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Ralathon

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Everything posted by Ralathon

  1. The difference between a reactionless drive and a conventional photon drive is the lack of an exhaust. Your kinetic energy depends on your frame of reference. A 1kg mass moving at 1ms has 0.5 J of energy. But if the mass is moving at 1000m/s it would have a whopping 0.5MJ. Now factor in this EM drive. Suppose it works as advertised and for simplicity converts 100% of the provided electrical energy into kinetic energy. Say we stand next to the drive and feed it 50 joulles. That's enough to speed up a 1kg drive by 10 m/s. So we see the drive accelerate from stationary to 10m/s. Now lets put it on a train moving 1000m/s. You again feed it 50 joulles. Now this is only enough to accelerate it by sqrt(2*500050J) - 1000m/s ~ 0.5 m/s. So now that same amount of energy is only enough to accelerate your drive by half a meter per second. This leads to a paradox. If you stand on the train next to the drive it's supposed to accelerate to 10m/s relative to you. But if you stand on the station it is supposed to accelerate by 0.5m/s. This is because the kinetic energy of an object is entirely dependant on the reference frame you choose to use. There is no preferential frame of reference, so the drive does not 'know' how much it is supposed to accelerate. This is why reactionless thrust breaks so much .... and why it's so easy to make infinite energy with it. In the case of that rotor it is assumed that the drive gives equal thrust regardless of the reference frame. This means you are generating more and more energy the faster it moves. So you can always pick some arbitrary speed where the drive is generating more energy than it consumes. If you have some form of exhaust you don't have this paradox since the missing energy is carried by the exhaust. Even with a photon drive the photons will redshift depending on the velocity, thus ensuring that no free energy is created.
  2. The san andreas fault moves about 40mm a year. Have fun with the earthquakes when the fault moves 2 meters back and forth every day (And days will be a lot shorter on a contact binary).
  3. Exactly, it will never be zero. Hence my surprise when you try to add an envelope 10 times as thick as needed for complete absorption. Read the words "Simplified model". I am fully aware that the greenhouse effect has nothing to do with the thermal conductivity of the atmosphere. But for the purposes of this model that's exactly what's happening. It becomes harder for the thermal radiation to escape so you can simulate that effect by decreasing the thermal conductivity. Just link the thermal conductivity to the greenhouse effect via some function f(P,d,T) and you have a simple 1 element model. Since you have a degree in thermodynamics I reckoned this would be obvious for you, but I guess not... conductivity vs resistance. Learn the difference. As far as I can see I never messed them up in my post. Increasing the greenhouse effect is equivalent to decreasing the thermal conductivity in the model, which is the way I used it. Good you figured it out on your own. Because your argument absolutely does not hold true if there is any energy transfer via visible light.
  4. Okay then mr degree-in-thermodynamics. Let's go through it step by step. Is the temperature of the isothermal gas equal to the surface temperature of the blackbody here, or are they different? So, infinite*10? That's not how heat transfer works and you should know this if you have a degree. You can simplify it to a trivially easy model. 2 boundary conditions and an element with a certain thermal conductivity. The thermal conductivity varies with pressure, volume, temperature etc, but unless the number of gas molecules goes to infinite the thermal conductivity will never be zero. And as I explained in my previous post, this has absolutely nothing to do with the greenhouse effect. The greenhouse effect is about decreasing the thermal conductivity of the atmosphere, it occurs even if the heatflow in equals the heatflow out. I can take a lump of molten lead and wrap it in insulation until it emits the same amount of heat as a mildly warm lump of lead. Kinda hard to argue that the lump of molten lead is only mildly warm, simply because it emits the same amount of thermal radiation. it DOES matter because the thermal conductivity of the atmosphere is lower than it would be without the greenhouse gasses. So this means the object needs to have a higher temperature to give the same outflow of heat. You've insulated the blackbody, so it needs a higher temperature to result in the same heat output. So the object will be hotter. Ah, I see what you're getting at. You're setting T(0)=T(end). You won't notice the greenhouse effect in that case. In the same vein, if you let the earth cool down to 2.76 kelvin and put it in an intergalactic void, you also wouldn't notice the greenhouse effect. But you're ignoring that not all incoming energy is in infrared. If you added a lamp that shone visible light onto your black body you would notice the greenhouse effect. Basically your whole argument is: If the sun was shining in infrared instead of visible we wouldn't have a greenhouse effect! Which is true, but kinda trivial...
  5. That's 2^(1/3). The 2 was inside the density ratio brackets. Derivation can be found here.
  6. It depends on the density ratio and the radi of the planets. The problem is the roche limit. The roche limit of the primary body needs to be smaller than the combined radi of the planets. So you need to solve the function 1.26*Rprimary*(rhoprimary/rhosecondary)^(1/3) < Rprimary+Rsecondary. Say we want 2 planets made of the same stuff. So rhoprimary = rhosecondary. 1.26*Rprimary < Rprimary+Rsecondary 0.26*Rprimary < Rsecondary So a contact binary between 2 bodies with the same density is possible provided that the secondary body is at least a quarter of the radius of the primary. It becomes harder if the secondary is less dense than the primary. The moon could never be a contact binary with earth because the density of the moon is only 60% that of earth.
  7. Then state your question more clearly. The way I read your post is that you're confused how the greenhouse effect works since an opaque sphere around a kerbal would radiate just as much heat as a naked kerbal. You then go off on some hypothesis about the temperature of the outer layer and that this causes less heat radiation. So my post explains how the same energy inflow and outflow can nevertheless result in higher surface temperatures. Is my reading comprehension completely off here?
  8. The outflow of energy will always be the same as the inflow (if you have a stable situation). But the greenhouse effect makes it harder for the heat to escape. It's important to make a distinction between heat and temperature here. An analogy I always like to use is a water bucket with a valve at the bottom. You pour water into the bucket at a constant rate (visible sunlight). The water fills up the bucket until the outflow via the valve (IR radiation) equals the inflow. If you now throttle the needle valve (increase the greenhouse effect) less water will escape the bucket. So the water level starts to rise until the outflow is once more equal to the inflow. The water level is equivalent to the temperature in this analogy. Even if we double the greenhouse effect we have just as much outflow as inflow. But the earth stores more joulles before radiating them and thus the surface temperature will be higher.
  9. That's exactly what they're saying there. They're discussing the thermal mass of the atmosphere dampening temperature swings. It's poorly worded though. But you're right, the dampening has nothing to do with the greenhouse effect. Again, that article is very poorly worded. What they mean is that when you look at the spectrum of the earth in infrared you don't see a perfect black body. You see this: As you can see the spectrum doesn't follow the perfect black body radiator. There are all sorts of gaps and valleys in the emission. This is where greenhouse gasses come in. These gasses absorb those spectra of light and scatter them. Because the energy keeps getting scattered the radiation only leaks out of the atmosphere very slowly. That's what's causing all those gaps. So what happens is that visible light from the sun comes down through the atmosphere. Hits the surface and gets converted into heat. Then the heat gets radiated as infrared, but because the atmosphere is opaque to infrared the heat can't escape. The more CO2 and water vapor the more opaque the atmosphere and the slower the heat escapes the atmosphere. And thus the higher the temperature on the surface. Yes. They detect more CO2 in the air and they see the valleys in the spectra due to CO2 and methane deepening. Water gets out of the atmosphere pretty quickly. All the water we inject by burning fossil fuels rains down pretty quickly. So we don't see any increase in the greenhouse effect due to water.
  10. I wouldn't call 100k cheap. Sure, it's cheap for spaceflight, but it's not exactly spare change. - - - Updated - - - Pretty much nonexistent. We don't know what dark matter is, but it seemingly does not interact via the electromagnetic force. If it did we could see it and our spacecraft would slow down when moving through it. Considering that microwaves are electromagnetic waves this shouldn't have any effect on dark matter. It is also an untestable hypothesis. We don't know what's causing the thrust of this EM drive. So its a bad idea to draw "all of physics is wrong" as a conclusion. Drastic changes like that are only warranted when all mundane causes are ruled out.
  11. Is there some way to determine the total torque a reaction wheel provides, the maximum thrust of an RCS thruster, the maximum gimbal of an engine and the total moment of inertia of the vessel? My steering script in RSS currently needs custom PID parameters per vessel. So I'm trying to build something that automatically tunes the PID. I could probably approximate the moment of inertia in the yaw and pitch direction by playing with the mass and position of each part, but that leaves the roll and it's not very elegant...
  12. Note how its a patent, not a build announcement. They're not going to build one, they're just locking down the patent.
  13. Even if we manage to build a magic von neumann machine that perfectly replicates itself in any situation, we'd still be in heavy contact with the thing. The entire point behind a Von Neumann swarm is to get a lot of data for a relatively small cost. Guiding the first few generations until the swarm is sufficiently large would likely be standard procedure. After those first few replications the swarm is so big that a single failed replication is no longer mission critical, and we can afford to just watch the data roll in. Also, I heavily contest your point that you always need a more complex machine to build another. Going by that logic you would never have a rise in complexity, as the child can never be more complex than the parent. Yet we see the exact opposite in nature and technology. We've build our current technological wonderland from sticks, stones and fire. Sure, it took us a few thousand years, but it is a clear sign that you can use simple machines to make more complex ones. The same thing can be observed in nature: 2 billion years ago we had pond scum, now we have cheetas and killer whales. If you give a probe a good 3d printer, an energy source, some resource extraction device and an arm to assemble stuff, it could rebuild itself. It would take a really long time, and probably take loads of intermittent steps, but provided your programming is smart enough it could be done.
  14. Very minor little bug: The scattering effect does not show up on the IVA RasterProp Monitors if you look at it via a camera.
  15. Light also gets weaker following the same inverse square law. This is why spaceprobes past Jupiter can't use solar panels for energy production. Or why a small light bulb isn't enough to light up your entire house. The reason this happens is a simple consequence of living in a 3 dimensional universe (with time as a 4th, but lets not get into GR here). Say you have a bubble with 100 small dots on its surface. The bubble has a surface area of 1m^2, so the density of the dots is 100 dots per square meter. Now say we inflate the bubble so its twice as big. The bubble now has 4m^2 of surface area, but the number of dots stays the same. So the number of dots per square meter is now 25. Twice the radius means the dot density becomes 4 times as low. Same thing happens in the universe. Gravity (or light) is equivalent to the dot density. The number of photons/gravitons is the same but they're smeared out over a larger area. And because that area increases by the square of the distance you end up with the familiar inverse square laws.
  16. That's because Newtonian gravity is a simplification of general relativity. If you take GR and stay away from relativistic masses or velocities you end up with Newton. That's not how things would work if this EM drive indeed produces reactionless thrust. It isn't "okay, we have this complex thing that simplifies to this stuff we know". Its "Okay, science says that this cloud is white but this EM drive says the cloud is purple...". They're simply not compatible.
  17. Sounds like your rocket is just aerodynamically unstable. So not a problem with FAR, but a problem with your design. Put some fins on the bottom and make sure the center of lift is close to the center of mass in the VAB
  18. Suppose that proxima centauri is about the size of the sun and we want at least 10 pixels of resolution. This means the telescope needs an angular resolution of arctan(70e3km/4.8ly)~1.56e-9 rad. We want to look at the star in visible light, so we have a wavelength of about 500nm. Plug these 2 numbers in the telescope equation and we find that we need a telescope with a main mirror 320 meters in diameter. The biggest telescope we have right now has a diameter of about 10 meters. The E-ELT that's in construction right now will have a diameter of almost 40 meters. So we're about an order of magnitude short in terms of mirror diameter to see stars as anything other than point sources.
  19. When using kOS in combination with an engine with alternative resource requirements, for example the LV-N or some mod, is it possible for kOS to figure out what resource the engine is using without actually activating the engine? For an autopilot I want the program to figure out how much dV the vehicle has. This means I need to figure out how much fuel each stage has and what the engines use. The first part works great, but I have no idea how to do the second.
  20. I knew the SLS was slow going, but 3018? Damn... They're lucky if there still are NEO asteroids left to explore by that point.
  21. Nah, it just has to be on a landed craft, no need for direct contact.
  22. Yea, it works just fine with the patch posted earlier and a swapped navball page. Here's the directory, just unpack it in gamedata like usual and it should work with the newest RPM. Few bugs in the resource panel, but I never use that one, so I haven't bothered fixing it.
  23. I just replaced the navball with the one that comes standard with RPM. All you need to do is swap the page definitions.
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