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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 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 Easy on Feb 28, 2017 3:36:50 GMT
I'd like to see a molten fission reactor. Where all the little bits can get as hot as they please and there is no worry of meltdown because it has already melted down.
In practice it would probably be like the hydrogen bomb molten salt power plants, where you detonate a nuclear bomb in metal salt and then use that latent heat source for however many months it lasts and you need to detonate a new bomb. That is is technically fusion power.
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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 Easy on Feb 28, 2017 0:58:23 GMT
I tend to have difficulty with armor penetration with flak weapons against capital ships.
Best results tend to be with lead or copper flak module with a osmium rad shield as a central penetrator.
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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 Easy on Feb 27, 2017 14:39:06 GMT
About Currency... The materials/ products here aren't important, just illustrative... ------------------------------------------------------------------------------------------------------------- Say Space Society A has Aluminum now and wants Beryllium 3 months from from now. Space Society B will have Beryllium 3 months from now, and wants Iron 6 months from now. Space Society C will have Iron 6 months from now, and wants Aluminum now. ------------------------------------------------------------------------------------------------------------ None of them can trade/barter directly (because none has exactly what the other wants, nor wants exactly what the other has, and none of them have the material to trade trade at the same time even if they did). Also all the materials are of various relative "values" and "worth". ------------------------------------------------------------------------------------------------------------ This kind of situation happens in real life all the time. Products needed, products available to trade, value of these products, and timing rarely match up perfectly. How is this solved without some type of currency/credit system? Even without a currency/credit system, would not a defacto one spring up out of necessity? If each colony has a local currency you can do a currency exchange much like FOREX. Colony A sells Aluminum in Acred, Colony C trades Ccred for Acred and buys the aluminum. C might even take a loan of Acred. Colony A now has Acred and Ccred to purchase Bcred to get Be from colony B. And it will continue. FOREX is a zero sum game where currency is not created nor destroyed and the exchange rate is based only on the willingness of the buyers and sellers. It is also how the world currency markets work.
Charlie on colony C wants Aluminum. Angela on colony A sells Aluminum for 3 Acred/kg. Angela doesn't accept Ccred, which is what Charlie has. Charlie negotiates with Alexander to purchase 3MAcred for 4MCcred. Charlie may pay Alexander the Ccred in lump or over an established time. Either way Charlie now has 3MAcred to purchase Aluminum from Angela. Charlie doesn't need to know Alexander personally, especially if he's paying immediately. Charlie logs onto the FOREX and bids to sell 4MCcred for 3MAcred. Alexander sees the bid and thinks it is a good deal and accepts the bid. Both exchange currency.
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Post by Easy on Feb 26, 2017 14:44:03 GMT
I think the game currency is a crpyto. In my headcanon, the Liberty Exchange is the only faction to use Cryptocurrency exclusively, but because of it's economic power everyone else's currency is sorta tied to it. I don't think factions would share currencies and may not choose to use the bitcoin model's method of inflation. Having administrate rights over a currency is a big deal, even if your only ability is controlling the injection rate of new credits. But cryptocurrency's major benefit which cannot be understated is it allows secure transactions that cannot be falsified or reversed unless the user's private keys are compromised. Data is a lot easier to transmit than it is to transport paper and metal.
I can see it very likely that each habitat would have its own cryptocurrency. Because it doesn't matter how rich you are on Mercury, your wealth there doesn't mean much to colonists on Saturn's moons.
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Post by Easy on Feb 24, 2017 21:08:10 GMT
Considering the tier 3, micronukes artillery shells with osmium flechettes works okay, but the silly fleet carrier doesn't like to die quickly.
I may switch to a pure chunk of metal for capship removal and build a specialized nuke spewer for missile defense.
As of now I am engaging everything between 40-30km. The Fleet carrier shoots back at 25km.
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Post by Easy on Feb 24, 2017 18:18:47 GMT
I don't see what you mean by 120 degree frontal protection, and how much armor is enough for this? 15kJ kinetic protection in a 120 degree cone. 15kJ means must be able to survive multiple hits of 1g projectiles at 5.48km/s. See image for 120 degree cone. Pick a vector, any vector and make sure it has an umbrella of armor that protects the crew module from attacks within 60° of that vector. A few centimeters of lithium backed up by S-glass works. The vector doesn't have to point along the ship's thrust axis but it should probably protect the crew module when broadside orders are issued.
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Post by Easy on Feb 22, 2017 5:06:14 GMT
When I saw the thread title, I imagined carefully painted realistic pictures of ships in the void of space. This isn't what I expected, but somehow I'm not disappointed. You have discovered art by engineers. Namely the great creative styles of schematic and concept.
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Post by Easy on Feb 22, 2017 4:51:55 GMT
Here's a submission. It is slightly less cheesy than I expected, because turns out that operating these drones still requires 177.1 people. If anyone asks why we can't spare any reactor techs in the Liberty thread I will direct them here.
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Post by Easy on Feb 22, 2017 0:50:35 GMT
Regarding economy of scale are thermocouples the reverse where multiple reactors with more relative surface area are more efficient? Because the theromocouple is getting big. The counter is that reactors require crew. The (cheesy) solution I'll use for this challenge is to leave the power management to computers. That is, built a battleship-drone with weapons, all the reactors I want, and a drone control module, and a low-powered human-crewed vehicle to launch it. That shouldn't work. Drones have a crew requirement based on the crew needs of the modules on the drone. Well I guess it does work because the combat drone itself doesn't care about the mass of the station that launched it. But you could use a bunch of simple ~100kW drones with low crew requirements without needing a large crew of drone operators.
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Post by Easy on Feb 21, 2017 23:15:30 GMT
Not a fuel cell, but here's a nuclear reactor that looks like a RTG. I hope you have depleted uranium to spare... A turbine would be much appreciated over a thermocouple. That sad yield strength of iron is disappointing. It doesn't afford much delta_T at all. I'm fighting just get into the MW range.
Regarding economy of scale are thermocouples the reverse where multiple reactors with more relative surface area are more efficient? Because the theromocouple is getting big. The counter is that reactors require crew.
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Post by Easy on Feb 21, 2017 20:55:55 GMT
That's a shame. The AI is unable to use the front laser and the side coils. If I just deactivate the laser when the enemy is in range, it's a massacre. Looks like I'll have to design around the dumb AI and put every weapon pointing the same direction.... it simply doesn't give the enemy much of a chance if they never get a shoot back. It also negates the armor part of the challenge. Realistically the tier 3 part is very difficult due to the fleet carrier's heavy hitting 286mm coilgun, but that ship also has the largest cross section. The devasator nuclear missiles are also much more resistant to point defenses than the drones. Tier 2 shouldn't be too bad since the cutter can be outranged by even a modest conventional cannon.
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