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C-14: Absurd range- how high would the muzzle velocity have to be?

by Galorian
Tags: absurd, muzzle, range, velocity
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Galorian
#1
Nov11-12, 01:34 PM
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I've been debating various Sci-Fi vs debates for a while now and I've decided to try and figure out how powerful the C-14 Gauss assault rifle from StarCraft needs to be in order to achieve the following feat:

"a complex mass of conflicting pressure systems and staggering
updrafts that have created what is, in simple terms, a stationary,
volatile vortex of immense size and unlimited life. It is four leagues
wide and over twenty high."
-Uprising
out of this predicament when suddenly the whole environment
darkened.
There wasn’t much light inside the funnel anyway,
but when the Wraith fighter and four dropships
appeared at the funnel’s top, it had the effect of eclipsing
what little light there was. Somo peered up and watched
the craft descend. Once they reached the funnel’s
halfway point, the turret came to life. The Wraith positioned
itself to take the entirety of the barrage of automatic
missile fire. Bright coronas erupted around the
fighter as several missiles struck the outer shields. There
were smaller impacts as well, and Somo realized that the
marines must have begun firing at the fighter . . . but at
least that meant the marines were distracted.
-Uprising
A league is 5.556km, so halfway up a 20 league vortex would be 55.56km from the ground.

The standard issue C-14 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 C-14 spike to reach 55.56km straight up?
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Simon Bridge
#2
Nov11-12, 07:30 PM
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what would be the minimal muzzle velocity of a C-14 spike to reach 55.56km straight up?
Ballistics problem - you will get an upper-bound by neglecting air resistance. After that it is down to your model for air.

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
Ibix
#3
Nov12-12, 05:58 PM
P: 378
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)mv2=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.

Simon Bridge
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Nov12-12, 10:12 PM
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C-14: Absurd range- how high would the muzzle velocity have to be?

I presume Simon means a lower bound.
Um - yes... we have the height, so sans air resistance gives the minimum initial velocity needed to get there grrrr.
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?
Galorian
#5
Nov12-12, 10:23 PM
P: 9
Quote Quote by Simon Bridge View Post
Um - yes... we have the height, so sans air resistance gives the minimum initial velocity needed to get there grrrr.
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?
The end result is probably going to be ridiculously high anyway so the minimal initial velocity required to reach the target would do.

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...
Simon Bridge
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Nov12-12, 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 v-t graph is, therefore, a triangle - with height v and base T, neither of which we know. The slope of the v-t 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.
Ibix
#7
Nov13-12, 02:03 AM
P: 378
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 10-ring 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 competition-quality 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?
Galorian
#8
Nov13-12, 10:56 AM
P: 9
Quote Quote by Ibix View Post
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 10-ring 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 competition-quality 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?
Another book has marines hold fire until the Zerglings were a km away to make every shot count and when they opened fire the first person perspective of one of them described the targeting system as "eerily effective" as he rapidly switched targets between skittering zerglings.
Mech_Engineer
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Nov13-12, 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 C-14 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 shoulder-fired weapon to me, maybe a vehicle-mounted weapon?
Attached Files
File Type: pdf Mathcad - DU Hypersonic Round Shot Straight Up.pdf (157.9 KB, 9 views)
chasrob
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Nov13-12, 10:58 PM
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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?
Galorian
#11
Nov14-12, 12:07 AM
P: 9
Quote Quote by Mech_Engineer View Post
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 C-14 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 shoulder-fired weapon to me, maybe a vehicle-mounted weapon?
It is meant to be fired by men in powered armor and there's probably inertial dampening involved (the C-14 rather blatantly delivers more impulse to the target than recoil to the shooter- there are cases where one marine "held down" another by shooting at him full auto, causing the impact force of the spikes to prevent him from moving despite the spikes' failure to penetrate the armor).
Mech_Engineer
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Nov14-12, 08:48 AM
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Quote Quote by chasrob View Post
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?
I agree it sounds crazy. I'm going to try comparing the calculations (in a horizontal configuration) to ballistic data for a couple of rounds and see if I can come up with some correlation. I kind of get the feeling the drag force isn't high enough in the calculation, artificially inflating the range estimates.
AlephZero
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Nov14-12, 09:55 AM
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Quote Quote by Mech_Engineer View Post
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.
The drag formula should be v squared, not v.

I haven't checked the arithmetic, but you have the same wrong formula on page 3 of your PDF attachment.
Mech_Engineer
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Nov14-12, 10:07 AM
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Quote Quote by AlephZero View Post
The drag formula should be v squared, not v.

I haven't checked the arithmetic, but you have the same wrong formula on page 3 of your PDF attachment.
There you go, that could be the difference!

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.
AlephZero
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Nov14-12, 05:57 PM
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Quote Quote by Mech_Engineer View Post
Well, I tried updating the equation in the calculation sheet, but now the ODE solver isn't converging to a solution.
Be careful with the signs. v squared is always positive, whether the bullet is moving up or down!

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.
Mech_Engineer
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Nov14-12, 06:25 PM
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Good call, I'll take a look at it on Friday if I can get to it.


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