
#1
Nov1112, 01:34 PM

P: 9

I've been debating various SciFi vs debates for a while now and I've decided to try and figure out how powerful the C14 Gauss assault rifle from StarCraft needs to be in order to achieve the following feat:
The standard issue C14 Gauss rifle fires 8mm DU spikes with a fairly aerodynamic shape (narrow rounded tip and slightly rounded rear) that is either 5cm or 6.4cm long. It is neosteel tipped so I'd assume that's how it avoids getting vaporized by the friction, neosteel being a super hard future alloy. The projectile mass should be somewhere between 43 and 66 grams. Assuming earthlike gravity and atmospheric pressure (it is an inhabited planet), ignoring the winds from the vortex and assuming the Marines were firing at a perfect 90 degrees angle, what would be the minimal muzzle velocity of a C14 spike to reach 55.56km straight up? 



#2
Nov1112, 07:30 PM

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See: http://www.physicsforums.com/showthread.php?t=281635 http://www.physicsforums.com/showthread.php?t=431557 http://www.physicsforums.com/showthread.php?t=214435 ... and: http://en.wikipedia.org/wiki/External_ballistics 



#3
Nov1212, 05:58 PM

P: 372

I presume Simon means a lower bound. You can do it by equating the kinetic energy at the muzzle with the gravitational potential energy at the target height  (1/2)mv^{2}=mgh. Back of the envelope, that's about 1km/s. Since that's about Mach 3, air resistance is in no way negligible, so the real answer is a lot higher (note that Wikipedia gives a muzzle velocity for the M16 of 948m/s and an effective range of about 800m). Serious aerodynamic knowledge is needed to give an accurate answer to your question.




#4
Nov1212, 10:12 PM

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C14: Absurd range how high would the muzzle velocity have to be?Also that would have the shards almost at rest when they hit the fighter  raising a "bright corona" off the shields. So  y'know, we cannot actually answer the question without knowing how strong the shields are... that would provide the upper bound. Did the marines expect to have some effect at that range? 



#5
Nov1212, 10:23 PM

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Problem is, I haven't learned how to solve equations where the variable is dependent on itself yet so I can't solve the problem at hand... 



#6
Nov1212, 10:48 PM

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... but... I gave you a bunch of links... which variable depends on itself?
The projectile starts out with speed v, it gets to height h, decelerating all the while at rate g. At h, it's speed is zero. the vt graph is, therefore, a triangle  with height v and base T, neither of which we know. The slope of the vt graph is the acceleration (g) which we know, and the area of the triangle is the height (h) which we know. 2 equations, 2 unknowns. You do it  builds character. 



#7
Nov1312, 02:03 AM

P: 372

By the way, isn't the Wraith a single seat fighter/bomber? An F15E is about 20m long. 55,000m away, that has an angular size of about 0.02 degrees. For comparison, the 10ring on a 10m olympic air rifle range has an angular size of 0.002 degrees. Scoring any hits at all on the Wraith under perfect conditions would be competitionquality shooting. These guys are getting hits in combat in a storm. I know very little about firearms, but I have my doubts about the quality of technical thought that went into this passage.
Also, I know marines are stereotypically dumb, but shooting at a vertically descending target that is right above you? Is that wise? 



#8
Nov1312, 10:56 AM

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#9
Nov1312, 03:34 PM

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I have a weak spot for hypersonic objects flying through the atmosphere (), so I tried adapting a calculation sheet I made several years ago to try and estimate the projectile range of a railgun. According to this StarCraft Wiki reference, the C14 Gauss rifle fire hypersonic rounds, so that's a good starting point.
Basically I model the projectile as a sphere, and using a basic aerodynamic drag model (F.d = 1/2*rho*v*A*C.d) and equations describing air density as a function of altitude (and gravity as a function of altitude although it is a smaller effect), I numerically solve for the trajectory. I've attached the sheet printout in case anyone is curious what it looks like. Note: I'm not sure the drag equation I used is valid for hypersonic speeds, oh well. From what I can tell from the sheet (unless there's a glaring error), a launch speed of around mach 3 (~33 kJ muzzle energy, about 65% more than a .50 BMG) would get you to a maximum altitude of about 50km, but that leaves no kinetic energy left over to do any damage. A launch speed of mach 4 (~60kJ muzzle energy, 3x that of a .50 BMG) gives a kinetic energy of around 25 kJ (about 25% higher than a .50 BMG's initial muzzle energy) at 50 KM altitude, which I would say is a pretty good hit but maybe not enough to damage heavy armor. Basically, the kind of muzzle energy you're looking at really doesn't seem like a shoulderfired weapon to me, maybe a vehiclemounted weapon? 



#10
Nov1312, 10:58 PM

P: 33

I don't understand. You're saying an M16 shooting straight up (muzzle velocity about Mach 3) will go 30 miles up before falling back?




#11
Nov1412, 12:07 AM

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#12
Nov1412, 08:48 AM

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#13
Nov1412, 09:55 AM

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I haven't checked the arithmetic, but you have the same wrong formula on page 3 of your PDF attachment. 



#14
Nov1412, 10:07 AM

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Edit Well, I tried updating the equation in the calculation sheet, but now the ODE solver isn't converging to a solution. I'll have to troubleshoot the problem later on and see what I can do, I'm guessing I might need to tweak the initial conditions of the solve. 



#15
Nov1412, 05:57 PM

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In one dimension using v * abs(v) should give you the right sign for the force. In 2D, ##v^2 = v_x^2 + v_y^2## and you have to resolve the force into the x and y components. 


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