From: chakatfirepaw@gmail.com   
      
   On Sat, 17 Sep 2016 06:17:20 -0700, Mikkel Haaheim wrote:   
      
   > 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.   
      
   When responding after a delay, it's a good idea to make note of it.   
      
   In my case this delay has to do with a visiting niece who watched a few   
   more videos on her tablet than we thought and burned though about a third   
   of our bandwidth for the month in under a week.   
      
   >> 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.   
      
   A radial burn results in a co-planar orbit to the original one that   
   passes across it at the point of the burn.  For obvious reasons, two co-   
   planar orbits that cross have to do so at two points.  If you still don't   
   understand:  Take a piece of paper and draw two ovals on it so that they   
   cross, (actually cross, not just be tangent), and count how many times   
   they cross.   
      
   Not that avoidance manoeuvres are likely to rely on radial burns, the   
   main ones are going to be (anti-)normal.   
      
   > For a point burn,   
   > there is only one point where the orbits intersect, and that is at the   
   > point of the burn.   
      
   That's only true of pro/retrograde burns, (which result in the orbits   
   being tangent to each other).  Radial and (anti-)normal burns result in   
   two intersections.   
      
   >> >> 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,   
      
   Oops, I did make a couple little errors there:  Calculating for ten   
   million square kilometres and forgetting to include the conversion to   
   metres.   
      
   So for a hundred million square kilometres that means:   
      
   100,000,000 km^2 * (1000 m/km)^2 = 1e22 m^2   
      
   Or about a quadrillion tonnes at 1g per 10 m^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.   
      
   You used both depending on whether you wanted to say "it's too wide to   
   dodge" or "you can reasonably saturate it with impactors."   
      
   It's a bit of dishonesty you were called on.   
      
   >> 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.   
      
   Hint:  Having one pellet every 10 m^2 does not mean you automatically hit   
   something larger than 10 m^2.  To give an extreme example, something 4m   
   wide and 10km long could fit into that perfectly spaced shot pattern   
   without getting hit.   
      
   > 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.   
      
   Um, _NO_ as has been pointed out before:  That's a diameter of only about   
   100km.   
      
   > That brings us down to 10^9 g, or 1000 tonnes.   
      
   Which almost certainly misses entirely.   
      
   >> > 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.   
      
   Your target is going to be somewhere in an area thousands of kilometres   
   wide.  You need a shot pattern of a similar size if you want a reasonable   
   chance of 'hitting' with your pattern, (a necessary prerequisite for any   
   individual shot connecting).   
      
   > However, the mass discussed is NOT unreasonable for important strategic   
   > targets.   
      
   Even with my erroneous calculation that got a very lowball figure you   
   were talking millions of tonnes _per target_.   
      
   > 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.   
      
   Um, no.  First of all, at the speeds we are talking non-nuclear warheads   
   are nothing but a sub-munition distribution technique due to the 1st law   
   of space combat, (at 3 km/s you pack your own mass in blam).  Second, you   
   have to drop the mass/area by multiple orders of magnitude, (using my   
   earlier, wrong, calculation:  To get to 'only' a kilotonne of stuff you   
   need to get down to 10 _micrograms_ per square metre).   
      
   > In war, "sane" amounts of material is quite subjective.   
      
   Masses measured in Zg are almost never sane.   
      
   >> 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.   
      
   Except that you aren't launching 100 of them, you are launching trillions   
   of them.   
      
   > BTW, your 10 km/s delta v estimate   
      
   [continued in next message]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)