From: mikkelhaaheim@gmail.com   
      
   Le lundi 26 septembre 2016 01:19:13 UTC+2, Rick Pikul/Chakat Firepaw a   
   écrit :   
      
   > 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.   
      
   Explanations don't matter to me. Nor does tardiness in reply. I am used to   
   dealing with people who have other priorities. When they DO reply, the topic   
   is bumbed to the top of the list, and it is marked as unread.   
      
      
      
   I will need to respond to this post in separate messages, as I don't have much   
   time.   
      
      
   > 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.   
   >    
      
   I will tentatively concede this point, given the assumption that we are   
   refering to punctual burns, and not the long duration "burns" (yes, I realise   
   that nothing is being burnt) that are required with solar sails. I still have   
   my reservations, but it    
   does make sense that a reversed vector applied at an appropriate point should   
   nullify an original vector. My reservations come from the fact that continuous   
   burns, even when achieving reversed vectors, continuusly spiral out, and do   
   not appear to nullify    
   any component of the previous burn... but I will admit that this fact might be   
   the result of the points of the original orbit being to far removed from the   
   modified "orbital" succession.    
      
      
   >   
   > >> 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   
      
   You are compounding your errors. 10^8 km^2 is already expressed as area. You   
   do not square this value. When converting from km to m, you do indeed have to   
   multiply the original area by (10^3 m/km)^2, or by 10^6 m^2/km^2. The result   
   is 10^14.   
      
   I will have to review my original statements. I do not know why I would have   
   chosen a figure such as 10^8 km^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.   
      
   As I said, I will go through my original posts. Keep in mind, that my response   
   was in the context of SeaWasp's suggestion of how easy it would be to avoid a   
   50 km asteroid. If you are willing to lob these around, even your erroneous   
   quadrillion tonnes    
   would be "reasonable".   
      
      
   > >> 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.   
      
   Assuming that your 5.77m spacing figure is accurate (I am assuming this is for   
   a triangular equidistant grid), then it won't pass through the spaces (10m is   
   just a few mm too large to pass... in any case, Hubble would be firmly hit at   
   13.2m x 4.2m,    
   assuming the larger cross section). Granted, you might get by with just a   
   graze, but be careful with shrapnel and spalling (but, again, that assumes a   
   shorter platform).   
   That said, I have already conceded a small possibility that a small platform   
   could be perfectly oriented to avoid a hit. This could easily be remedied with   
   slightly smaller pellets spread out more closely. OTOH, we've been talking   
   about Hubble sized    
   platforms, and such platforms are TOO SMALL to detect the small amount of   
   energy being discussed here.    
      
      
   >    
   > > 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.   
      
   Yes... and that was the original escape proposal suggested by SeaWasp, in   
   order to escape the 50km asteroid.   
      
   >    
   > > That brings us down to 10^9 g, or 1000 tonnes.   
   >    
   > Which almost certainly misses entirely.   
      
   Not at all, especially when you consider that a useful platform makes a nice,   
   big target... especially when it is pointed at you. This is especially true if   
   ou are trying to manoeuvre with a light sail.   
      
      
   >    
   > 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).   
      
   This area is easily reducible through patient observation   
      
   >    
   > > 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_.   
      
   Yes. A little less than 100m x 100m x 100m of excavation. Assuming you want to   
   take the overkill route.   
      
   >    
   > > 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).     
      
      
   [continued in next message]   
      
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