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Post by Easy on Feb 28, 2017 0:47:12 GMT
This thread is about observibility, heat, spectra, and telescopes.
So lets start with those big glowing radiators. The example ship is the Corvette. 66.4MW engines off, 575MW engines on. The radiators are mostly arranged in sets of four, perpendicular to the thrust axis. For two radiators it is pretty simple, each side of the radiator emits 50% of the energy but you'll only see the full strength if you're normal to the radiator. The brightness will decrease with the Cosine of the angle until you are 90 degrees (edge on) where you won't see much of that radiator at all. But we have four radiators, so that brightness is spread into a cylindrical distribution. But nose on or tail on you'll have the same decrease, you could use the same equation but if you measured the angle from the nose you would use the sine. These equations would assume that a Corvette pointing its nose at you would not emit any light, which is a bad assumption, but you're getting the minimal light from the radiator.
Okay the math. First bit is to spread that heat power P, 66.4MW or 575MW, over a cylinder, so we can simply divide it by 2pi. Next we add in the inverse square law to get the equation. Flux = P*sin(angle from nose)/2pi*r²
So lets be 109m away, broadside (90°), with 66.4MW. We're simplifying the spectrum in that we can see really well. Flux = Radiator_Heat*sin(angle_from_nose)/2*pi*(distance_to_observer²) Flux = 66.4MW*sin(90°)/2*pi*(109m)² Flux = 1.05*10-11 W/m²
Now how bright is that? The sun is 1361 W/m² in Earth Orbit. ~1100-1000W/m² at the surface. Lets use 1000W/m² as what most people would think the sun's brightness is. Divide them out. 1000 / 1.05*10-11 = 9.46*1013
Now to get the apparent magnitude. (Sun from Earth's Surface is magnitude -26.74) Apparent Magnitude = 2.5*log(1000W/m² / Flux_of_Target) + -26.74 Apparent Magnitude = 2.5*log(9.46*1013) + -26.74 = 8.20. The brightness of Titan when seen from the Earth.
Play with the equations, at 1 AU the Corvette is dimmer than Pluto.
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Post by Enderminion on Feb 28, 2017 2:03:06 GMT
but brighter then a whole bunch of other stuff, like backround stars, also a suprise corvette shaped shadow on jupider is very suspisious
Edit: one AU is really effing far to
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Post by Easy on Feb 28, 2017 2:40:35 GMT
but brighter then a whole bunch of other stuff, like backround stars, also a suprise corvette shaped shadow on jupider is very suspisious Edit: one AU is really effing far to You won't be able to see shape at long distances. Just a point source. The Hubble Telescope with its 2.4m mirror only sees .05 arcseconds. or 10 -5 degrees. The Corvette is 105m long so 105m/tan(10-5)= 6*108m = 600,000 km.So you wouldn't be able to measure the length of a corvette much further than six hundred thousand kilometers. The average distance of the Moon from the Earth is 384,400 km. Remember this is a 2.4m diameter mirror, the biggest space telescope was the Herschel Space Observatory at 3.5m wide, but it ran out of coolant in 2013 and is now useless. qswitched talked about 10cm or 0.1m telescopes in the blog.
In regards to reflected light we can assume the ships is roughly a hemisphere (1/2*pi) of rough (not mirrored) highly reflective surface. Because I'm lazy we'll multiply it by the solar irradiance and cross section for the advanced equation: Flux = (Cross_Section*Solar_Irradiance + Radiator_Heat*sin(angle_from_nose))/2*pi*(distance_to_observer²)1080m² * 1361W/m² = 1.47MW which is peanuts compared to the radiator heat or engines on heat. However the reflected light has a spectrum based on the Sun and whatever material is reflecting, not a black body profile. Remember that the 1.47MW is all the energy reflecting off the hull, you need to divide it by 2*pi because it reflects across a single hemisphere. If you wanted to be fancy and had both a broadside cross section and a nose-on cross section you would hook those numbers up to sine and cosine respectively. Crosssection = Broadside_Crosssection*sin(angle_from_nose) + NoseOn_Crosssection*cos(angle_from_nose)Hint: nose cross section is pi*hull_radius² Hint2: broadside cross section is somewhere between Length*hull_radius for conical ships and 2*length*hull_radius for cylinder ships
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Post by Enderminion on Feb 28, 2017 3:16:18 GMT
600k km or 600Mmm, 600 times the max engament range for lasers or about 1.8 times the distance from the earth to the moon, also qusitched follows the "microrecondrones everywhere" train of thought, there may or may not be stratgic stealth but tactical steatlh is out of the question
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Post by lieste on Feb 28, 2017 3:55:39 GMT
Not at all sure of your numbers.
I get around -8 Apparent at 1000km, and 0 Apparent at 41,260km for 575MW over the surface of a sphere of observation.
This is brighter than the ISS as seen from earth at a higher distance (-5.9 Apparent at 330km), but isn't 'brighter than the full moon', by a long shot (-12.74 at 38,400km)
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Post by Easy on Feb 28, 2017 4:07:56 GMT
lieste you are correct. I have updated the original post with magnitude 8.20. For 575MW at 41,260km I got -1.07 which agrees with your 0 due to your choice to use a sphere. It took me two tries because I didn't convert from km to m the first time. Also for an exhaust plume you would want to use the spherical equation. So you're more correct if 575MW is talking about an exhaust plume.
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Post by lieste on Feb 28, 2017 4:24:52 GMT
You are out by a factor of 10^8 in the flux calculation (too high - I guess you failed to raise the distance to the second power, or used a different distance than you wrote). I also used slightly different distribution of flux (over a sphere vs cylinder, but difference in magnitude is relatively minor for the 0.5x flux density this causes).
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Post by Easy on Feb 28, 2017 4:45:36 GMT
You are out by a factor of 10^8 in the flux calculation (too high - I guess you failed to raise the distance to the second power, or used a different distance than you wrote). I also used slightly different distribution of flux (over a sphere vs cylinder, but difference in magnitude is relatively minor for the 0.5x flux density this causes). The sphere is 1/4*pi so your result should be half as bright, flux-wise but will result in a similar absolute magnitude due to the logarithmic scale. Thank you for checking my math.
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Post by n2maniac on Feb 28, 2017 6:54:37 GMT
Small point in observability (sorry, not math related): the ship's spectrum is that of a ~1200K blackbody, which is actually quite far from ordinary around the sun (solar reflection is based on ~5700K visible, planetary blackbody is typically under 300K). Searching the surroundings for the 1200K - 2400K range would find very few other candidates (red giants are 3000K and above).
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Post by Easy on Feb 28, 2017 12:29:20 GMT
Small point in observability (sorry, not math related): the ship's spectrum is that of a ~1200K blackbody, which is actually quite far from ordinary around the sun (solar reflection is based on ~5700K visible, planetary blackbody is typically under 300K). Searching the surroundings for the 1200K - 2400K range would find very few other candidates (red giants are 3000K and above). But how easy is that really? The 5700K blackbody still emits in the infrared and near infrared range. Don't forget about red and brown dwarfs. Luhman 16 is the nearest brown dwarf at 2 Parsecs and an apparent magnitude of 8.87. I think it would be very interesting to run the numbers for what kind of telescope and sensor you would need. And then compare the required data processing in order to subtract the known starchart, when you have realistic diffraction limited and noisy data. Could you do it with a bunch of guys with really nice amateurs telescopes who know star charts really well? What ranges would the amateurs be effective? What about fancier computerized telescopes. Obviously we don't want exposure time to be too long since expired data is useless.
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Post by Enderminion on Feb 28, 2017 13:26:14 GMT
micro-recon-drones everywhere, I don't need a 10m telescope when I have 50 1m telescopes
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Post by caiaphas on Feb 28, 2017 18:15:34 GMT
Small point in observability (sorry, not math related): the ship's spectrum is that of a ~1200K blackbody, which is actually quite far from ordinary around the sun (solar reflection is based on ~5700K visible, planetary blackbody is typically under 300K). Searching the surroundings for the 1200K - 2400K range would find very few other candidates (red giants are 3000K and above). But how easy is that really? The 5700K blackbody still emits in the infrared and near infrared range. Don't forget about red and brown dwarfs. Luhman 16 is the nearest brown dwarf at 2 Parsecs and an apparent magnitude of 8.87. I think it would be very interesting to run the numbers for what kind of telescope and sensor you would need. And then compare the required data processing in order to subtract the known starchart, when you have realistic diffraction limited and noisy data. Could you do it with a bunch of guys with really nice amateurs telescopes who know star charts really well? What ranges would the amateurs be effective? What about fancier computerized telescopes. Obviously we don't want exposure time to be too long since expired data is useless. The issue there is that the sun takes up a very small section of the sky, as do any of the other celestial bodies, and all of the major ones are going to be mapped (for traffic control purposes) anyways. The discussion concerning this is much more nuanced at the source, but basically what it boils down to is the argument that if you have any sort of major space-faring civilization it's more or less trivial to set up the logistics for a major orbital telescope observation network (for example, Hubble is around 11 tons; we throw around fleets of drones that weigh that much), and some basic calcs tell me that, for example, a network in orbit around Luna (diameter of around 3500 km) observing in the infrared (800 nm to 1 mm) can distinguish two objects with an angular separation of anywhere between 1.31*10 -11 to 1.64*10 -8 degrees. To put this into useful terms, if we have this array around Luna and we're looking at something at Jupiter 500 million km away, you could resolve drone launches from a fleet carrier if you're observing at 800 nm (can resolve objects down to around 6.55 m) and could resolve fleets on maneuver at the high end of 1 mm (can resolve objects down to 8.19 km). Realistically I figure that you'd structure a detection web like this; of course you'd place them in orbit around the largest available celestial body to get the largest effective mirror diameter. - Early-warning infrared satellites. These don't need to have fantastic resolution, they're just looking for large infrared sources where there wasn't anything before. Once they do see what's going on they send an alert signal to the other satellites in the network, progressively focusing more and more scopes on this weird new infrared source until they can confirm through parallax that yeah, this isn't a supernova or something, it's in-system.
- Trajectory acquisition satellites. I don't know what wavelength they'd operate in but I figure that the role of these is to determine the direction in which the target is moving, and at what velocity. Once you've determined whether it's heading towards someplace you don't want them to go, ping them with active radar and lidar to get better measurements.
- Several different dedicated tracking satellites operating in near-infrared to the high end of the visual spectrum. These stay focused on the acquired targets so that you can figure out what the heck they actually are; the range of wavelengths is to make it harder for your enemy to hide from you.
EDIT: I'm going to quote one of the guys on the discussion I linked above, who lays out the sensor strategy in broader and more technically relevant terms. In addition, I wasn't entirely clear on a point that I wanted to make above that addresses one of the basic assumptions in the OP. It doesn't matter if your ship has the same apparent magnitude as stars in the background, because of two reasons. One, you know where most of the background stars/planets/asteroids are; pete's sake, leaving spectral analysis aside, which would make this even more useful for spotting enemy warships, this technique is actually how astronomers find new asteroids, because they can look at the sky for a couple of nights running and observe where the little dots of lights are and which have moved measurably from their position. Anything that moves significantly is something in-system. By analogy, you know where all the stars and major asteroids and where all the planets are relative to your satellite's position, and all it'll take is one or two orbits of constant observation to know how all of that background looks at any position along your orbit. You then subtract that data digitally, and look for any dots of light that appear against that background. Two, you can use parallax to determine if that thing that magically has the same emission profile and apparent magnitude of, say, Proxima Centauri, is actually Proxima Centauri or whether it's a ship that just looks like Proxima Centauri.
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Post by n2maniac on Mar 1, 2017 5:18:27 GMT
Hold on, aperture synthesis in the IR/vis range? Is that yet cutting edge, particularly at that long of distances? If so, why aren't we getting images of extrasolar planets crossing their parent star?
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Post by caiaphas on Mar 1, 2017 7:40:16 GMT
Just a quick correction to my statement above; I forgot to account for the sensitivities of the CCD chips we'll be using. Those will set a lower limit on the radiance that any given object would need before it gets picked up by our sensors.
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Post by Enderminion on Mar 1, 2017 14:46:00 GMT
Thermal batteries and delivery system, the size of the battery determines how long the thermal sensor operates and the delivery systems, how low temp the enemy has to be in order to not be detected. At least this is how modern A2A and SAM missiles work.
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