Remedial science: Force and Energy
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- Currald
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Remedial science: Force and Energy
Allright, you were all so caring and understanding about my little math problem earlier, so I thought that I'd share a little science problem I'm having. I'm having a bit of a problem with the relationship between force and energy here. Here is the specific problem.
I'm trying to do some weapons calcs, and I ran across an energy construct which is able to withstand the temperature and pressure inside the core of Sirius B. Additionally, the weapon who capabilities I am trying to calculate is capable of puncturing said construct. The weapon is a beam with a firing duration of 0.60 seconds.
Sirius B stats:
Effective Temperature: (K) 24,790
Density: (kg/m3) 2,467,598,176
and
Pressure= density * temperature (right?)
pressure=61,171,758,783,040 N/m^2 (right?)
So if we assume (for simplicity's sake) that the weapon aperature happens to be 1 m^2 and it fires for 0.60 seconds, how can I determine the minimum energy involved?
Thanks in advance for what promises to be re-educational!
I'm trying to do some weapons calcs, and I ran across an energy construct which is able to withstand the temperature and pressure inside the core of Sirius B. Additionally, the weapon who capabilities I am trying to calculate is capable of puncturing said construct. The weapon is a beam with a firing duration of 0.60 seconds.
Sirius B stats:
Effective Temperature: (K) 24,790
Density: (kg/m3) 2,467,598,176
and
Pressure= density * temperature (right?)
pressure=61,171,758,783,040 N/m^2 (right?)
So if we assume (for simplicity's sake) that the weapon aperature happens to be 1 m^2 and it fires for 0.60 seconds, how can I determine the minimum energy involved?
Thanks in advance for what promises to be re-educational!
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Right...so you have something which is capable of surviving inside the core of the star, and you want to do calcs for a weapon that could penetrate this thing. Of course, for said calculation you need to assume that being able to survive inside the core is at the maximum of the thing's capabilities.
First of all, the formula you used for pressure is completely wrong. Dimensionally it isn't even accurate (i.e. the units don't add up).
The fact that it is capable of withstanding the temperature inside the star gives you an idea of how much heat the thing can take (i.e. the heat capacity), and can be factored into the amount of energy needed to penetrate the thing (i.e. the energy needed to liquify or melt it), unless the thing is a kinetic energy weapon, in which case the heat tolerance of the thing is irrelevant (asbestos can take way more heat energy than ice, but it's easier to penetrate).
If the thing is stuck immobile inside the star, then the pressure on its surface is:
pressure = height*density of star*local gravity
For height, in this case it should probably be defined as the height above the absolute centre of the star...as for density of the star (i.e. the density of the atmosphere of the star...a star is basically a ball of gas), you need you need the exponential atmosphere equation...I could do it for you, but you will need fairly advanced calculus and thermodynamics to understand it. It's also pretty complicated with the gravity, since gravity obviously falls off as you go further from the star, and you need the right gravity to calculate the total effect of the pressure.
If you do manage to get the pressure the thing is capable of withstanding, then it's a fairly simple thing to calculate the effect of kinetic energy weapons. But since your "weapon" seems to be an energy weapon, this seems to be a moot point.
Basically, I think you are biting off way more than you can chew with this...unless you want to simplify things to a point where it has little relevance with the scenario, or get more "ready made" data.
First of all, the formula you used for pressure is completely wrong. Dimensionally it isn't even accurate (i.e. the units don't add up).
The fact that it is capable of withstanding the temperature inside the star gives you an idea of how much heat the thing can take (i.e. the heat capacity), and can be factored into the amount of energy needed to penetrate the thing (i.e. the energy needed to liquify or melt it), unless the thing is a kinetic energy weapon, in which case the heat tolerance of the thing is irrelevant (asbestos can take way more heat energy than ice, but it's easier to penetrate).
If the thing is stuck immobile inside the star, then the pressure on its surface is:
pressure = height*density of star*local gravity
For height, in this case it should probably be defined as the height above the absolute centre of the star...as for density of the star (i.e. the density of the atmosphere of the star...a star is basically a ball of gas), you need you need the exponential atmosphere equation...I could do it for you, but you will need fairly advanced calculus and thermodynamics to understand it. It's also pretty complicated with the gravity, since gravity obviously falls off as you go further from the star, and you need the right gravity to calculate the total effect of the pressure.
If you do manage to get the pressure the thing is capable of withstanding, then it's a fairly simple thing to calculate the effect of kinetic energy weapons. But since your "weapon" seems to be an energy weapon, this seems to be a moot point.
Basically, I think you are biting off way more than you can chew with this...unless you want to simplify things to a point where it has little relevance with the scenario, or get more "ready made" data.
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Re: Remedial science: Force and Energy
The relationship between force and energy is very simple and given by the relationshipCurrald wrote:Allright, you were all so caring and understanding about my little math problem earlier, so I thought that I'd share a little science problem I'm having. I'm having a bit of a problem with the relationship between force and energy here. Here is the specific problem.
W = Fd*cos (theta),
where W is work, F is the amount of force applied, d is distance and (theta) is the angle at which the force was applied.
No. Your output unit for your equation would be K*m^3, which is completely meaningless. Pressure has units of N/m^2, as you have above. Therefore,I'm trying to do some weapons calcs, and I ran across an energy construct which is able to withstand the temperature and pressure inside the core of Sirius B. Additionally, the weapon who capabilities I am trying to calculate is capable of puncturing said construct. The weapon is a beam with a firing duration of 0.60 seconds.
Sirius B stats:
Effective Temperature: (K) 24,790
Density: (kg/m3) 2,467,598,176
and
Pressure= density * temperature (right?)
pressure=61,171,758,783,040 N/m^2 (right?)
P = F/A,
or pressure is force per unit of area.
The density isn't really needed. The relationship between temperature and energy is well-defined asSo if we assume (for simplicity's sake) that the weapon aperature happens to be 1 m^2 and it fires for 0.60 seconds, how can I determine the minimum energy involved?
Thanks in advance for what promises to be re-educational!
E = kT,
where k is the Boltzmann constant, which is 1.38E-23 J/K. So every particle in the core of the star has 3.5E-19 J of energy, or about 2.2 eV. What you need that you don't have is the particle density of the core.
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Some of the base facts are either vague or incorrect, so you'll have trouble resolving this. For one thing, the listed temperature for Sirius B cannot possibly be its core temperature. Stellar core temperatures are measured in the millions of K in order to create fusion conditions, not a few tens of thousands. This sounds more like the surface temperature, not the core temperature.
Also, you don't know how long this target construct is capable of sitting in there. The fact that it can "withstand" those temperatures is somewhat meaningless if you don't know how long it can do so.
PS. What is an "energy construct"?
Also, you don't know how long this target construct is capable of sitting in there. The fact that it can "withstand" those temperatures is somewhat meaningless if you don't know how long it can do so.
PS. What is an "energy construct"?
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Oh, yeah, of course. So that's not very helpful, since we need the core temperature.Darth Wong wrote:Some of the base facts are either vague or incorrect, so you'll have trouble resolving this. For one thing, the listed temperature for Sirius B cannot possibly be its core temperature. Stellar core temperatures are measured in the millions of K in order to create fusion conditions, not a few tens of thousands. This sounds more like the surface temperature, not the core temperature.
Hummmm... How about, "as long as its projector is supplied with power."Also, you don't know how long this target construct is capable of sitting in there. The fact that it can "withstand" those temperatures is somewhat meaningless if you don't know how long it can do so.
Shit, man. I dunno. I don't have my source material handy at the moment, but as I recall it was a cylinder of force, open at both ends. Sort of a cylindrical force field that could in effect extend a gun barrel quite a distance... 10km, if memory serves... all the way to the target.PS. What is an "energy construct"?
So given: the q-type helix can survive at the core of Sirius B for as long as power is supplied to it (assuming the ship generating it could somehow get within 10 km of the core of Sirius B), what what information do I need to determine the lower limit of a weapon yield that would exceed those conditions. How can I determine how much energy is battering the q-type helix? What data do I need?
Isn't the core of a star hydrogen? Or is it helium? Couldn't I determine the particle density by taking the (known) density and dividing it by the atomic mass of hydrogen or something along those lines?Durandel wrote:What you need that you don't have is the particle density of the core.
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Whaaaaat?!? That means that the pressure would be zero at the center of the star, since the height would be zero! I may have forgotten everything I learned in high school, but I remember that anything times zero equals zero!kheegan wrote: If the thing is stuck immobile inside the star, then the pressure on its surface is:
pressure = height*density of star*local gravity
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No, that's essentially correct. The pressure at the gravitational center of a star will be zero by definition, because that's the point where all the forces cancel each other out. It goes the same way for the Earth.Currald wrote:Whaaaaat?!? That means that the pressure would be zero at the center of the star, since the height would be zero! I may have forgotten everything I learned in high school, but I remember that anything times zero equals zero!kheegan wrote: If the thing is stuck immobile inside the star, then the pressure on its surface is:
pressure = height*density of star*local gravity
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The pressure will be at its highest since you are essentially equidistant from both ends of the star if you will. Gravity will be a non factor for the above reason by Durandal.Currald wrote:While I can accept that gravitational effects will be zero, pressure increases with depth, in my SCUBA-diving experience. Or is there some definition of pressure which I am not aquainted with?