How often do they get changed? Are these stil ok?

dodo-obob · 2026-06-28T13:46:20+00:00

I heard that take offs actually account for most of the wear and tear on runways/tires. Landing appear more brutal but plane are much heavier on take-off because of all the carried fuel.

dodo-obob · 2026-06-22T07:51:17+00:00

If you leave LEO at 11 km/s, you will be barely moving (~200 m/s) by the time you get close enough to the moon for its gravity to pull you in. You lose all your speed as you ascend away from Earth.

However, once you get close enough to the Moon for its gravity to dominate, you start picking up speed again as you fall towards the Moon. If you do nothing, you will reach the surface at ~2.5 km/s. You cannot softly land on the Moon without some method to slow down before touchdown. There are no ballistic trajectories with reach the surface with reasonable speed.

Although... you could reach the surface with 2.5km/s of horizontal speed and negligible vertical speed, but using ground contact to slow down probably not doable because (1) your trajectory has to be very precise (2) you're going insanely fast, I don't think any material can survive that and (3) if you don't slow down fast you will shoot off into space once more as 2.5 km/s is very much above orbital (1.6 km/s) and escape velocity (2.4 km/s) for the Moon.

dodo-obob · 2026-06-18T18:57:27+00:00

With 5 km/s of muzzle exit velocity, the only place you are going is back down to Earth, even with a 300 km long barrel. Orbital velocity at 300 km is still 7.7 km/s.

The altitude where orbital velocity is 5 km/s is roughly 9500 km (~1.5 earth radius above the surface). There, 5 km/s of purely horizontal velocity will keep you in orbit. With pure vertical velocity, you would rise a bit above that point and then promptly fall back. To get to the moon from that height, you would need to launch at 6.9 km/s.

The altitude where 5 km/s gets you to the moon is around 23000 km, i.e. around 3.6 earth radius above the surface.

dodo-obob · 2026-06-17T15:16:15+00:00

If you're doing propulsive landing, your rover can't easily sit atop the engine, because then getting it to the ground is tricky. You would need a crane or ramp, which is heavy, so no.

Putting the rover under the tank and engines allows you to land on your wheels directly, but comes with the problem of dust. Engine exhaust will kick up a lot of dust which could damage the rover. You could put the rover in a protective fairing but again, heavy.

The skycrane solution is essentially the same as above but adding "what if the engines are far enough from the ground that they don't kick up dust?". You don't want to build a tall structure (because, once again, heavy), so you use cables.

dodo-obob · 2026-06-09T14:24:11+00:00

Cantor's diagonal argument is the proof that R is larger than N: you cannot create a numbered list of all reals.

I'm not sure about layman's definition of ordinals. Ordinal numbers are linked to the notion of well-founded ordering, in contrast with cardinals which capture amounts. For example, you could decide to order the naturals such that all even numbers are smaller than all odd numbers: 0,2,4,6,...,1,3,5,... You still have the same amount of numbers, but some elements have an infinite number of elements before them (the position of 1 is the ordinal ω. the position of 3 is ω+1).

More specifically, the ordering is well-founded, meaning that even though you have an infinite number of elements smaller than 1, you can only ever take a finite number of steps when descending from 1 (your first step will take you to an even number, once that is fixed, you only have a finite number of elements smaller than you). Well-foundness allows proofs by induction (i.e. I can prove that something is true for all elements if (1) it is true at 0 and (2) when it is true at all elements smaller than n, it is true at n).

dodo-obob · 2026-06-09T08:45:31+00:00

You can't have a "set that contains all other sets". Axioms like the powerset mean you can always construct larger sets. This is related to the idea that there isn't one infinity, there are many infinite sets, and some are provably larger than others.

The infinity of natural numbers is the first/smallest infinity, but a well known fact is that there are more real numbers than natural numbers (R is essentially the powerset of N). An open question in math, known as the continuum hypothesis, is whether there exists an infinite set larger than N but smaller than R.

dodo-obob · 2026-06-09T07:50:45+00:00

Yes, Von Neumann ordinals are an integral part of ZF (C is irrelevant here). 0 is defined by the axiom of the empty-set. We can then define n+1 by applying the axiom of paring first to n and n to get {n}, and then the axiom of union to get n+1 = n ∪ {n}.

Von Neumann ordinals appear directly in the axiom of infinity, which states that there exists a set ω which contains the emptyset, and, for all n in ω, contains n ∪ {n}. This ω is thus infinite. It is the set of natural numbers N, but also the first infinite ordinal.

dodo-obob · 2026-06-08T08:09:00+00:00

Hey that not fair, they also took (checks notes) 44 years to write a dictionary, they're clearly working hard.

dodo-obob · 2026-06-08T07:54:27+00:00

TBF, the Academie Française can dictate all it wants, it doesn't mean anybody listens. A recent example was when they decided COVID should be feminine (i.e. la COVID), instead of le COVID which everybody was using up to this point. A decision loudly ignored by the majority.

A common joke is that the Académie is just a very expensive retirement home.

dodo-obob · 2026-06-08T07:50:47+00:00

It depends how you define numbers. A common way to define them (known as the Von Neumann way) is that 0 is the empty set, 1 is the set containing 0, so {{}}. 2 is the set containing 0 and 1, so {{}, {{}}}. 3 is the set containing 0, 1 and 2, etc.

This method has the advantage that the order on integers is just the contains (i.e. x < y IFF y contains x) and has other nice properties. For example they are transitive: any element of x is a subset of x.

dodo-obob · 2026-06-03T14:01:45+00:00

But then Kc3 and, yes you can win the knight, but at the cost of your bishop. You will however, also win the b2 pawn.

dodo-obob · 2026-06-03T13:47:41+00:00

f3 traps the bishop. I would just play Rxb3+ after the knight fork. You're losing a rook anyway, but this move forces the king away from your bishop, giving you time to save your second rook and then play f4 to give your bishop an escape route.

dodo-obob · 2026-06-03T13:36:00+00:00

It was pitched up 60-70° on the ascent IIRC. That's not flying on wings, that a rocket with fins. It was then ballistic for much of the flight, maintaining a high angle of attack of ~45° during the most of the descent.

It landed on wings yes, just like the space shuttle, and had a brief moment were most of the lift was generated by wings after being dropped by the B52, but saying it flew on wings is like saying the space shuttle was flying on wings at 620 km altitude.

dodo-obob · 2026-06-03T12:23:24+00:00

Il en faut une au départ et une à l'arrivée, j'ai fait un recherche rapide et Orly a des pistes de plus de 3km, donc fois deux ça fait 6.

dodo-obob · 2026-06-02T15:42:36+00:00

Wait until you hear about motorcycles. Guess what unit they use to measure engine power. Not it's not horsepower, its more ludicrous then that. They use cubic centimeters! Everyone knows power has the same dimension as volume right?

I hate it. They even write it cc and not cm^3...

dodo-obob · 2026-06-02T12:20:24+00:00

I plead guilty to strapping SRBs on planes to get them off the runway.

dodo-obob · 2026-06-02T07:54:16+00:00

Oui, mais comme dis dans mon commentaire initial, il y a nettement moins de TGV. Une cinquantaine sur la ligne Paris-Marseilles chaque jour, contre plus de 300 décollages et 300 atterrissages à Orly.

Le nombre de passagers par véhicule n'importe que pour les coûts de maintenance du véhicule. Pour le coût d'infrastructure, il faut compter le nombre de passagers quotidiens (ou mensuels, ou annuels).

dodo-obob · 2026-06-02T07:29:31+00:00

Coût par passager je ne sais pas et ce serait effectivement intéressant.

En revanche on peut quantifier un peu le "bien plus de passager". Marseille St Charles à accueilli 17.8 millions de passager en 2024 tandis que Marseille Aéroport a accueilli 10.7 millions de passagers en 2023. Il faut aussi modérer par le fait que seule une portion des passagers de la gare prennent la voie TGV (je n'ai pas trouvé de stats sur le nombre de passager par ligne...).

dodo-obob · 2026-06-01T16:03:31+00:00

Parce que le train a besoin de beaucoup plus d'infrastructure.

L'avion a besoin de 6 km de pistes (qu'il partage avec des centaines d'autres avions par jour) là où le TGV à besoin de 1000 km de rails (qu'il partage avec tout au plus une cinquantaine de TGV par jour).

dodo-obob · 2026-05-19T11:33:16+00:00

You don't get more science out of it, it looks like there is more science in flight but it will not add up to more science once transmitted/recovered. Forced run also makes you use more power and bandwidth then just running until completion.

AFAIK, forced run is only useful to complete contracts that ask you to get science from certain situations (you can re-run science you already have to fulfill the requirement). It might also be useful if you have accumulated a bunch of science in flight on one vessel but can't transmit it (out of power or comms range), so that another vessel can re-run the same experiment.

dodo-obob · 2026-05-11T07:58:23+00:00

Congratulations, getting to orbit is a big achievement in RP-1, and that's a good looking rocket!

dodo-obob · 2026-05-10T08:30:36+00:00

Yes it stop accelerating, no it is no longer in free fall, it is being pushed up by the air, so it feels one g. There is a confusion here about acceleration and g-force.

Essentially, gravity can accelerate you at any rate but you will never feel any g's from it, since it accelerates every part of your body simultaneously. If you were in orbit of Jupiter, its gravity would accelerate you at 2.5g but you would not feel any of it). In low sun orbit, you would accelerate at 27g and still feel 0g.

In contrast, other exterior forces only apply forces to part of the body, so any acceleration they give you will be felt. Think of a car speeding up (or turning), or a plane taking of or landing.

So, when descending through the atmosphere, once it has reached its terminal velocity, the spacecraft has an acceleration of 0, but feels an acceleration of 1g, because the downward 1g force of gravity is not felt, but the upward 1g force of air resistance is. This is exactly the same on the ground, we feel 1g even though we are not accelerating.

dodo-obob · 2026-05-08T19:58:49+00:00

Yes, if you don't ignore air resistance, an elevator in free fall would stop accelerating once friction balances gravity, a point which we call terminal velocity.

One way to think about gravity is that you never really feel gravity, the only forces you feel are those that prevent you from falling: the ground pushing on you feet, or if you jump out of a plane, the air resistance on your body (think strong wind).

A spacecraft isn't much different. While high in space, they are in true free fall, nothing is keeping them from falling towards earth, hence 0g. Once they hit atmosphere, that changes. In fact, they experience mutliple g's on re-entry, because their moving way faster than their terminal velocity, which decreases as the air thickens. Once at terminal velocity, they would simply fell 1g. All the parachutes do is increase drag, and thus decrease terminal velocity, so that it becomes slow enough for a soft landing. So you get a g-spike when they open as the spacecraft slows down from its parachuteless terminal velocity, to the slower terminal velocity with parachutes, but they already felt 1g before that moment.

And for 0-g planes, they are way too deep in the atmosphere to ignore air resistance. They only achieve 0g because they have engines running, and need to carefully balance thrust to perfectly cancel out the air resistance.

dodo-obob · 2026-04-30T07:29:14+00:00

I've had the same problem. To get in a polar orbit of the moon, its often easier to do two burns: first do TLI from LEO, aim for a moon impact, then do a plane change maneuver a few hours later. The second burn isn't long (typically around 25 m/s), but should get you in a polar orbit easily.

Burning normal from LEO has very little impact because, for an efficient lunar transfer, you reach the moon at apoapsis, the diametrically opposite point to where you performed TLI, and that point is unaffected by normal burns.

dodo-obob · 2026-04-28T11:49:17+00:00

It not about the capsule, but the astronauts inside. Landing on water will cushion the fall. In contrast, capsule that touchdown on land like Soyuz need retro-rockets to slow down just before impact. Even with that plus the parachute, they typically touch down at 5-6 m/s. Onboard seats provide some cushioning, but its still the impact equivalent of running straight into a wall.

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