32

In Star Trek, we have warp drive. In Star Wars, there was hyperdrive. Is there any canon information on how the ships in Firefly travel around the 'verse?

I understand that the planets are all in one solar system with several stars, but still, it took seven months for the Mars Reconnaissance Orbiter to reach Mars using current technology, and that's the nearest planet in a solar system with one sun. It seems there are dozens of worlds in the Firefly universe, so you'd think it would take quite a while to travel between them unless they had faster-than-light technology. Are there any details on the speed of the ships, or the technology used?

  • Pluto can be as far as 49 AU from the sun, according to Wikipedia, and is a pretty cold and barren place. I don't know the Firefly universe (which is why I'm posting this as a comment), but let's use that real-world number as an example. Travelling 49 AU at .5c would take 24.5 million seconds, or about 283 days, roughly twice the length of a long Atlantic crossing in the 1600s. Definitely a long trip, but certainly feasible even without FTL drive. – a CVn Nov 30 '12 at 8:48
  • 6
    @MichaelKjörling Your math is off. 1 AU is about 500 light seconds, so at .5c the trip would take under 14 hours. It would certainly be feasible to traverse a large solar system at even a small fraction of light speed, say .01c. – Travis Christian Nov 30 '12 at 15:40
  • 3
    @TravisChristian Oops, you're right. I see my mistake; I almost certainly used 300k m/s rather than 300M m/s for c. That would explain the error. Too bad comments can't be edited after the short grace period. It doesn't negate my point though; quite the opposite: inside a system or even a group of closely spaced solar systems, you don't need FTL drive to get reasonable or semi-reasonable travel times. – a CVn Nov 30 '12 at 21:06
  • 5
    And wasn't the MRO transfer orbit chosen for energy efficiency rather than speed? It's perfectly possible to haul something off to Mars much faster using current, Earth technology, particularly if you can choose a good launch window. But if you don't really care about the time it takes to get there, it makes a lot of sense to optimize for transfer orbit energy efficiency instead. en.wikipedia.org/wiki/Low_energy_transfers – a CVn Nov 30 '12 at 21:15
  • 3
    This is the first time I have heard of the 'Verse being that small. Looking into it, I see that according to various show sources it indeed is. My mind is blown at the ridiculousness of this. – Broklynite Oct 26 '15 at 11:58
41

The Firefly class transport ships have an acceleration of 5.5g. At that rate it will take about a week to cross the entire verse from end to end which is around 400 AU across in diameter. About 55 light hours (for comparison, the current most distant know object in our solar system is Eris at 96 AU). There is no need for a faster-than light drive.

Acceleration source:
https://firefly.fandom.com/wiki/Firefly-class_transport_ship

Size source: https://store.qmxonline.com/assets/images/serenity/atlasoftheverse/the_verse_in_numbers_v11.pdf

*Approximate numbers.

  • 14
    Two weeks if you want to stop once you get to the other side :) – OrangeDog Mar 10 '16 at 18:44
14

it is entirely plausible to go from end of the Solar System to another in two weeks without approaching the speed of light. Of course, you can go 99.999999% (any arbitrary number of 9s) of the speed of light without violating Relativity, as long as you don't actually go 100%. This is Hard SF. :)

Now, if your ship accelerated at 1G, flipped over halfway through, and then decelerated at 1G for the rest of the trip, you could go from Earth orbit to Mars orbit in less than 2 days. This would also cause you to feel normal Earth gravity aboard the ship, without requiring Star Trek/Star Wars artificial gravity. Presumably you would use an antimatter or fusion rocket to maintain this acceleration.

http://www.johndcook.com/blog/2012/08/30/flying-to-mars-in-three-days/

Where Wikipedia says that the Serenity can accelerate at 5.5G, obviously that would be even faster, but I doubt she cruises at that speed normally; walking around at that acceleration would be hard, and if you slip and fall, you'd get hurt pretty bad, I imagine.

The Star Trek/Star Wars magic gravity floors -- even if they are just orbiting a planet, or traveling at what seems like a very low acceleration, they still have gravity -- are obviously not Hard SF, however! :-)

  • 2
    No it's not hard scifi. If you approach 1c close enough your own mass will crush you. Never mind the structural integrity of your spacecraft or the ridiculous amount of energy (i.e. fuel) you need to invest. But otherwise I agree, if you disregard your first paragraph. I am, however, too lazy to calculate the maximal speed you would have (classically) in your 1g-Earth-Mars scenario. I'd say it's plausible if it's below .9c. – bitmask May 26 '13 at 12:13
  • Okay, I lied. Turns out I wasn't too lazy; If you go at 1g for a week you would merely reach .02c, for a day only .003c. Not enough for relativistic effects to kick in. Not by a long shot. – bitmask May 26 '13 at 12:21
  • 12
    @bitmask Why should you be crushed by your own mass or have a problem with structural integrity? In relativity frames moving with a constant velocity to each others are equally valid. This means that the frame where the spaceship is at rest is valid, so no problems there. However, collisions with dust particles will be a major limitation. – Erik Sep 7 '14 at 16:27
  • 6
    Yes, bitmask's comment suggests a lack of understanding of the principle of relativity which says the laws of physics work identically in all inertial reference frames, so if I am traveling at 0.99999c relative to the Earth, I am free to choose a frame where I am at rest and the Earth is moving at 0.99999c, the predictions about physical events made using this frame should be identical to those made using the Earth's rest frame. – Hypnosifl Oct 25 '15 at 1:45
  • Note that gravity in Firefly is not due to acceleration: scifi.stackexchange.com/questions/13621/… – Wolfie Inu Oct 26 '15 at 12:00
12

Bharat B's answer covers the essentials well (though for the 5.5g acceleration figure, I think it would be better not to cite the wiki page but rather the source it got this info from, which is revealed at the bottom to be the Serenity Blueprints Reference Pack). But I wanted to add more exact numbers for Serenity's time and final velocity to cross the 'verse (assuming the 'verse is about 400 AU or 55 light-hours wide as stated in The Verse in Numbers), and also figure out what this implies about the amount of fuel that would be needed for a full trip across the 'verse. I also wanted to include the detailed calculations, though most people probably won't be interested in them so I'll put them in bold so you'll know what to skip if you just want to see the results.

For the actual time, I find that if Serenity accelerates continuously across the 'verse at 5.5g (presumably using artificial gravity to keep from killing the crew), and the 'verse is 55 light-hours across, then it should take 141.5 hours, or about 5.9 days, as measured by people at rest relative to the center of the 'verse. Calculations:

A lot of useful formulas for a rocket accelerating at a fixed rate to a significant fraction of the speed of light can be found on the relativistic rocket page here. If we use units of hours for time and light-hours for distance, 5.5g works out to 0.00647 light-hours/hour², so if you plug that in as the "a" into the formula t = sqrt((d/c)² + (2*d/a)) (where t is time in hours, d is the distance of 55 light-hours, and c is the speed of light of 1 light-hour/hour), the time would be sqrt((55/1)² + (2*55/0.00647)) = 141.515 hours.

On the other hand, if you want to accelerate as you travel the first 27.5 light-hours and then decelerate during the second half so you arrive at low speed, plugging 27.5 into the same equation shows the first half takes a time of 96.2 hours and the second takes the same amount, for a total of 192.4 hours or 8 days.

As for speed, in the first case where Serenity just accelerates continuously at 5.5g (which minimizes fuel), its final velocity after having travelled for those 141.5 hours will be around 0.675c. In the second case where it accelerates for 96.2 hours until the midpoint of the journey and then decelerates, its speed at the midpoint will be 0.528c. Calculations:

Another formula on the relativistic rocket page says that if I accelerate continuously, my final change in velocity at the end should be v = (a * t) / (sqrt(1 + (a * t/c)²)), in this case (0.00647*141.515)/(sqrt(1 + (0.00647*141.515)²)) = 0.675298 light-hours/hour, i.e. 0.675c. And for the second case of 96.2 hours, we have (0.00647*96.2)/(sqrt(1 + (0.00647*96.2)²)) = 0.528c.

Using this change in velocity, the Tsiolkovsky rocket equation can tell you the ratio of initial mass m₀ (both fuel and payload) to final mass m₁ (just the payload, fuel used up) needed to get such a large change in velocity. The formula depends on the velocity of the exhaust, and I don't think that's stated directly in any Firefly material, but this answer says some material on the DVD indicated that Firefly-class ships light up their tails when they travel due to a nuclear fusion reaction. So Serenity is probably a type of fusion rocket, perhaps similar to Project Daedalus in which some real-world engineers tried to draw up a rough plan for an interstellar space probe, and which used a fusion reaction involving deuterium and helium-3. The chart on this page indicates that this is one of the best fusion reactions in terms of exhaust velocity, with a v_e of about 8.9% the speed of light or 0.089c (which is incidentally much, much faster than any chemical rocket like the ones we have today), so I assumed that in my calculations.

When I worked it out, I found that the total mass of the ship including fuel at the beginning of the trip would have to be about 10,000 times more than the mass of the ship at the end of the trip when it used up its fuel! This suggests it's not very realistic that a ship that seems to have as much non-fuel mass as Serenity could make the trip without constant refuelling. Calculations:

For the relativistic case where the velocities are a significant fraction of light speed, the formula they give on the Tsiolkovsky rocket equation page is that the change in velocity Delta-v of the rocket is equal to:

Delta-v = c * tanh((v_e/c) * ln(m₀/m₁))

where tanh is the hyperbolic tangent function, ln is the natural logarithm function, and v_e is the "effective exhaust velocity" (and we're still using the units of hours and light-hours, so c=1).

With a little algebra you can solve this for the ratio (m₀/m₁), giving:

(m₀/m₁) = e^((c/v_e) * atanh(Delta-v/c))

where e is a mathematical constant, and atanh is the inverse hyperbolic tangent function (e^x is the inverse of ln(x), and atanh(x) is the inverse of tanh(x)). Both functions are recognized by this online calculator, so I just plugged in e^((1/0.089)*atanh(0.675298)), and got an answer of 10079 for the ratio of m₀ to m₁.

If anyone wants to play around with different distances (in case Serenity makes shorter hops and refuels) and different exhaust velocities, I'll put all the above equations together into one long one that can just be copy and pasted into the the online calculator to find the ratio of masses before and after using up the fuel, just substitute whatever distance you want for the four D's in the equation (measured in light hours), whatever acceleration you want for the four A's in the equation (where 5.5g = 0.00647, and 1g = 0.00011776), and whatever exhaust velocity you want for the single V in the equation (measured as a fraction of light speed):

e^((1/V) * atanh((A * sqrt((D)² + (2 * D/A)))/(sqrt(1 + (A * sqrt((D)² + (2 * D/A)))²))))

For example, plugging in D=55, A=0.00647, and V=0.089, the calculator gives the same mass ratio of 10079 found earlier.

On the other hand, I suppose you could imagine they use some exotic technology (perhaps the same technology they use to make artificial gravity) to accelerate the exhaust to speeds much faster than real-world fusion reactions, or you could ignore the material on the DVD saying it was fusion and imagine it was something like a matter/antimatter reaction, where the effective exhaust velocity can be close to light speed. For example, if we use the same equation but with an exhaust velocity of 0.9c rather than 0.089c, the ratio of initial mass with fuel to final mass without fuel would just be around 2.5.

Note: That last formula was assuming continuous steady acceleration all the way from departure point to destination, which would result in reaching the destination at a very high speed; a more natural assumption might be to accelerate for the first half of the trip, then decelerate at the same rate for the second half, so you'd come to rest at your destination (you can also imagine a constant-velocity coasting phase in between the end of acceleration and the beginning of deceleration, this wouldn't affect the amount of fuel used). In that case, all you have to do is set the D to be half the total distance crossed during both acceleration and deceleration--i.e. the distance traveled during the deceleration phase alone--then find the fuel/payload ratio using the above formula, then square the resulting ratio to get the ratio for the whole trip. (Similarly, if you have the final velocity Delta-v at the end of the acceleration phase/beginning of the deceleration phase, then you can use the formula e^((c/v_e) * atanh(Delta-v/c)) I mentioned earlier, which gives the mass ratio as a function of Delta-v and the exhaust velocity v_e, then square the result to get the ratio for the whole trip including both acceleration and deceleration phases.)

If you want to know the rationale for this, start by noting that the formula I posted should work fine for the deceleration phase alone (decelerating at A meters/second² from an initial velocity V to rest requires exactly the same fuel ratio as accelerating at A meters/second² from an initial state of rest to a final velocity of V), so it can tell you how many kilograms of fuel+payload will be needed at the beginning of the deceleration for every kilogram of payload at the end of deceleration. Then the number of kilograms at the beginning of deceleration can be treated as the "payload" mass that needs to be brought up to speed during the first acceleration phase, so you'll have the same ratio as before for fuel+payload to final payload over the course of the acceleration. Thus if the mass at the beginning of deceleration is R times bigger than the final mass at the end of deceleration, and the mass at the beginning of acceleration is also R times bigger than the mass at the end of acceleration (which is also the mass at the beginning of deceleration), then this implies the mass at the beginning of acceleration is R*R = R² times bigger than the mass at the end of deceleration.

  • 1
    Phew, nice calculations +1 – Zommuter Sep 26 '16 at 10:53
6

The Serenity and other ships (not sure if it's all of them) use what is essentially controlled fusion explosions at the tail end of the ship to propel the ship at greater than normal speeds. It is never made clear how fast the ships do move at top speeds, but it is still at sub-light speeds. But even so, they appear to achieve speeds much greater than what is capable of our technology today.

There are production documents from the movie that indicate the top speed of Serenity was in the range of 400,000 mph, which would allow it to travel 1 AU (distance from Earth to the Sun) in about 16 days. Whether or not that is canon is highly debated on many Firefly forums however. There does not appear to be any official canon from JW on what the actual top speeds are.

  • 4
    "top speed" for a spaceship does not mean much. More important is the maximum acceleration developed. Applying this acceleration over time results in a velocity, the more time, the higher the velocity. There is nothing to limit the speed achievable except for the amount of fuel carried. – Organic Marble Sep 6 '14 at 16:20
  • 2
    That's a good point Organic Marble. Maybe when they say top speed they do mean the total velocity achievable with a full load of fuel at maximum acceleration. But I doubt it :) – PhysicalEd Sep 6 '14 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.