Some help with math. :)
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Some help with math. :)
In a fanfic I'm writing a weapon is developed that uses a ball of energy that when in contact with a ship's hull\shields it pulls it towards the ball of energy. Think of it as a black hole gun. But in this case it is a ball of energy and the gravitational forces is equaled to 100,000 times Earth standard. So form poking around the internet, the gravitational force on Earth is around 9.8meters per second squared. Would this mean that the weapons gravitational force is 980km per seconds squared?
Here is the potentially stupid question. How much energy would a ship's shield need to resist the weapon without collapsing? Another potential stupid question, is there a way to equate the weapon's gravitational force to a yield?
Before anyone asks, this is mainly for myself with some generalization used for the fic.
Here is the potentially stupid question. How much energy would a ship's shield need to resist the weapon without collapsing? Another potential stupid question, is there a way to equate the weapon's gravitational force to a yield?
Before anyone asks, this is mainly for myself with some generalization used for the fic.
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Re: Some help with math. :)
Can't be done without knowing how shields work.
Nothing material is going to stand up well under that kind of force, and if you're using handwavium force fields then there's no way to tell how much power they consume. The answer might be "nothing," it might be "a jillion."
And since the gravity weapon doesn't have an explosive yield, does not melt or tear apart materials and disperse them by pumping energy into them, trying to express a yield in tons of TNT equivalency is useless- there's nothing TNT-equivalent about the weapon.
Nothing material is going to stand up well under that kind of force, and if you're using handwavium force fields then there's no way to tell how much power they consume. The answer might be "nothing," it might be "a jillion."
And since the gravity weapon doesn't have an explosive yield, does not melt or tear apart materials and disperse them by pumping energy into them, trying to express a yield in tons of TNT equivalency is useless- there's nothing TNT-equivalent about the weapon.
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Re: Some help with math. :)
You know, the gravity of an object is proportional to its mass and inversely proportional to the distance from the center-of-mass. If the Earth were squeezed into a ball one half its current radius, then the surface gravity would be four times as great. At a distance from the center-of-mass equal to a planet's radius, the gravity becomes 1/4 of what it is at the surface (since you're twice as far away from the center). This is how black holes have such a huge surface gravity: since they collapse in on themselves, you're dividing by a very small number. The "surface" of the black hole is simply the point at which the gravity becomes high enough to trap light. OTOH, their gravity drops off to "normal" levels very quickly, and at further distances the gravity is the same as it was prior to its collapse into a black hole (not counting the mass that got blown away in the supernova). Thus, if the Sun were magically squeezed into a black hole now (without a supernova), the planets would continue orbiting it like nothing had happened, except that they wouldn't receive any more light from it.
You can't really compare gravity with yield rates directly since an object doesn't retain it's gravity by expending energy. You might conceivably calculate the gravitational potential energy an object receives from being in the artificial gravity field and inferring that the device somehow generates this. Then the energy requirement would become a function of the mass of nearby objects affected by the device. Of course, this would be a "soft" sci-fi solution. Alternatively, assert that the device actually increases the energy density of a region in space to form an actual black hole. Of course, once you do that, it is probably going to stay there until it evaporates in Hawking radiation (and if it's a low-mass black hole, this happens very quickly and explosively).
You can't really compare gravity with yield rates directly since an object doesn't retain it's gravity by expending energy. You might conceivably calculate the gravitational potential energy an object receives from being in the artificial gravity field and inferring that the device somehow generates this. Then the energy requirement would become a function of the mass of nearby objects affected by the device. Of course, this would be a "soft" sci-fi solution. Alternatively, assert that the device actually increases the energy density of a region in space to form an actual black hole. Of course, once you do that, it is probably going to stay there until it evaporates in Hawking radiation (and if it's a low-mass black hole, this happens very quickly and explosively).
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Re: Some help with math. :)
You're mixing around your term and units a bit. The 9.8 m/s^2 is actually the acceleration due to the gravitational force. The force would be that acceleration times the mass of the object the force is acting on. So your weapon would cause an acceleration of 980km/s^2, or 100,000 g's. The actual force would depend on the mass of the ship being affected.
Also, it take no energy at all to resist a force as long as you are not moving away from that force. For instance, standing still I am still experiencing earth's full gravitational force, but I expend no energy to do so. If I jump, climb up stairs, a hill, etc then I do expend energy. So providing that the shield can hold against the force, you ship will not expend any energy, unless it attempts to pull away from the energy ball.
Force does not equal energy, two different things, so you can't really equate the weapon's gravitational force to a yield. You could take the mass of the black hole thingy and calculate the energy needed to disperse it so every particle is an infinite distance but I'm not sure what use that would be to you.
What you could do I guess, is say the ships shields generate an opposing force to counteract the black hole and then assign an inefficiency in the process to generate the shields, so keeping the shields up does consume energy. This allows you keep the shields active as long as the plot requires.
Also, it take no energy at all to resist a force as long as you are not moving away from that force. For instance, standing still I am still experiencing earth's full gravitational force, but I expend no energy to do so. If I jump, climb up stairs, a hill, etc then I do expend energy. So providing that the shield can hold against the force, you ship will not expend any energy, unless it attempts to pull away from the energy ball.
Force does not equal energy, two different things, so you can't really equate the weapon's gravitational force to a yield. You could take the mass of the black hole thingy and calculate the energy needed to disperse it so every particle is an infinite distance but I'm not sure what use that would be to you.
What you could do I guess, is say the ships shields generate an opposing force to counteract the black hole and then assign an inefficiency in the process to generate the shields, so keeping the shields up does consume energy. This allows you keep the shields active as long as the plot requires.
Re: Some help with math. :)
You're coming at this the wrong way for elf magic weapons. You don't start with a figure and then try to work out effects from it, because you're using elf magic and it is by its very nature absurd. You start with what you want it to do, then make some general ballpark estimates to make sure that it matches the rest of what your universe can do. For instance, if your moon-sized battlestation needs to blow up planets, your star destroyers need to be able to vaporize asteroids and make a serious mess of the surface of a planet. You don't say that your Death Star needs to have a firepower of 1.21 jiggawatts, then try to figure out what its effect will be on the secret rebel base.
Re: Some help with math. :)
If you really must have some concrete description of an absurd magical item like this ("ball of energy"? like Dragonball Z or some shit like that?), you might want to consider the damage done by tidal forces. Presumably this "energy ball" has some concrete radius; then you can model a gravitational acceleration field as rapidly going from 0 just outside the ball to -gr/|r| just inside the ball, where r is the position vector in a coordinate system centered on the center of the ball and g is whatever big number you want. So what you'd really have going on is when this energy ball comes in "contact" with any object is the gigantic gradient on its edge would produce enormous tidal forces in the object.
If you want the tidal forces to not just happen on the edge of the ball, scale the gravitational field by some nice function like
|r|^2/(1+|r|^2).
If you want the tidal forces to not just happen on the edge of the ball, scale the gravitational field by some nice function like
|r|^2/(1+|r|^2).
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Re: Some help with math. :)
Ach! This makes my head hurt. Would it help if the "ball of energy" is 50cm in diameter and is fired at 10x the speed of light.Surlethe wrote:If you really must have some concrete description of an absurd magical item like this ("ball of energy"? like Dragonball Z or some shit like that?), you might want to consider the damage done by tidal forces. Presumably this "energy ball" has some concrete radius; then you can model a gravitational acceleration field as rapidly going from 0 just outside the ball to -gr/|r| just inside the ball, where r is the position vector in a coordinate system centered on the center of the ball and g is whatever big number you want. So what you'd really have going on is when this energy ball comes in "contact" with any object is the gigantic gradient on its edge would produce enormous tidal forces in the object.
If you want the tidal forces to not just happen on the edge of the ball, scale the gravitational field by some nice function like
|r|^2/(1+|r|^2).
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Re: Some help with math. :)
For that you'll have to specify how your in-universe FTL mechanism works, otherwise it has no meaning because you would need 10*infinity = infinity amount of energy to overcome relativistic effects. Could become complicated if your FTL is sensitive to gravity. And if you have a "stuttering" FTL, where 10x light speed is actually a "pseudovelocity" i.e. a result of many short jumps in sequence, then you don't have any momentum other than the one you had before activating FTL.Enigma wrote:is fired at 10x the speed of light.
edit: typo
Re: Some help with math. :)
something fired at 10x the speed of the light?
the planet behind the target explodes, one week ago.
seriously, you're breaking too many rules to even start with a numerical working.
the planet behind the target explodes, one week ago.
seriously, you're breaking too many rules to even start with a numerical working.
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Re: Some help with math. :)
That's not really relevant to what I said, so no.Would it help if the "ball of energy" is 50cm in diameter
You should probably keep your sci-fi non-retarded if you want to apply math to it.is fired at 10x the speed of light.
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Re: Some help with math. :)
Dumping it then. Still thinking of a light speed weapon (as in 1x speed of light. Think of it as a ls railgun.). During lunch, I had an about using for a projectile a torpedo casing filled with 25 kilograms of deuterium and anti-deuterium so when it hits a target, the m/am mixes and goes kaboom. At least something like that.
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Re: Some help with math. :)
What it comes down to is that the damage mechanism, the "at the speed of light stuff," and so on are so... wacky by the standards of known physics, that applying the equations of known physics to them produces nonsensical results.
Frankly, I think you should start by figuring out how much damage you want the weapon to do (not "TNT equivalent yield," damage in the practical sense of what size hole it leaves in things). Then, if you really want to, start waving in figures like "ball is X centimeters wide and travels at the speed blah blah blah."
Because there's no way we're going to be able to construct a useful description of the effects of such a weapon from a handful of arbitrary parameters you plunked down on a page.
Frankly, I think you should start by figuring out how much damage you want the weapon to do (not "TNT equivalent yield," damage in the practical sense of what size hole it leaves in things). Then, if you really want to, start waving in figures like "ball is X centimeters wide and travels at the speed blah blah blah."
Because there's no way we're going to be able to construct a useful description of the effects of such a weapon from a handful of arbitrary parameters you plunked down on a page.
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Re: Some help with math. :)
That might be a near light speed weapon, but it won't get all the way to the speed of light.Enigma wrote:Dumping it then. Still thinking of a light speed weapon (as in 1x speed of light. Think of it as a ls railgun.). During lunch, I had an about using for a projectile a torpedo casing filled with 25 kilograms of deuterium and anti-deuterium so when it hits a target, the m/am mixes and goes kaboom. At least something like that.
Depending on your target, it might be redundant to include the matter, since presumably the target is composed of matter and will annihilate with the anti-deuterium quite nicely. Unless your ships use shields of some kind.
25 kilograms total will result in about 535 megatons of energy being released, though not all of that will be useful (half of the total energy will be blasted outwards from the target, and some of the energy that goes in the direction of the target will be released as neutrinos).
If it's 25 kg of antimatter and the same amount of matter on the target, then double the yield.
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And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
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Re: Some help with math. :)
Depending on just how fast you fire the projectile, the AM would be redundant. If you fired it at c-1 m/s (299792457 m/s) you'll get ~13.9 gigatons from the impact. Once you hit ~86.6% of c as your muzzle velocity don't bother with a warhead, it just makes the projectile more hazardous to you than to the enemy.
EDIT: This assumes a 25kg projectile, because that is the only mass number provided.
EDIT: This assumes a 25kg projectile, because that is the only mass number provided.
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Re: Some help with math. :)
^ Good point, there. If you want to use AM, it makes more sense if you don't have railguns which can achieve near light speed projectiles.
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TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
Re: Some help with math. :)
Gotcha. 0.5c or 0.25c instead? Projectile configured so that it'll be shape charged in order to deliver as much destruction to the opponent as possible?Lord Zentei wrote:^ Good point, there. If you want to use AM, it makes more sense if you don't have railguns which can achieve near light speed projectiles.
I want the weapon to be able to basically easily smash through a (unshielded) ship. Also the projectile can hold various amounts of AM from a kilo to 25 kilograms.
Thanks for the help.
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Re: Some help with math. :)
The difficulty of smashing through an unshielded ship depends on the ship's density, any armor, and the momentum per cross-sectional area of your projectile. I wouldn't be surprised, though, if even in a moderately soft sci-fi setting you don't need more than 10^1-10^2 km/s speeds on reasonably not-massive ( < 10^1 tons) projectiles to seriously fuck up an unshielded ship. (Obviously, you need to take into account things like structural integrity fields, which will strengthen hull armor as well.)
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Re: Some help with math. :)
OK, it depends on the nature and overall strength of the ship's armour. Keep in mind that armour penetration doesn't depend on yield rates alone: a bomb has much less penetration than a bullet with the same energy, so if it's armour penetration you want, the high speed regular railgun is the way to go. If you want to dish out a lot of high-energy radiation over a wide area, frying people inside ships for example, then the AM bomb works nicely. Good for planetary bombardment too.
But as far as yield rates go, it's easier than armour penetration. Here's some more stuff:
In space, almost all of the energy of the AM bomb (or a nuclear bomb, for that matter) will be in the form of high-energy ionizing radiation (specifically, gamma rays); the shockwave and fireball only arise if there's an atmosphere. What happens in the atmosphere is that there is that the radiation ionizes the surrounding air turning it into a fireball which rapidly expands due to its temperature and pressure going through the roof as all that energy is dumped into it. All of this is absent in the case of bombs exploding in space, though it might fry a ship it explodes right next to.
Here's a quick calculator for the yield of an bomb that converts mass totally into energy: 21.48 megatons per kilogram. So, for a device that converts two kilograms into energy, that's about 43 megatons and so on. Keep in mind that this includes both the matter and antimatter that gets converted.
This is on account of the E=mc^2 equation: just take the speed of light in meters per second, roughly 300 million m/s (precisely 299792458 m/s), square it and multiply by the mass, and you've got the energy in Joules (you can use other units of course, but the important thing is to make sure you've got the right units - in metric it's meters, seconds, Joules and kilograms). The energy of a kilogram of mass, fully converted is therefore 8.98755 x10^16 Joules.
Then you need to know the yield of a kiloton of explosive energy, which is defined to be equal to one trillion calories of energy, or 4.184 x10^12 Joules. Thus a megaton is 4.184 x10^15 joules. It used to be that it was just 1000 tonnes of TNT as the name implies, but since not all TNT is created equal, they defined this figure for the sake of consistency (I forget which year they did it), which happens to be pretty close to what most types of TNT manage.
As to what effects this has: the main site has an atomic explosion calculator here. Keep in mind that this is for a planet surface.
An antimatter device involving only electrons and positrons might convert all the mass into gamma rays, but that's going to be impossible AFAIK to direct in any way, since the gamma ray distribution is pretty random. So that's a bomb that goes in all directions. The only way you could direct it is if your civilization has some kind of gamma-ray mirror which would be placed behind the device. Incidentally, this would also lead to a very efficient STL rocket known as a Photon Drive rocked (since it uses gamma rays for propulsion) - though low thrust and thus low acceleration, it would require very little of its mass as fuel, and could reach near lightspeed velocities if you're willing to take a journey of a few years.
The second type of antimatter bomb would involve protons and antiprotons (or hydrogen and antihydrogen), IIRC you would lose about a third of the energy as neutrinos, but most of the rest could be directed to some extent - I'd have to look it up. Directing it would be a bit of a bother for a bomb, though. Most - NOT all - of the energy gets converted into mesons, the majority - NOT all - of which are electrically charged, and thus subject to magnetic fields. You just need to keep the magnetic field alive while the mesons last - they decay very quickly into more neutrinos and gamma rays. By then, hopefully the blast has been directed at the target. As for whether this is feasible for a bomb is another matter. Frankly, it's better for a less-efficient-than-photon-drive antimatter rocket. Don't expect lightspeed, but it sure beats most other realistic STL out there.
Overall, unless you have a gamma ray reflector, the bomb probably won't be shaped-charge. BTW, a gamma ray reflector would also permit a gamma ray laser - a pretty fearsome deathray.
As an aside, you can look at the theoretical optimal yield rates for fusion bombs and such by looking at this. The atomic mass numbers of the various atoms is in the top right hand corner of the entries in the periodic table. Hydrogen is 1.00794, while Helium is 4.002602. Since four hydrogen atoms fuse into one helium atom, the helium "should" have had a mass of 4.03176, but the difference (about 0.7%) was lost as energy. Therefore, a device which converts 1 kilo of mass using fusion would only be about 0.7% as efficient as total conversion, and thus a kilogram of hydrogen fusing completely into helium would deliver about 0.15 megatons, or about 150 kilotons (for comparison, Fat Man and Big Boy of WW2 fame were about 15 kilotons or so, and the Russian Tzar Bomba of Cold War infamy were from 57 megatons and up. Keep in mind that a bomb will be more massive than the actual mass that gets used as fuel.
FAIR WARNING: the above calculations assume that you have 100% efficient use of fuel, which is virtually impossible in practice.
Then we move to the railguns... the energy it delivers to the target is the kinetic energy of the projectile, obviously. So to calculate the kinetic energy, you need to know the kinetic energy equation for relativity - the classical Newtonian equation of 1/2 mv^2 doesn't work, even if you use the relativistic mass increase. What you have to do is jump through a few more hoops. Fortunately you don't need to do that, though, since Darth Wong also has a calculator for that on his page: here. Just plug in the mass and speed of the projectile in kilograms and meters per second, and you get the "relativistic KE" on the right for the kinetic energy. The result is in joules.
But as far as yield rates go, it's easier than armour penetration. Here's some more stuff:
In space, almost all of the energy of the AM bomb (or a nuclear bomb, for that matter) will be in the form of high-energy ionizing radiation (specifically, gamma rays); the shockwave and fireball only arise if there's an atmosphere. What happens in the atmosphere is that there is that the radiation ionizes the surrounding air turning it into a fireball which rapidly expands due to its temperature and pressure going through the roof as all that energy is dumped into it. All of this is absent in the case of bombs exploding in space, though it might fry a ship it explodes right next to.
Here's a quick calculator for the yield of an bomb that converts mass totally into energy: 21.48 megatons per kilogram. So, for a device that converts two kilograms into energy, that's about 43 megatons and so on. Keep in mind that this includes both the matter and antimatter that gets converted.
This is on account of the E=mc^2 equation: just take the speed of light in meters per second, roughly 300 million m/s (precisely 299792458 m/s), square it and multiply by the mass, and you've got the energy in Joules (you can use other units of course, but the important thing is to make sure you've got the right units - in metric it's meters, seconds, Joules and kilograms). The energy of a kilogram of mass, fully converted is therefore 8.98755 x10^16 Joules.
Then you need to know the yield of a kiloton of explosive energy, which is defined to be equal to one trillion calories of energy, or 4.184 x10^12 Joules. Thus a megaton is 4.184 x10^15 joules. It used to be that it was just 1000 tonnes of TNT as the name implies, but since not all TNT is created equal, they defined this figure for the sake of consistency (I forget which year they did it), which happens to be pretty close to what most types of TNT manage.
As to what effects this has: the main site has an atomic explosion calculator here. Keep in mind that this is for a planet surface.
An antimatter device involving only electrons and positrons might convert all the mass into gamma rays, but that's going to be impossible AFAIK to direct in any way, since the gamma ray distribution is pretty random. So that's a bomb that goes in all directions. The only way you could direct it is if your civilization has some kind of gamma-ray mirror which would be placed behind the device. Incidentally, this would also lead to a very efficient STL rocket known as a Photon Drive rocked (since it uses gamma rays for propulsion) - though low thrust and thus low acceleration, it would require very little of its mass as fuel, and could reach near lightspeed velocities if you're willing to take a journey of a few years.
The second type of antimatter bomb would involve protons and antiprotons (or hydrogen and antihydrogen), IIRC you would lose about a third of the energy as neutrinos, but most of the rest could be directed to some extent - I'd have to look it up. Directing it would be a bit of a bother for a bomb, though. Most - NOT all - of the energy gets converted into mesons, the majority - NOT all - of which are electrically charged, and thus subject to magnetic fields. You just need to keep the magnetic field alive while the mesons last - they decay very quickly into more neutrinos and gamma rays. By then, hopefully the blast has been directed at the target. As for whether this is feasible for a bomb is another matter. Frankly, it's better for a less-efficient-than-photon-drive antimatter rocket. Don't expect lightspeed, but it sure beats most other realistic STL out there.
Overall, unless you have a gamma ray reflector, the bomb probably won't be shaped-charge. BTW, a gamma ray reflector would also permit a gamma ray laser - a pretty fearsome deathray.
As an aside, you can look at the theoretical optimal yield rates for fusion bombs and such by looking at this. The atomic mass numbers of the various atoms is in the top right hand corner of the entries in the periodic table. Hydrogen is 1.00794, while Helium is 4.002602. Since four hydrogen atoms fuse into one helium atom, the helium "should" have had a mass of 4.03176, but the difference (about 0.7%) was lost as energy. Therefore, a device which converts 1 kilo of mass using fusion would only be about 0.7% as efficient as total conversion, and thus a kilogram of hydrogen fusing completely into helium would deliver about 0.15 megatons, or about 150 kilotons (for comparison, Fat Man and Big Boy of WW2 fame were about 15 kilotons or so, and the Russian Tzar Bomba of Cold War infamy were from 57 megatons and up. Keep in mind that a bomb will be more massive than the actual mass that gets used as fuel.
FAIR WARNING: the above calculations assume that you have 100% efficient use of fuel, which is virtually impossible in practice.
Then we move to the railguns... the energy it delivers to the target is the kinetic energy of the projectile, obviously. So to calculate the kinetic energy, you need to know the kinetic energy equation for relativity - the classical Newtonian equation of 1/2 mv^2 doesn't work, even if you use the relativistic mass increase. What you have to do is jump through a few more hoops. Fortunately you don't need to do that, though, since Darth Wong also has a calculator for that on his page: here. Just plug in the mass and speed of the projectile in kilograms and meters per second, and you get the "relativistic KE" on the right for the kinetic energy. The result is in joules.
CotK <mew> | HAB | JL | MM | TTC | Cybertron
TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
Re: Some help with math. :)
What I was going for is a Star Trek version of a Bolo's Hellbore, but a bit fancied up, so to speak. I'm at the point of my fic that one of the secondary characters is giving the admiral in charge of the Bolo project, ideas for how to go about creating their version of the Hellbores\infinite repeaters\etc...
I've got the hull chosen along with what type of propulsion\hull plating\shielding and armor. Those were easy.
For infinite repeaters, I'm going for a gatling gun style pulse phaser cannon. Rate of fire between 100 to 200 shots per second.
The sticky part has been the Hellbore itself. I hoped to beef it up and modify it but I got carried away. lol I guess I'll go for a 100kg neutronium projectile (in the fic the Feds have begun to feasibly manufacture it in reasonable quantities.) that can be fired at 0.5c (my understanding is that the Bolo's Hellbore can fire it's ammo at a high fraction of c). Or I might just use some handwavium metal that is dense and easy to manufacture.
That should give me a 1.123E+18J yield or roughly 268MT?
Smaller Hellbores would use smaller projectiles between 20 to 50kg.
Thanks.
I've got the hull chosen along with what type of propulsion\hull plating\shielding and armor. Those were easy.
For infinite repeaters, I'm going for a gatling gun style pulse phaser cannon. Rate of fire between 100 to 200 shots per second.
The sticky part has been the Hellbore itself. I hoped to beef it up and modify it but I got carried away. lol I guess I'll go for a 100kg neutronium projectile (in the fic the Feds have begun to feasibly manufacture it in reasonable quantities.) that can be fired at 0.5c (my understanding is that the Bolo's Hellbore can fire it's ammo at a high fraction of c). Or I might just use some handwavium metal that is dense and easy to manufacture.
That should give me a 1.123E+18J yield or roughly 268MT?
Smaller Hellbores would use smaller projectiles between 20 to 50kg.
Thanks.
ASVS('97)/SDN('03)
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
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Re: Some help with math. :)
Use the relativistic KE, that's 1.392E+18J = 332.7 MT, that's in the right hand column - DW presumably just included the Newtonian calculations to show how they differed from the relativistic ones. Otherwise, yes. BTW, remember that this is the energy that the cannon has to pump into the bullet in the time it takes for it to leave the barrel. You might want to include capacitor banks that supply the energy and which then have to be recharged after use, at least if you're looking for some kind of limiting factor.
I'm not sure that neutronium can easily be fired from a railgun (since neutrons aren't affected by electromagnetic forces). That is, unless you include the neutronium pellet in some kind of capsule that is made of ordinary matter which can contain the neutronium and transfer the acceleration onto it by some handwavium means until it impacts on the target. Incidentally, neutronium is a liquid since unlike ordinary matter it doesn't include any electron shells and thus can't form chemical bonds to keep the stuff in place.
BTW, we have experimental railguns today that can fire projectiles that can punch through multiple tanks, all lined up (and they certainly don't manage a significant fraction of c). Presumably, the armour of these ships will be rather tougher than the depleted-uranium laced steel plate used today.
I'm not sure that neutronium can easily be fired from a railgun (since neutrons aren't affected by electromagnetic forces). That is, unless you include the neutronium pellet in some kind of capsule that is made of ordinary matter which can contain the neutronium and transfer the acceleration onto it by some handwavium means until it impacts on the target. Incidentally, neutronium is a liquid since unlike ordinary matter it doesn't include any electron shells and thus can't form chemical bonds to keep the stuff in place.
BTW, we have experimental railguns today that can fire projectiles that can punch through multiple tanks, all lined up (and they certainly don't manage a significant fraction of c). Presumably, the armour of these ships will be rather tougher than the depleted-uranium laced steel plate used today.
CotK <mew> | HAB | JL | MM | TTC | Cybertron
TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
TAX THE CHURCHES! - Lord Zentei TTC Supreme Grand Prophet
And the LORD said, Let there be Bosons! Yea and let there be Bosoms too!
I'd rather be the great great grandson of a demon ninja than some jackass who grew potatos. -- Covenant
Dead cows don't fart. -- CJvR
...and I like strudel! -- Asuka
Re: Some help with math. :)
Well, neutronium in ST seems to be just a super dense metal.
I've made some more refinements after rubbing more than two brain cells together.
The infinite repeaters and the secondary and main Hellbores are all railguns of various sizes. The infinite repeaters fire off 1cm diameter spherical neutronium pellets and has a rate of fire of 10 per second. The main and secondary Hellbores use 10cm and 5cm diameter spherical neutronium pellets and their rate of fire is one per second (sustained) or two per second (10 seconds sustained fire, 10 seconds to cool off).
The Bolo will have a dedicated warp core and fusion reactors for the Hellbores, infinite repeaters and any other weapons, shielding and so forth.
I've made some more refinements after rubbing more than two brain cells together.
The infinite repeaters and the secondary and main Hellbores are all railguns of various sizes. The infinite repeaters fire off 1cm diameter spherical neutronium pellets and has a rate of fire of 10 per second. The main and secondary Hellbores use 10cm and 5cm diameter spherical neutronium pellets and their rate of fire is one per second (sustained) or two per second (10 seconds sustained fire, 10 seconds to cool off).
The Bolo will have a dedicated warp core and fusion reactors for the Hellbores, infinite repeaters and any other weapons, shielding and so forth.
ASVS('97)/SDN('03)
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
Re: Some help with math. :)
Enigma, try this:
How many Enterprise Ds would it take to beat one of your spaceships (or vice versa, if the spaceship is less powerful than the Enterprise)?
Described in general terms of what you would see on the screen if this were a movie, what do you want your weapons to do, look, and sound?
What is the peak rate of fire of these weapons and what is the time-averaged rate of fire of these weapons for an actual engagement?*
What kind of general ideas do you have about the kind of technology the weapons use?
Give us those, and we can give you some ideas that will fit the bill and won't be horribly out of balance with the rest of Trek.
Mightn't this thread be better off in SciFi?
*for instance, a submachinegun might have a rate of fire of 900 rounds per minute, but in a real fight you might use 60 per minute, or something.
How many Enterprise Ds would it take to beat one of your spaceships (or vice versa, if the spaceship is less powerful than the Enterprise)?
Described in general terms of what you would see on the screen if this were a movie, what do you want your weapons to do, look, and sound?
What is the peak rate of fire of these weapons and what is the time-averaged rate of fire of these weapons for an actual engagement?*
What kind of general ideas do you have about the kind of technology the weapons use?
Give us those, and we can give you some ideas that will fit the bill and won't be horribly out of balance with the rest of Trek.
Mightn't this thread be better off in SciFi?
*for instance, a submachinegun might have a rate of fire of 900 rounds per minute, but in a real fight you might use 60 per minute, or something.
Re: Some help with math. :)
Probably best in sci-fi.Feil wrote:Enigma, try this:
How many Enterprise Ds would it take to beat one of your spaceships (or vice versa, if the spaceship is less powerful than the Enterprise)?
Described in general terms of what you would see on the screen if this were a movie, what do you want your weapons to do, look, and sound?
What is the peak rate of fire of these weapons and what is the time-averaged rate of fire of these weapons for an actual engagement?*
What kind of general ideas do you have about the kind of technology the weapons use?
Give us those, and we can give you some ideas that will fit the bill and won't be horribly out of balance with the rest of Trek.
Mightn't this thread be better off in SciFi?
*for instance, a submachinegun might have a rate of fire of 900 rounds per minute, but in a real fight you might use 60 per minute, or something.
The fic takes place about 80 years or so after TNG's All Good Things. So the Galaxy Class is an obsolete relic even by the mid 25th century Starfleet with their own starships let alone a Bolo.
I can't describe how it would sound since I'm hard of hearing I don't know how to compare it to. Possibly something like a muted electronic whooshing sound. The Hellbores and the infinite repeaters are dual barreled turret mounted. They can be fired individually or twin linked. The barrels are the typical long narrow tube but squarish on both sides where light blue light would pulse along the side as the pellet is being fired.
The Bolo itself will be built from an old starship even by TNG's standards for various reasons which will be explained in the story if some of it hasn't been already mentioned. (BTW, the first few chapters of it has been posted in Fanfic forum for a while. Need to update it with new chapters. )
The Bolo prototype will be overpowered since it will only have a max crew capacity of three so over 95% of available space will be dedicated to propulsion systems, energy production, armor, shield emitters, weapons and automated repair systems. I may be missing something but oh well.
The Bolo prototype will retain it's torpedo launchers and phasers (though the phasers will be modified and relocated to a better suited area.)
The rate of fire is fluid for the infinite repeaters from anywhere from one per second to ten per second. While the secondary and main Hellbores are as previously stated. It can go two per second but only for ten seconds before having to stop for another ten seconds to cool down or else the barrels will overheat and warp.
I've did my best to crunch some numbers then compared it to Wong's calculator. A 1cm infinite repeater pellet will have a KE of around 2.67petajoules (~639KT) (assuming the metal's mass is 0.454kg per centimeter squared or one pound per centimeter squared). Velocity being 0.5c, the mass being 0.238kg and the volume 0.52cm2.
A 5cm Hellbore spherical projectile would have a KE of 333.78petajoules (~79.77MT)(metal's composition being the same). The volume is 65.45cm2 and the mass is 29.71kg with the velocity being the same.
Finally a 10cm Hellbore spherical projectile would have a KE of 2.67exajoules (~638MT). The volume is 523.6cm2, the mass is 237.71kg and the velocity remaining the same as the other two.
ASVS('97)/SDN('03)
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
Re: Some help with math. :)
I'd recommend strapping the guns onto the warp nacelles, or otherwise arranging for them to go the full length of the space ship (if not farther) for firing c-fractional projectiles that have significant mass. The longer the accelerator, the less force and torque gets put on weapon to achieve the same muzzle velocity.
Re: Some help with math. :)
I've thought about the accelerator length. Just haven't decided on how long it would be.Feil wrote:I'd recommend strapping the guns onto the warp nacelles, or otherwise arranging for them to go the full length of the space ship (if not farther) for firing c-fractional projectiles that have significant mass. The longer the accelerator, the less force and torque gets put on weapon to achieve the same muzzle velocity.
I know exactly where I'll be placing the main Hellbores but I'm less certain about the exact placement of the secondary Hellbores and a vague idea where to put the infinite repeaters.
I'm thinking three main Hellbore turrets on top of the saucer section and two below but forward facing only (the last part is a necessity due to the ship's design. If you are starting to get an idea as to what type of ship, please don't mention but if you wish, you can PM me and I'll let you know if you are right or not. ) Two more would be installed at the rear of the secondary hull.
Secondary Hellbores, I'm guessing four turrets per nacelle (three above and one below). A few placed on and underneath the saucer but not in the way of the mains. Three turrets possibly on back end of the saucer. Six on the secondary hull (four at the back and two at the front [one above and below the forward section of the secondary hull])
The infinite repeaters would be spread out in order to act as a PD\ anti-fighter\gunboat type weapon.
Where possible a phaser strip would be placed along the circumference of the saucer as much as possible and I'm thinking of another strip wrapping around the sides of the secondary hull.
Any torp launchers will remain as is.
I was thinking of using a missile delivery system similar to the Bolo's VLMs but I figure that the torp launchers are better suited to the task.
ASVS('97)/SDN('03)
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!
"Whilst human alchemists refer to the combustion triangle, some of their orcish counterparts see it as more of a hexagon: heat, fuel, air, laughter, screaming, fun." Dawn of the Dragons
ASSCRAVATS!