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Post by The Astronomer on Oct 28, 2017 15:32:02 GMT
This can be defeated with very thin whipple shields. You need penetrators of decent sectional density and overall momentum to get through whipple shields. Wave tactics. Obviously, one could use rotating armor schemes/roll ship or carry additional Whipple layers; but swarms of cheap tiny micromissiles boosted in at high relative velocity are less mass/cost-intensive than the structural requirements for armor, especially if your ship has significant acceleration (even rotational acceleration!) AKA 'hit it 'til it dies' tactic. No matter how weak you are, just do it.
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Post by matterbeam on Oct 28, 2017 16:18:54 GMT
Kerr: We must establish a standard basis for comparison. We will establish the missile as having 12km/s of deltaV, powered by a 9km/s exhaust velocity nuclear thermal rocket. It accelerates to 10km/s at the target and uses 2km/s of deltaV to match lateral accelerations. The dry mass is assumed to be 100kg and the mass ratio is 3.8. The majority of the dry mass will be the fragmentation warhead. We want dense, cheap fragments, so let's use iron. The target is a cylinder 30m long and 10m wide. It has a 10cm layer of carbon as its final layer of armor. This masses 161.8 tons. The RCC has a density of 1700kg/m^3 and a yield strength of 700MPa. Let's aim for a projectile that can penetrate 10cm of RCC armor. This can be a 2.5cm long iron nail or a 6.6g projectile at 10km/s. We'll use the latter as it is lighter. Let's say the missile follows the target spaceship up to point-defense ranges and releases the fragments. The fragments must reach the target before the target can evade their dispersion area. So, what's the target's acceleration? Let's calculate a table of values for the release ranges and the accelerations. Modifying the closing velocity too would lead to matrices I'm unwilling to attempt to display. Anyhow, we know that the target spaceship will accelerate laterally. Accelerating towards or away from the fragments would have little to no effect. For this same reason, there is no need for a three-dimensional cloud, only a disk of projectiles. Lateral acceleration can be very low (RCS thrusters) or an emergency thrust of the combat engines. Let's set the acceleration from 0.01g to 3g. The point defense range depends on the effective range of high rotation and tracking rate weaponry. Railguns, reactive armor, lasers, phased arrays, electronically steered particle beams... there's quite a selection! We'll set the ranges from 10 to 1000km. Finally, the biggest challenge to high velocity projectiles is thin sheets of armor spaced far from the main hull. These Whipple shields turn the projectiles' kinetic energy against them. How thick does a whipple shield have to be to destroy the projectiles? Let's aim for straight up vaporization upon impact. This requires that the iron be excited with 7.8MJ/kg from cold. For our 6.6g projectile, we need an impact that releases 47.4kJ. This means it has to intersect a section of Whipple shield massive enough to release 47.4kJ of energy. A 6.6g iron projectile can be a sphere 6mm in diameter and have a cross-section of 0.00002826m^2. At 10km/s, it needs to hit 0.95g of material. An aluminium foil of depth 12.45mm is sufficient. A 0.5m separation between this whipple shield and the spacecraft's main hull means that the whipple shield covering the flanks would mass 34.8 tons. You suggest staging the fragments so that the second stage would pass through the holes in the whipple shield created by the first stage. A stage would be a disk of fragments just behind the first disk. One stage table: The values are in kg for 1 fragment/m^2. In red are the values where the fragmentation warhead is more massive than the entire Whipple shield. With two stages, the fragmentation warhead doubles in mass while the Whipple shield adds another layer of 38 tons for a total 72.8 tons. Two stage table: No change? That's because the fragmentation warhead mass increases linearly while the whipple shield mass increases at a rate of (R^2 - r^2). My conclusion is that the mass of the missile becomes incredibly high as the Whipple shields are added. If the Whipple shields are rotated, at least double the mass of the warhead required. For example, a two-stage fragmentation cloud released 100km from a target accelerating a 10m/s^2 would mass 10.36 tons. The missile that delivers it has a total mass at launch of 39 tons. This is a large fraction of the Whipple shield mass required to perfectly defend against the missile. The attacking ships must bring along roughly as much mass as the defenders do! When we consider the deltaV requirements for bringing all the missiles up to the target's position, and the addition of other types of defenses, the ratio is strongly in the defender's favour. For example, if the target ship's defenses can handle 100 missiles, can roll its shields between attacks, and the attacker needs to make a 7km/s trip to reach the battlefield plus 2km/s of maneuvers to set up an intercept, then the mass ratio between missiles and replaceable Whipple shields is at least 294:1. This is assuming that one single missile getting through the Whipple shields is enough to kill a target, and that the two ships use the same 9km/s exhaust velocity rockets as the missile.
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Post by matterbeam on Oct 28, 2017 16:28:13 GMT
bigbombr: Laser fire at very long range has a maximum effectiveness - the intensity at which the target's armor does not heat up. This is strongly affected by the target's reflectivity and slightly by the thermal conductivity of the armor material. Laser turrets aren't a good idea when facing literal millions of tiny fragments. You'll want something like a phased array that can electronically steer a beam in milliseconds. The pellet gun is an offensive weapon but percentage C velocities are not necessary - just hitting the target should be sufficient. It become a highly efficient pulsed laser that doesn't disperse at range, with the output of your weapon arriving in packets at your target. EshaNas: Pretty much, yes. Kerr: The plasma released by a laser strike is usually too cold to absorb the laser wavelengths. Plasma transparency decreases with increasing temperature, but unless you are creating million K plasmas, you can assume that they are quite transparent.
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Post by Kerr on Oct 28, 2017 17:14:20 GMT
Kerr: We must establish a standard basis for comparison. We will establish the missile as having 12km/s of deltaV, powered by a 9km/s exhaust velocity nuclear thermal rocket. It accelerates to 10km/s at the target and uses 2km/s of deltaV to match lateral accelerations. The dry mass is assumed to be 100kg and the mass ratio is 3.8. The majority of the dry mass will be the fragmentation warhead. We want dense, cheap fragments, so let's use iron. The target is a cylinder 30m long and 10m wide. It has a 10cm layer of carbon as its final layer of armor. This masses 161.8 tons. The RCC has a density of 1700kg/m^3 and a yield strength of 700MPa. Let's aim for a projectile that can penetrate 10cm of RCC armor. This can be a 2.5cm long iron nail or a 6.6g projectile at 10km/s. We'll use the latter as it is lighter. Let's say the missile follows the target spaceship up to point-defense ranges and releases the fragments. The fragments must reach the target before the target can evade their dispersion area. So, what's the target's acceleration? Let's calculate a table of values for the release ranges and the accelerations. Modifying the closing velocity too would lead to matrices I'm unwilling to attempt to display. Anyhow, we know that the target spaceship will accelerate laterally. Accelerating towards or away from the fragments would have little to no effect. For this same reason, there is no need for a three-dimensional cloud, only a disk of projectiles. Lateral acceleration can be very low (RCS thrusters) or an emergency thrust of the combat engines. Let's set the acceleration from 0.01g to 3g. The point defense range depends on the effective range of high rotation and tracking rate weaponry. Railguns, reactive armor, lasers, phased arrays, electronically steered particle beams... there's quite a selection! We'll set the ranges from 10 to 1000km. Finally, the biggest challenge to high velocity projectiles is thin sheets of armor spaced far from the main hull. These Whipple shields turn the projectiles' kinetic energy against them. How thick does a whipple shield have to be to destroy the projectiles? Let's aim for straight up vaporization upon impact. This requires that the iron be excited with 7.8MJ/kg from cold. For our 6.6g projectile, we need an impact that releases 47.4kJ. This means it has to intersect a section of Whipple shield massive enough to release 47.4kJ of energy. A 6.6g iron projectile can be a sphere 6mm in diameter and have a cross-section of 0.00002826m^2. At 10km/s, it needs to hit 0.95g of material. An aluminium foil of depth 12.45mm is sufficient. A 0.5m separation between this whipple shield and the spacecraft's main hull means that the whipple shield covering the flanks would mass 34.8 tons. You suggest staging the fragments so that the second stage would pass through the holes in the whipple shield created by the first stage. A stage would be a disk of fragments just behind the first disk. One stage table: The values are in kg for 1 fragment/m^2. In red are the values where the fragmentation warhead is more massive than the entire Whipple shield. With two stages, the fragmentation warhead doubles in mass while the Whipple shield adds another layer of 38 tons for a total 72.8 tons. Two stage table: No change? That's because the fragmentation warhead mass increases linearly while the whipple shield mass increases at a rate of (R^2 - r^2). My conclusion is that the mass of the missile becomes incredibly high as the Whipple shields are added. If the Whipple shields are rotated, at least double the mass of the warhead required. For example, a two-stage fragmentation cloud released 100km from a target accelerating a 10m/s^2 would mass 10.36 tons. The missile that delivers it has a total mass at launch of 39 tons. This is a large fraction of the Whipple shield mass required to perfectly defend against the missile. The attacking ships must bring along roughly as much mass as the defenders do! When we consider the deltaV requirements for bringing all the missiles up to the target's position, and the addition of other types of defenses, the ratio is strongly in the defender's favour. For example, if the target ship's defenses can handle 100 missiles, can roll its shields between attacks, and the attacker needs to make a 7km/s trip to reach the battlefield plus 2km/s of maneuvers to set up an intercept, then the mass ratio between missiles and replaceable Whipple shields is at least 294:1. This is assuming that one single missile getting through the Whipple shields is enough to kill a target, and that the two ships use the same 9km/s exhaust velocity rockets as the missile. Great! That's some blog post quality stuff! How well does this work against tandem/spearhead missile formations? What if there are breacher submunition used? Small Nukes/Guided heavy fragments. How can it be avoided that missiles hide behind the shields shadow? What if the nails have impact driven seperation charges? Also 12km/s is pretty good in CDE and near future standards, but my thinking was in terms of MicroAIM's. Few hundred km/s delta-v, and in quantities of a thiusand.
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Post by Kerr on Oct 28, 2017 17:18:19 GMT
bigbombr: Laser fire at very long range has a maximum effectiveness - the intensity at which the target's armor does not heat up. This is strongly affected by the target's reflectivity and slightly by the thermal conductivity of the armor material. Laser turrets aren't a good idea when facing literal millions of tiny fragments. You'll want something like a phased array that can electronically steer a beam in milliseconds. The pellet gun is an offensive weapon but percentage C velocities are not necessary - just hitting the target should be sufficient. It become a highly efficient pulsed laser that doesn't disperse at range, with the output of your weapon arriving in packets at your target. EshaNas: Pretty much, yes. Kerr: The plasma released by a laser strike is usually too cold to absorb the laser wavelengths. Plasma transparency decreases with increasing temperature, but unless you are creating million K plasmas, you can assume that they are quite transparent. Matterbeam, the plasma would also have many liquid/solid fragments of armor in it which will absorb most of the frequencies. PS: You seem to like this overall discussion here, literally.
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Post by newageofpower on Oct 28, 2017 21:18:26 GMT
matterbeam Rather than a large 9km/s missile, consider a 5-6 km/s missile package being boosted in on a nuclear electric bus with above 20km/s. ( For example, my clone of the Tukuro missile) You require more than double the fuel to double the dV for high thrust propulsion.
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Post by Kerr on Oct 28, 2017 21:47:30 GMT
matterbeam Rather than a large 9km/s missile, consider a 5-6 km/s missile package being boosted in on a nuclear electric bus with above 20km/s. ( For example, my clone of the Tukuro missile) You require more than double the fuel to double the dV for high thrust propulsion. Wouldn't a nuclear closed-gas core make more sense? Higher acceleration and comparable Dv.
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Post by matterbeam on Oct 28, 2017 23:27:55 GMT
Great! That's some blog post quality stuff! How well does this work against tandem/spearhead missile formations? What if there are breacher submunition used? Small Nukes/Guided heavy fragments. How can it be avoided that missiles hide behind the shields shadow? What if the nails have impact driven seperation charges? Also 12km/s is pretty good in CDE and near future standards, but my thinking was in terms of MicroAIM's. Few hundred km/s delta-v, and in quantities of a thiusand. Thanks. The staged fragmentation warhead is technically a tandem warhead, but sub-optimal. You could improve your penetration efficiency through multiple whipple shields by dividing the projectiles into stacks of thin disks that precede the main projectile. Each disk blows a hole through the whipple shield large enough to let the rest of the stack through. Instead of multiplying the fragmentation warhead's mass by each Whipple shield layer, you only increase it by a small fraction. However, the enemy can defeat this tactics by using paper-thin layers of Whipple shield on the outside, and thick anti-main-projectile layers on the inside. Of course, these thinner layers become very vulnerable to laser strikes preceding the missile wave... and so on. I'm not sure what you mean by 'spearhead missile formations'. A directed explosive charge preceding a bunch of missiles is a very effective tactic. It would create a large hole through multiple whipple shields by using explosive force to push each layer down and on top of each other. The problem comes from the poor coupling between the explosive charge and the first Whipple shield layer - there is no air to transmit the explosive force, so it can only rely on the momentum transfer of high velocity gasses. These are more likely to bounce off then push the Whipple shield efficiently. Impact driven separation charges aren't terribly reliable when the impact has enough energy to vaporize all of the missile's components. At a few hundred km/s, there's an interesting tactic for defeating incoming kinetic projectiles. A thin plasma can be held a few dozen meters away from the target spaceship. This plasma is enough to vaporize incoming projectiles and then ionize the gasses. These ionized gasses are strongly affected by magnetic fields - they can be deflected. Here's an example: 300km/s incoming projectile, made of iron and massing 10kg. It can be a cylinder 10cm wide and 18cm long. At one atmospheric pressure, a room-temperature plasma will contain 1.2kg/m^3. The projectile traversing this plasma would encounter 9.42 grams of plasma per meter depth. This releases 423.9MJ of energy. Since the projectile crosses this meter in about 3.3 microseconds, it is safe to say that it absorbs all of this energy before dispersing. 423.9MJ of energy is enough to vaporize the iron projectile (-78MJ) and then heat up the rest to (very roughly) 67000K. There's about 1.07e26 atoms in that projectile. They gain an energy of 24.5eV. That's enough to ionize iron twice over and then some. I don't know the magnetic field equations nor can I quickly determine the answer to 'what magnetic field strength is needed to contain a plasma of temperature A of conductivity B and composition C to a pressure D at a distance E?' nor can I determine 'how is a particle of mass F with a charge G travelling at a velocity H at angle I to a magnetic field of strength J for a distance K displaced by a distance L and a final lateral velocity M?'.
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Post by matterbeam on Oct 28, 2017 23:32:05 GMT
Kerr: The energy being absorbed from a laser by a target material is constant. It either goes to vaporizing the material (lowest penetration rate per energy absorbed) or physically deforming and ejecting chunks of the material (highest penetration rate of energy absorbed). Any liquid or solid bits escaping the area under laser attack means that some of the laser's energy was not spent vaporizing the target. This can only improve the penetration rate, even if non-intuitively the chunks absorb part of the beam for a fraction of a millisecond as they escape at hypersonic velocity. newageofpower: Switching around the propulsion types adds more variables, but would help reduce the mass disadvantage on the attacker's side.
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Post by samchiu2000 on Oct 29, 2017 1:21:44 GMT
Glad to see my thread led to a such high quality conversation Just one question: how heavy will these pellet gun are? Either the 12 km/s or 300km/s one. For me micro missile horde is still my favorite, followed by laser.
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Post by matterbeam on Oct 29, 2017 1:45:38 GMT
Glad to see my thread led to a such high quality conversation Just one question: how heavy will these pellet gun are? Either the 12 km/s or 300km/s one. For my micro missile horde is still my favorite, followed by laser. That depends on at least 50 different factors I can think of right now, one of which is 'how big do you think big is?' and another is 'what's the median salary in the future?'
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Post by samchiu2000 on Oct 29, 2017 1:48:25 GMT
Glad to see my thread led to a such high quality conversation Just one question: how heavy will these pellet gun are? Either the 12 km/s or 300km/s one. For my micro missile horde is still my favorite, followed by laser. That depends on at least 50 different factors I can think of right now, one of which is 'how big do you think big is?' and another is 'what's the median salary in the future?' I mean the mass of the gun itself.
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Post by matterbeam on Oct 29, 2017 2:30:34 GMT
That depends on at least 50 different factors I can think of right now, one of which is 'how big do you think big is?' and another is 'what's the median salary in the future?' I mean the mass of the gun itself. I mean the same. Think about how you would answer the question today: what is the mass of a gun? Not any gun, mind you, but specifically the one where a chemical propellant encased in a cartridge and accelerates bullets from a firing chamber. That's pretty specific. How much does it mass?
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Post by samchiu2000 on Oct 29, 2017 3:10:10 GMT
I mean the mass of the gun itself. I mean the same. Think about how you would answer the question today: what is the mass of a gun? Not any gun, mind you, but specifically the one where a chemical propellant encased in a cartridge and accelerates bullets from a firing chamber. That's pretty specific. How much does it mass? Alright fine and i don't know. I just wanna know the cost effectiveness of the pellet gun when compare with missile and laser.
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Post by EshaNas on Oct 29, 2017 4:30:54 GMT
All off these qualms is why I'm leaning back towards missiles.
Think of it this way: if we absolutely needed a weapon in space *now*, for whatever reason, we're stuck with either huge laserdrones, missile swarms, BEAR neutron beam drones and the like, or gunpowder. And of all those, the missile swarm seems the most versatile and implementable enmasse.
This will, of course, change with time and new technology, but that thereof becomes a question of mathematics clashing with the theoretical.
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