# Break-even distance for sending data at c versus sending an FTL hard drive?

In the mid-21st century, humanity has an interplanetary colony system principally spanning the Moon, Mars, and Titan (as per usual). The furthest established colony is on Proxima b. Humans have FTL drives that can reach 12,500 times c without time dilation effects, and which can accelerate to this speed in under a minute. However, we are still reliant on light-speed signal transmission. We send data back and forth very frequently between our colonies.

Some people on Proxima b, while waiting for the previous Summer Olympics results, figure that beyond a certain distance, it's more efficient to just strap an FTL drive to a hard drive and lob it upwards into space than it is to send a signal and wait for it to arrive at c. A hard drive will need time to accelerate and decelerate to avoid damage, and it will need to be processed at its destination to access the data. This processing could take less than a minute if, say, a satellite net of some sort catches it, plugs it in, and then relays the data at c to the colony below, or several hours if the hard drive has to actually land. The FTL drive can launch from the ground. But let's say they've already got the former system set up.

This is a two-part logistical problem. Firstly, what is the break-even distance for sending a light-speed signal versus sending an FTL hard drive? It's not simply a matter of the distance from the source at which an FTL hard drive overtakes a signal at c, it also involves what people would find more convenient, and it might involve signal decay by the inverse square law (I don't know). For example, I don't think an FTL hard drive would be worth it between the Earth and the Moon - 3 seconds of ping time beats the processing time, even if it gives Lunar Dota 2 players a headache.

Secondly, given the colonies I've noted, do any of them have differences in their maximum and minimum distances from each other such that they would pass break-even at different times, and one method would be more convenient than the other for part of the time? In such a case, would one method of data transmission still be preferable to the other?

(I'd like to credit xkcd for partial inspiration.)

• I think you need to tell us how long it takes to copy the data onto the hard drive, launch it into space, accelarate to top speed, decelarate back to the colony's orbital speed, land it (or put it on a satellite), and copy from the hard drive to their system. And also, how much FTL the Hard drive can move at. – colmde Oct 13 '16 at 16:45
• how much data you can pack per distance unit is going to be pretty important here for at least some cases. I recall a race between the internet and a carrier pigeon with a flash drive, The Pigeon won. Today though, The pigeon could only win on very short distances at best (or in certain slow internet only areas), and at some internet speeds it may be impossible for it to win. Its the exact same scenario here. The faster the speed (not travel time, but in Bytes/sec kind of deal), the longer the distance before a set size hard drive becomes better. – Ryan Oct 13 '16 at 16:47
• This feels like an algebra problem, not a world building problem. The only part that seems world-building based is the question of how people may keep using the same transmission medium, even when the other is faster. – Cort Ammon Oct 13 '16 at 17:48
• It's hard to tell, since I don't see a date on it, but I suspect that this xkcd (File Transfer) might have come first. OK, the Wayback Machine says that the cartoon goes back to 23 Sep 2011, while the FedEx Bandwidth what if? originated in Feb 2013. – Peregrine Rook Oct 13 '16 at 19:12
• You may be able to draw inspiration from reality. We really have to worry about these latency vs. bandwidth questions in real life, so there's plenty of examples of how each individual problem calls for a different balance. It also points to other tradeoffs like security or flexibility. – Cort Ammon Oct 13 '16 at 23:43

The simple answer: for anything further than the moon, you should use FTL.

The slightly more technical answer: 18 million km is the break-even point.

Obviously, price is going to have a large affect on what you actually use FTL for, but if speed is your worry, FTL is already 3 times faster by Mars (at closest approach), and just gets better from there. By the time you're sending stuff to Proxima, there's a 12300 times speed increase (it doesn't quite hit 12500 because of the acceleration times).

Short Derivation

The closest planet-like object beyond the moon is Mars. At its closest approach, it's 56 million km from Earth. That's 3 minutes, 7 seconds by light.

Our hard drives can be processed in less than a minute, hit max speed in less than a minute, then decelerate in less than a minute. That's less than 3 minutes. Ignoring the fact that the drive is moving during acceleration (reducing the time even more), at 12.5k c, the 56 million km takes 15 ms.

Less than 3 minutes + 15 milliseconds is less than the 3 minutes and 7 seconds it takes light at closest approach. For any other part of the orbit, the difference is even greater. For any other planet or moon, the difference isn't even close.

Longer Derivation

I'll use exactly 60 seconds for processing time since "less than" is annoying. Then, let's assume constant acceleration to/from 12500 c.

$A(t)=a$

$V(t)=\int{A(t)}dt$$=at+v_0$$=at$

$D(t)=\int{V(t)}dt=\frac{a}{2}t^2+v_0t+d_0$$=\frac{a}{2}t^2 Where a is constant acceleration, v_0 is initial velocity (zero here), d_0 is initial position (which we can call zero), and A, V, D are functions of acceleration, velocity, and distance. First, let's solve for a: V(t)=12500c$$=3.747\cdot10^{12} \frac{m}{s}$$=a\cdot60s a=\frac{1}{60s}\cdot3.747\cdot10^{12}\frac{m}{s}=$$6.245\cdot10^{10}\frac{m}{s^2}$

Now, how far does the drive go before hitting max velocity?

$d=\frac{6.245\cdot10^{10}}{2}\frac{m}{s^2}(60s)^2=$$1.124\cdot10^{14}m Which is about 8 times the distance to the edge of the solar system. Since your furthest colony, Proxima b, is 4.2 ly (3.978\cdot10^{16}m) away, we need to use piecewise functions for it, but for everything else, we'll never see max velocity. We can solve for the time taken to get to any specific planet. Since (I presume) we have to accelerate then decelerate, take the half-distance time and multiply by 2. \frac{d}{2}=\frac{a}{2}t_\text{half}^2$$\rightarrow t_\text{half}=\sqrt{\frac{d}{a}}$$\rightarrow t=2t_\text{half}=2\sqrt{\frac{d}{a}} Specific Travel Times t_\text{MarsClose}=2\sqrt{\frac{56\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$1.894 s$
(plus 60 seconds for processing is 62ish seconds compared to 182 seconds at light speed)
$t_\text{MarsAverage}=2\sqrt{\frac{225\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$3.796 s (64 seconds compared to 751 seconds) t_\text{MarsFar}=2\sqrt{\frac{401\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$5.068 s$
(65 seconds compared to 1342 seconds)

$t_\text{JupiterNear}=2\sqrt{\frac{588\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$6.137 s (66 seconds compared to 1961 seconds) t_\text{JupiterFar}=2\sqrt{\frac{968\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$7.874 s$
(68 seconds compared to 3229 seconds)
$t_\text{Proxima}=2\sqrt{\frac{1.124\cdot10^{14}m}{6.245\cdot10^{10}\frac{m}{s^2}}}+\frac{3.978\cdot10^{16}m-1.124\cdot10^{14}m}{3.747\cdot10^{12}\frac{m}{s}}=$$10671 s (about 3 hours compared to about 4.2 years) Exact Point of Equality We can calculate the exact distance where the two times are equal. p is the 60 seconds of processing time. \frac{d}{c}=2\sqrt{\frac{d}{a}}+p (Mumble mumble stupid sign errors doing it by hand, just plug it into WolframAlpha /mumble mumble.) d=\frac{1}{a}[acp+2c^2\pm2\sqrt{ac^3p+c^4}]$$=cp+\frac{2c^2}{a}\pm\frac{2}{a}\sqrt{ac^3p+c^4}$

$cp+\frac{2c^2}{a}$$=3\cdot10^8\frac{m}{s}\cdot60s+\frac{2(3\cdot10^8\frac{m}{s})^2}{6.245\cdot10^{10}\frac{m}{s^2}}=$$1.8\cdot10^{10}m$

$\frac{2}{a}\sqrt{ac^3p+c^4}$$=\frac{2}{6.245\cdot10^{10}\frac{m}{s^2}}\sqrt{6.245\cdot10^{10}\frac{m}{s^2}\cdot60s\cdot(3\cdot10^8\frac{m}{s})^3+(3\cdot10^8\frac{m}{s})^4}=$$3.221\cdot10^8m$

Since the second term is tiny compared to the first term, we can pretty much ignore it, but the exact distances where the two meet are:

$d_+=$$1.832\cdot10^{10}m d_-=$$1.768\cdot10^{10}m$

Plotting the curves on a graph, I only see an intersection at $d_+$, so I'd go with that one.

$1.8\cdot10^{10}m$ is $18$ $\text{million}$ $km$, or $1.019$ light minutes, which makes sense, given the 60-second processing time and how ridiculously fast our FTL accelerates.

Conclusion

For anything further than 18 million km, FTL is faster. That's a lot further than the moon (about 0.39 million km), but nowhere near Mars (56 million km at closest approach) or Jupiter (588 million km at closest approach).

• I haven't fully checked your maths but I'm pretty sure you haven't added the time required to courier the disk from my desktop to the spaceport, loading times, waiting for a launch window, holding patterns waiting to land, unloading time and courier to destination. In practice you can add a minimum of several hours on at either end, meaning it only breaks even once you're going interstellar. – Separatrix Oct 14 '16 at 8:01
• @Separatrix That's not how it would work though. You would send the signal over the internet to an address that belongs to the other planet. The "router" at the space port would package your data along with loads of others into a hard drive and then be firing the hard drives at regular intervals as soon as they are full. They would also be processing return packages from incoming hard drives and sending those off the same way. – Tim B Oct 14 '16 at 8:45
• @TimB, DHL Internet as a service, there's definitely a business idea there. There are still the same bandwidth considerations that the original xkcd link refers to which you don't have on a full Fed-Ex service, you've also now included a volume based time delay, if traffic is too low, it's now also much slower. – Separatrix Oct 14 '16 at 10:01
• @Separatrix True. However it's likely you would have "low priority sync" items going over as well, for example wikipedia or stack exchange synchronizing between star systems. For cheaper rates they would just use spare capacity during cheaper periods and in doing so keep the liquidity up. – Tim B Oct 14 '16 at 10:11

First, lets list the variables

• Bandwith

• Bit's per second per stream
• Number of streams per signal
• Time delay = Distance / c
• Time = (bits/(seconds) * streams) + Time Delay
• Harddrive

• Write bits per second
• Setup/catch time/rate (how fast to set up for transit and rate for it to get to FTL speed)
• Time in FTL
• Travel time = Time in FTL + time entering FTL + time leaving FTL
• Time = Setup time + write time + Travel Time + read time + process time

And of course the distance.

And then once you have everything but distance find Bandwith_Time = HardDrive_Time, So...

(bits/(seconds) * streams) + (Distance / c) = Setup time + write time + Travel Time + read time + process time

Than solve for the break distance (you can replace c with speed of bandwidth)

Distance = (Setup time + write time + Travel Time + read time + process time - (bits/(seconds) * streams)) * c

And just plug your numbers into that.

• Many of these numbers need to be added by OP, since we are talking more SciFi than anything, so they can choose how powerful the computers are and such. – Ryan Oct 13 '16 at 16:54

The time to deliver a FTL hard drive from Earth to Proxima Centauri b is 2.88 hours. The time it takes for it takes for the Summer Olympics to arrive at lightspeed is 4.243 years. The FTL information delivery system is simply faster.

If the Proxima Centauri colonists were waiting to see the Olympics coming to them at lightspeed and someone recorded the next Olympics, four years later than the previous event, and the current Olympics came via FTL hard drive delivery, they would see the current year's Olympics first.

The question may be ultimately economic. It could depend on the cost of delivering information at different speeds.

There is popular joke among Russian programmers, that Ministry of Defence has world record of data transfer rate between 2 cities - they use military trucks full of HDD.

Assume, that this track carries 10 tons of HDD. It is a very big data. I have 500Gb HDD in my PC, it has weight of about 250 g. So, for 10 tons there will be about 40.000 of HDDs, and about 20.000.000 Gbytes of data per military truck.

And, if this truck can move between 2 cities in one hour, the rate will be 20.000 Tb per hour or 5 Tb per second!

And nothing prevents us to use more than one truck. So we can get much better speed!

Also there is Amazon Snowball service - they send you some sort of big reinforced and rugged HDD, you upload data on it, and it is delivered back to Amazon and uploaded to their services. If i were a geologist from distant lands, and i have 2 Tb of geological data i need to send to my boss, i'd use this service.

So, i think, FTL signal transfer is ok for some high importance data with small size - email messages, news, RSS feed contents.

For big, and less important data - AVIs, MP3, Olympics, porn, games and so on, it is easier to use HDD, and upload and distribute data using the local Content delivery networks.

So, i think both methods of transfer - FTL signals and FTL HDDs will be used.

• The US equivalent is "never underestimate the bandwidth of a FedEx truck" – John Feltz Oct 14 '16 at 12:20