From: mikkelhaaheim@gmail.com   
      
   Le vendredi 16 septembre 2016 02:33:04 UTC+2, Rick Pikul/Chakat Firepaw a    
   > You did notice that this discussion was a month and a half ago, right?   
       
   Sorry. My wife imposed an incommunicado during our vacation in order for us to   
   do some work on our house and rental properties.   
      
      
   >    
   > You do not understand because I never said anything about the opposing    
   > orbital point:  "...the other point the initial and final orbits    
   > intersect."   
      
   Okay, you will have to explain this a little, then. For a point burn, there is   
   only one point where the orbits intersect, and that is at the point of the   
   burn. If you are using a continuous sail manoeuvre, there IS no intersection,   
   because you are either    
   spiralling in or spiralling out. Once a manoeuvre is completed, the only point   
   of intersection is with the point where you withdrew the sail, unless you were   
   constantly changing the angle of the sail in order to circularise the final   
   orbit... but, again,    
   there would be no common point with the initial orbit. You CAN NOT oppose a   
   radial out burn with another radial out burn.   
      
   >    
   > >> Not really, hitting the platforms involves firing shots that saturate a   
   > >> target area at least one hundred _million_ square kilometres in size.   
   > >> Cheap, unguided, shot means a minimum of about one 'pellet' per square   
   > >> metre.   
   > >    
   > > This is not a problem. Yes, I have suggested 1 pellet/m^2, but you   
   > > actually only need 1 pellet per sensor. The sensors will have to be   
   > > large in order to obtain useful info. 1 pellet/10m^2 would likely be   
   > > more than sufficient.   
   > > Furthermore, pellets can be small. The impact of even a single gram will   
   > > likely sufficiently damage the sensors.   
   >    
   > And what will be the total mass of those pellets?   
   >    
   > 10,000,000 km^2 = 1e14 m^2   
      
   10^13, actually... However, I think you might have meant 100 000 000 km^2,   
   which would roughly match the 10 000 km spread I meantioned; although I no   
   longer remember if I intended that "spread" to refer to diameter or area.   
      
   >    
   > For a 1g pellet every 10 m^2 that's:   
   >    
   > 1e14 m^2 * 0.1 g/m^2 = 1e13g   
   >    
   > That's ten _million_ _TONNES_.  For a shot that isn't even certain to    
   > connect, (perfectly spaced they're 5.77m apart).   
      
   Actually, the shot is pretty certain to connect. The Hubble has a minimal   
   cross section of over 17m^2. But I suppose that I can concede an insignificant   
   probability that all the pellets will miss.   
   You have a point that that is a lot of mass (and energy) to dedicate to a   
   single target, although it might still be worthwhile for strategic purposes.   
   based on the original discussion, 10 000 km^2 should be more than sufficient.   
   That brings us down to 10^   
   9 g, or 1000 tonnes. This would be about the size of a backyard pool of   
   material. This would also be the mass equivalent of the bomb load of an   
   average WWII sortie of B-17s (6 combat boxes of 12 aircraft each), generally   
   dedicated against a single target    
   (actually, this count is small... typical sorties used anywhere from 100 to   
   600 bombers, not counting support aircraft).   
      
   >    
   > > You don't even need pellets...   
   > > sensors are VERY sensitive to any kind of impact. Even high velocity   
   > > dust will scour the collector dishes and optics, rendering them   
   > > unusable. Given the size of the pellets or dust in question, the mass   
   > > required to do sufficient damage, results in high arial coverage   
   > > actually yielding fairly low actual masses or volumes.   
   >    
   > You obviously either didn't do the math or made rather unrealistic    
   > assumptions.   
   >    
   > Or are you jumping around between units again:  Using one figure as the    
   > radius of your shot cloud when claiming it will hit, then using the same    
   > number as the area when claiming you can connect using a sane amount of    
   > material.   
      
   The area originally discussed was probably overkill. However, the mass   
   discussed is NOT unreasonable for important strategic targets. Still, I have   
   demonstrated that the mass can be significantly reduced for more reasonable   
   strikes. You can even decrease    
   the mass below 1g, especially if you decide to use an explosive shell or   
   armour piercing design.    
   In war, "sane" amounts of material is quite subjective.   
      
      
   >    
   > Um, no.  You need on the order of 10km/s for escape trajectories, and    
   > that's if you eject prograde from the body orbiting the sun.  For these    
   > kinds of shots think 15-20km/at least.   
      
   Again, easily achievable with small pellets. Existing rail gun design will   
   fire a 3.5 kg projectile at over 2.5 km/s. The same energy will fire 100 1g   
   pellets at 15 km/s. BTW, your 10 km/s delta v estimate applies to launch from   
   Earth orbit. From the    
   asteroid belt, you only need on the order of 3 km/s delta v.   
      
   >    
   > > Also, dust and pellets will continue to disperse, leaving them no more a   
   > > hazard than currently existing meteoroids.   
   >    
   > Actually, they don't.  They just transform into a stream, meteor showers    
   > are annual for a reason.   
   >    
      
   Actually, relative to each other, they DO. Even if they are locked into   
   respective orbits, they will disperse within that orbit (actually, a band of   
   orbits). But, yes, you now have another region of meteor showers.   
   Largely irrelevant... it would not be difficult to impart sufficient momentum   
   for solar escape velocity; or, alternatively, to have a programmed detonation   
   to reduce the shells to dust.    
      
      
   >    
   > Actually, yes you do:  All of the stations can see almost everywhere, you    
   > have no horizon to hide behind.   
      
   If you scan, yes. However, if you scan, you leave open large windows where you   
   can sneak manoeuvres in.   
      
      
   >    
   > >> Remember that 'cold-running' means turning everything off, including   
   > >> things like life support and any computer systems.  If you have   
   > >> anything running you will have waste heat to get rid of.   
   > >    
   > > Incorrect. Cold running simply means that waste heat will have to be   
   > > limited to what can be safely absorbed by the internal heat sink   
   > > (cryogenic supply)   
   >    
   > Which isn't that much, expect no more than a few months under ideal    
   > conditions.   
      
   This depends upon the object, the amount of heat sources, the amount of   
   cryogenic supply or other heat sink material, and the amount of surface area   
   for low level radiation of waste heat.   
   If I have time, I will try to draw up some scenarios.   
      
   >    
   > > or what can be emitted without allowing for detection.   
   >    
   > Which also isn't that much   
      
      
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