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Intruder at the top of the 20 meter amateur band?

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Re: Intruder at the top of the 20 meter amateur band?

#161

Post by caymanjim »

filmfact wrote: post #66
caymanjim wrote: post #49

Earlier quoted context omitted.

You'd need to be in orbit. Edit: actually it might not be possible at all from a single point anywhere in the universe.

I just stared at a globe, and it’s possible, but not from a geostationary orbit. Think about a hemisphere projected onto a 2d circle - you can see all points on the surface of the hemisphere on the projection. Logically it must be possible to select a hemisphere, such that any two random points on a sphere lie on that hemisphere, and would thus be visible in the projection. The only edge cases is two points that are…

In this case I was considering three points, but yeah, they're all in the northern hemisphere, so your scenario applies.

Re: Intruder at the top of the 20 meter amateur band?

#162

Post by londons_explore »

scatters wrote: post #158
KirillPanov wrote: post #154

Earlier quoted context omitted.

Heh. Yeah I have no idea why everybody who isn't doing HFT hasn't decided to stop letting these guys rip them off and moved to investor sex change er, sorry, put the spaces in the wrong place there, I meant: investors exchange They basically encoded a bunch of HFT-thwarting principles into their clearing policies: https://en.wikipedia.org/wiki/Investors_Exchange#Operating_p... It's a great idea. But every time I see…

Yeah, why? Maybe because HFT actually do provide a service valuable to retail.

Or because retail have no choice but to lose a few tenths of a percent of every trade to HFT leeches..

Re: Intruder at the top of the 20 meter amateur band?

#163

Post by tlrobinson »

vgrigorescu wrote: post #130

Think I found them. Using the update from the bottom of the page: > … and M-Wave is authorized the 14-14.99 MHz at 16 kW. I found this petition from M-Wave Networks, LLC for permission to build 4 towers in Kane County, Illinois: https://www.countyofkane.org/FDER/Zoning%20Petitions%20Docum... The location in the petition is approx. 41.82907908782928, -88.4940281199101, which is about 16 miles (27 km) from the center c…

> If it was approved, it seems remarkably fast to build a tower given that the article was from Dec 15th.

It makes sense they moved quickly, since there's a limited time period where they'd have this advantage over other firms which set up similar systems.

Re: Intruder at the top of the 20 meter amateur band?

#164

Post by rlt »

There was speculation Starlink could be used for HFT once it gets laser links, but this would even beat that since you don't need to travel up and down an extra 500 miles, plus any buffering in the satellites. Starlink would have more bandwidth, but you probably don't need much for HFT.

Re: Intruder at the top of the 20 meter amateur band?

#165

Post by tlrobinson »

KirillPanov wrote: post #153
baobrien wrote: post #150

Earlier quoted context omitted.

Stream ciphers can operate without adding any extra delay and are considered plenty strong.

At HF bitrates they can use a one-time pad. The only unbreakable cipher. A single hard drive shipped (or hand-delivered) to each transmitter location ought to last a decade.

No need to ship a hard drive, just send it over the regular internet.

It would be an interesting exercise to design an encryption and compression protocol that has a high bandwidth/high latency link and a low bandwidth/low latency link working together.

Re: Intruder at the top of the 20 meter amateur band?

#166

Post by widforss »

WarOnPrivacy wrote: post #9

The inexact coordinates in the image lead to the front yard of some farm. Now I can't stop thinking of Kash Hill's Maxmind story ref: https://www.google.com/maps/place/41°36'00.0"N+88°36'00.0"W/ ref: https://splinternews.com/how-an-internet-mapping-glitch-turn...

https://goo.gl/maps/aWdYVhVLbBfuSKvq9

Re: Intruder at the top of the 20 meter amateur band?

#167

Post by welterde »

kreelman wrote: post #142
souprock wrote: post #140

Someday the traders will be beaming neutrinos through the Earth.

...Maybe one day they'll have quantum entangled particles and they'll be able to do spooky things at a distance and communicate instantly? I don't pretend to understand the physics, but sounds like it would be interesting... if it could work. Perhaps then Radio Amateurs would become Amateur Tanglers?

While it is spooky interaction at a distance it can't be used to transfer information faster than light. In simple terms, the states are correlated, but you have no way to influence what they are. You get a random number and know instantly what the other random number of the remote particle is, but that doesn't transfer any information from you to the remote location or vice-versa.

The most obvious explanation would have been that there is some kind of hidden state and they don't interact at all anymore, but that doesn't appear to be the case.

Re: Intruder at the top of the 20 meter amateur band?

#168

Post by beervirus »

jtsiskin wrote: post #147
beervirus wrote: post #82

Earlier quoted context omitted.

At least arbitrage provides some value by tightening bid/ask spreads. What about all the smart nerds who work on adtech instead of important problems? That's entirely irredeemable.

Targeted advertising seems to benefit everyone compared to untargeted advertising? My google ads are mostly saas software and tech gadgets, which I prefer over random ads. If a lip gloss company wanted to advertise, is it not better that they can target people who like lip gloss instead of a random audience?

The cost of targeted ads, as they are currently implemented, is privacy. The juice ain’t worth the squeeze.

Re: Intruder at the top of the 20 meter amateur band?

#169

Post by [deleted] »

CraigJPerry wrote: post #38

Chicago to London one way is around 35ms over fibre but that could be down to 21ms over the air like this. That signal is only about 10-15khz wide[1] from that diagram. That’s not a whole lot of data transfer ability, measured in the low kbps range rather than mbps. Very low kbps when you add in error correction. We are moving out of / have left the solar minimum so the MUF is usually well above the 20 metre band eac…

[deleted]

Re: Intruder at the top of the 20 meter amateur band?

#170

Post by zaarn »

filmfact wrote: post #66
caymanjim wrote: post #49

Earlier quoted context omitted.

You'd need to be in orbit. Edit: actually it might not be possible at all from a single point anywhere in the universe.

I just stared at a globe, and it’s possible, but not from a geostationary orbit. Think about a hemisphere projected onto a 2d circle - you can see all points on the surface of the hemisphere on the projection. Logically it must be possible to select a hemisphere, such that any two random points on a sphere lie on that hemisphere, and would thus be visible in the projection. The only edge cases is two points that are…

If the two points you select on the sphere are poles of eachother (ie, their distance through the sphere is equal to the sphere's diameter, then you need to have an infinite distance from the sphere.

What you're looking for is a triangle between the two points and a third point forming your observer. The two points are visible at the same time if the derivative of the sphere's surface and the side of the triangle between a point form a positive angle (the derivative intersects the triangle at a point other than the points).

So the question then becomes "if the two points are poles, what does that mean"; easy, if the two points are poles and they're only visible if the surface derivative is within the triangle, the angle between the observer, the point and the other point must be 90°. This must hold for both points as the solution is symmetric. However, there is no well defined triangle that has two angles of 90° inside.

So the solution is "doesn't work" (or possibly infinite distance).

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