Blast radii and orbital weaponry
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Blast radii and orbital weaponry
I need some help (don't I always)
I'm trying to work out the blast radius of some weaponry from an RPG I'm running.
One of my players fired a missile battery at a surface target from a mid-earth orbit, and while I have already dealt with it in game, I wanted to work out exactly what effect such an impactor would have.
Now, with some basic mathematics (Suvats, kinetic energy equations, and the mass of the projectile) I worked out it would be impacting with a kinetic energy of 40 gigajoules, which is something like the equivalent of ten tons of TNT. Now, I might be completely wrong, but I'm pretty sure I'm ok up too there.
I'm assuming the following:
1) missile has a mass of 500 pounds/227 kilograms
2) accelerates at 9.81m/s, for 20,000 kms (yes, I know gravity changes, but I used that as a standard, because a rocket motor is involved as well)
3)initial velocity is 0.
Note I have no idea what force the rocket motor applies, hence the continuous acceleration assumption (once the motor gives out, gravity should take over)
I got a speed of some 19,000 m/s by the time of impact
Also, a friend of mine came up with a blast radius of about 150 metres, but I have no idea how he did it.
Can anyone give me any help or point out mistakes I have made?
I'm trying to work out the blast radius of some weaponry from an RPG I'm running.
One of my players fired a missile battery at a surface target from a mid-earth orbit, and while I have already dealt with it in game, I wanted to work out exactly what effect such an impactor would have.
Now, with some basic mathematics (Suvats, kinetic energy equations, and the mass of the projectile) I worked out it would be impacting with a kinetic energy of 40 gigajoules, which is something like the equivalent of ten tons of TNT. Now, I might be completely wrong, but I'm pretty sure I'm ok up too there.
I'm assuming the following:
1) missile has a mass of 500 pounds/227 kilograms
2) accelerates at 9.81m/s, for 20,000 kms (yes, I know gravity changes, but I used that as a standard, because a rocket motor is involved as well)
3)initial velocity is 0.
Note I have no idea what force the rocket motor applies, hence the continuous acceleration assumption (once the motor gives out, gravity should take over)
I got a speed of some 19,000 m/s by the time of impact
Also, a friend of mine came up with a blast radius of about 150 metres, but I have no idea how he did it.
Can anyone give me any help or point out mistakes I have made?
"Seriously though, every time I see something like this I think 'Ooo, I'm living in the future'. Unfortunately it increasingly looks like it's going to be a cyberpunkish dystopia, where the poor eat recycled shit and the rich eat the poor." Evilsoup, on the future
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Re: Blast radii and orbital weaponry
Being 20,000 km out means you're pretty much at your final velocity by the time your projectile hits atmosphere. At 19,000 m/s more or less, I'd be pretty doubtful of the missile surviving reentry unless you're giving it some sort of handwavium heat shield.
From here...
The drag force on your missile would can be calculated using:
drag = 1/2 * (mass density of atmosphere) * (velocity2)*(cross-sectional area) * (drag coefficient)
Since the density of the atmosphere is not constant, we'll see what happens near the outer edge at say 1% of sea level density, = .01 * 1.2 kg/m3
velocity is of course 19,000 m/s
you didn't mention what the cross-sectional area of your missile was, but I'll go with a circular area with radius of 5 cm.
for drag coefficient we'll go with 0.01 which is 1/4 of the value for a general streamlined body.
so, drag = 0.5*(0.012kg/m3)*((19000m/s2)2)*((0.05m)2*pi)*0.01 = 170 N which doesn't seem to bad, but's that just the force when the atmosphere is only 1%. At 50%, it will be 8000N.
If we assuming a depth of 100 km in total and a linear distribution of density and further neglecting any deacceleration on the missile we can calculate the work done on the missile's by the atmosphere.
Work = force * distance. Mass density of atmosphere = (1.2 E-05 x)kg/m3, x is distance into the atmosphere
leaving out units, dW = force * dx = 0.5*(((1.2 E-05 x) 3)*(190002)*((0.05)2*pi)*0.01* dx
integrating we get W = (0.25*1.2 E-05)x2*(190002)*((0.05)2*pi)*0.01 = 8.51 x2
evaluating from 0 to 100,000 metres, we get W = 8.51(1000002) = 85GJ
Now this is more energy than the missile's kinetic energy before you hit the atmosphere, which can't be so, but that is the result of ignoring the deacceleration of the missile by the atmosphere. Factoring in the deacceleration really complicates things, but the above calculation gives us an idea of what the missile is going to run into when it hits the atmosphere.
From here...
That's about the speed of your missile and a hell of a lot heavier, but it lost a lot more that 227 kgs. Now it certainly started off with a cross-sectional area a lot more than your missile, but planetary atmospheres are just really good at blowing up hypervelocity impactors.The object that excavated the crater was a nickel-iron meteorite about 50 meters (54 yards) across, which impacted the plain at a speed of several kilometers per second. The speed of the impact has been a subject of some debate. Modeling initially suggested that the meteorite struck at a speed of up to 20 kilometers per second (45,000 mph), but more recent research suggests the impact was substantially slower, at 12.8 kilometers per second (28,600 mph). It is believed that about half of the impactor's 300,000 metric tons (330,000 short tons) bulk was vaporized during its descent, before it hit the ground.
The drag force on your missile would can be calculated using:
drag = 1/2 * (mass density of atmosphere) * (velocity2)*(cross-sectional area) * (drag coefficient)
Since the density of the atmosphere is not constant, we'll see what happens near the outer edge at say 1% of sea level density, = .01 * 1.2 kg/m3
velocity is of course 19,000 m/s
you didn't mention what the cross-sectional area of your missile was, but I'll go with a circular area with radius of 5 cm.
for drag coefficient we'll go with 0.01 which is 1/4 of the value for a general streamlined body.
so, drag = 0.5*(0.012kg/m3)*((19000m/s2)2)*((0.05m)2*pi)*0.01 = 170 N which doesn't seem to bad, but's that just the force when the atmosphere is only 1%. At 50%, it will be 8000N.
If we assuming a depth of 100 km in total and a linear distribution of density and further neglecting any deacceleration on the missile we can calculate the work done on the missile's by the atmosphere.
Work = force * distance. Mass density of atmosphere = (1.2 E-05 x)kg/m3, x is distance into the atmosphere
leaving out units, dW = force * dx = 0.5*(((1.2 E-05 x) 3)*(190002)*((0.05)2*pi)*0.01* dx
integrating we get W = (0.25*1.2 E-05)x2*(190002)*((0.05)2*pi)*0.01 = 8.51 x2
evaluating from 0 to 100,000 metres, we get W = 8.51(1000002) = 85GJ
Now this is more energy than the missile's kinetic energy before you hit the atmosphere, which can't be so, but that is the result of ignoring the deacceleration of the missile by the atmosphere. Factoring in the deacceleration really complicates things, but the above calculation gives us an idea of what the missile is going to run into when it hits the atmosphere.
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Re: Blast radii and orbital weaponry
That's cool.
I thought it might burn up, but I was handwaving that away to make the attack at all useful
does anyone know how to calc the damage it would do?
I thought it might burn up, but I was handwaving that away to make the attack at all useful
does anyone know how to calc the damage it would do?
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Re: Blast radii and orbital weaponry
@Korvan, I guess the numbers showing that if the "handwavium heatshield" would keep it from melting, drag would slow it down to terminal velocity. (The most kinetic damage they would do is the terminal velocity of a object of the rocket diameter, times the mass. The explosives on board would do the lion's share of damage...)
1.28 MJ/kg for a 227 kg impactor means a lousy 300 MJ would be enough to completely melt it - if it were a solid block of iron.
Since it has components - probably a warhead, and definitely a rocket engine with hollow tanks, it will break apart much faster...
These missiles won't do anything but pretty fireworks...
1.28 MJ/kg for a 227 kg impactor means a lousy 300 MJ would be enough to completely melt it - if it were a solid block of iron.
Since it has components - probably a warhead, and definitely a rocket engine with hollow tanks, it will break apart much faster...
These missiles won't do anything but pretty fireworks...
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Re: Blast radii and orbital weaponry
I may as well explain the handwavium then.
The pbp is based on the battle-star galactica series (the new one). During one episode, the BSG threatens to fire nuclear missiles at the surface of a planet. They seem to be relatively sure of their missile being able to get into the atmosphere. Therefore... I'm ignoring the air resistance, at least in terms of whether the missiles will get too the surface or not...
The pbp is based on the battle-star galactica series (the new one). During one episode, the BSG threatens to fire nuclear missiles at the surface of a planet. They seem to be relatively sure of their missile being able to get into the atmosphere. Therefore... I'm ignoring the air resistance, at least in terms of whether the missiles will get too the surface or not...
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Re: Blast radii and orbital weaponry
An ICBM usually only reaches around 7km/s maximum velocity. As velocity and drag are a squared relationship. Also, that missile is MUCH larger, and constructed to withstand re-entry (which a normal ship-to-ship missile wouldn't be) which means it can absorb much more heat before it goes pear-shaped.
I suppose that BSG missiles are much bigger, and probably much slower than your stated impactor.
I suppose that BSG missiles are much bigger, and probably much slower than your stated impactor.
A minute's thought suggests that the very idea of this is stupid. A more detailed examination raises the possibility that it might be an answer to the question "how could the Germans win the war after the US gets involved?" - Captain Seafort, in a thread proposing a 1942 'D-Day' in Quiberon Bay
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Re: Blast radii and orbital weaponry
It's kinda rules light unfortunately. All I know is that the ship carries 500 pound or 1000 pound missiles, for ship-to-ship, which can be used in ground bombardment....I suppose that BSG missiles are much bigger, and probably much slower than your stated impactor.
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Re: Blast radii and orbital weaponry
Wait a sec. You are in a mid-earth orbit, remeber? you orbital speed is somewhere between 7.7 km/s and 10 km/s (max 14 km/s since that's the Earth orbital escape speed).3)initial velocity is 0.
That's the missile's initial velocity (note that those above are kilometers per second, it's a significant speed).
Orbital bombardment FTW.
Too fast. Its main concern is resisting the impact with the atmosphere, since at those speeds it starts to look very solid.I thought it might burn up
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Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
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Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
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Re: Blast radii and orbital weaponry
Sorry. I meant that it's initial velocity in a downwards direction is 0, no? Because all of that velocity is horizontal rather than vertical... Unless I am really really bad at physics these daysThat's the missile's initial velocity (note that those above are kilometers per second, it's a significant speed).
Ah, very interesting...Too fast. Its main concern is resisting the impact with the atmosphere, since at those speeds it starts to look very solid.
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Re: Blast radii and orbital weaponry
Hell, I can imagine the missile's engine burn including a long phase of deliberately braking just to slow down for reentry. Depending on how much delta-v it has to play with, it could enter the atmosphere at much less than orbital velocity.
And useful ship to ship missiles in a BSG setting would need a lot more than 20 km/s worth of delta-v, I think.
But in that case yes, the main source of damage will be the explosive warhead. Is this missile chemical or nuclear?
Is it designed for bombardment? Does the firing missile battery have a ground attack mode? If so, there will be missiles designed to survive hypervelocity reentry; they'll probably have less delta-v than the ship to ship missiles, and use some of the extra mass for an armored nosecone that can survive reentry relatively intact.
And useful ship to ship missiles in a BSG setting would need a lot more than 20 km/s worth of delta-v, I think.
But in that case yes, the main source of damage will be the explosive warhead. Is this missile chemical or nuclear?
Is it designed for bombardment? Does the firing missile battery have a ground attack mode? If so, there will be missiles designed to survive hypervelocity reentry; they'll probably have less delta-v than the ship to ship missiles, and use some of the extra mass for an armored nosecone that can survive reentry relatively intact.
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Re: Blast radii and orbital weaponry
IIRC there were once a probe sent into Jupiter that hit the atmosphere at something like 50 km/s and performed a succesful entry. If 20th century humans can build a heatshield that can whithstand such extreme velocity then for a much more advanced civilization building a heathshield that can protect from reentry at mere 20 km/s should be trivial.
Re: Blast radii and orbital weaponry
The probe was destroyed by the entry-it was 'successful' in that destroying the probe was the intended affect, in order to to avoid possible contamination of the system by any microbes on the probe.Sky Captain wrote:IIRC there were once a probe sent into Jupiter that hit the atmosphere at something like 50 km/s and performed a succesful entry. If 20th century humans can build a heatshield that can whithstand such extreme velocity then for a much more advanced civilization building a heathshield that can protect from reentry at mere 20 km/s should be trivial.
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Re: Blast radii and orbital weaponry
The probe was destroyed, but it was intended to get data back, and it did. The mere fact that it survived long enough to get usable data back, rather than being smashed or vaporized, is suggestive.
Design of heat shields for 20 km/s reentries really shouldn't be impossible, I'd think, just difficult as an engineering challenge. The kind of thing you need specialized equipment for, and you have to accept that a lot of heat shield mass will vaporize away during entry.
Design of heat shields for 20 km/s reentries really shouldn't be impossible, I'd think, just difficult as an engineering challenge. The kind of thing you need specialized equipment for, and you have to accept that a lot of heat shield mass will vaporize away during entry.
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Re: Blast radii and orbital weaponry
I think you cofused between Galileo spacecraft and atmospheric entry probe released from it. Entry probe made a succesful entry, it slowed down to subsonic speed - a succesful atmospheric entry. t stopped transmitting data only after it had descended under parachute 150 km into Jupiter.Kryten wrote:The probe was destroyed by the entry-it was 'successful' in that destroying the probe was the intended affect, in order to to avoid possible contamination of the system by any microbes on the probe.Sky Captain wrote:IIRC there were once a probe sent into Jupiter that hit the atmosphere at something like 50 km/s and performed a succesful entry. If 20th century humans can build a heatshield that can whithstand such extreme velocity then for a much more advanced civilization building a heathshield that can protect from reentry at mere 20 km/s should be trivial.
Re: Blast radii and orbital weaponry
There's an interesting discussion we had quite awhile ago about orbital bombardment and the effect of atmosphere. One of the things that came out of that if I recall correctly, is that increasing the speed of the projectile just makes things worse and actually decreases the distance it penetrates into the atmosphere. Which does make sense as the drag is proportional to the square of the velocity. The best way to make sure you hit the surface is to just add mass, and a lot of it.
As for blast radius, I can do a very rough estimation. Lets say your impactor is 227 kgs of solid iron and it all makes it through to ground where it impacts with your value of 40 gigjolues. We're going to put all that energy into the iron. 227 kgs is about 4000 mols and the heat capacity at constant pressure for iron is about 25 J/k mol.
So the iron gets vapourized and ignoring the phase change enthalpies we get a cloud of iron vapour at about 400000 K. Now at these temperatures, a whole lot of crazy stuff that I can't begin to deal with happens, so we're just going to pretend the iron vapour is an ideal gas and go from there.
Ideal gas Law, PV = nRT, P = 1 atm, n = 4000 mols, T = 400000 K, R = 0.08206 L atm/K mol
V = 4000 * 400000 * 0.08296 = 131 Mega L = 131,000 m3 which is a hemisphere of radius = 40m.
Now this is not the blast radius, this is a hemisphere of super heated iron vapour, a zone of more or less instant annihilation. I don't know enough to calculate how this would equate into a blast radius I'm afraid, but your friends estimate of 150 seems reasonable at least.
As for blast radius, I can do a very rough estimation. Lets say your impactor is 227 kgs of solid iron and it all makes it through to ground where it impacts with your value of 40 gigjolues. We're going to put all that energy into the iron. 227 kgs is about 4000 mols and the heat capacity at constant pressure for iron is about 25 J/k mol.
So the iron gets vapourized and ignoring the phase change enthalpies we get a cloud of iron vapour at about 400000 K. Now at these temperatures, a whole lot of crazy stuff that I can't begin to deal with happens, so we're just going to pretend the iron vapour is an ideal gas and go from there.
Ideal gas Law, PV = nRT, P = 1 atm, n = 4000 mols, T = 400000 K, R = 0.08206 L atm/K mol
V = 4000 * 400000 * 0.08296 = 131 Mega L = 131,000 m3 which is a hemisphere of radius = 40m.
Now this is not the blast radius, this is a hemisphere of super heated iron vapour, a zone of more or less instant annihilation. I don't know enough to calculate how this would equate into a blast radius I'm afraid, but your friends estimate of 150 seems reasonable at least.
Re: Blast radii and orbital weaponry
The battlestar wiki implies that the Galactica does carry a handful of nuclear-tipped missiles designed for orbit-to-surface use. I seem to remember hearing that the physical dimensions of the missiles placed an upper yield limit of 150 KT, but I don't have a source for that.Simon_Jester wrote:Hell, I can imagine the missile's engine burn including a long phase of deliberately braking just to slow down for reentry. Depending on how much delta-v it has to play with, it could enter the atmosphere at much less than orbital velocity.
And useful ship to ship missiles in a BSG setting would need a lot more than 20 km/s worth of delta-v, I think.
But in that case yes, the main source of damage will be the explosive warhead. Is this missile chemical or nuclear?
Is it designed for bombardment? Does the firing missile battery have a ground attack mode? If so, there will be missiles designed to survive hypervelocity reentry; they'll probably have less delta-v than the ship to ship missiles, and use some of the extra mass for an armored nosecone that can survive reentry relatively intact.
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Re: Blast radii and orbital weaponry
Yup. But by thrusting perpendicularly to the surface of the planet (either up or down, irrelevant) you make your orbit more elliptical, until it becomes so eccentric that the orbit would carry it through the Earth's surface from a more or less perpendicular direction. That's a pretty easy feat.Sorry. I meant that it's initial velocity in a downwards direction is 0, no? Because all of that velocity is horizontal rather than vertical...
Uh? In this case the problem isn't successful reentry, but reaching the ground at a fast enough speed to deal significant damage.IIRC there were once a probe sent into Jupiter that hit the atmosphere at something like 50 km/s and performed a succesful entry.
That probe entered atmosphere at a very small angle, so that the "descent" is actually a spiral inward as the atmosphere brakes the craft. It was intended to use the atmosphere to brake the craft. I mean, Apollo anyone?
You cannot compare it to a thing supposed to go down at 19 km/s + some more from orbital speed.
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Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
--
Stereotypical spacecraft are pressurized.
Less realistic spacecraft are pressurized to hold breathing atmosphere.
Realistic spacecraft are pressurized because they are flying propellant tanks. -Isaac Kuo
--
Good art has function as well as form. I hesitate to spend more than $50 on decorations of any kind unless they can be used to pummel an intruder into submission. -Sriad
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Re: Blast radii and orbital weaponry
In some respects, going down straight through the atmosphere might actually be easier; the heat shield only has to stop you from breaking up during a few seconds' atmospheric friction, rather than for hundreds of seconds.
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Re: Blast radii and orbital weaponry
The system is certainly designed for it. The battery has a ground attack mode. It's a chemical explosive, the Admiralty decided not to let a destroyer captain play with nukes just yet, especially since the current mission is a rescue oneSimon_Jester wrote:And useful ship to ship missiles in a BSG setting would need a lot more than 20 km/s worth of delta-v, I think.
But in that case yes, the main source of damage will be the explosive warhead. Is this missile chemical or nuclear?
Is it designed for bombardment? Does the firing missile battery have a ground attack mode? If so, there will be missiles designed to survive hypervelocity reentry; they'll probably have less delta-v than the ship to ship missiles, and use some of the extra mass for an armored nosecone that can survive reentry relatively intact.
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Re: Blast radii and orbital weaponry
If the battery has dedicated ground-attack missiles, then it's fair to assume that those missiles are designed to enter planetary atmosphere at speeds on the order of 10 km/s and penetrate without burning up entirely- though loss of mass due to ablation on the nose could be significant. So I'd expect them to hit with a large fraction of their 'possible' kinetic energy, somewhere between 50% and 100% depending on how generous you want to be about the tech level that went into designing the missiles.
The chemical warhead, given the missile's stated weight, will be an afterthought.
The chemical warhead, given the missile's stated weight, will be an afterthought.
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Re: Blast radii and orbital weaponry
That sounds about right
So I'm good with the blast radius?
Does anyone know the formula he used?
So I'm good with the blast radius?
Does anyone know the formula he used?
"Seriously though, every time I see something like this I think 'Ooo, I'm living in the future'. Unfortunately it increasingly looks like it's going to be a cyberpunkish dystopia, where the poor eat recycled shit and the rich eat the poor." Evilsoup, on the future
StarGazer, an experiment in RPG creation
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