Space Speeds and Jump Gate near Io (NEED HELP!!!)

[lastbesthope]

Say the freighter has a deceleration rate of 10, then one round before reaching its destination it's travelling at 10, so is 10 away, the round before it was travelling 20 and so was 30 away, 60 away the turn before that and 100 the turn before that so you have 5 rounds notice on the fine tuned sensors (and considerably before that in reality - the heat bloom from the thrusters, which would have to be firing towards you, would be damn obvious against the cold of interplanetary space).

I think you're distances are off there frobisher, but your speeds are right.

If the freighter has a deceleration of 10 then 1 time unit before the target is reached it will have a speed of 10, but it will not be 10 distance units away as the distance travelled in that last time period will be less than 10 due to deceleration. Remember from physics:

s=ut+1/2*a*t^2

s=distance
u=initial velocity
t=time over which distance travelled is being calculated
a=accel/decel (depending on sign, accel= +ve)
^2=squared

so if at time T the target is reached then the distance from target at time T=0.

At time T-1, u=10, a=-10 (decelerating), t=1 (time we are decelerating for) so:

s=10*1 +1/2*-10*1^2
s=10 -5
s=5

so at time T-1 the freighter would be 5 distance units away from the target assuming constant deceleration.

At time T-2, u=20, a=-10, t=2 so
s=20*2+1/2*-10*2^2
s=40-20
s=20, so 20 distance units away at time T-2.

It's easier to work out the other way round, using the same formula and assume the object starts at the target and accelerates away from it, the distances travelled will be the same as under the same level of decel. It simplifies though as

s=ut+1/2at^2 becomes
s=1/2*a*t^2
as u=0 (Object at rest at target)

I hope I got my numbers right.

LBH[/code]

 

[lastbesthope]

Havinfg said all that I've realised the rules abstract the physics and frobisher is right in game I think. But for real world calculations my numbers are right.

Sorry

LBH

 

[frobisher]

Havinfg said all that I've realised the rules abstract the physics and frobisher is right in game I think. But for real world calculations my numbers are right.

Absolutely

I probably should have made that clear in my example though - too many years of flipping between real physics and games programs simulated physics (programming such) that I tend to treat examples contextually IYSWIM.

 

[lastbesthope]

If I'd been more awake I would have spotted it before taking 5 mins typing all the above. but at least it'll give people something to work out real world estimates for travel. Accel to the mid point then decel all the way in for max speed travel.

LBH

 

[ShadowScout]

The average speed would increase the longer your accel phase.

I know - see answer to point "1".
I was talking "quick and dirty" game terms here, as many a GM would go mad if he had to do real-life physical calculations of acceleration, distance and speed every time his players do some travelling... :wink:

It should be noted that most civilian ships (freighters) won't keep under constant acceleration/deceleration from start to finish; they'll likely accelerate to a "fast enough" speed and coast until they need to slow down again.

Quite likely as long as speed and time is no issue. THE thing for low-priority freight, especially for cheap transport companies. But as one can see from the calculations above, it would then take many Weeks if not months to get from earth to the jumpgate at io (as it almost takes a week at constant 1G accel). While that may be nice for tramp freighters, it wouldn't do for passengers and priority cargo (and not for Marcus and Dr.Franklin, as their total travel time was supposed to be a mere week - and considering the urgency of their mission they probably took the best mix between speed and anonymity...)

If you want evidence for this in the show, witness the first leg of Marcus and Franklin's trip to Mars that we saw. They were clearly in a near weightless environment aboard the freighter (floating, unsecured cargo!!!) so the ship had no significant acceleration at that point (say less than .05g). The ship was coasting, and given how "settled in" they were, it had been for some time.

Possibly. Likely even. Though problematic for the cargo shipper - once he decelerates, as he has to, the unsecured cargo will do some banging...

Military ships would probably have a greater fuel reserve, and are less wary of "running costs". It's also likely (for the EA ships) they would run under a slight acceleration when ever possible so that the crew got some "heavy time" so as not to lose muscle tone etc.

True.

 

[Kizarvexis]

Two things.

First, IIRC, the Excalibut was moving at 7.5% of light speed in normal space. I remember thinking that he used a lower, more realistic value than other 'so called' space shows on TV. But it has been awhile since I read the unfilmed Crusade scripts, so I could be wrong.

Second, the real life asteroid belt between Mars and Jupiter is very widely spaced apart. It is roughly on the plane of the rest of the planets and it is 2 to 4 AU wide. On average, there is about 2 MILLION kilometers (over a million miles) between asteroids. So if you are near one, you probably can't see any others.

'Tons of Asteroids' article from Bad Astronomy.com

Kizarvexis

 

[ShadowScout]

First, IIRC, the Excalibut was moving at 7.5% of light speed in normal space. I remember thinking that he used a lower, more realistic value than other 'so called' space shows on TV. But it has been awhile since I read the unfilmed Crusade scripts, so I could be wrong.

I remember 75%... but it's been a while for me too, so I could be of course wrong. I don't think so, because I was wonderintg about that figure back then... and "75%" is more likely to be picked then "7.5%" ("nice" numbers and all that).
Personally I'd be more happy with figures below 10% LS for real-space travel. That'd still be 0.72 AU per hour - over 112 mio Km/h; or 1.33 hours per AU.
The game-suggested average speed would be about 0.5% of LS BtW - 24 hours per AU. Just for those who want to know.
And all that of course supports if not your claim that it was 7.5% in the unfilmed scripts, but certainly that it should be that.

In that spirit... for any physically calculating people here... anyone wants to do the math and see how long you'd have to accelerate at 1G until you reach 75% of LS? I don't have the formulae, but I'm sure you'd be looong out of our solar system by the time you reack that velocity... and most likely many months older besides.

Second, the real life asteroid belt between Mars and Jupiter is very widely spaced apart. It is roughly on the plane of the rest of the planets and it is 2 to 4 AU wide. On average, there is about 2 MILLION kilometers (over a million miles) between asteroids. So if you are near one, you probably can't see any others.

Well, that's basically why I wrote: ...now, your usual asteroid belt doesn't look like in "Star Wars", with rocks floating within meters of each other -that's more the kind of thing you find withing a gas giant's rings BtW-, it usually means at least several klicks of space between the rocks...
You are of course completely right that in Our belt there usually are thousands or missions of klicks between rocks (at least between big rocks, so far we're not completely sure if there aren't small rocks floating between those - but we'd be talking real small here - I think the observatories are certain they found everything larger then a couple of meters)
But even a denser belt that the one in our system is pretty easy to fly through - if you know the vectors of the asteroids (one good astronomical scan would do it for an explorer). As I wrote, even if it's relatively much, much denser then ours, there'll still be kilometers between asteroids.
Of course, if you do go through at an high enough speed, it becomes a gamble - as I mentioned, the higher the speed, the more thrust needed to alter your vector. The more thrust needed, the more burn-time that means, especially for ships that have no AG. Meaning if you go fgast enough you can't turn in time, even if you see the obstacle.
Not to mention that as I also mentioned, one never knows if there aren't small rocks floating with the big asteroids. And at a high enough speed a fist-sized rock can ruin many a spaceship's day...

 

[frobisher]

Meaning if you go fgast enough you can't turn in time, even if you see the obstacle.

Actually not entirely true (depending upon speed).

You just need to be able to impart to yourself a tangential velocity that is sufficent to clear the presented width of the object in the time it would take you to hit the impact point. In relative terms, this might not need to be much;

For instance, assuming the ship is capable of a 1g acceleration, then to avoid a 300km wide rock you only need a 3 minute warning (before impact) to be able to avoid it, and only 30 seconds if you've got a 2g acceleration, regardless of what speed you're travelling at.

 

[lastbesthope]

Where are these speeds as factors of c coming from, I don't remember them being mentioned anywhere?

In that spirit... for any physically calculating people here... anyone wants to do the math and see how long you'd have to accelerate at 1G until you reach 75% of LS? I don't have the formulae, but I'm sure you'd be looong out of our solar system by the time you reack that velocity... and most likely many months older besides.

Ok, using this site as my source of constants:

http://www.astro.wisc.edu/~dolan/constants.html
and assuming a value for g of 9.8065 m s^-2 (IIRC)

I get that from an absolute 0 velocity standing start accelerating at 1G constantly, compensating engine thrust for relativistic effects) it would take you about 0.484248782 Light Years to accelerate to c, so about 0.36 ly to get to 0.75 C. That is about 30631 AU for c, and about 22650 Au or do for 0.75 C.

That puts you well outside the Solar System and about 7 or 8% of the way to Alpha Centauri.

LBH

 

[frobisher]

I get that from an absolute 0 velocity standing start accelerating at 1G constantly, compensating engine thrust for relativistic effects) it would take you about 0.484248782 Light Years to accelerate to c, so about 0.36 ly to get to 0.75 C. That is about 30631 AU for c, and about 22650 Au or do for 0.75 C.

That puts you well outside the Solar System and about 7 or 8% of the way to Alpha Centauri.

Though of course at this point you should be discarding Newton and going with General Relativity

Effective mass increases with velocity, plus time and length dilation etc, would play merry hell with those figures. As far as a static observer is concerned, the ship would take greater than 0.36 ly to get to 0.75c, and would never achieve c. After that it gets much more complicated...

 

[Kizarvexis]

I found my notes on 'To the Ends of the Earth". After Gideon finds out where the enemy ship is, he says the following.

Gideon - Give the coordinates to Lieutenant Matheson. We're going after them. Full burn.

There is a screen cut with the Excalibur racing past the point of view of the camera.

Gideon - What's our current speed?

Matheson - Point seven five percent of light speed. We can't go any faster than that in normal space.

Gideon - Assuming they have the same limits we do, the laws of physics being what they are, their maximum speed should be roughly the same as ours.

Matheson and Gideon go on to discuss how to catch the enemy ship using hyperspace. So I was wrong and Matheson says .75% C, which means that the Excalibur is limited to moving less than 3,000 km per sec in normal space.

Kizarvexis

 

[frobisher]

Matheson and Gideon go on to discuss how to catch the enemy ship using hyperspace. So I was wrong and Matheson says .75% C, which means that the Excalibur is limited to moving less than 3,000 km per sec in normal space.

Which is kinda refreshing to know

At a rough guess, that figure came from the guys at JPL that the production team were working with (it could be the threshold at which the relativistic effects kick in with a vengeance with particular drive types - though given that the Exclaibur has gravitic drive, "real world physics" might not apply).

 

[ShadowScout]

Actually not entirely true (depending upon speed).
You just need to be able to impart to yourself a tangential velocity that is sufficent to clear the presented width of the object in the time it would take you to hit the impact point. In relative terms, this might not need to be much;
For instance, assuming the ship is capable of a 1g acceleration, then to avoid a 300km wide rock you only need a 3 minute warning (before impact) to be able to avoid it, and only 30 seconds if you've got a 2g acceleration, regardless of what speed you're travelling at.

Hmmm... You know, I don't think that is true.

Correct me if I am mistaken, but isnt it that the more speed your ship has, the more thrust it needs to change it's vector? And that the acceleration capability is just a result of available thrust and to-accelerated mass? (which BtW is why LS cannot be reached while playing fair with Einstein - as the speed gets closer to LS, the mass starts to increase, thus requiring more and more thrust for the same acceleration - and the increase of mass grows exponentially as the speed gets closer to LS - so you can't reach LS without unlimited - and I mean UNlimited, as the mass reaches "infinite" at lightspeed - power)

It's just as with cars - the faster they go, the bigger a turn radius they have. Everyone can try this (though I wouldn't recommend it on public roads... actually I wouldn't recommend it at all, just take my word for it or ask someone whose word you trust :wink: ) - a narrow turn you can make at 25 Km/h will be mostly impossible to do at 150 Km/h. People die every year because drivers forget that.

Now, to get back to our ships, if the obstacle appears at the start of your burn cycle, where speed is low, you need only little warning, because a short blast of thrust from your side thrusters would be enough to change the ship's vector to avoid collision. But if the same obstacle appears near turnover, where speed is high, you need a much, much longer period of thrust to shift your vector away from the impact zone. Meaning you might even need to pivot your ship and use the big main thrusters. And the longer a distance travelled, the more speed at turnover, so the less maneuvering capability available to the ship.

Remember, the distance we're all here talking about is Io jumpgate to mars - average distance ca. 5 AU. The edge of our solar system on the other hand (Neptune, as Pluto has an skewed orbit) is somewhere around 30 AU. And an exploration vessel couldn't risk jumping into a solar system, they'd have to jupm in from hyperspace elsewhere and work their way in through normal space (though I don't expect them to stay in the ecliptic - the hypotethical explorer would most likely try to jump in, say, 5-10 AU above/below the ecliptic and get a good look at the system before it starts the more serious exploring)

Of course, all that is talking about constant accel trips - a freighter slowly coasting in free fall doesn't chage it's "cruising speed".

And since we DO have a top speed now (see below), the question becomes more like: "...how much warning does a ship need at 8 Mio. Km/h if it can sustain 0.5, 1, 1.5 or 2 G thurst; and how much for short 3-5 G emergency thrust?"

Though of course at this point you should be discarding Newton and going with General Relativity

For real life calculations - sure. But before that point, newton is good to take for what his formulae really are - an easy to work with approximation. :wink:

I found my notes on 'To the Ends of the Earth". After Gideon finds out where the enemy ship is, he says the following.
-snip-
So I was wrong and Matheson says .75% C, which means that the Excalibur is limited to moving less than 3,000 km per sec in normal space.

([muttering...] I knew I should have ignored JMS and got a copy of these scripts too...)
You know, that is Really, really good to know. Never was I more happy to be wrong, for it saves the reputation of my favorite SF show in that regard.

ALL RIGHT!

We now have a good number!

So, 0.75% of LS... 2250 km (or roughly 1500 miles for those who are not yet evolved enough to use the metric system :wink: ) per sec. as maximum real-space speed. That'd be about 8 Mio. km/h (5 Mio. MPH). And that means about 19 hours per AU, or 1.3 AU per day at that speed.
Good.
Now... if we had a calculation how long (time & distance) it take to reach that speed at 1G, 1.5G and 2G (I unfortunately haven't found my physics formula book yet, and it's been a too long time since I did such stuff to still know it on my own) we would have a rough (therefore game-usable) formula for calculating long-distance realspace travel. Though it starts to look as if the values mongoose wrote in their GG are basically good enough to run with even for real-life physics enthusiasts. :wink:

 

[lastbesthope]

@Frobisher

I di mention adjusting engine thrust for relativistic effects to maintain 1G accel.

@ Shadow Scout

To keep the sums easy, make 1G accel 10 ms^-2. So 2250 km/s would take 225000 seconds to reach at 1G (correcting for relativity) This equates to 62.5 hours if I got my numbers right.

LBH

EDITED my numbers as Shadow Scout pointed out

 

[ShadowScout]

To keep the sums easy, make 1G accel 10 ms^-2. So 22500 km/s would take 2250000 seconds to reach at 1G (correcting for relativity) This equates to 625 days if I got my numbers right.

I see two errors again... for one there is a zero here that doesn't belong (2250 km/s = 0.75% LS; not 22500 km/s, that'd be 7.5% LS...), for another you say "days" where it's supposed to be hours...

OK, let's do the numbers again...

at 1G = 10 ms²...
...speed 2250 km/s...
...100 sec to reach 1000 m/s...
...means 225000 sec to reach that 0.75% LS...
...1 hour = 3600 sec...
...225000 sec = 62.5 hours...
...62.5 hours = 2 days, 14 hours, 30 minutes!

I may not have all the info, but at least my basic math skills seem to be intact :wink:

OK, roughly two and a half days burntime for maximum realspace speed at 1G. Means almost two days at 1.5G, or one day and six hours at 2G (maximum sustainable accel for your typical humanoid non AG-race).

AG races will be at hat spaad much quicker - I suppose a Minbari gravity generator can at least absob 5G, so they'd be at this speed in 12 hours. Maybe they can do even 10G (at least on ships that are supposed to be "fast", like the WhiteStar), that'd mean 6 hours to 0.75% LS.

 

[lastbesthope]

You're right Shadow Scout, my bad.

Note to self, do not attempt to do sums in your head when tired late at night, while watching TV, surfing the web and talking to your mother on the phone.

(It never used to happen in my dial up days)

LBH

 

[frobisher]

Actually not entirely true (depending upon speed).
You just need to be able to impart to yourself a tangential velocity that is sufficent to clear the presented width of the object in the time it would take you to hit the impact point. In relative terms, this might not need to be much;
For instance, assuming the ship is capable of a 1g acceleration, then to avoid a 300km wide rock you only need a 3 minute warning (before impact) to be able to avoid it, and only 30 seconds if you've got a 2g acceleration, regardless of what speed you're travelling at.

Hmmm... You know, I don't think that is true.

Correct me if I am mistaken, but isnt it that the more speed your ship has, the more thrust it needs to change it's vector? And that the acceleration capability is just a result of available thrust and to-accelerated mass?

Oops - just spotted I said "tangential" there when I meant "perpendicular", which does kinda change the meaning a bit

But, regardless of current velocity (ignoring relativistic considerations...) a 1m/s perpendicular acceleration will always result in same change to the velocity vector.

Remember that the X, Y and Z components of a vector are operated upon entirely independantly.

For instance (I'll write the vector triplets as (X, Y, Z)) if your ship has a current velocity vector of (20,0,0) and you impart an acceleration of (0,1,0) for 10 seconds then at that point its velocity becomes (20,10,0). That 10 in the Y position would give you the drift required to miss the asteroid completely Yet you're still heading down the X axis at the same old speed you were.

It's just as with cars - the faster they go, the bigger a turn radius they have.

If you were turning, yes. Good old F = M v^2 /r applies there;

But the ship isn't looking to turn (change its velocity vector and keeping its heading vector pointing the same way), but merely change its velocity vector (it will be imparting a sideways drift).

Now, to get back to our ships, if the obstacle appears at the start of your burn cycle, where speed is low, you need only little warning, because a short blast of thrust from your side thrusters would be enough to change the ship's vector to avoid collision. But if the same obstacle appears near turnover, where speed is high, you need a much, much longer period of thrust to shift your vector away from the impact zone. Meaning you might even need to pivot your ship and use the big main thrusters. And the longer a distance travelled, the more speed at turnover, so the less maneuvering capability available to the ship.

Timewise, the warning you need is the same, regardless of your velocity. The distance at which you need to react is greater at greater velocities...

Obviously you want to be travelling at a velocity such that your detection radius gives you sufficent time to avoid the objects you're most likely to encounter.

 

[ShadowScout]

For instance (I'll write the vector triplets as (X, Y, Z)) if your ship has a current velocity vector of (20,0,0) and you impart an acceleration of (0,1,0) for 10 seconds then at that point its velocity becomes (20,10,0). That 10 in the Y position would give you the drift required to miss the asteroid completely...

Exapmles! Yes, that I can work with.

What I meant that if one takes your example, and adds that the asteroid needs these 10 in the Y position to be cleared by the ship, and is, say, 300 units away (X position).

Now, if the ship has a low speed (vector X=20 / sec) - no problem as in your example. Ten sec burn, then it has travelled 200 units, and attained the vector change it needs to clear the asteroid, with a bit more as safety margin by the time it gets there.

But if it has a high speed (X=200), it will crash into the asteroid before it's altered it's vector enough to get clear (at impact time it'll hav an vector of 200,1½,0 - the "1½" in the Y not being enough to avoid hitting the obstacle). So at the high speed it either needs a lot more thrust (acceleration change of 0,10,0), or a lot more room to maneuver (start accel at 0,1,0 at 2100 units so you can get your ten sec. of burn before intersecting vectors with the obstacle)

I hope what I meant is clear now, yes? :wink:

Timewise, the warning you need is the same, regardless of your velocity. The distance at which you need to react is greater at greater velocities...

Which is basically the same if one assumes limited sensors. If you can detect the obstacle at 5000 units (to continue the above example), the warning time is quite enough in both cases. But if your sensors can't detect smaller obstances above a distance of, say, 1000 units, then you need to slow down so your warning time is enough to give you the reaction time you'd need to change your vector (see above - and include a margin for the human factor, as I haven't seen any ships with emergency autopilot in B5 yet; all the automatics did was tell the pilot to change his couse to avoid collision - see Ivanova at "Severed Dreams")

Obviously you want to be travelling at a velocity such that your detection radius gives you sufficent time to avoid the objects you're most likely to encounter.

And that was all I wanted to state! :wink:
Slow speed - no problem as the maneuvering distance is enough to avoid obstacles
High speed - dangerous if there may be obstacles floating through your course, because yo might not be able to change your vactor fast enough.
See examples above.