3 ms·
Avoid the pesky curvature of the Earth's surface by going directly along a chord through the mantle rather than along a great circle? Using https://planetcalc.
by schoen 6y ago
Avoid the pesky curvature of the Earth's surface by going directly along a chord through the mantle rather than along a great circle?
Using https://planetcalc.com/7725/ https://planetcalc.com/7725/ and https://planetcalc.com/73/ https://planetcalc.com/73/ I just concluded that the straight-line chord distance between Tokyo and San Francisco is 1552 km whereas the great-circle distance is 8270 km. (Neither of these is very precise because it's unclear where in each city you should measure from, and unclear whether either calculator uses data about the irregularity of the Earth's curvature.)
It does make a noticeable difference for HFT applications, I guess: 1552 km/c is about 5 ms while 8270 km/c is about 28 ms. (The neutrinos might do better in another way because I guess light in a fiber doesn't directly follow the curve of the fiber itself, since it's getting repeatedly reflected off of the inside surface of the glass.)
- hinkley 6y agoSpeed of light in a fiber is slower than the speed of light in a vacuum, so it's the straight line speed of light versus the great circle speed of light / index of refraction. So you're looking at somewhere north of 35 milliseconds (because the fiber is NOT great circle, and repeaters and routers will add little snips of time every time they touch the signal). But can you decode a neutrino signal in real time?
- DaiPlusPlus 6y agoPardon my ignorance, but what would a neutrino signal look like? Can’t it be pulsed on/off like an electron beam - does it have a wavelength/frequency like a photon?
- sbierwagen 6y agoPulses, yes. Making a neutrino beam is remarkably convoluted. https://en.wikipedia.org/wiki/Accelerator_neutrino https://en.wikipedia.org/wiki/Accelerator_neutrino >The process of the muon neutrino or muon antineutrino beam production consists of the following steps[1][2]: >Acceleration of a primary proton beam in a particle accelerator. >Proton beam collision with a fixed target. In such a collision secondary particles, mainly pions and kaons, are produced. >Focusing, by a set of magnetic horns, the secondary particles with a selected charge: positive to produce the muon neutrino beam, negative to produce the muon anti-neutrino beam. >Decay of the secondary particles in flight in a long (of the order of hundreds meters) decay tunnel. Charged pions decay[3] in more than 99.98% into a muon and the corresponding neutrino according to the principle of preserving electric charge and lepton number... There's several different places in this chain where you could modulate the beam by turning various magnets on or off. Probably not in the proton accelerator. Unfortunately, "long string of expensive experimental equipment" means "not efficient". From the paper: >A neutrino source delivering muons at a rate of 10^14 s^−1 with an energy of 150 GeV would require about 4 MW in proton beam power and 2.4 MW acceleration power, which for a 10% electrical efficiency translates into a total power consumption of roughly 65 MW. Yow. Something like ten times more power than the NuMI neutrino beam. He concludes this transmitter could do something like 100 bits/s to a stationary detector string anchored at the ocean bed. Deeper the better, for shielding against cosmic rays and solar radiation. Would be tough to put a neutrino detector close enough to a financial hub and still get useful bandwidth.
- londons_explore 6y ago65 MW is only $6M/year at the cheapest available electricity prices. Even less if you turn it off at night or when markets aren't as volatile. 100 bits/sec is plenty to make lots of money... If you consider 100 bits/sec is 1 bit per 10 milliseconds, with say a 5% bit error rate (can't do error correction without introducing delay). Each end of the connection can make two candidate investment strategies based on globally available data. The 'bit' decides which strategy to go for. Easy money for anyone who can turn the science into reality... Easy money that will dry up as soon as a few other people start doing the same...
- DaiPlusPlus 6y ago> can't do error correction without introducing delay Hamming Codes don’t, unless I’m missing something?
- londons_explore 6y agoIn hamming codes, the correction can only be done when an entire block has been received. With a block length more than 1, you therefore introduce latency of 10s of milliseconds. With a block length of 1, a hamming code doesn't do any error correction. Hamming codes would let you access the uncorrected data immediately, and do error correction later (after the block length has passed), but at that point you have already used it to make trading decisions, so it's too late.
- sbierwagen 6y agoNuMI cost $139 million to construct, including the detectors. It was driven by the Tevatron, a proton accelerator 2km across that itself cost $197 million to build in 1991. 150 GeV particle accelerators are not cheap, and are going to be impossible to build secretly. If everyone knows you have a phone to the future, finding counterparties is going to be hard. Who's going to take the other side of a trade you know you're going to lose?
- femto 6y agoAre those numbers correct? In the worst case, the chord would be the diameter and the great circle would be half the circumference, so the ratio would be (pi.d/2)/d = pi/2 = 1.57. For fun the cos rule can be used to derive a general formula for the ratio of great circle to chord, given the angle (theta) in radians subtended at the centre of the Earth: theta / sqrt( 2-2.cos(theta) ) where: theta = (great circle distance) / radius (Assuming I haven't mucked up my algebra.) -- Not sure why my comment is rendering in italics. No emphasis is meant. Figured it out, it was the asterisk I used for a multiplication symbol.
- schoen 6y agoThank you for the sanity check. It looks like the web site has a bug!
- deleted 6y ago[deleted]
- ianhorn 6y agoThe app you linked has some issues with data entry. If you re-enter the minutes field for SF, it will give a straight line distance of about 7800km, instead of the 1500km you listed. As another commenter pointed out the worse case is ratio is about 1.5, so I'm guessing either it's totally buggy or 7800km is the correct one.
- rocqua 6y agoLight in fiber travels at about 2/3 s of the speed of light in a vacuum. Not because of bouncing (the fiber creates a waveguide) but just because it is inside glass. This different speed of light is what gives glass its refractive index, which makes it optically useful.
- schoen 6y agoThanks, I feel like I've just experienced Feynman's complaint about his Brazilian students not applying their knowledge of the concept of "index of refraction" to an actual physical object. That is, I didn't think about how the whole idea of the index of refraction of a material is directly derived from the different speed of light in that material. I'm still confused because I thought I had learned that fiber optical cables work by total internal reflection, in which the light inside is repeatedly reflected by the surface of the fiber. Is this an overly simplistic view for thinking about the path that the light will follow in the fiber?
- rocqua 6y agoIt is a somewhat simplistic view because light isn't just a particle bouncing off walls, its a wave. What happens is that the fiber creates a 'waveguide'. This will gently curve the wavefront around the fiber. I barely understand this myself. But the way I imagine this is essentially as a water wave running through a trough. Lets consider a simple sinusoidal wave originating at one end of the trough. Lets call a wavefront any line across the trough that follows the peak of a wave (really it can be any fixed phase, but peaks are nicer to visualize). What happens to these wavefronts as time goes on? They bend around the corners. Essentially the water just in front of the wavefront is going to be pushed up next, and when this happens has a lot more to do with distance from the current wavefront than whether the actual water particles are bouncing off the wall or not. My explanation of wavefronts moving forward doesn't quite explain why the wavefront 'rotates' in a curved waveguide. I would guess something about path interference being different on the inside of a corner than the outside. Someone who actually studied this stuff probably knows a lot better.