From: chakatfirepaw@gmail.com   
      
   On Wed, 28 Sep 2016 03:19:39 -0700, Mikkel Haaheim wrote:   
      
   > Le lundi 26 septembre 2016 01:19:13 UTC+2, Rick Pikul/Chakat Firepaw a   
   > écrit :   
   >   
   > 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.   
      
   While an explanation is not required, it is still good form to at least   
   acknowledge long delays.   
      
   Necroposting to get the last word really is a thing.  If you are silent   
   regarding a long delay between posts it comes across as if you were   
   trying to slip a final post past those you are arguing with so that those   
   who look at it later see you as having an unchallenged final shot.   
      
   >> 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.   
      
   The core point was that you don't need to be able to do a radial in burn   
   to undo the effect of a radial out burn.  Yes, in reality it is more   
   complicated than just a pair of point burns but it can still be done.   
      
   >> >> 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.   
      
   Oh, yes you are right.  Forget the totally ridiculous number I came up   
   with.  So 'just' ten million tonnes per target and I only made the one   
   error the first time, (which looks like it was just a typo since I got   
   the right result).   
      
   > 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.   
      
   It's what you need to hit when using unguided projectiles, you need to   
   saturate an area thousands of kilometres across.   
      
   >> > 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".   
      
   Avoiding a 50km asteroid and avoiding a 50km wide cloud of shot are of   
   equal difficulty.  The only way the cloud of shot can be harder to avoid   
   through blind jinking is if it covers a larger area.   
      
   >> >> 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...   
      
   You misunderstood:  That was 4m wide, 10_km_ long.  If it was lined up   
   just right a perfectly spaced pattern could have a 10km long string of   
   near misses.   
      
   (Assuming a triangular grid would be right, although I approached it as   
   hexagonal packing.)   
      
   > 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).   
      
   Remember that that spacing was if you managed to distribute things   
   perfectly evenly.  That isn't going to happen, I suspect that you will   
   get something a lot closer to a fully random distribution with all sorts   
   of clumps and holes.   
      
   > 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.   
      
   Your energy assumptions are, put simply, wrong.   
      
   >> > 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.   
      
   Using shot covering the same area doesn't help, it still hits the same   
   area.   
      
   >> > That brings us down to 10^9 g, or 1000 tonnes.   
   >>   
      
   [continued in next message]   
      
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