sevenperforce

There were obviously going to be comparisons. It looks like NG has a dedicated actuated-fin section independent of the interstage? Seems like extra weight. Falcon 9 uses grid fins, which require less torque to operate and are useable at a broader range of regimes and thus can be stuffed inside the ordinary interstage.

Right, it's not a regular tetrahedron. It does, however, describe what I believe is a single mirrored sphenoid that grows and shrinks periodically. This appears to have been beaten to death now, haha.

I spent entirely too long playing around with the Alt-12 menu, haha. Here's a 1986 article by Draim but unfortunately I can only see the first page: https://arc.aiaa.org/doi/pdf/10.2514/3.20244 Of note: "Throughout the 1970s, it was generally thought that the minimum number of satellites required to give continuous global coverage (including the polar regions) was five. A credible proof of this minimum number...is valid only for circular-orbit constellations. "In 1984, the author presented the concept of a four-satellite elliptical-orbit constellation giving continuous global coverage. It was based on a three-satellite elliptic-orbit continuous hemispherical coverage constellation of 78 h or longer, with a fourth circuit orbit satellite having a period one-half that of the other three. "The new four-satellite constellation described in this paper uses common period orbits and can maintain continuous global coverage at approximately one-half the altitude of the earlier four-satellite constellation. [T]he orbital periods, inclinations, and eccentricities of the satellites in the new constellation are identical." And here's a 2012 article which DOES have full text: https://www.researchgate.net/publication/286375838_Common-period_four-satellite_continuous_global_coverage_constellations_revisited "Draim’s analysis of the optimal, four-satellite, continuous single coverage satellite constellation is based on a tetrahedron formed by four planes–each plane contains three of the four satellites in the constellation (see Fig. 1). The basic requirement is for the planes of the tetrahedron to always encompass the Earth, without ever intersecting it, as the tetrahedron changes shape or warps during the constellation repeat groundtrack period." Bingo. There it is. Here's the design: This is essentially the same thing I proposed upthread -- two pairs of satellites, each describing a line segment which orbits the planet while remaining at right angles to the corresponding line segment. He describes it: "Both pairs of satellites drift to the West. One pair of satellites moves in a clockwise fashion. The other pair moves in a counterclockwise fashion."

I wonder if the uplook camera has enough dynamic range to set a low long exposure and really get a good skyview as well a surface lit by starlight....

Tangentially, I wonder what the BE-4 uses for startups. I believe the BE-3 uses a set of sacrificial solid igniters that have to be replaced after each flight.

Yep, two extra cores would be the same stress as about four and a half SRBs, so there wouldn't be any problems on launch. There might be issues before burnout, though. The SRBs have a nominal burn time of 90 seconds, while an additional Vulcan core at full throttle the whole way would burn for much, much longer. I don't know what the propellant load of the first stage is but I'm guessing that the gee-loading on the core at side-booster burnout would be ridiculous.

Ooooooh, fascinating! If the center of the triangle maintains an essentially constant orbit, you could absolutely make that triangle bigger, pull it closer to the planet, and then have a corresponding satellite on the opposite end, creating a triangular pyramid. Not a regular tetrahedron but a stretched one. Total, nearly equidistant coverage.

Someone on Facebook made an edit of one of the 360 images from Perseverance's navcam to include the sky. This is not the actual sky as seen from the surface of Mars but it is really cool nonetheless. https://www.facebook.com/360creator/posts/751879702136518

Tory comments (not very convincingly) on a true Vulcan Heavy render: Of course I had to make an observation:

There's more of a difference between "crazy" and "so crazy it just might work" than people think.

So I think I have a partial solution. I hopped into KSP and played around with a couple of satellites for a bit to get an idea of how the line segment connecting two satellites (let's call them A and B) with equal period and inclination rotates in 3-space. I drew the lines in Paint for simplicity. Take a look: As you can see, the line segment (let's call it AB) "tumbles" around the planet, returning to the exact same position with each circuit. Its midpoint essentially follows an orbit equidistant between the two other orbits but at a slightly lower inclination and it rotates 360 degrees with every orbit, lengthening as it approaches its AN/DN and contracting as it approaches its highest and lowest latitudes. Trivially, we can place an equivalent pair of satellites (let's call them C and D) in an orbit such that their corresponding line segment CD is always at a right angle to AB. The two line segments are equal in length at two points in the orbit; otherwise one is always smaller while the other is always larger. The four satellites ABCD thus continuously form a tumbling irregular tetrahedron with an insphere centered on the planet, which oscillates between two different tetrahedral isometries that I believe are a mirrored sphenoid and digonal disphenoid. While this solution is not a rotating regular tetrahedron as requested by the OP, it is a tetrahedron which rotates and oscillates non-chaotically around the planet and can provide 100% coverage of the surface.

That's anyone's guess.

More on this... Cost Breakdown Two-launch campaign A (single_reuse) Two-launch campaign B (expendable) Costs One F9 booster F9 booster recovery hardware Two F9 upper stages Prop for two launches Two launch operations Booster recovery operation Booster refurb operation Two F9 boosters Two F9 upper stages Prop for two launches Two launch operations Revenue 1x price for 5.5 tonne payload to GTO 1x price for 8.3 tonne payload to GTO 2x price for 8.3 tonne payload to GTO That's a lot of numbers. Fortunately for us, we don't have to know most of these numbers directly because all we have to look at is the arithmetic difference in the profits between the two types of operations; lots of things can just cancel out. First, let's talk about revenue. SpaceX prices its reusable GTO launch at $62M but doesn't say what its expendable launch price is. Elon has thrown numbers around but I don't find them particularly reliable, especially because SpaceX can price in significantly lower than the competition, which muddies the water a little. It's very rare that the entire expendable payload to GTO is actually needed. However, ULA has dial-a-rocket options on their Rocket Builder, so we can take a look at that. Remember, we only need to look at the difference in revenue, not the gross revenue. RocketBuilder removed its actual pricing values a couple of years ago, but when it launched (no pun intended) it priced an Atlas V 401 at $109M and an Atlas V 551 at $155M. The site says that you'll need one SRB for a 5.5-tonne payload to GTO and five (plus the 5-meter fairing) for an 8.3-tonne payload to GTO. Setting aside the cost of the different fairing, we can estimate that the market price charged to customers for those five SRBs is $46M or $9.2M each. A little bit of math, and we can estimate that SpaceX could conceivably charge $36.8M more for an 8.3-tonne launch than for a 5.5-tonne launch. So that's going to be the difference in revenue between Campaign A and Campaign B. Now, on to costs. Most of the things cancel out: upper stages, propellants, launch operations. You can also cancel the price of the booster for Campaign A. So you're left with the following simplification: Cost Breakdown Two-launch campaign A (single_reuse) Two-launch campaign B (expendable) Costs Recovery hardware Booster recovery Booster refurb One F9 booster Revenue $0 $36.8M For the costs, I don't think it's at all conceivable that recovery hardware, recovery operations, and refurb costs MORE than a single booster. It's just not reasonable. There's no way that landing legs, grid fins, boat operations, and additional work-hours can actually exceed the cost of a brand new booster. However, the added revenue from sending a heavier payload to GTO certainly makes up the difference...IF that revenue is guaranteed. But it isn't. There's simply not a market for that many gigantic comsats. So without the added revenue, I think they break even after a single reuse. These aren't independent variables though.

Well, I think we have to break out development costs separately. If we are talking about operational reuse, we can’t try to amortize in the costs of developing Falcon 9‘S reusability, especially considering that the process of reaching reuse also involved a lot of upgrades that made even an expendable Falcon 9 a lot better of a vehicle. We can talk about how long it will take for operational reuse to pay back the development investment, but for comparing expendable to reusable it needs to be a head-to-head comparison. We also need to specify what is meant by a single reuse. Are we are comparing two expendable flights to two reusable flights, or to a single booster that is reused once and then expended? I think an accurate head-to-head comparison requires us to use the latter approach; otherwise we are adding in additional recovery costs that aren’t going to convert into profits until the next launch. We also have to factor in profit margins. SpaceX benefits from a world where dial-a-rocket capability exists but is fairly imprecise. Falcon 9 has so much margin that it can deliver most commercial payloads to GTO without needing to go expendable, meaning a commercial reusable launch will usually earn the same gross revenue as a commercial expendable launch would.

Per my source anything they want to record will need to be programmed in advance. It won't be just sort of casually/passively listening.

Fantastic work. That's exactly what I was missing.

I believe the audio worked just fine; they didn't have it downloaded yet. Unless there's been some other announcement.

I'll ask around. The only person I know inside JPL works on Curiosity, not Perseverance, but I'm sure she can find out.

My contact at JPL says the audio from the landing just finished downloading from the DSN today and so they should be able to release it publicly in a day or two.

Absolutely breathtaking. It's so strange having absolutely no relative scale for the ground. Until the surface starts getting blasted by the thrusters, it's impossible to tell whether Perseverance is 400 meters up or 40 meters up. Almost feels fractal.

The answer is always yes.

As discussed upthread (I believe), this is trivially approximable with a series of very eccentric Molniya-style orbits. Take a spherical non-rotating Earth and define two congruent tetrahedrons centered on Earth, one much larger than the other, and then rotate the smaller tetrahedron 180 degrees such that each of its vertices are opposite the center of earth from a corresponding vertex of the larger tetrahedron. Set each orbit with the perigee at the lesser vertex and the apogee at the corresponding greater vertex. Most of the time, the four satellites very nearly form a projected tetrahedron. But it is not a perfect tetrahedron and I believe the whole thing falls apart at perigee. The reason I'm interested in pushing this a little farther is that while the OP specifically asked for the perfect regular tetrahedral solution, there's a slightly more relaxed solution possible. The real need expressed by the OP is to maintain perfect coverage over the surface of a sphere with only four observation points. A two-point solution leaves an unobserved ring (and leaves the satellites unable to communicate with each other) while a three-point solution leaves the poles unobserved. A four-point solution, on the other hand, works perfectly: each satellite can communicate with every other satellite and they have perfectly-overlapping symmetric coverage of the surface. Let's suppose we imagine statites rather than satellites, so that the positions can be maintained. It can be noted that the symmetric coverage is independent of the altitude of the statites, assuming they are above a certain minimum altitude. If we imagined statites that could alter their altitude, they wouldn't even have to alter their altitude in any regular coordination. Their altitudes could be completely chaotic and the underlying coverage remains completely fixed. We still may be able to find a solution by solving it piecewise. Let's take the limited set of solutions where the four satellites are in two pairs, with each pair sharing an orbital period. Each pair defines a line segment which follows some regular quasi-orbit around the planet. The problem then becomes whether two such line segments can maintain symmetry with each other such that their endpoints remain fixed on projections from the vertices of a free regular tetrahedron. The tetrahedron need not be rotating around a single axis at all; in fact, its rotation/tumble doesn't even need to be periodic. One could start by considering the prior example of two circularly-orbiting satellites with the same inclination but different AN longitudes, positioned such that they both cross their ANs simultaneously. This defines a line segment which remains parallel to the equatorial plane at all times, rotates around 360 degrees during each quasi-orbit, and goes up and down in latitude. You can then change the length and rotation of the line segment by changing the orbital parameters of the two satellites. As long as you do not change their orbital period, the line segment's quasi-orbit remains regular and periodic. I believe that the two line segments would need to also have the same periodicity which may further restrict the solution space.

Yes, I am accepting @K^2's proof for the regular tetrahedron. But I'm not so sure that @HebaruSan's non-equidistance conjecture is accurate. If English is your third language you are a LOT smarter than me...but what do you mean by standing face? I'm sorry; I must just not be following. Okay, so let's look at the reoriented system. Let's take satellite A as being the equatorially-orbiting satellite and B as being the non-equatorial one. Both are circular and have an orbital radius R from the center of Earth. Define time t0 as the time when A is positioned at the intersection of the two orbits. Suppose that at time t0, B is some angular distance Φ away from the point of intersection. The shortest distance between the two satellites is the line segment ℓ0 which lies in the orbital plane of B. Clearly, the length of ℓ0 = 2*sine(Φ/2)*R. Now, define time t1 as the time when B is positioned at the intersection of the two orbits. At this point, A has traveled some angular distance along the equator from the point of intersection. Since the two orbits have the same period, the time for B to traverse an angular distance of Φ is the same amount of time for A to traverse an angular distance of Φ, so the distance between A and B is the line segment ℓ1 and trivially the length of ℓ1 = 2*sine(Φ/2)*R. So the two satellites are the same distance apart at t0 as they are at t1. The question, then: is there a Φ such that ℓ = f(Φ) remains constant?

I agree, you have proven it impossible for a regular tetrahedron. I am musing that it might be possible for a regular tetrahedral projection. Wait, why is it the hypotenuse of a right (I assume that's what you mean by rectangular) triangle? For two equal-inclination orbits with only a slight difference in the longitude of the ascending node, it is the shortest leg of a very very acute triangle. it does though, at four moments (or two if they coincide - and somehow phase through each other): when they pass through the AN/DN of their respective orbits. Which is the moment of least distance between them. If the two satellites pass through their respective ascending nodes at the exact same time, then the straight line between them is definitely not in either of their orbital planes.