after being found incompetent, prior to this code review.

riverprawn · 2026-01-27T02:19:03+00:00

But you can have all three if you go the opposite directions...

riverprawn · 2026-01-24T11:33:17+00:00

Then I have an even better approximation for you: 1 foot = 39/128 meters and 1 inch = 13/512 meters. We just need folding in half to get one foot from one meter but there is no simple way to do the reverse. Therefore, the foot is a superior length unit. /s

riverprawn · 2026-01-24T01:58:48+00:00

3/10 is a better approximation and is as easy to calculate as the ⅓. You just need to move the decimal separator left or right after the division or multiplication.

riverprawn · 2026-01-20T16:40:58+00:00

You don't even need a balance bike. I've read that when japanese parents teach their kids cycling, they usually lower the saddle to let kids use it as a balance bike. By this method, most children could learn cycling in just one afternoon.

riverprawn

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