Long Distance Supercruise times - Anyone got rough numbers?

mossfoot

mossfoot
Has anyone compiled a rough estimate of times for various long distances to other stars? Like if a star is 50,000LS, 100,000LS, 200,000LS, etc? Obviously the ETA given is useless since you're steadily accellerating and then decellerating as you get close.

A rough guide so if I'm exploring and want to scan a distant star I have an idea of how long I can walk away and then check back in on my progress?
 

Yomamas Fett

Yomamas Fett
Did 250k today in the time it took to make a pot of tea and roll a smoke, so about 6 or 7 minutes, did 60k today and arrived just before the tea was on the table, so maybe 3minutes?

It's nowhere near as long as some folks like to complain, thing is at 800c+ you really zip to the destination, but it can take 3-4mins to get up to speed, and if you're facing a binary with gas giants you start slowing down earlier, so results WILL vary.
 

CMDR iLikeCookiesQ

C
ignoring the masslock range of the starting star and the target star/planet (large objects reduce your acceleration and can help you slow down faster when going very close to them)

These observations are mere speculation, but I think you could make some kind of model with it:
Assuming your speed grows linearly. We get that the distance travelled as a function of time is: x(t) = c * t^2. Where c is a constant. With this function, we get that the time to get to a destination as a function of the distance is: t(x) = k * sqrt(x). (Where k is a constant equal to 1/sqrt(c).)

Assuming your speed grows exponentially (you do seem to accelerate faster the higher your speed is) Then we get x(t) = c * e^t and subsequently t(x) = k * ln(x) (where k = -log(c))

So depending on what your acceleration type is (linear or exponential) (it seems to be more the former case than the latter. If it really is exponential, then it is very slow exponential growth) You can pre calculate the time. Just fill in the data of real examples and you can calculate the constants. A model is likely to be accurate if the constants are roughly equal for different real examples (examples of distance and time taken).

I think it is actually a combination of linear and exponential growth, with exponential acceleration near large objects and linear when you are a bit of a distance from it.
 
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Ian Doncaster

Ian Doncaster
For star-to-star journeys (star-to-planet or star-to-empty-space would be faster, of course)

65,000 Ls ~= 8 minutes
360,000 Ls ~= 15 minutes
 

mossfoot

mossfoot
Ian Doncaster said:
For star-to-star journeys (star-to-planet or star-to-empty-space would be faster, of course)

65,000 Ls ~= 8 minutes
360,000 Ls ~= 15 minutes
Click to expand...

That's a good start, thanks!

Maybe some other people can give their own experiences while exploring and we can work out a rough table?
 

kafolarbear

K
200,000ly as of last week in my asp took me about 10 minutes trying to get to another star in the same system so I could scan it. In case you are wondering, yes I did actually time it and I started a bit late and get a bit under ten minutes for 203,000 ly distance

-CMDR Kbear
 

TerrorTrooper

T
As mentioned above, it varies depending on the masses in the solar system.

On a side note, I did .22 Ly in just over 46 minutes. (I bet there's many CMDR's who will recognise that infamous number.) ;D

Oh., and I can do the Kessel run in under 12 parsecs.
 

Zoy

Zoy
I did some work in Excel fitting a regression line to a scatter chart of time (minutes) and distance (ls) and it derived the following formula -

time_in_minutes = 8.6692*LN(distance_in_ls)-92.826

Seems to be pretty reasonable, but I could do with a few more data points for my scatter chart - if anyone has collected some times vs distances, please reply here with your data and I will try and refine the formula

Values for ls > 1,000,000 would be most useful, but anything will help

I would prefer the timings to be as follows -

Hyperspace into target system, select distant target and start supercruise at max speed = Start Time
Supercruise at max speed until time to target < 15 seconds = End Time

Thx

timevsdistance.png
 
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The Tick

The Tick
TerrorTrooper said:
As mentioned above, it varies depending on the masses in the solar system.

On a side note, I did .22 Ly in just over 46 minutes. (I bet there's many CMDR's who will recognise that infamous number.) ;D

Oh., and I can do the Kessel run in under 12 parsecs.
Click to expand...

Oh, the Hutton run! But TerrorTrooper seems to have some secret trick up his sleeve, as everybody else needs approx. 80-85 mins.
Willing to share your trick, TerrorTrooper :-D?
 

Lupin

Lupin
I have listed on my note pad to remind me when looking for viable trade routes and if it's worth traveling to explore:

Aprox.
100,000 = 10mins
233,000 = 11mins (Odd but true, as you speed up more over longer distances).
633,000 = 18mins

These times taken in Asp, shorter now as running most places in Python.

Hope this helps.
 

shadragon

shadragon
mossfoot said:
You'd think some industrious player would be actively catologing that sort of data ;)
Click to expand...
Which ship, what loadout, what class engine, which system? All of those things are variables and to get consistent data you have to establish that first.

An A Class Python versus a stock Sidey would give you different numbers. I like Cargobane's answer personally. It takes as long as it takes.
 

mossfoot

mossfoot
shadragon said:
Which ship, what loadout, what class engine, which system? All of those things are variables and to get consistent data you have to establish that first.

An A Class Python versus a stock Sidey would give you different numbers. I like Cargobane's answer personally. It takes as long as it takes.
Click to expand...

I don't think a python and Sidey accelerate differently in super cruise, and even if they do no doubt it probably evens out in the long run when going 100,000ls or more. The numbers at the top of this Page are the kind I'm looking for
 
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