Fuel efficiency in Elite Dangerous

timwglass

Greetings, commanders.

One of my wing-mates recently attempted the record distance from Sol, so we got talking about fuel efficiency and max distance. I saw the thread that listed Cmdr. Mcgrew's 65,804.4 LY accomplishment. I began to wonder if my little DBX could ever go that far. So, I started to attempt to calculate the math. But I ran into problems. Either my math isn't working, or there's something funny going on. Now, I'm an English Language Arts teacher - not a mathematician, so I'm asking for your help in double checking my math and reasoning.

Let's start with the speed of light, our constant. Google tells me that's 299,792,458 meters/second. (Which is odd. I always thought it was meters/second^2).

So, I know from my wing-mate that max speed is 2001c, which equals 5.99885E+11 meters/second.

Fine.

I really need the speed in meters/hour, though, because my fuel burn rate is listed in tons / hour.

So, 5.99885E+11 meters/second * 60minutes/hour *60second/minute = 2.15958E+15 meters/hour.

So far, so good (I assume).

I need to know how many light years I can travel per hour to know the maximum number of light years I can get away from the star.

So, Google tells me 1 light year = 9.461E+15 meters. So far, so good.

Now, (1LY /9.461E+15 meters)/(2.15958E+15 meters/hour) = .228262 LY/hr

Okay. So, my wing-mate's ship has 800 tons of fuel. At 1.54 tons/hour (not what he's getting, it's actually what I'm getting - but it serves to make the point because his fuel efficiency isn't much different), he can burn for roughly 519 hours.

Now, 519 hours (burn time) * .228262 LY/hr = 118.4679 LY max distance.

This doesn't account for the possibility of a return trip. I'm just trying to figure out the math.

If the main star in Oevasy SG-Y d0 is 65,647.34, then: 65,647.34 + 118.4679 = 65,765.81 maximum distance from the star.

Correct?

So, if I want to get to 65,804.4 LY from Oevasy SG-Y d0, I need to travel an additional 157.06 LY from the star.

That means, traveling at .228262 LY/hr, I need 688.0696 hours of burn time. (157.06 LY / .228262 LY/hr = 688.0696 hours).

Now, knowing I burn at 1.54 tons / hour, I need 105.9627 tons of fuel (1.54 tons / hour * 688.0696 hours = 1,059.627 tons).

Phew.

Does this math seem correct?

~Tim

1

Trollymctrollerson

I'm not sure I fully understand your question. Your math seems to be based on fuel consumption in supercruise. That all goes out the window when you jump between systems in hyperspace. The fuel consumption of inter system jumping depends on the FSD, the optimized mass and any GFSB . Fuel consumption per jump is a non linear formula where the current jumps % of maximum range is weighted with an exponent. The exact formula is unknown to me.

timwglass

@troyllymctrollerson, you're correct. I'm trying to figure out how far I can travel in supercruise, not how far I can jump.

SpeakerToAliens

I think the OP is saying that he wants to get the record for being as far from Sol as possible and to do this he'll fly to the furthest star "Oevasy SG-Y d0" and then refuel at the star and fly in SC on a heading away from Sol to increase that distance still further. Of course, he'll die out there in the black as his power fails when his fuel is exhausted, but he'll hold the record - until somebody builds a better ship.

Trollymctrollerson

So you are telling me somebody flew 65,000+ LY from sol in supercruise? What would possess somebody to expend so much electricity to power a PC for something so pointless?

Trollymctrollerson

Oh ok. I keep forgetting that supercruising out of the galaxy is actually a thing people try to do for bragging rights.

timwglass

@trolleymctrollerson, no, people fly to Oevasy SG-Y d0 by jumping, then fly away from the star there in supercruise for max distance from Sol.

As usual, I'm sorry for asking.

~Tim

Factabulous

You'd be better asking in the Exploration forum - too many trolls in here.

If you want this thread moved then hit 'report post' on your first post & ask the mods to move it - gl!

iain666

Your maths seems plausible - I know the record breakers do SC for several weeks to get there.

Deluvian

Focus on T/h, that's the key

Orvidius

Let's start with the speed of light, our constant. Google tells me that's 299,792,458 meters/second. (Which is odd. I always thought it was meters/second^2).
Squaring the time (such as meter/second^2, or also sometimes described as "meters per second per second") indicates an acceleration rate, not a speed. For the speed of light, you're just looking at a straight speed, so "meters/second" is correct.

Braandlin

Your maths is basically OK. You'll find it easier, and possible to do it mostly in your head if you make some simplifications however.

Speed of light is (near enough) 3x10^8 m/s

Assume top SC velocity is 2000c. So your top velocity is 6x10^11 m/s

3600 seconds in an hour. So Top velocity is 2.16x10^15 m/h

One Light year is approximately 9.5x10^15 m. So 0.23 Ly/h

You want to travel 157 Ly. Will take you 157/0.23 = 682 hours.

Burn rate of 1.54 tons per hour. Means you need 1,050 tons of fuel.

Now add a few % to make up for aproximations... say 1,100 tons to be sure.

You need to remember that you can do these calculations to several orders of accuracy, based on real life accurate numbers. What we don't know for sure though is what numbers FDev have used in the game. eg, have they used 3x10^8m/s for speed of light, or have they used 299,792,458 m/s ... likewise have they used 9,460,730,472,580,800 or 9.5 x10^15 metres in a light year. so an estimation to 2 significant figures is almost certainly accurate enough.

Be aware that additional distance caused by course corrections can become significant over these distances.

• BaldEagle

Thatchinho

Greetings, commanders.

One of my wing-mates recently attempted the record distance from Sol, so we got talking about fuel efficiency and max distance. I saw the thread that listed Cmdr. Mcgrew's 65,804.4 LY accomplishment. I began to wonder if my little DBX could ever go that far. So, I started to attempt to calculate the math. But I ran into problems. Either my math isn't working, or there's something funny going on. Now, I'm an English Language Arts teacher - not a mathematician, so I'm asking for your help in double checking my math and reasoning.

Let's start with the speed of light, our constant. Google tells me that's 299,792,458 meters/second. (Which is odd. I always thought it was meters/second^2).

So, I know from my wing-mate that max speed is 2001c, which equals 5.99885E+11 meters/second.

Fine.

I really need the speed in meters/hour, though, because my fuel burn rate is listed in tons / hour.

So, 5.99885E+11 meters/second * 60minutes/hour *60second/minute = 2.15958E+15 meters/hour.

So far, so good (I assume).

I need to know how many light years I can travel per hour to know the maximum number of light years I can get away from the star.

So, Google tells me 1 light year = 9.461E+15 meters. So far, so good.

Now, (1LY /9.461E+15 meters)/(2.15958E+15 meters/hour) = .228262 LY/hr

Okay. So, my wing-mate's ship has 800 tons of fuel. At 1.54 tons/hour (not what he's getting, it's actually what I'm getting - but it serves to make the point because his fuel efficiency isn't much different), he can burn for roughly 519 hours.

Now, 519 hours (burn time) * .228262 LY/hr = 118.4679 LY max distance.

This doesn't account for the possibility of a return trip. I'm just trying to figure out the math.

If the main star in Oevasy SG-Y d0 is 65,647.34, then: 65,647.34 + 118.4679 = 65,765.81 maximum distance from the star.

Correct?

So, if I want to get to 65,804.4 LY from Oevasy SG-Y d0, I need to travel an additional 157.06 LY from the star.

That means, traveling at .228262 LY/hr, I need 688.0696 hours of burn time. (157.06 LY / .228262 LY/hr = 688.0696 hours).

Now, knowing I burn at 1.54 tons / hour, I need 105.9627 tons of fuel (1.54 tons / hour * 688.0696 hours = 1,059.627 tons).

Phew.

Does this math seem correct?

~Tim

1
You've kind of made the maths a bit more difficult than you need to.

Just bear in mind that a light year is the distance light travels in one year. So, travelling at c means it will take 1 year to travel 1 light year. Measured in c, your speed is light years per year.

So 2,001 c = 2,001 ly per year.

So to work out your hourly speed in ly, you just need to convert from years to hours, which you do by dividing by the number of hours in a year.

So 2,001 c = 2,001 ly per year = 2,001 / (365 * 24) = 2,001 / 8,760 = 0.2284 ly per hour.

In terms of the rest of the journey you just need factor in a bit of time and distance for the acceleration period. It's not linear so can't be worked out simply. But say 6 hours at 0.12 ly per hour average. (Just a stab in the dark on this bit.) This would then effectively take 0.72 ly to get up to full speed.

So to work out the fuel you need the calculation is then fuel needed = (6 * hourly fuel rate) + ((total distance - 0.72ly)/0.2284) * hourly fuel rate).

o7

• LifesAJourney

Qohen Leth

• Arnoldo

Kenneth Mcgrew

Focus on T/h, that's the key

That's the key to your quest.

1.54T/Hr is a terrible fuel consumption rate if you're aiming to beat me p.s your math is correct.

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timwglass

Thanks everyone for the comments. @Kenneth Mcgrew - I can see that 1.54 tons/hour isn't going to work. I did some preliminary tests on my beta account switching various modules in a DBX and couldn't get the burn rate any lower than about 1.12 tons / hour. Obviously, a DBX isn't going to work anyway. I was just trying to figure out what module combination effects fuel efficiency. It's going to take a lot of testing and engineering to get that figured out. I wouldn't ask you to share that, discovery is half the fun. I think my first test will be to see if engineering the thrusters effects efficiency.

Again, I appreciate y'all taking the time to comment and confirm my thinking (plus offer simpler methods).

cercata

Focus on T/h, that's the key
Should focus on Deposit size/ (T/h), that will give you the hours you can run, and since the max supercruise speed is constant, the number of hours is the key

He should try all the ships, DBX is best in FSD Jumps, that's based on Thargoid technology incorporated to human ships in the last 10 years, but supercruise is another older technology, I think ... so maybe another ship is more efficient.

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Allitnil

So to work out your hourly speed in ly, you just need to convert from years to hours, which you do by dividing by the number of hours in a year.

So 2,001 c = 2,001 ly per year = 2,001 / (365 * 24) = 2,001 / 8,760 = 0.2284 ly per hour.
A small correction, but the definition of "year" used in light year is 365.25 days so there are an extra 6 hours. Very slightly reduces the per hour figure to 0.22827.

• Thatchinho

Thatchinho

A small correction, but the definition of "year" used in light year is 365.25 days so there are an extra 6 hours. Very slightly reduces the per hour figure to 0.22827.
Top info, ta! And now you’ve said it, it’s obvious why, and I’m kicking myself. Bob Lighthouse

Banned
Greetings, commanders.

One of my wing-mates recently attempted the record distance from Sol, so we got talking about fuel efficiency and max distance. I saw the thread that listed Cmdr. Mcgrew's 65,804.4 LY accomplishment. I began to wonder if my little DBX could ever go that far. So, I started to attempt to calculate the math. But I ran into problems. Either my math isn't working, or there's something funny going on. Now, I'm an English Language Arts teacher - not a mathematician, so I'm asking for your help in double checking my math and reasoning.

Let's start with the speed of light, our constant. Google tells me that's 299,792,458 meters/second. (Which is odd. I always thought it was meters/second^2).

So, I know from my wing-mate that max speed is 2001c, which equals 5.99885E+11 meters/second.

Fine.

I really need the speed in meters/hour, though, because my fuel burn rate is listed in tons / hour.

So, 5.99885E+11 meters/second * 60minutes/hour *60second/minute = 2.15958E+15 meters/hour.

So far, so good (I assume).

I need to know how many light years I can travel per hour to know the maximum number of light years I can get away from the star.

So, Google tells me 1 light year = 9.461E+15 meters. So far, so good.

Now, (1LY /9.461E+15 meters)/(2.15958E+15 meters/hour) = .228262 LY/hr

Okay. So, my wing-mate's ship has 800 tons of fuel. At 1.54 tons/hour (not what he's getting, it's actually what I'm getting - but it serves to make the point because his fuel efficiency isn't much different), he can burn for roughly 519 hours.

Now, 519 hours (burn time) * .228262 LY/hr = 118.4679 LY max distance.

This doesn't account for the possibility of a return trip. I'm just trying to figure out the math.

If the main star in Oevasy SG-Y d0 is 65,647.34, then: 65,647.34 + 118.4679 = 65,765.81 maximum distance from the star.

Correct?

So, if I want to get to 65,804.4 LY from Oevasy SG-Y d0, I need to travel an additional 157.06 LY from the star.

That means, traveling at .228262 LY/hr, I need 688.0696 hours of burn time. (157.06 LY / .228262 LY/hr = 688.0696 hours).

Now, knowing I burn at 1.54 tons / hour, I need 105.9627 tons of fuel (1.54 tons / hour * 688.0696 hours = 1,059.627 tons).

Phew.

Does this math seem correct?