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Astronomy and/or scale question.

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Astronomy and/or scale question.
Post by Ensign Re-read   » Tue Mar 21, 2017 1:47 pm

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#1:
Generally speaking, roughly how large would an Honorverse ship have to be in order to be visible from earth, when parked at Lagrange point “L1”?

NOTE: Choose your own scale. In other words, select if you want the ship’s shadow and/or visible profile to be detected by the biggest of telescopes (yea, I know about the brightness issue) all the way down to the human eyeball (presumably using a “Pinhole Projection Method”).


#2:
Same question, but for “L4” and/or “L5”?


FFI, see URLs:
http://www.space.com/30302-lagrange-points.html
https://en.wikipedia.org/wiki/Lagrangian_point
.
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The Celestia "addon" for the Planet Safehold as well as the Kau-zhi and Manticore A-B star systems, are at URL:
http://www.lepp.cornell.edu/~seb/celestia/weber/.
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http://www.flickr.com/photos/68506297@N ... 740128635/
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Re: Astronomy and/or scale question.
Post by kzt   » Tue Mar 21, 2017 2:37 pm

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What's the skin temperature?

http://www.projectrho.com/public_html/r ... detect.php
A Russian Oscar submarine is a cylinder 154 meters long and has a beam of 18 meters, which would be a good ballpark estimate of the size of an interplanetary warship. If it was nose on to you the surface area would be 250 square meters. If it was broadside the surface area would be approximately 2770. So on average the projected area would be 1510 square meters ([250 + 2770] / 2).

If the Oscar's crew was shivering at the freezing point, the maximum detection range of the frigid submarine would be 13.4 * sqrt(1510) * 2732 = 38,800,000 kilometers, about one hundred times the distance between the Earth and the Moon, or about 129 light-seconds. If the crew had a more comfortable room temperature, the Oscar could be seen from even farther away.

Sorry, on my phone so I can't link deeper.
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Re: Astronomy and/or scale question.
Post by MuonNeutrino   » Thu Mar 23, 2017 2:10 pm

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Kzt's link is about detectability to infrared sensors, which is a reasonable enough way to go about it (while I do have the occasional quibble with their calculations, that site is usually on the right track). However, it sounds like you're asking about optical visibility, which is a somewhat different kettle of fish. A precise answer would require lots of annoying details, but we can make some estimates reasonably easily.

In this case you're observing reflected light from the sun rather than blackbody radiation emitted by the object. Both L1 and L4/5 are roughly 1 AU from the sun (L1 is only 0.01 AU from earth, so still about the same distance), and at this distance the illumination from the sun is about 1360 W/m^2. Of that, something like 45% is visible light, so roughly 610 W/m^2. If we assume that the ship has a cross-sectional area of A and reflects all the light that hits it evenly into a half-sphere (crude assumptions but close enough for estimation), then the effective 'luminosity' of the ship in terms of the reflected visible light is very roughly L = 1220 W * A/m^2.

We're interested in the flux that we observe, so we need to factor in the distance from earth to the lagrange points. This gives F = (1220 W * A/m^2)/(4*pi*r^2) = 97.1 W/m^2 * (A/r^2). I'm going to convert this to apparent visual magnitude, because that'll make it easier to compare to the detection limits of various astronomical instruments later on. Skipping all the gory details, this ends up giving m = -2.5*log(32.4 * A/r^2) - 21.1. (Note that the magnitude system is backwards, smaller numbers = brighter.) For L1 the distance is about r = 1.5e9 m, and for L4/5 the distance is 1 AU, r = 1.5e11 m. Plugging those in and rearranging a bit, we end up with the following:

m(L1) = 21.0 - 2.5 log(A/m^2)
m(L4/5) = 31.0 - 2.5 log(A/m^2)

The limiting magnitude of the human eye is about 6. If you substitute that into the above and solve for A, you get that you'd need a surface area of about 1e6 m^2 to be visible at L1 and about 1e10 m^2 to be visible at L4/5. Assuming the usual honorverse proportions, that'd be a ship about 2650x375 m in size for L1 and 265x37.5 km in size for L4/5, assuming we're seeing it broadside on.

Now, we made a lot of assumptions to get here, so these numbers aren't necessarily exact, but they're probably good to less than an order of magnitude, so we're still talking sizes of a couple km for L1 and a couple hundred km for L4/5. Those are pretty big - the former is twice the dimensions (and thus 8 times the mass) of an invictus, while the latter is even bigger than Hephaestus - but that's what you get if you want something to be visible to the human eye over astronomically significant distances. Of course in reality you can't see something at L1 via reflected light at all because it's inbetween you and the sun, but the numbers would work about the same for something at the exterior L2 point.

The limiting magnitude of telescopic observations varies based on a very large number of factors relating to the properties of the telescope, camera, and observing parameters, so it's not really practical to try to give a definitive answer here. As an example, though, the limiting magnitude of the Sloan Digital Sky Survey for point sources in the g band is 22.2. If you put that into the above, you find that you'd need a surface area of only 0.33 m^2 at L1 and 3300 m^2 at L4/5. The former figure is obviously tiny, while the latter figure happens to be a bit more than the surface area of a Chanson class DD viewed broadside-on.

The SDSS is an important astronomical survey (which is why I used it as an example), but it only used a 2.5 meter telescope and its sensitivity is good but not particularly outstanding compared to observations using the really big instruments. Given that SDSS can see destroyers at L4/5, it's pretty safe to say that any larger instruments could easily detect all honorverse ships, probably even a LAC, at the 1 AU distance of that point. And given that you only need 0.33 m of surface area to be visible to SDSS at the distance of L1, honorverse ships at those ranges could likely be seen even in relatively small scopes.
Last edited by MuonNeutrino on Thu Mar 23, 2017 6:58 pm, edited 1 time in total.
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Re: Astronomy and/or scale question.
Post by Jonathan_S   » Thu Mar 23, 2017 2:55 pm

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MuonNeutrino wrote:Kzt's link is about detectability to infrared sensors, which is a reasonable enough way to go about it (while I do have the occasional quibble with their calculations, that site is usually on the right track). However, it sounds like you're asking about optical visibility, which is a somewhat different kettle of fish. A precise answer would require lots of annoying details, but we can make some estimates reasonably easily.

In this case you're observing reflected light from the sun rather than blackbody radiation emitted by the object. Both L1 and L4/5 are roughly 1 AU from the sun (L1 is only 0.01 AU from earth, so still about the same distance), and at this distance the illumination from the sun is about 1360 W/m^2. Of that, something like 45% is visible light, so roughly 610 W/m^2. If we assume that the ship has a cross-sectional area of A and reflects all the light that hits it evenly into a half-sphere (crude assumptions but close enough for estimation), then the effective 'luminosity' of the ship in terms of the reflected visible light is very roughly L = 1220 W * A/m^2.

We're interested in the flux that we observe, so we need to factor in the distance from earth to the lagrange points. This gives F = (1220 W * A/m^2)/(4*pi*r^2) = 12.6 W/m^2 * (A/r^2). I'm going to convert this to apparent visual magnitude, because that'll make it easier to compare to the detection limits of various astronomical instruments later on. Skipping all the gory details, this ends up giving m = -2.5*log(4.19 * A/r^2) - 21.1. (Note that the magnitude system is backwards, smaller numbers = brighter.) For L1 the distance is about r = 1.5e9 m, and for L4/5 the distance is 1 AU, r = 1.5e11 m. Plugging those in and rearranging a bit, we end up with the following:

m(L1) = 23.2 - 2.5 log(A/m^2)
m(L4/5) = 33.2 - 2.5 log(A/m^2)

The limiting magnitude of the human eye is about 6. If you substitute that into the above and solve for A, you get that you'd need a surface area of about 7.6e6 m^2 to be visible at L1 and about 7.6e10 m^2 to be visible at L4/5. Assuming the usual honorverse proportions, that'd be a ship about 7x1 km in size for L1 and 700x100 km in size for L4/5, assuming we're seeing it broadside on.

Now, we made a lot of assumptions to get here, so these numbers aren't necessarily exact, but they're probably good to less than an order of magnitude, so we're still talking sizes of a few km for L1 and a few hundred km for L4/5. Those are pretty dang big, much larger than any mobile objects the honorverse builds, but that's what you get if you want something to be visible to the human eye over astronomically significant distances. (They're on the size scales of astronomical objects rather than ships!) Of course in reality you can't see something at L1 via reflected light at all because it's inbetween you and the sun, but the numbers would work about the same for something at the exterior L2 point.

The limiting magnitude of telescopic observations varies based on a very large number of factors relating to the properties of the telescope, camera, and observing parameters, so it's not really practical to try to give a definitive answer here. As an example, though, the limiting magnitude of the Sloan Digital Sky Survey for point sources in the g band is 22.2. If you put that into the above, you find that you'd need a surface area of only 2.5 m^2 at L1 and 2.5e4 m^2 at L4/5. The former figure is obviously tiny, while the latter figure happens to be a bit less than the surface area of a Star Knight class CA viewed broadside-on.

The SDSS is an important astronomical survey (which is why I used it as an example), but it only used a 2.5 meter telescope and its sensitivity is good but not particularly outstanding compared to observations using the really big instruments. Given that SDSS can see heavy cruisers at L4/5, it's pretty safe to say that any larger instruments could comfortably detect all honorverse ships at the 1 AU distance of that point. And given that you only need 2.5m of surface area to be visible to SDSS at the distance of L1, honorverse ships at those ranges could probably be seen even in relatively small scopes.

Thank you for doing all the number crunching.

And that probably applies if the Honorverse ships have their smartpaint set to bright white. If they were playing games and had it set to a far less reflective color you'd need a bigger (more light collecting) scope to see them at those same ranges. (Though looping back to kzt's IR detection point if you're reflecting less visible light it's going into warming you up so the hull temp should be higher and therefore more visible to an IR search).


Hmm, how visually impressive would it be for an SD to fire up it's wedge at the L1 point? (Assuming it's not edge on to you)
You'd have a 300x300 km square of high enough gravity to majorly distort the photons coming from the sun...
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Re: Astronomy and/or scale question.
Post by robert132   » Thu Mar 23, 2017 3:23 pm

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Ensign Re-read wrote:#1:
Generally speaking, roughly how large would an Honorverse ship have to be in order to be visible from earth, when parked at Lagrange point “L1”?

NOTE: Choose your own scale. In other words, select if you want the ship’s shadow and/or visible profile to be detected by the biggest of telescopes (yea, I know about the brightness issue) all the way down to the human eyeball (presumably using a “Pinhole Projection Method”).


#2:
Same question, but for “L4” and/or “L5”?


FFI, see URLs:
http://www.space.com/30302-lagrange-points.html
https://en.wikipedia.org/wiki/Lagrangian_point
.


Naked eye visible at the nearest LaGrange point? Death Star.

Any further away? Dahak.
****

Just my opinion of course and probably not worth the paper it's not written on.
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Re: Astronomy and/or scale question.
Post by MuonNeutrino   » Thu Mar 23, 2017 7:04 pm

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Just FYI, I fucked up the math a bit in my original post, turns out it helps if you actually write the correct numbers when you carry them from one line of math to the next. :lol:

I edited the original post, upshot is that things are a bit easier to see than I thought. Honorverse ships are still too small to see with the naked eye at those distances, but not by quite as much as I thought, and the SDSS can actually see down to destroyer sizes at 1 AU instead of just CA sizes.

As far as wedge effects, unfortunately we don't know *exactly* what a wedge does to light - we know it bends photons into a pretzel, but not in exactly what manner. So it's hard to say exactly what it'd look like, but it's probably safe to assume it'd be pretty impressive looking in *some* way!
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Re: Astronomy and/or scale question.
Post by kzt   » Thu Mar 23, 2017 8:03 pm

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At honor on they talked about seeing honorverse ships. The damn things are black in action, and the wedge scatters light, so they are pretty much invisible optically if you see the wedge. You'd see the blob of missing stars with an automated surveillance network.
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Re: Astronomy and/or scale question.
Post by Jonathan_S   » Thu Mar 23, 2017 10:24 pm

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kzt wrote:At honor on they talked about seeing honorverse ships. The damn things are black in action, and the wedge scatters light, so they are pretty much invisible optically if you see the wedge. You'd see the blob of missing stars with an automated surveillance network.

Sure, but that's focusing on looking for them sneaking into your system. I was wondering about the look of a ship passing between the occupied planet and it' sun at astronomically close range. The light scattering of 90,000 km^2 worth of solar photon on their way to the planet should be pretty noticeable...
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Re: Astronomy and/or scale question.
Post by Annachie   » Fri Mar 24, 2017 4:00 am

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You'd think an SD with it's wedge up passing through L1 would make quite a show.

Let alone closer.

In fact, I suspect that on a good cloudless day you could detect it in shadow variances on a planet.
Maybe not by eye, but not with great difficulty either.

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