Please, allow us to visualize fuel tank info (total tonnage remaining in tank)

[Koben]

I'm out there exploring and need to do some calculations in order to see if I can reach a target by flying in super cruise.
Unfortunately I have to use a small scale on my monitor screen to try to figure that out, once there's no place I can get the information on the actual total tonnage of fuel remaining in my ship.

Also, there was before a lot of information about my gameplay that is now gone, like total hours spent in game, profit from several roles... etc... please, bring those back!


Thanks in advance!

 

[Winterwalker]

I'm out there exploring and need to do some calculations in order to see if I can reach a target by flying in super cruise.
Unfortunately I have to use a small scale on my monitor screen to try to figure that out, once there's no place I can get the information on the actual total tonnage of fuel remaining in my ship.

That would be a very useful stat.

You can do some estimating of course; With a ruler against the screen you can work out the size of your fuel supply tank by seeing how much a refill takes off your main tank, which has a capacity known from outfitting. It can be fun doing stuff like this, I did some calculations on the efficency of (fusion) powerplant and conversion rate of hydrogen into MW.

Nice request though.

Also, there was before a lot of information about my gameplay that is now gone, like total hours spent in game, profit from several roles... etc... please, bring those back!
Thanks in advance!

You can still find this. Look up your commander stats in codex, right panel.

 

[Kelster87]

Try looking in the Codex, that's where most of your data is..

 

[Koben]

Thanks for the tips on the CODEX tab for the game stats!

 

[MK Regular]

... I did some calculations on the efficency of (fusion) powerplant and conversion rate of hydrogen into MW.

Doesn't 1 ton of fuel generate something like 45000 MJ of energy (plus however much heat the power plant makes) in-game?

1 ton of hydrogen certainly doesn't generate a measly 45 GJ of energy in a star. Knowing that hydrogen fusing into helium loses 0.8% of it's mass (8 kg for every metric ton), Einstien's equation (e = mc²) says that a single ton of hydrogen should generate 720 PJ of energy, which is 16 million times higher than the amount of usable energy 1 ton of hydrogen generates in ED.

Unless I missed something, our power plants in ED are ridiculously inefficient (less than 0.00000625% of the energy released by the fusion reaction is usable).






Anyways, back to OP's idea. I like it. It seems useful, although I am not too sure where to put it in the UI.

 

[Winterwalker]

Doesn't 1 ton of fuel generate something like 45000 MJ of energy (plus however much heat the power plant makes) in-game?

I used a pair of small mine lasers on a sidewinder with everything else powered down. Mining lasers consume 0.5MW each for easy maths so by running one then both you can isolate the fuel consumption rate from the powerplant. If I recall, as I don't have the record anymore, 22.2 grams hydrogen per joule I think it got?

That is inefficiant but as there's no medium to conduct heat away in space, you can only radiate and heat is an issue, eg. go silent running. 100% efficient conversion (E=mc2) might cook you so it could be 'inefficient' by design. In fact it might also be practical to dump hot hydrogen in your vapour trail, with propellant conducting waste fusion heat away.

 

[Shiro Akai]

although I am not too sure where to put it in the UI.

The most logical place would be next to fuel gauge, before "fuel/hour". Or slightly re-arrange to:
FUEL: 1.6t/h 3.14/8t.
Should be relatively easy since we got lights indicator in a similar way some time ago. After years of begging but still...

Fantasy-ing further - if we had proper UI that support mouse input (sorry console users, your inferior pads are heavily limiting overall performance) we could have simple toggle-on-click display so you could alter between fuel consumption and fuel remaining.

 

[metatheurgist]

Not 100%, but I'm pretty sure the reservoir is always 1T, and it always steals 1T from your main tank. You have to eyeball the Res but you should be able to work out how long you can fly with the fuel/h reading. Your bigger problem will be working out how long it will take you to get somewhere since there's no way to account for slowdowns and the time to destination is elastic.

 

[Morbad]

Unless I missed something, our power plants in ED are ridiculously inefficient (less than 0.00000625% of the energy released by the fusion reaction is usable).

I'd assume most of the fuel was being spent as reaction mass, with little of it actually fused, except for hyperspace jumps.

At the other end of the spectrum, there are weapons that appear to be more than 100% efficient at turning energy input into damage.

In fact it might also be practical to dump hot hydrogen in your vapour trail, with propellant conducting waste fusion heat away.

Reaction mass, sure. A heat dump doesn't seem to jive so much as a silent running ship has no more of a heat signature while boosting than it does while stationary. If anything, our exhaust seems to be deliberately cooled.

Not 100%, but I'm pretty sure the reservoir is always 1T, and it always steals 1T from your main tank. You have to eyeball the Res but you should be able to work out how long you can fly with the fuel/h reading.

It's not. It varies heavily per ship and only the largest ships have a ~1T reserve.

Ships - Reserve tank sizes for all ships

Fuel tanks The main tank We all know that our ships have one or more fuel tanks from which the frame shift drive draws fuel when jumping between systems. The amount of fuel available in these tanks is shown in a thick bar on the right of the HUD. When we plot a jump, a portion of that bar...

forums.frontier.co.uk

 

[MK Regular]

I used a pair of small mine lasers on a sidewinder with everything else powered down. Mining lasers consume 0.5MW each for easy maths so by running one then both you can isolate the fuel consumption rate from the powerplant. If I recall, as I don't have the record anymore, 22.2 grams hydrogen per joule I think it got?

That is inefficiant but as there's no medium to conduct heat away in space, you can only radiate and heat is an issue, eg. go silent running. 100% efficient conversion (E=mc2) might cook you so it could be 'inefficient' by design. In fact it might also be practical to dump hot hydrogen in your vapour trail, with propellant conducting waste fusion heat away.

22.2 grams per megajoule equates to 45045 megajoules per ton, which is quite close to what I got.

I'm not sure if venting heated hydrogen would be enough to get rid of enough heat to explain why our reactors are so inefficent, so I'll do some quick^TM math:

Spoiler

The specific heat capacity of Hydrogen is ~14.32 kJ/kg K, meaning that you would need to raise the temperature of a single kilogram oh hydrogen by 70 K in order to get rid of 1 MJ of thermal energy. However, in order for our ships to actually be losing heat, we would need to increase the temperature of the hydrogen from what it was scooped at. Thankfully, this actually works in our favour, since the surface of the sun is about 6000 K, at which point hydrogen has a specific heat capacity of ~20 kJ/kg K and we will only need to heat a single kilogram of hydrogen by 50 K in order to dump 1 MJ of thermal energy.

Making the following assumptions:

1. A stock power plant is capable of converting 5% of the energy generated by the fusion reaction into usable energy (this leaves some wiggle room for engineer mods)
2. All remaining energy generated by the fusion reaction is in the form of thermal energy
3. The thermal rating of our power plant denotes the number of MJ of thermal energy retained for every MJ of usable energy generated
4. All thermal energy that is not retained is vented with the waste gasses from the reaction
5. Waste hydrogen has a specific heat capacity of 20 kJ/kg K
6. Waste helium has a specific heat capacity of 5 kJ/kg K
7. The fusion reaction of hydrogen to helium loses 8 kg per metric ton of hydrogen

We can determine an approximate temperate of the waste gasses as follows:

​

45 GJ of usable energy generated per ton means that a total of 900 GJ of total energy is generated per ton​

​

For a stock A-rated power plant, 18 GJ of thermal energy is retained, meaning 837 GJ of energy needs to be released​

​

900 GJ of energy generated means that 0.00001 kg of mass was lost in the reaction​

​

0.00001 kg of lost mass indicates that 0.00125 kg of hydrogen was consumed for a return of 0.00124 kg of hydrogen​

​

We now have 999.99875 kg of hydrogen and 0.00124 kg of helium and 837 GJ of waste heat to dump into it (for a stock A-rated power plant with a thermal rating of 0.40)​

​

There is so little helium that it can be ignored entirely​

​

999.99875 kg of hydrogen at a specific heat capacity of 20 kJ/kg K means that we can dump 19.999975 MJ (let's just call it ~20 MJ) of heat for every 1 K we increase the exhaust temperature​

​

In order to dump 837 GJ of thermal energy using nothing but hydrogen, we would need to heat up the exhaust gas by 41850 K​

Yeah, that's a lot of heat. How about I up the efficiency of the reactor in assumption 1 from 5% to 10% efficiency:

45 GJ of usable energy generated per ton means that a total of 450 GJ of total energy is generated per ton​

​

For a stock A-rated power plant, 18 GJ of thermal energy is retained, meaning 387 GJ of energy needs to be released​

​

450 GJ of energy generated means that 0.000005 kg of mass was lost in the reaction​

​

The amount of hydrogen converted to helium is so small we'll just say that we have 1000 kg of hydrogen​

​

1000 kg of hydrogen at a specific heat capacity of 20 kJ/kg K means that we can dump 20 MJ of heat for every 1 K we increase the exhaust temperature​

​

In order to dump 387 GJ of thermal energy using nothing but hydrogen, we would need to heat up the exhaust gas by 19350 K​

This is much more manageable, but still quite high. What if we increase the efficiency to 20% (which would be fairly good efficiency)?

45 GJ of usable energy generated per ton means that a total of 225 GJ of total energy is generated per ton​

​

For a stock A-rated power plant, 18 GJ of thermal energy is retained, meaning 162 GJ of energy needs to be released​

​

The amount of hydrogen converted to helium is so small we'll just say that we have 1000 kg of hydrogen​

​

1000 kg of hydrogen at a specific heat capacity of 20 kJ/kg K means that we can dump 20 MJ of heat for every 1 K we increase the exhaust temperature​

​

In order to dump 162 GJ of thermal energy using nothing but hydrogen, we would need to heat up the exhaust gas by 8100 K​

That's still a lot of heat, but it at least somewhat reasonable considering that we are already dealing with temperatures in the range of 6000 K when we are scooping our for to begin with.

Also, yes, this could explain why we get so little energy from running an entire ton of hydrogen through our power plants.

Oh, and using an E-rated power plant (with 1.00 thermal efficiency) instead of an A-rated power plant decreases your exhaust temperature by 1350 K, in case you were wondering.

Yeah, this could explain why we get so little energy from running an entire ton of hydrogen through our power plants.

Edit: Moved all of the quick^TM math into a spoiler

 

[Winterwalker]

Also, yes, this could explain why we get so little energy from running an entire ton of hydrogen through our power plants.

Temps look pretty good but if there's a concern they're a bit high, the estimations can possibly come down if you imagine fuel scooping as a ram? If the hydrogen is stored under pressure then released (and depressurised) into coolant, you get a reduction in temp from the gas expansion? (The opposite effect to pressuration temperature rises in the ram, when you're scooping?).

The more I think about this OP the more I would like a fuel remaining tonnage on the readout. Hours remaining might be another way to go, basically the same info without the calculation involved. Be nice if there's still some thinking to do though, so one or the other I think (tons or hours remaining). These are spaceships, not Audi's !

In the absence of data on ship model feed tank capacities I'm definitely going to do this on my RSV (Chieftain) with a stopwatch next week. Not a bad community project in fact, to collate feed tank sizes for all the ships in game. Useful stat, as meanwhile, graduation marks on the main tank gauge would also really help.

 

[MK Regular]

Temps look pretty good but if there's a concern they're a bit high, the estimations can possibly come down if you imagine fuel scooping as a ram? If the hydrogen is stored under pressure then released (and depressurised) into coolant, you get a reduction in temp from the gas expansion? (The opposite effect to pressuration temperature rises in the ram, when you're scooping?).

Hmmmm, that might work, let me figure it out....

Spoiler

Th ideal gas law states that the Pressure (P), Volume (V), and Temperature (T) of an ideal gas are bound to each other by the equation PV = nRT, where n represents the amount of the gas present (which we will treat as being constant) and R is the gas constant (which is also constant). We can rearrange the equation to take the form of:

P1 V1 / T1 = P2 V2 / T2

Allowing us to determine the pressure, temperate, or volume of the gas in a given state provided that we have the other 2 paramaters for the new state and all 3 parameters for the initial state. Starting with the parameters for the initial state:

• The surface of the sun is ~6000 K
• The surface pressure of the sun is ~87 Pa
• The density of the surface of the sun is about 0.2 mol/m³
• Hydrogen has a molar mass of 1.008 g/mol
• The density of hydrogen scooped from the surface of the sun is therefore about 0.2016 g/m³
• Therefore, a single ton of hydrogen takes up 4.96 * 10⁶ m³
• Hydrogen makes up ~75% of the sun's mass, with the next most common element being helium at ~24%
• Helium has a molar mass of 4.003 g/mol
• Therefore hydrogen makes up ~92% of the sun's volume, rounding down
• Given that the surface pressure of the sun is ~87 Pa, the partial pressure of Hydrogen is ~80 Pa, rounding down

Plugging these values into the equation, we get:

(80 Pa) (4.96 * 10⁶ m³) / (6000 K) = P2 V2 / T2
66.138 kPa m³ / K = P2 V2 / T2

Knowing that we need to shove all of this hydrogen into our ship, we need to massively decrease the amount of volume it takes up. Let's say we compress the hydrogen to take up 4 cubic meters per ton and simplify the equation:

66.138 kPa m³ / K = P2 (4 m³) / T2
16.534 kPa / K = P2 / T2

Now, if we assume that the scoop acts like an isothermal process (temperature is constant), we can calculate the pressure inside the tank:

16.534 kPa / K = P2 / (6000 K)
99.206 MPa = P2

Oh yeah, this fuel tank is fine. It's only pressurized to nearly 1000 atmospheres with an internal temperature of 6000 K and filled with a highly volatile gas, nothing bad will happen if it gets damaged in combat....

Spoiler

Now, let's decompress the hydrogen to use as coolant. Let's say our coolant system runs at 1000 kPa (about 10 atmospheres). For this one, we actually need to use a variation of the ideal gas law for an isentropic process (entropy is constant) that uses the specific heat ratio (k). Knowing that the specific heat ratio for hydrogen is about 1.6:

T2 = T1 (P2 / P1)^{(k − 1)/k}
T2 = (6000 K) (1000 kPa / 99.206 MPa)^{(1.6 − 1)/(1.6)}
T2 = (6000 K) (0.01008)^0.375
T2 = (6000 K) (0.1784)
T2 = 1070 K

Assuming that this is the hydrogen that gets heated by 8100 K when it cools the reactor, the exhaust gases have a total temperature of 9170 K.

Also, for the love of god, DO NOT try to assume that the fuel scoop is an isentropic process unless you want to have a fuel tank pressurized to over 12 GPa with a temperature exceeding 6 million degrees Celsius. I repeat, DO NOT.

Yeah, it works (kinda). If you assume that:

• The fuel scoop is an isothermal process
• The fuel in the tank has a density of 250 kg/m³
• The expansion into the cooling system is an isentropic process
• The cooling system runs at 1 MPa

The exhaust temperature is reduced by just under 5000 K (from 14100 K to 9170 K). It's an improvement, but it's still extremely hot (for reference, tungsten will melt at about 3700 K and graphite will melt at about 3800 K).

Note that it is possible to reduce the heat even more by lowering the operating pressure of the cooling system. Lowering the operating pressure to 100 kPa would have the effect of reducing the exhaust temperature by another ~500 K, but it won't make a massive difference as most of the heat comes from the cooling process instead of residual from the scooping process.