The ships of Elite: Dangerous are known to have fusion power plants capable of producing enormous amounts of power. Testament to this is that shield strength is rated in megajoules. However, whilst thinking about the relationship between the power plant and the shields, I've become confused. I'll use my Corvette as an example, with stats given by coriolis.io. Specifically, here's my ship at the moment:
It has a 36 megawatt fusion power plant, which we can assume is quite small and efficient. In comparison, Arkansas Nuclear One (fission) power plant, which I live relatively close to, has a total nameplate capacity of 1,776 megawatts. Assuming these measurements of power are taken over the same amount of time (I don't know-feel free to correct me), the power plants can generate 36 MJ/s and 1,776 MJ/s, respectively. I found this conversion at
. In short, one MWe is equal to one MJ every second. Here's the problem: My ship's power plant produces 36 MJ/s, but my shields require 1,481 MJ (a ridiculous amount of energy). Is the lack of "per second" an important distinction here? Or is there something else entirely that I'm missing here? My only theory is that the 1,481 MJ is the pool, and the 36 MJ/s would be the available power to recharge the shields, but that would mean the shields would be able to recharge from empty in about 40 seconds (1,481/36=41.14), and from experience I know it takes far longer than this. Someone with a better understanding of electricity, physics, and mathematics, please enlighten me. :S
I need a scientist...or a mathematician
- Thread starter HereticPilot101
- Start date
Well, I think you're assuming 100% efficiency in your calculation for shield recharge time (1,481/36=41.14). 41 secs would be right if there's no loss at all in the Shield Generator especially in the Power Distributor, possibly in the Power Plant too.
You can probably assume the Shield Gen (1,481 MJ) is rated to include losses really, as <power in> is the easiest point of access to make any measurement, eg, it's power "drawn".
The Power Distributor would heat up, as it transfers power through its various coils and capacitors. Heat is a direct loss that means the (Power Plant) output has to provide something over 1,481 MJ (IN) through the Power Distributor, for 1,481 MJ (OUT) to reach the (Shields) input. The difference is the efficiency of the distributor which is highly unlikely to be 100%.
The rating of the power plant might depend .. You'd really want to know how much power the "reactor" (subsystem) gives out, compared to the power out available, overall, to decide if 36MW is a "power plant" rating or a "reactor" rating, as it really could be either. You should probably assume 36MW includes internal Power Plant loss (and 36MW is useable power, out) but it would depend, very realistically, on how the Power Plant has been rated (that is, either the powerplant stats give you a measured power out, or they might easily show an 'expected' power out, according to fuel consumption, which is basically a rating of the reactor inside. Yes, in this case the manufacturer is over selling the powerplant but it's also true, it's a 36MW (IN) powerplant!).
If you measure the shield recharge time (seconds) and compare to your 41.14 seconds ideal, you should be able to come up with a figure for Power Distribution efficiency. This definitely indicates Power Distributor efficiency but might include an error given by the Power Plant (vs Reactor) internal efficiency. However, if you already know your Power Distributor % efficiency (maybe it's given in the stats?) if that equals your own stopwatch (distributor efficiency) calculation you can for interest, work out how the Power Plant was rated.
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You can probably assume the Shield Gen (1,481 MJ) is rated to include losses really, as <power in> is the easiest point of access to make any measurement, eg, it's power "drawn".
The Power Distributor would heat up, as it transfers power through its various coils and capacitors. Heat is a direct loss that means the (Power Plant) output has to provide something over 1,481 MJ (IN) through the Power Distributor, for 1,481 MJ (OUT) to reach the (Shields) input. The difference is the efficiency of the distributor which is highly unlikely to be 100%.
The rating of the power plant might depend .. You'd really want to know how much power the "reactor" (subsystem) gives out, compared to the power out available, overall, to decide if 36MW is a "power plant" rating or a "reactor" rating, as it really could be either. You should probably assume 36MW includes internal Power Plant loss (and 36MW is useable power, out) but it would depend, very realistically, on how the Power Plant has been rated (that is, either the powerplant stats give you a measured power out, or they might easily show an 'expected' power out, according to fuel consumption, which is basically a rating of the reactor inside. Yes, in this case the manufacturer is over selling the powerplant but it's also true, it's a 36MW (IN) powerplant!).
If you measure the shield recharge time (seconds) and compare to your 41.14 seconds ideal, you should be able to come up with a figure for Power Distribution efficiency. This definitely indicates Power Distributor efficiency but might include an error given by the Power Plant (vs Reactor) internal efficiency. However, if you already know your Power Distributor % efficiency (maybe it's given in the stats?) if that equals your own stopwatch (distributor efficiency) calculation you can for interest, work out how the Power Plant was rated.
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Or, you know, you can also just reasonably figure that your ship isn't putting any where near 36mj into your shields per second. A good chunk of that goes into thrusters, and every other subsystem on the ship, like life support and sensors and such. Plus the shield generator works off of a capacitor in the generator that stores power fed to it by the power distributor, thus further gating how many mjs a sec your shields get in charge.
Valid point, but shield recharge time is not the only issue here. If what you say is true, and 36MW is a maximum useable capacity (this makes sense, the Nuclear One power plant I mentioned has a certain nameplate capacity, but that's the maximum capacity possible, not what it's giving out at all times), this may actually solve the issue. Let's do more math. Adding Divine Cerinian's input, let's say that since we have 6 pips available, and can allocate a maximum of 4 to systems, that we're only giving 66% of that 36MJ/s to the shields, but let's also account for heat loss due to inefficiency. In a modern nuclear power plant, the number is pretty low. 28-32%, but we can take our current predictions for fusion energy efficiency (48%) and multiply it by 1.5 (72%). This is pure conjecture-there may be a maximum limit to fusion power plant efficiency that I may have just surpassed. Either way, 72% of 66% is approximately 48%. That's the amount of power allocated to ALL systems other than the subspace engines and the weapons. All other systems power percentages add up to roughly 65%, but the shields only take roughly 14%, so they are taking 14%/65%= 22% of the systems allocated power at 4 pips. 22% of 48% is about 11%. 11% of 36MJ/s is 3.96MJ/s. 1,481MJ/3.96MJ/s= 374 seconds, or about 6.2 minutes. This is much closer to the actual time it takes for my shields to recharge at standard (I'm going to get some Engineer upgrades that will increase that time later). So Frontier must've put a lot of thought into power allocation in ships. It's actually extremely impressive. I wonder what efficiency they actually used for their fusion reactors.
I wonder if you could gain any kind of estimate off a field trial. The energy avialble from fusion does depend on the fuels used, but I think you can rule out any Lithium 6 reaction (as we fuel scoop, hydrogen). The most obvious use of Hydrogen might be to use Deuterium Cycling ..
From that page, and from the bottom up;
6 x Hydrogen [2,1] (deuterium) atoms, output the Deuterium cycle as follows ...
The helium, neutrons and hydrogen [1.1] (protium) are waste because they can't be used in any further reactions, so that's exhaust.
Now, 1 gram of hydrogen contains 6.022 x 10 23 hydrogen atoms
(6.022 x 10 23 is avogadro's number and gives the number of atoms in one mol.
And, also by definition, the molar mass of hydrogen, atomic mass 1 is 1 gram, per mol)
6 atoms used, per deuterium cycle reaction above, you have approx 1 x 10 23 reactions in 1 gram,
.. yielding 43.2 x 10 6 eV (electron volts) per reaction.
that's a yield of 4.32 x 10 30 eV per gram
(for conversion to watts, 1 electron volt, 1eV = 1.60 x 10 -19 Joules)
So how much energy (in joules) potentially, from one gram of hydrogen? 6.9 x 10 11 J
And how much, per tonne? 6.9 x 10 17 J
(that is when 1 metric ton (tonne) is 1 million or 1 x 10 6 .. 1,000,000 grams)
and that also is, pretty hot! (so check me, but I also think you may need to do pre-processing of pure hydrogen to make deuterium, for the cycling process .. this is difficult to guess about but maybe you filter the scooped hydrogen, for example, on a mass basis, deuterium accounts for approximately 0.0312% of all the naturally occurring hydrogen in Earth's oceans wiki .. dropping the tonne yield to 2.07 x 10 14 J)
With a calculation on the energy made available from deuterium cycle, fusion of one tonne of fuel, the field trial might involve working out how much energy your ship expends in example isolating the engines and 'burning' one tonne, or I think (maybe) weapons fire (???) while using one ton of fuel ...
So, in a super simple example without checking if you can get this number, a 1 MW weapon would use 1 x 10 6 Joules per sec (1W = 1 J/s) so if the energy available from one tonnne of fuel was one million joules, that will provide enough energy for 1 second of weapons fire before your tonne of fuel runs out? So, suggest, turn everything else off .. If you now fire your (eg. 1MW) weapon (1) how long to drain the fuel tank? and (2) how big is your fuel tank? (though I suggest using no more that 50% of your fuel tank before returning to base unless you're good mates with a fuel rat!)
PM me if you want assistance in the game for a field trials, or over to you
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Try turning off each subsystem to see what how the available power changes (based on percentage of 36MW).
This OP is pretty good. Don't over-complicate things as ED is just a game. The only way to solve this would be for FD to reveal their formula to us.
Obv. any system has efficiency losses due to 'laws o'physics'. Assume whatever you want, but energy conversion and transfer will cause big losses regardless of theory. Just like a 1 Gb wireless router in theory will probably never transfer data at 1 Gbps due to dependency on other parts of the data flow, protocol overhead, etc.
My guess is FD put some thought into this but trying to make "realistic" calculations for a possible fusion reactor 2,200 years in the future, not so much.
IMHO, Fusion reactors will never be practical. The power density of fusion is actually quite low. To initiate the reaction takes pretty much the energy as we can get out, barring huge leaps in containment and initiation and conversion through heat exchangers. Then there is the problem of very nasty radioactivity of the torus, for most designs.
Nice theory-crafting though, good fun! +1 for the maths effort.
So .. some telemetry testing done. Hope you enjoy.
--- --- elite pilots federation eyes only --- --- --- --- ---
TEST ONE : STANDARD LOAD, CALIBRATION
Sidewinder (Research Vessel, Rsv Avogadro)
Chassis pre-equipped with 2 off standard 0.5MW (1D) mining laser, 1MW total draw .. specs
Method: Weapons free, mining lasers fired at asteroid in an endurance experiment that show identical power being drawn by active hardpoints, whether trigger is being fired, or not. Measurement was made by comparing fuel use (tonnes/hr, see standard cockpit HUD, fuel gauge) and under firing/non-firing conditions.
Conclusions:
Hardpoint weapons draw (and therefore must vent) energy while not being fired. However "Distributor Draw" (see specs, 1.50MW per 1D unit) seems to be the mechanism causing heat to build up in ship. Perhaps acting as an auxilliary, when firing lasers, heat is no longer dissipated but contained inside the ship as the firing coils run hot. To save fuel overall, retract harpoints! However, mining lasers have now been pre-calibrated ready for a fusion test. Laser power to output efficiency is not yet known but (drawing 0.5MW whether firing or not) may be used to test the fusion powerplant as follows.
--- --- elite pilots federation eyes only --- --- --- --- ---
TEST TWO : FUSION FUEL TEST
RSV equipped with two different powerplants, 2A and 2C, for comparative testing.
Method: ( caeritis three facility ) All ship systems were powered down with hardpoints retracted. (1) On emergency life support, ship fuel use was observed at 0.00 tonnes/hr ie. datum. (2) Hardpoints deployed, fuel use for 1 x 0.5MW mining laser (1D) measurements give 0.04/hr (see fuel calculations below). For confirmation, a second, identical mining laser also deployed, observed similar (doubled) fuel use, of 0.08/hr.
CONCLUSIONS: see fuel calulations
(1) fusion systems supply power on demand. ie. if no load on powerplant it powers down to use 0 fuel. Coupled with fact that no modules aboard are generating any heat at 0 power, causes ship / cabin temps to drop off!
(2) Fule Calculations (above) show that 2A and 2C powerplants both consume 22.2 grams of hydrogen fuel per second, and per mega watt of power out.
This experiment showed that <fusion> efficiencies are identical when comparing the fusion process between powerplants tested. Standard ship outfitting figures (mass, overall power, heat etc.) are therefore practical, as there is no gain from using a smaller or larger powerplant, on fusion efficiency alone .
Calculations allow ship range to be developed, as follows; If 22.2 grams hydrogen yields 1MW, 1 gram hydrogen provides 44kJ energy (and, equivalent, one tonne thus yields 45,000MJ). This might lead to an expectation, that a ship loadout drawing 12.5MJ has an operational "endurance" of one hour per ton of fuel. Casual observation however suggests that endurance is less than this (to be expected as systems are not 100% efficient) but no module draw efficiency conclusions can be drawn from initial experiments carried out at Avogadro Outpost, Caeritis 3, on Oct 17 3302. All that is known, is that the power plant regulates its output to suit the load demand and has a 22.2 gms/MW nominal yield on fuel hydrogen.
--- --- elite pilots federation eyes only --- --- --- --- ---
General Experimental Observations
Hardpoints are deployable in proximity to station without alerting security but weapons should NOT be fired(!). Mining lasers do not operate in NFZ's however so testing can be safely carried out under security protection of station. Recommended use of docking computer is for rapid return to base on low power and in any emergency (eg. a tiny powerplant requires shields to be disabled) and as automated DC flight allows pilot monitoring and simultaneous testing of ship power systems during station approach. A high quality off-power life support system is also recommended as extended survival time allows pilot confidence for a low heart rate, allowing careful scientific measurement and without risk of CO2, asphyxia, death.
Winterwalker RTB / o7 out.
--- --- elite pilots federation eyes only --- --- --- --- ---
TEST ONE : STANDARD LOAD, CALIBRATION
Sidewinder (Research Vessel, Rsv Avogadro)
Chassis pre-equipped with 2 off standard 0.5MW (1D) mining laser, 1MW total draw .. specs
Method: Weapons free, mining lasers fired at asteroid in an endurance experiment that show identical power being drawn by active hardpoints, whether trigger is being fired, or not. Measurement was made by comparing fuel use (tonnes/hr, see standard cockpit HUD, fuel gauge) and under firing/non-firing conditions.
Conclusions:
Hardpoint weapons draw (and therefore must vent) energy while not being fired. However "Distributor Draw" (see specs, 1.50MW per 1D unit) seems to be the mechanism causing heat to build up in ship. Perhaps acting as an auxilliary, when firing lasers, heat is no longer dissipated but contained inside the ship as the firing coils run hot. To save fuel overall, retract harpoints! However, mining lasers have now been pre-calibrated ready for a fusion test. Laser power to output efficiency is not yet known but (drawing 0.5MW whether firing or not) may be used to test the fusion powerplant as follows.
--- --- elite pilots federation eyes only --- --- --- --- ---
TEST TWO : FUSION FUEL TEST
RSV equipped with two different powerplants, 2A and 2C, for comparative testing.
Method: ( caeritis three facility ) All ship systems were powered down with hardpoints retracted. (1) On emergency life support, ship fuel use was observed at 0.00 tonnes/hr ie. datum. (2) Hardpoints deployed, fuel use for 1 x 0.5MW mining laser (1D) measurements give 0.04/hr (see fuel calculations below). For confirmation, a second, identical mining laser also deployed, observed similar (doubled) fuel use, of 0.08/hr.
CONCLUSIONS: see fuel calulations
(1) fusion systems supply power on demand. ie. if no load on powerplant it powers down to use 0 fuel. Coupled with fact that no modules aboard are generating any heat at 0 power, causes ship / cabin temps to drop off!
(2) Fule Calculations (above) show that 2A and 2C powerplants both consume 22.2 grams of hydrogen fuel per second, and per mega watt of power out.
This experiment showed that <fusion> efficiencies are identical when comparing the fusion process between powerplants tested. Standard ship outfitting figures (mass, overall power, heat etc.) are therefore practical, as there is no gain from using a smaller or larger powerplant, on fusion efficiency alone .
Calculations allow ship range to be developed, as follows; If 22.2 grams hydrogen yields 1MW, 1 gram hydrogen provides 44kJ energy (and, equivalent, one tonne thus yields 45,000MJ). This might lead to an expectation, that a ship loadout drawing 12.5MJ has an operational "endurance" of one hour per ton of fuel. Casual observation however suggests that endurance is less than this (to be expected as systems are not 100% efficient) but no module draw efficiency conclusions can be drawn from initial experiments carried out at Avogadro Outpost, Caeritis 3, on Oct 17 3302. All that is known, is that the power plant regulates its output to suit the load demand and has a 22.2 gms/MW nominal yield on fuel hydrogen.
--- --- elite pilots federation eyes only --- --- --- --- ---
General Experimental Observations
Hardpoints are deployable in proximity to station without alerting security but weapons should NOT be fired(!). Mining lasers do not operate in NFZ's however so testing can be safely carried out under security protection of station. Recommended use of docking computer is for rapid return to base on low power and in any emergency (eg. a tiny powerplant requires shields to be disabled) and as automated DC flight allows pilot monitoring and simultaneous testing of ship power systems during station approach. A high quality off-power life support system is also recommended as extended survival time allows pilot confidence for a low heart rate, allowing careful scientific measurement and without risk of CO2, asphyxia, death.
Winterwalker RTB / o7 out.
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electrical losses
Your ships systems do not run directly from the powerplant. Everything in your ship excepting the FSD runs on electricity stored in capacitors labeled WEPS, ENG and SYS. The power plant simply charges the capacitors. Assume losses.
Your ships systems do not run directly from the powerplant. Everything in your ship excepting the FSD runs on electricity stored in capacitors labeled WEPS, ENG and SYS. The power plant simply charges the capacitors. Assume losses.
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Awesome job, CMDR. Sorry I couldn't be present for the experiment. Time zone differences...sigh. Someone should move the UK closer to the US.