FSD jump speed

Noogillim

Noogillim
What is the Averaged Jump speed of the FSD

Given I timed a 40 lyr jump at 13-14 seconds (I know it's game loading etc)

I have been trying to calculate it but my result seems way off of around 263c this can't be right I think I may have dropped a few zeros somewhere[haha]

Given we can do 2000c in supercruise it seems very wrong:x

Anyone know? be nice for FD to give us a figure.
 

777Driver

777Driver
I am fairly certain the concept of time and velocity doesn't exist in witchspace. SC is more like star trek warp (bending spacetime) witchspace is more like punching a hole in normal space.
 

Dagwit

Dagwit
AFAIK the time taken by a jump is determined by the game mechanics - loading the next instance - and isn't really affected by the distance travelled. One light year or eighty will take more or less the same amount of time, with any minor variation caused by the time taken to construct the new instance, rather than the distance travelled.

Edit: come to think of it, I've had jumps where the variation wasn't so minor - but there's no lore reason for that, something's just gone wrong.
 
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Chris Simon

Chris Simon
The hyperspace (witchspace) jump doesn't have "speed". Every jump, no matter how long, takes the same amount of time.
Witchspace isn't a place where time or place would matter.

edit: reps for ninjas...
 

moorison

M
Keep in mind there is at least one pair of systems less than 1ly apart. Jump time would still be the same for out of game reasons.
 

ibmalone

I
Noogillim said:
What is the Averaged Jump speed of the FSD

Given I timed a 40 lyr jump at 13-14 seconds (I know it's game loading etc)

I have been trying to calculate it but my result seems way off of around 263c this can't be right I think I may have dropped a few zeros somewhere[haha]

Given we can do 2000c in supercruise it seems very wrong:x

Anyone know? be nice for FD to give us a figure.
Click to expand...

As already noted, supposedly this is about warping space and so you are not going that fast. By this token though, supercruise is also not at >c velocities, and yet the speedo reads it out anyway...

So what speed will be recorded if we set up speed cameras in Sol and Mulag (exactly 40ly), and they send you a ticket based on timing? Light travels at c, and 1ly is the distance light travels in 1 year. So if you do that 1ly in 1 second then you are going as many times the speed of light as there are seconds in a year. If you do 40ly in one second you are going 40 times that speed. If it instead takes you 14 seconds, you are going 14 times slower. So, how many seconds are there in a year? For this purpose 365.25 days is used, 31,557,600 seconds (I'm told geologists approximate this to π*107, but then, they are horribly imprecise). So, we've got 3.15576*107*40/14=~ 9.016*107 times the speed of light, about 90Mc.

Of course travelling at this speed you should end up with relativistic effects, including going back in time, but representing this in any computer game, let alone a multiplayer one, is simply not possible.
 

Valorin

Valorin
Noogillim said:
What is the Averaged Jump speed of the FSD

Given I timed a 40 lyr jump at 13-14 seconds (I know it's game loading etc)

I have been trying to calculate it but my result seems way off of around 263c this can't be right I think I may have dropped a few zeros somewhere[haha]

Given we can do 2000c in supercruise it seems very wrong:x

Anyone know? be nice for FD to give us a figure.
Click to expand...

Wibbly wobbly, timey whimey. What has been said above. The FSD bends spacetime and shortens the way instead of making us travel at ridiculous speeds. Same with supercruise by the way. In that case we don't really warp out of our dimensions into witchspace, but space in front of our ship is being shortened. The speed indicator is kinda relative depending on the time the ship needs to get from one point to another and doesn't measure real speed. Also why thrusters don't do anything while supercruising. They really don't do anything.

It's roughly based on this: https://en.wikipedia.org/wiki/Alcubierre_drive
 

Robert Maynard

Volunteer Moderator
Robert Maynard
Noogillim said:
Given I timed a 40 lyr jump at 13-14 seconds (I know it's game loading etc)

I have been trying to calculate it but my result seems way off of around 263c this can't be right I think I may have dropped a few zeros somewhere[haha]
Click to expand...

40 LY equates to around 1,262,304,000 Ls, so a 40 LY jump in 13-14 seconds would be somewhere between 90,164,571c and 97,100,308c.

.... except that's not how hyperjumps work in E: D.... :)
 
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Shiro Akai

Shiro Akai
Dagwit said:
AFAIK the time taken by a jump is determined by the game mechanics - loading the next instance - and isn't really affected by the distance travelled. One light year or eighty will take more or less the same amount of time, with any minor variation caused by the time taken to construct the new instance, rather than the distance travelled.
Click to expand...

Confirm. One day it hit me. I decided to buy new ship - discount sale was 200 LYs away from my position. In my engineered ship it was 6 or 7 jumps. made there quickly, bought new ship and made my way home. Now it was over 20 jumps due to low grade FSD.

The clue is: each jump took the same amount of time. It didn't mattered whether it was 35 LYs jump or 8 LYs jump. It took equal time to perform it.

The other day internet had a hiccup (later it was reported it was due to heavy traffic made by some DDOS attack somewhere) and all my jumps took 4x the usual. To the point I was wondering if I still have a connection.

In general - your jump speed relies on your internet connection (the faster the better) and your machine's ability to build jump destination. With good internet and fast machine your times will be shorter.
 

Kenneth McGrew

Kenneth McGrew
Noogillim said:
What is the Averaged Jump speed of the FSD

Given I timed a 40 lyr jump at 13-14 seconds (I know it's game loading etc)

I have been trying to calculate it but my result seems way off of around 263c this can't be right I think I may have dropped a few zeros somewhere[haha]

Given we can do 2000c in supercruise it seems very wrong:x

Anyone know? be nice for FD to give us a figure.
Click to expand...

well travelling beyond semotus beacon at 2001C is roughly 0.24LY/Hr (approx)

So for 40LY in 14 seconds, thats 2.85712~ LY per second or 10,285.7LY/hr


Edit... so you need 10,285LY/hr..

2001C = 0.24LY
therefore as a relationship
if you X 0.24 x 4.1667 you get 1 so

8,337.5C = 1ly/hr (x 10,285)

85,751,873C to cover 40ly in 14 seconds.
 
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ibmalone

I
Ted Striker said:
"Jumping" is just a cleverly disguised loading screen really, so there's no speed.
Click to expand...

Distance over time is speed. It might depend on loading time, may not be constant for a given ship (different length jumps in the same ship take the same amount of time all other things being equal, but it can also take an hour to drive across London, while the same time on a motorway will be 70 miles), and we might say that it's about space warping (and of course it's not, it's about updating some numbers in a computer, but so is normal flight in ED), but we can still talk about how fast an external observer sees you transit that distance, hence using speed cameras in my example. How speed cameras 40ly apart are supposed to synchronise their clocks is where things get sticky.
 

ibmalone

I
Kenneth Mcgrew said:
well travelling beyond semotus beacon at 2001C is roughly 0.24LY/Hr (approx)

So for 40LY in 14 seconds, thats 2.85712~ LY per second or 10,285.7LY/hr

10,285/0.24 = 42,857C?

its early i may have made an error...

but you can jump 40LY in the same time as 5LY it makes no sense... So yeah.. erm 42,857C over the 14 seconds...

^^ actually no that's wrong... my brains pickled atm will try again later
Click to expand...

Nice alternative take (though since ly are defined in terms of c it's sort of the long way around), missed step is that 10285 ly/hr / 0.24 ly/hr is going to be the ratio of the 0.24 ly/hr speed to the 10285ly/hr one, so you still need to multiply by 2001c, to get ~86Mc, which is right ballpark for the equivalent travel speed.
 

Kenneth McGrew

Kenneth McGrew
ibmalone said:
Nice alternative take (though since ly are defined in terms of c it's sort of the long way around), missed step is that 10285 ly/hr / 0.24 ly/hr is going to be the ratio of the 0.24 ly/hr speed to the 10285ly/hr one, so you still need to multiply by 2001c, to get ~86Mc, which is right ballpark for the equivalent travel speed.
Click to expand...

aye, early morning, and got to the same figure in the end but by using a relationship of 4.1667 to get to speed required for 1ly/hr and multiplied up from there. good to know we got the same figure though so must be about right!
 
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