whh@PacBell.COM (Wilson Heydt) (06/05/90)
From: whh@PacBell.COM (Wilson Heydt) In article <16052@cbnews.ATT.COM> eshop3@ma.ecn.purdue.edu (Electronic Shop Part-Timers) writes: > >I have a quiz question for you all to figure out. While actual information >is of course classified, see if you can fathom how quickly the Stinger >moves. If it self-destructs after *14* seconds of flight with no kill, and >is intended to hit *ANY* aircraft flying below 10,000 feet Top of the head calculations from the data given (assuming that "any aircraft" may be traveling up to about 2000 mph), the *minimum* *average* velocity has to be aroud 2700 mph. Given that the missile has to accelerate from zero, this implies a terminal velocity somewhere in the neighborhood of 4000 to 5000 mph. While we're on the subject of missile performance figures-- For those who are Science Fiction fans (and have read Hal Clement's "Mission of Gravity"), consider the performance of the 1960's Sprint missile. It was designed to kill incoming warheads below 10 miles. Takeoff acceleration was around 600 to 700 Gs, with steering thrust at around 100 Gs. Within 2 seconds of launch, the outside of the nosecone was hotter than the inside of the engine. . . --Hal ======================================================================= Hal Heydt | An earthquake is Mother Nature's Analyst, Pacific*Bell | "silent" pager going off . . . 415-823-5447 | whh@pbhya.PacBell.COM |
timk@xenitec.on.ca (Tim Kuehn) (06/05/90)
From: timk@xenitec.on.ca (Tim Kuehn) In article <16052@cbnews.ATT.COM> eshop3@ma.ecn.purdue.edu (Electronic Shop Part-Timers) writes: > > >From: eshop3@ma.ecn.purdue.edu (Electronic Shop Part-Timers) > >I have a quiz question for you all to figure out. While actual information >is of course classified, see if you can fathom how quickly the Stinger >moves. If it self-destructs after *14* seconds of flight with no kill, and >is intended to hit *ANY* aircraft flying below 10,000 feet (as well as some >slower ones up higher, though I only know the upper ceiling by sight and not >measurement) at *ANY* angle of approach (unlike Redeye, where you could not >engage aircraft heading right for you head-on), how fast do you think that >little momma is moving? (The responses should be interesting...) Hmmm - assuming the missle has constant acceleration until detonation, and that the target being fired on is directly above the launch point (ie the missle is launched vertically) and stationary then equation comes out to: 0.5* a * t^2 + c1 * t + c2 = 10,000 Since the initial velocity = 0, the c1*t term goes away, the initial height = 0, so the c2 term is gone too, leaving to be solved: 0.5 * a * t^2 = 10,000 or 0.5 * a * (14) ^ 2 = 10,000 sorting terms around, the result is an acceleration of 102 ft / sec^2 Assuming the accelleration is constant, by definition velocity = accelleration * time yielding v = 14 * 102 = 1428 ft / second. The standard speed of sound is 1129 ft / second, so this means the missle has a peak speed of Mach 1.25 when attacking a vertical target. (This would vary slightly depending on atmospheric conditions) But wait - the quiz question said *any* angle of approach. So we're not dealing with an ideal case (stationary target vertically above the the firing point). Assuming then, that the we're firing at a fixed point 10,000 ft away and at 10,000 ft (or a 45 degree angle). The missle would then have to traval a distance of 1.414 times the distance it traveled in our original scenario. This means that accelleration would be 144 ft / sec^2 and reach a velocity of 2020 ft / sec or Mach 1.8. If the target's any further away than 10,000 ft, the accelleration, and hence final velocity would have to be correspondingly higher. For example, say the terminal velocity of the missle at detonation (no kill) was Mach 3. This is a velocity of 3387 ft / sec, which translates into an accelleration of 241 ft / sec^2. Going backwards through the previous equations, this means that it could reach a target that's 23,709 ft away, or 21,496 ft downrange at 10,000 ft, which means it could hit a stationary target 3.78 miles away, and at 10,000 ft. If the aircraft is at a lower altitude (say 2500 ft), then we can expect to hit a target on 23,576 ft downrange, or 4.46 miles away. Of course, our target's an aircraft, which means that it's *apparent* velocity (relative to the missle) will vary depending on whether it's flying towards, away from, or transverse to the missle flight path. If this was taken into account we may come up with an even higher terminal speed to catch up with a plane flying away from the launch point. It could be slower, though, if the aircraft was approaching the launch point (which is why it can hit a higher-flying but slower aircraft). There's a whole lot of other things left out of these equations - like the drag of the missle, it's *actual* terminal velocity (which gets factored into how far it can fly to catch the target), wind conditions (max headwind allowed for, etc.) and the like. That makes these numbers entirely suspect - but good for purposes of discussion. Do I pass? :-) ------------------------------------------------------------------------------ Timothy D. Kuehn TDK Consulting Services 871 Victoria St. North, Kitchener, voice: (519)-741-3623 Ontario, Canada N2B 3S4 DOS/Xenix - SW/HW. uC uP RDBMS !watmath!maytag!xenitec!timk timk@xenitec.on.ca ------------------------------------------------------------------------------