Sci-fi Weapons, Vehicles and Equipment

De marquis: And no one said they would. The brief interlude on mobility was pointing out there would likely be additional thrusters that were not depicted in a particular work of fiction. Even then I have yet to hear of atmospheric fighter craft needing or having vernier thrusters for use in atmo. Most of the rest discussion was revolving around types and roles of fighter like craft in role and thinking in terms of scales for vehicle.

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The Old-School Dogfighting Space Fighter is actually realistic. There's nothing engineering wise preventing us from actually making such craft.

The better question is: is it advantageous to have craft that behave solely that way? Have a spacecraft that acts too much like a terrestrial aircraft? That's where the debate should focus on.

Drones and missiles are better for Space fighting. I was just commenting on Gundam.

That depends entirely on range. Remember the travel distance and speed charts? You have to get going pretty damn fast to make big far reaching space battles a thing with anything that is kinetic. For the amount of energy that takes you may as well find a way to make viable energy weapons to cross the gap.

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^ And in the same token, enough energy production to make a means of stopping said long range attack. If you have the power output available on a ship to accelerate a ferrous slug to significant fractions of the speed of light you probably have the technology level to implement things like Deflector Shields or amazing point defenses and evasion systems.

The charts highest projectile velocity was 50 km/s only a very small fraction of a percent of C.

A quick exercise here. KW Calculator Includes penetration of chosen armor type and the cost in power. At 50 km/s you are cooking at a very high velocity and whatever you hit will be having a bad day. However the cost is high.

A quick example Using this I picked the Bofors L/70 We are assuming this is a solid shot type round pure kinetic impactor. No guidance kits, no in flight fragmentation, no special warheads etc.

The average shell weight is 4.6 kg. We are gong to fire it at 50 km/s. The power required is about 16 TW or Terrawatts. The equivalent of a years worth of power expenditure of our planet in 2010. That is just 50 km/s. The Speed of light is 299,792 km/s. It is a fractional of a percentage point of the speed of C. It is roughly .02 percent the speed of light. The energy output for a single shot is our modern planets power expenditure for a single year.

At that speed it takes a kinetic round nearly 12 minutes to travel the distance from earth to HEO.

If you can put that much into an EM weapon you likely have the power for energy weapons with some serious reach and power.

edited 28th Mar '15 10:44:37 PM by TuefelHundenIV

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2 Terawatts are a measure of power, not energy. A 4.6kg projectile moving at 50km/s has a kinetic energy of about 5.8GJ. A 100000W car engine produces that much energy for every 16 hours of operation. 16TW would let you continuously fire over 2000 projectiles every second.

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Melkar: That is the power plant requirements for a single shot. Non-Repeating. You are not going to get that easily.

Your 2000 round 1 second burst is 49 Terrawatts three times the power requirements for a single shot.

Oh and for some perspective your 100,000 w car engine is putting only 1.0 × 10-7 Terrawatts .0000001 Terrawatts. I think your math is a wee bit off. Well that is unless your car engine has higher power output then the nuclear generator of the Nimitz.

Here the 16 Terrawatt for a single shot is 16,000,000,000,000 watts of power. Lets assume the car engine is putting out 100,000 watts a second. It would take that car 160,000,000 seconds/2666666.66 minutes/44444.44 hours/1851 days/61 months/5 years to generate 16 terrawatts equivalent of power.

If that was watts per hour the car is only putting out only about 27 watts per second. You would be waiting a very long time.

If you wanted to charge a single shot like that in 16 hours you need to generate a 1 Terrawatt an hour. You would need something generating 277 MW a second which is more power output at peak levels then the nuclear reactor of the USS Nimitz.

If you were comparing engine out put to impact energies they don't quite translate like that. The impact forces are exerted in a fraction of a second.

edited 29th Mar '15 12:35:31 AM by TuefelHundenIV

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And as mentioned if you have that kind of power levels your technology level is sufficient enough to render that speed and range advantage moot. You might have shields, you might have impressive evasion capabilities, Hell you might even possess the means to vaporize a solid slug as point defense.

Point being for as much power as it takes to throw something that fast or faster, you can use just as much power defeating or evading the incoming shot. In your example 12 minutes in point defense is a relative eternity. Plenty of time to get out of the way or shoot it down many times over.

Is there some part of the distinction between power and energy that you don't understand? A projectile of a given mass moving at a constant velocity has a certain fixed kinetic energy (measured in joules, or an equivalent measure such as TNT yield). Any given constant supply of power (measured in watts, which are joules per second, or an equivalent measure such as horsepower) thus corresponds to some particular rate of fire, given the mass and desired velocity of the projectile. 'Watts per second', or equivalent measures, are some sort of bizarre unit that I don't think has any relevance here; it would imply that the power, itself, is increasing linearly over time (and energy scaling with the square of elapsed time).

Well, the longer a barrel you have in which to accelerate the projectile, the lower power you need. The car engine just produces an equivalent amount of energy across 16 hours. Unless the barrel is ridiculously long, you would need to deliver that energy to the projectile in a shorter span of time, implying that the engine would charge up some form of energy storage such as an electrical capacitor or a flywheel.

A given barrel length, combined with the original mass and velocity of the projectile, would imply a level of power necessary to be imparted to the projectile while it is actually being accelerated down the barrel. Assuming that the level of power delivered to the projectile is constant, your 16TW would correspond to a barrel length of about 18 meters.

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A Key thing to remember about Kinetics is that the projectile's kinetic energy is applied very fast. The time frame is between the time the projectile first make contact with the target to the time the shot has either penetrated the ship entirely or has been caught within the ship.

If we assume the shot is moving at 10 Km/s and is shooting through a cross section 100 meters wide then the kinetic energy of the shot will be transfered in one tenth of a second at most. If the shot had 1 MJ of kinetic energy then it applies it's forces as 100 MW.

bel: Exactly. What makes the kinetic strike so damaging is that energy is being transferred into the target over that short time span which exerts forces on the materials .

Melkar: The last bit of your post was kind of odd. This bit. 16TW would let you continuously fire over 2000 projectiles every second. It made it sound like you were talking about power generation in relation to the car.

No you are not going be able to fire 2,000 shots a second at 16TW at 50km/s. The huge difference between one shot and 2,000 in one second is time projectiles spend accelerating to max speed in the barrel. The more shots you want to squeeze out in a burst of one second the faster you have to force the projectile accelerate to it's max velocity through the barrel. That means more power expended per shot. The more shots you fire in a given time frame the more the power requirements are going to be to achieve it. Lengthening the barrel does not solve your problem you still have expend energy to accelerate the projectile down the full length of the barrel.

No matter how you try and look at it your power requirements for more shots in a given moment go up with the number of shots in that moment.

The power requirements are not linear. If it is 16 TW for single shot a linear power requirement for 2,000 would be 32,000 TW. The total required for 2,000 of the same projectile in one second is 46TW bit of a difference there. Not linear but still a lot of juice.

Take the calculator and make all the values except relative target velocity and set them to all 1.

1 kg shot moving at 1m/s with a radius of 1cm burst duration of 1 second firing 1 shot a second. No big deal right?

Power plant needs a rough out put of about 0.0000021 MW. Now jump that shot per second to 2,000 per second. Your power plant now needs 0.0043 MW. Why? Because you are having to accelerate individual rounds more quickly to achieve the rate of fire.

If you want to improve energy consumption improving weapon efficiency helps a lot. Remember most weapons are not very energy efficient and a lot of that energy is wasted.

edited 29th Mar '15 3:57:42 PM by TuefelHundenIV

Who watches the watchmen?

That's the rate at which kinetic energy (attached to the projectiles) actually leaves the gun. Whether the gun imparts energy to the projectiles at a constant rate, or uses some form of short-term energy storage to apply that energy in bursts, is a matter of how the gun itself is engineered to accelerate the projectiles. Treating the gun as a black box, 16TW is the power requirement you would see from the outside for firing about 2000 shots per second.

Again, what a single shot carries is energy, not power. 16TW gives you something like 2000 shots per second, if the gun operates at near 100% efficiency.

You seem to be getting your 16TW figure from that calculator you linked to earlier. I looked at the source code, and it seems that that figure is the average power applied to a given projectile while it is in the gun barrel, based on two extra parameters: The energy efficiency of the weapon, which is set at 0.23 by default, and the time the projectile takes to traverse the length of the barrel, which is always set at 1/650 of a second. I'm not sure why these two parameters are given those particular values. 0.23 might be a fairly realistic efficiency, but the 1/650 seems pretty arbitrary, and at a constant acceleration would translate to a barrel length of about 38 meters for our 50km/s projectile. Even if we assume those parameters are accurate, though, the figure labeled 'Per Round Power Output' is still not what the power plant sees (that's the other figure, 'Average Power Output', and for my own calculations I was assuming an efficiency of near 100%).

In any case, so long as the gun is engineered such that its energy efficiency doesn't change with its rate of fire, the power requirement for continuous firing will scale linearly with the rate of fire. Certainly it can't scale any lower than linearly, or else you could get free energy just by firing fast enough.

Sounds like a linear increase to me.

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Going back to the initial hypotheses : a 4.6 kg projectile launched at 50 km/s has a kinetic energy equal to 5.75 GJ (about 1.3 T-TNT, for reference). Naval nuclear reactors (an acceptable benchmark for a starship's power plant) can have a power output of more than 100 MW. This gives us an order of magnitude of one shot per minute.

More futuristic power generators (e.g. a fusion power plant) could bring us closer to one shot per second.

EDIT : screwed up the TNT-equivalent conversion.

edited 30th Mar '15 6:34:12 PM by Aetol

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It's a bit less than the burning of an oil barrel, or a bit more than 1.3 tons of TNT.

edited 30th Mar '15 6:26:50 PM by ManInGray

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What the actual fuck are you talking about? Rate of fire has nothing to do with the kinetic energy of the round.

Major Tom : it seems like you're confusing power and energy. Watts measure power, which is energy over time. So, for a given energy per round, the power input determines the rate of fire (i.e. energy/second = energy/round * round/second).

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The math is a little long and confusing for me but I'd like to point out there are actual physical constraints on the fire rate of a weapon in the form of the mechanism. How fast the gun can rechamber, the size of the projectiles, things like that will affect the fire rate.

Furthermore, modern machine guns struggle to deal with overheating. How would you cope with that in space where there is only radiation to vent the heat with?

Pixel: Same is true for pretty much any weapon system out there in terms of both physical limitations of the mechanisms and heat restraints. For the rail guns some have speculated the naval rail gun may have cooling channels running the length of the barrel along the rails pumping coolant to deal with heat. Though keep in mind there are weapons capable of some impressive rates of fire that use various solutions to the assorted problems.

There are some systems like the Metal Storm that take the short cut of the projectiles already being in the barrel when fired. They have the buffer room given by the propellant and increasing barrel length from the first round fired to the last round. They could fire all rounds at once and none should collide by the time the last round leaves the barrel unless there was excessive variation in the propellants. The weapon system had trouble though with excessive pressures in some of their designs and they tended to burst or explode when that happened.

Aeotol: Efficiency is also rather important. I mention it in the wall of text below. If you want to make it easier to fire that powerful a round there are a few options. The question that should be asked is what do you want the weapon to do? Do you want or need high impact forces, rates of fire, something more modest and efficient, point defense by single shot volume or by multi-round volume, you need a small but very fast projectile, etc. Same is really true of all possible weapon systems like lasers, rockets, missiles, and chemically propelled gun/cannon type weapon systems.

Man: And like was pointed out earlier. The difference is the amount of time that energy is delivered in. That burning barrel of oil will deliver that energy over a much longer time frame then the energy delivered by kinetic impact or an explosive. Given enough time to disperse or distribute it in you can make any amount of energy seem trivial by comparison in terms of energy over time.

Melkar: No it is not the kinetic energy at all, that is the individual drain on the power plant for a single shot. You are grossly confusing power and energy. Power as in power generated by a power plant to be consumed by the weapon on firing. The power per shot is 16TW of power to fire that one individual shot. That is why that particular result is under the section Power Plant under Per Shot Power Output. That is charging up and firing the weapon. For the calculation only the projectile values were altered nothing else. The power plant requirement is to power the gun firing one shot in one second. That is again the total charge for the weapon and firing of the projectile inside of a certain time frame. The faster it needs to fire the shot(s)in that same time frame the more power it will need from the power plant. The inverse is true if you have more time between shots. That per shot power requirement for energy is to put the projectile from breach to muzzle with the projectiles max velocity at exiting the muzzle. It is not the exterior view of kinetic energy it is the total power required to get the shot out of the gun supplied by a power plant. That is the amount of electrical energy needed to generate the forces to propel the round not the amount of kinetic force on the round as it is accelerated. The actual forces exerted on the round are measured differently and expressed differently.

The way these weapons work is a power plant of certain output charges some form of power storage device like a capacitor. When the weapon is charged that stored energy is dumped all at once into firing the projectile. That is why you can have such disparity in power plant requirements to the power of a single shot. The more rounds you put through a barrel at a given point the faster every one of those rounds has to accelerate to clear the barrel. The shorter the time frame and/or the more rounds out of a given time period the more power you need to dump into the weapon to achieve it.

We can even test the calculator to see if it is reasonable.

We can plug in the numbers for the Naval Rail Gun that are known or reasonably estimated for what they want out of the weapon. That means an average velocity of 2.5km/s firing 20kg saboted [[note:complete]] projectile at a rate of .2 rounds per second or about 12 rounds a minute with a estimated 37-40% efficiency. I have seen some say the rate of efficiency seems a bit generous as is the rate of fire but that is ultimate goal for the weapon. Whether or not that is what we get in the fielded weapon has yet to be seen.

The various estimates put the average minim power plant output at about 30MW depending on variations of efficiency and rates of fire. Plug all of that into the calculator. The calculator gives us an estimate of average power plant output of 31MW at 40% efficiency and 33 at 37% efficiency. It matches the estimated specs of the Naval Rail Guns target goal rather closely. Without exact specs of the weapon system that is about as good as we are going to get. Other things to consider those numbers are only good for the first shot through the rails. Efficiency degrades as the rails degrade with successive shots because of high wear and tear. How quickly the rate of degradation will be in the final product is unknown as of yet.

That means the calculator is reasonably accurate for our purposes.

You are not getting your 2k shots in one second on 16 TW without upping efficiency or altering other parameters like projectile mass and velocity. As it is the original calc firing the single shot is using the same parameters only with the 2k rounds in 1/s burst. The more efficient the weapon is the less power it takes to fire that round. The far easier alternative is to take longer time periods between shots. Even if you take two or three minutes to charge the total power for a shot remains the same. If you change how long you take between shots you need to generate less power from your power plant to charge the same shot. Given enough time between shots you can use rather weak power generation sources but the time between shots will be rather high.

Lets get back to the 4.5kg slug again to drive the point home. Same speed only instead of firing one shot every second or faster I plugged in .002 rounds per second or a round every 7 minutes. The per shot power requirement stays the same no surprise there. However the power plant output requirements are no more than 48 MW. Lets look at the impact of efficiency. Same round same rate of fire but up the efficiency to 50% and you get 7.3 TW per shot and the minimum power plant output required drops as well. What about a slug with less mass? Using the same efficiency, velocity, and ROF as the original but firing a 3 kg slug you need 10.6 TW of power per shot and the power plant requirements drop as well.

There you have it.

edited 30th Mar '15 11:29:28 PM by TuefelHundenIV

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It does if we're assuming a particular power input and projectile mass.

Inside of a certain timeframe, yes. That is the difference between power and energy. This mystical measure of 'power per shot' is irrelevant. Energy per shot is what's relevant, and in this case it comes to about 5.8GJ.

As I mentioned before, my original calculations were assuming a near 100% efficiency. Using the calculator's default efficiency of 23%, we get 640 rounds per second. At the 40% efficiency you suggested for the railgun, we get 1113 rounds per second.

EDIT: Forum wouldn't parse my Wolfram Alpha link. :(

edited 30th Mar '15 9:53:27 PM by Meklar

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Yup...maybe I shouldn't have slept through Physics class.

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If the fora is eating links run it through tiny url or similar service. Usually fixes it especially the long urls or ones with percent signs in it. It has been one of those ever pervasive never fixed oddities of the fora.

The power per shot is very much relevant. That is what is one of the key things determining your power plant requirements in the first place. It is also why more time between rounds means the minimum requirements for a power plant can be lower. It is not measured in joules it is measured in Watts because that is the total stored electrical energy the weapon needs in total for any individual shot either as one round at a time or part of a burst. The 46 TW power plant requirement comes in because it needs that output to maintain the 16TW power charge for the duration of the burst.

Look at the Naval Rail Gun again. The joules for the propulsive energy is rated at 32-33MJ at the muzzle. That is the kinetic force of the round as it leaves the barrel. The total power per shot of the US Naval Rail gun is 101.59 GW converted to joules/s is the same. That however does not match the at muzzle joules. In this case the joule measurements of 32-33 MJ in question are not representative of the power needed to fire but an example of the muzzle energy as the round leaves the barrel. The power per shot of the calculator is the total electrical energy needed to propel the round stored in capacitor or similar device.

I forgot this bit The reason the energy in joules is different is from firing to projectile exiting some of that energy is used up propelling the round the rest winds up as waste and some of it is in the muzzle blast.

Now depending on the situation it might be more efficient to fire multiple shots in that one second than to fire them across a longer period of time. Like say you need an immediate threat neutralized ASAP it would be perfectly ok to hose it with that kind of spray. The difference would be say firing that burst to shred an incoming flight of missiles or other threats vs a series of aimed shots across a 30 minute engagement period.

edited 31st Mar '15 1:35:30 AM by TuefelHundenIV

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No it's not. The burst power needed during a shot only determines what capacitors you'll be using. The power plant requirement depends on the overall power input of the weapon — how much energy it consumes each second. It only depends on the kinetic energy of each round (i.e. their mass and their muzzle speed), the rate of fire, and the efficiency of the gun system.

This power is fed continuously to an array of capacitors, which deliver their energy in a high-power burst each time a shot is fired. You can't have your power plant work in bursts.

On the other hand, if the rate of fire is such that there is always one, or even several rounds inside the barrel at any given time, then the "power per shot" (which may vary as the round gains speed, but whatever) is indeed relevant. But then the power requirements are truly ridiculous.

As an aside, the "power per shot" given by the calculator is meaningless. The "burst duration" field is never used — I suspect whoever created it forgot to replace the arbitrary "residentTime" value (seems to be the duration spent by a round in the barrel) with an appropriate formula. What you need to look at as far as power plant requirement are concerned is "Average Power Output". I also suggest we stick to using the relevant formulas directly — they're not that complicated anyway.

edited 31st Mar '15 4:10:42 AM by Aetol

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Yes power per shot is relevant. You can't take it out of the formula if you do there is no formula. No shot per second means no power plant requirements because there is nothing to base the calculation off of. What bizarre idea are you two sharing that cuts out a key factor of a formula?

The power plant requirement does not determine the capacitors. The power per shot does. The power per shot is the electrical power stored in the capacitors in total that it takes to fire the given mass to reach a given velocity. The power plant requirement is the overall power input into the capacitors at any given moment not the capacitor requirements itself. You have it completely backwards.

The amount of power per shot has a direct impact on power plant requirements. Power per shot is determined by round mass, velocity, and efficiency of the weapons system. You can't cut it out of the formula. The power plant is not working in bursts it is continuously feeding the capacitors. That is why if you have longer charge times between shots you can use lower output power plants. If we tried to use the same power plant output from the single shot for the 2k 1 second burst the power plant could not do it. It could not generate the power fast enough to keep the capacitor banks filled and it is the capacitor banks that fire in bursts.

The power per shot remains constant unless you change individual round velocity, efficiency, and/or mass. Those three factors determine the power needed to provide energy for a single shot as long as those three variables are consistent. Power plant requirements is not only determined by those three initial factors but also how much time it takes to fire the weapon. The more time before or between shots the lower the minimum average power plant output is. How much power is needed in the capacitors is Power Per Shot, Power Plant requirements reflects the needed power output for the duration of the shot. Again from the 4.5kg shot this is why if you just fire one shot at one second you expend 16 TW per shot. The per shot doesn't change unless you change the values of the shot to begin with. The power plant requirement is determined by four factors instead of just three.

As for the burst duration box it is irrelevant in general. Considering it is trivially easy to translate burst duration into rounds per second the lack of its functioning is a non-issue.

Here are a few examples. The USS Iowa Main guns for a start. We use a burst duration of 60 seconds which is a common measurement for weapons. In 60 seconds it fires about two rounds this varies because crew have a direct impact on fire rates due to manual handling. 2 divided by 60 is .03 or .03 rounds per second. That value will work with burst duration left as one second and shots per second as .03 because you have already converted it. As long as you know that and account for it the rest of the calculations work. That is how I got it to give us nearly the same estimate as the proposed goals for the US Naval Rail gun which has about 12 RPM or about .2 rounds a second.

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