 January 23, 2014 - 10:59pm | After seeing Terl Obar's estimate of the energy of an SEU, I decided to do a few of my own. This is all just for fun and I doubt if there's a single value that will come out of this and I don't think one of them is going to be “right.” I just want to see what the numbers come out to be. It would be really cool to have a single value that could be used as a scale but I think that's unlikely. Please note that there are other ways of doing things and that all of these estimates have error bars you could fly a Star Destroyer through. Sideways. Some quick definitions: Joule (J) metric unit of energy (1 J = the amount of energy that raises a cubic centimeter of water 1 degree Celsius) Calorie: Here, these are food calories, not the smaller scientific ones. 1 Cal = 4184 J Watt (w) a measure of power. 1 w = 1 J per second.Kilowatt-hour (kwh) Confusingly, this is a measure of energy, not power. 1Kwh = 3600 kj On to the estimates: Terl Obar's Laser 100 J per SEU Project Rho's laser 1-10kJ per shot depending on laser efficiency and target composition. I tend to think that energy weapons are less efficient than kinetic weapons at actually inflicting damage. Here's a table of a few common rounds and their kinetic energy according to Wikipedia. I'm giving a range because there are different loads that each have their own energy levels.
What does this tell us? NOTHING! :) Very powerful rounds can cause tremendous damage but might overpenetrate, meaning that not all of the energy gets used killing the target. Even 22 short has caused deaths. Damage is strongly dependent on hit location, shooter skill, and most of all luck. But let's ignore all that. Let's start with the fact that a 1 SEU laser shot does the same damage as a autopistol/rifle. Also note that a burst does an extra 4d10. I'm going to call this the burst bonus. So without that bonus, a machine gun does 6d10 of damage. Let's pretend that works. If we assume that an autopistol is about equal to a .380 (250 J), that would mean a machine gun is about 1.5 kJ, so it would be a 5.56 or 7.62 Soviet type round. If we went with 500 J, about the same as a 9mm, the machine gun would be like the M-60 used today. Neat. So a laser pistol would have an energy output from 250 J to 2500 J, about the same range from a 380 ACP to a .500 S&W and an SEU could be 250 J. But lasers aren't 100% efficient. Today's lasers are generally 25-30% although there are systems up to 60% efficient. Frontier tech would be better but let me give Blankbeard's laser input power estimate 1SEU = 250J – Minimum, 100% efficiency 1 SEU = 2 kJ – 9mm damage, 25% efficiency 1SEU = 1000 J or 1 kJ |
January 23, 2014 - 11:01pm |
Powerclip size A powerclip is about the size of a pack of cigarette pack. How big is that? (Oh how times change, eh?) http://www.dimensionsinfo.com/malboro-pack-dimensions/ gives a Marlboro pack as 8.5 x 5 x 2 cm, 85 cubic centimeters or 0.085 liters. They are non-rechargeable so I chose to use a zinc-air battery as a stand-in with an energy density of 35 MJ per liter. That would imply 2.9 MJ per power clip. A lithium battery (not lithium ion) would have 367 kJ capacity. We actually need packaging and lasers need a way to get rid of waste heat. 50 cc of water would actually work fairly well to hold most of the heat in conjunction with a small radiator on the pistol. If an SEU is about 1kJ and we drain the battery to 50% so we can get the power out quickly, our 20 SEU clip needs 40 kJ of energy. So we need about 10 cc of lithium battery to make powerclip with a small radiator. Since the pistol is limited to 10 SEU per shot, we don't have a problem with vaporization. But a 20SEU shot from a rifle needs to vaporize a few cc of water to cool itself so we can assume the rifle has ports that release a small amount of steam when a 15 SEU or higher shot is made. It's not a problem because each new clip has a fresh supply of coolant. This could be a way of finding that beam weapon user who's trying to cook you alive. Belts and power backpacks use lithium ion chemistry, meaning a power belt would the size of about 5 clips, the backpack would be 10 times the size of a clip. So a powerclip is happy at 1 SEU = 1000 J 1SEU = 1 kJ From here on out, the estimates are much larger. |
January 23, 2014 - 11:05pm |
Ground Cars Typical efficient cars today use .250-.350 kwh per km traveled. The Nissan Leaf uses .212. The Alpha Dawn ground car is actually lower performance than the Leaf but the one in SFMan #8 is similar to the Leaf so let's use the low value. The ground car has a 1000 SEU battery and a 1000 km range so it conveniently uses half a SEU per km traveled. Using the Leaf's value, that means 1 SEU = 0.212 kwh or 1472 kJ. The .350 value gives us an SEU value of 2.5 MJ. 1 SEU = 2.5 MJ Aircars There is a formula that lets you calculate the power a jet-like aircraft is using if you know its top speed and a few other things that I'm going to list below.
Top speed – 900 kph air density – 0.41 kg/cubic meter (This is about earth's atmosphere at 10 km. This means an aircar has similar performance to a personal jet or airliner.)drag coefficient – 0.15 (I chose to use car style coefficient because the air car lacks large wings. I assume it has a lifting body design at speed because you're not using fans to lift at 900 kph. The Mercedes S class has a .23 coefficient so the air car is pretty slippery.) Reference area – 5 square meters (A bit more than a regular car would have, I assumed the aircar is 2.5 meters wide, 2 meters tall, and 3 meters long, about the same as an old fullsize car) This works out to be about 112 MW of power or 124 kilonewtons of thrust. Just for comparison, one F-100-229 engine from a F-15E develops 129 kilonewtons. The aircar would be an amazing piece of technology. Conveniently, with a range of 1000km and a top speed of 900kph, an aircar has an endurance of 4000 seconds so it uses 1 SEU per second. So it uses 112 MW or 112 MJ per second. 1 SEU = 100 MJ Lifestyle This one is a bit strange, so please bear with me. The US uses about 105 exajoules per year. Assuming all of that energy is used to provide all of the needs and wants (bear with me) of the US population of 300 million, that works out to 954 megajoules per person per day, about 11kwh continuous. This includes gas, oil, manufacturing, and anything else. In addition, we eat about 1500-3000 calories a day including waste. Adding that on means we use about 1 GJ per day to provide our lifestyle. Future populations will probably use even more but I don't get the impression that the average Frontier citizen is that much better off than we are. We know that a parabattery costs 1 credit per SEU and that in general that exchange rate holds since the credit originally was the price of an SEU. This fits well with future societies being extremely energy dependent. If hireling wages are typical of what an NPC can earn (and they should be if hirelings take those jobs), middle class is probably 50-100 cr per day in earnings. Since we're interested in total lifestyle, not just living expenses, we use the whole value. That would mean a typical Frontier middle class being uses 50-100 SEU per day. If they are using energy at the same rate we are, an SEU would be 10-20 MJ. I think a higher rate (hover cars and robots) is appropriate so I'd say 50-200 MJ per SEU. For the estimate, let's match the aircar. 1 SEU = 100 MJ
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January 23, 2014 - 11:10pm |
Parabatteries Parabatteries range from a few hundred cubic centimeters up to one cubic meter. Assume that the type 4 parabattery is one cubic meter. That's 1000 liters. Assuming high end lithium ion chemistry (2.4 MJ per liter) and that about half the volume is cell with the rest being packaging and cooling, the type 4 has 1.2 gigajoules of energy. With a capacity of 4000 SEU, that would mean 1 SEU = 300 kJ. 1 SEU = 300 kJ To match vehicles, we'd need much better technology. Zinc-air fuel cells have an energy density of about 35 MJ per liter. Gasoline is about the same, 35 MJ per liter. A type 4 battery holds 4-8 SEU per liter depending on the percentage that is actually energy storage so using ground car SEU you'd need 10-20 MJ per liter. Aircar SEU would be 40 times that. The type one is going to be the densest one. It holds 1/8 the energy of a type 4 in less than 1/1000 of the volume. Using the aircar value for SEU as the worst case as you would need several hundred gigajoules per liter. While that's much lower than the energy density of a nuclear reaction, I'm not aware of any non-nuclear method to store that much power in such a small volume easily. Maybe really efficient superconductors could do it. If you don't care how a parabattery works, just ignore the last two paragraphs. Generators Generators can produce from 500-4000 SEU per hour. Using the aircar SEU, the largest produces 400 GJ per hour, a bit more than 100 megawatts. The type 4 generator is thus capable of serving a fairly large powerplant. You'd need about 1.6 square kilometers of solar panels, or a small nuclear reactor to provide the power for your generator. Of course, you can always assume one of the smaller values but I think this fits the generators as not being something you casually move around. I think you can actually come up with something useful out of all of this but you need 3 energy units, essentially 1 kJ, 1 MJ, and 1 GJ. I'll stop posting and wait for some comments before I write up what I think could be done to make this work.
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 January 23, 2014 - 11:14pm | So your last part of your essay, the laser input power estimate is a range from 25% efficient laster to 100% efficient laser. Thus the choice of 1SEU = 1 kJ would equate to delta(efficiency)/delta(energy)*1kJ => 43% efficiency for a 1SEU, 1kJ shot. Correct? -iggy |
January 23, 2014 - 11:32pm | I don't think I'm following your math. I was thinking in terms of overall efficiency. If we assume 1000J input energy and 250 J of laser energy out, that's 25% efficiency. My honest reason for chosing 1 kJ over any of the other possible values was to have a perfectly round number. And when the groundcar and air car came out near 1 MJ and 1 GJ I thought some fudging was in order. :) Also, sorry about the text size. That's readable on my screen in openoffice but it's small here. I don't seem to be able to change it with the editing controls. |
January 25, 2014 - 7:20pm | OK, I've reread your post. Are you setting your energy for the ammunition SEU by a 9mm bullet energy imparted to the target? What I was trying to calculate is the effeciency of the SF laser for 1kJ = 1SEU. This would be greater than the 25% efficency I understand to be the 1SEU = 2kJ modern day equivelent of a 9mm bullet energy imparted to the target. Concerning the inconsistancy of the rules where SEU in vehicles is needing a much higher energy than ammunition SEUs I think this may be easily adjusted by using your 1MJ an 1GJ numbers. Frontier residents may easily say their car takes 10SEU to get to work and back in common speach when the scientist and engineer know better that it is actually 10kiloSEU. The same would be true of the guy driving the air car stating that he used 2SEU when in fact he used 2MegSEU. I'll go over this again and run the numbers myself. This could make a great Frontier Explorer article and rules update. What would be fun is to lay it out in a science brief format. -iggy |
January 25, 2014 - 10:25pm | Since both the energy imparted by the laser and the efficiency would be arbitrary, I didn't work that number out. I just came up with 1 kJ as being a reasonable number for the range and for being an "attractive" number. I think you can assign a laser pistol any efficiency between 10 and 80% and still justify the 1d10 damage. There might be high efficiency lasers that do 2d10 per SEU but have a lower SEU limit and are much more expensive, possibly requiring regular maintenance. I like the idea of a 357 and 44 equivalent handguns that do 2d10 and 3d10 a shot. An assault rifle might be 6d10/10d10 burst, while a .50 cal type sniper rifle would probably be 20d10. You're certainly right about the different SEU sizes and the way they could be blended in everyday speech. We do that with calories and kilocalories aka food calories and in general there's no end of confusion about watts and kilowatts. There's a bit of weirdness when you charge a power backpack from a parabattery, but you can either ignore it or simply say that a parabattery type 1 holds 500 kSEU and can convert 1 kSEU into 500 SEU for charging small equipment each day. It might have a small battery system in the same way that modern generators have separate load capacities for 110 and 220 volt loads. That's not completely realistic - electricity is electricity - but it does keep a parabattery from being infinite ammo while still letting you have your vehicle gassed up. Can you point me to an example of the science brief format? It's funny, this is the one I didn't format as an article and it's the only post that's been mentioned as article-worthy :) Please keep up with the observations. I appreciate them! |
January 26, 2014 - 12:30am | I think we take the general format of the UPF planetary brief and merge it with the typlical published article format for science and engineering, say an IEEE article, and make the first ever UPF science brief. I'd be glad to help you out with it. And if we use 357, 44, and other balistic weapon calibres as a reference we can also introduce stats for them into the game. We may even be able to borrow or brush up specs from an old SFman article or two. I believe there was one on various handgun stats. -iggy |
 January 26, 2014 - 8:55am | It's funny, this is the one I didn't format as an article and it's the only post that's been mentioned as article-worthy :) They are all article worthy! This one was called out probably because it wasn't "article formatted" already. You should submit any and all of them as articles when you're happy with them.Ad Astra Per Ardua! My blog - Expanding Frontier Webmaster - The Star Frontiers Network & this site Founding Editor - The Frontier Explorer Magazine Managing Editor - The Star Frontiersman Magazine |
January 28, 2014 - 7:05pm | Thanks! I'll work on this next weekend and try to get some of the rest looked over to see what can be submitted. I really want to get the Vectorish movement hammered out soon too. |
 March 8, 2019 - 5:51pm | Did this ever get published? Seems pretty interesting. Joe Cabadas |
 March 9, 2019 - 8:51am | Did this ever get published? Seems pretty interesting. The answer would be no. Wonder if blankbeard is still around. EDIT: just sent Blankbeard an email have to wait and see if he is still around and willing to pull this together as an article. I might not be a dralasite, vrusk or yazirian but I do play one in Star Frontiers! |
March 9, 2019 - 9:37am | Did this ever get published? Seems pretty interesting. The answer would be no. Wonder if blankbeard is still around. EDIT: just sent Blankbeard an email have to wait and see if he is still around and willing to pull this together as an article. Yes, this is good, but a bit beyond me to do fact checking on the science... at least not without a lot of time and effort that I don't have at the moment. Joe Cabadas |
March 9, 2019 - 9:41am | Did this ever get published? Seems pretty interesting. The answer would be no. Wonder if blankbeard is still around. EDIT: just sent Blankbeard an email have to wait and see if he is still around and willing to pull this together as an article. Exactly, and for this reason we'll send a email to original poster and see if he'll pull this together. I think there isn't much to do to publish it. Yes, this is good, but a bit beyond me to do fact checking on the science... at least not without a lot of time and effort that I don't have at the moment. I might not be a dralasite, vrusk or yazirian but I do play one in Star Frontiers! |
