can be configured per application

wireguard can too using network namespaces

not exclusively as an interface in kernel mode

Which devices are you people running where you want VPN/proxy but don’t have kernel permissions to run wireguard? Firefox on iPhone? Porn on wifi washing machine?

Curious, what is SOCKS5 used for that regular wireguard cannot do? I’m only familiar with the use case of telling Firefox to connect through a SOCKS5 proxy, which may be convenient as a form of split tunneling - only firefox traffic goes through the VPN and everything else through clearnet - but wireguard can be configured into a split tunnel form as well with a bit more work, and works for all software not just the ones aware of SOCKS proxies. Is it for use on a system where your permissions are too limited to turn on wireguard but not so limited that you cannot change Firefox proxy settings?

Talescale is a VPN, “private network” is what P and N stand for. It’s just one with only forwarded ports and no outbound traffic. The question was are forwarded ports important, and yes they are. So important that some users pay for a VPN twice! Once for something like Mullvad with no port forwarding, and once for Talescale for port forwarding. It’s true it has benefits like static IP, but even on my commercial VPN I get the same forwarded IP and port when connecting to the same server, so I don’t want to pay twice.

I agree that OP is in the best position to report the crime to the police - they are closest to the police station, they have video evidence, they literally know who the thief is - but it should not be their responsibility! OP has done nothing wrong and there are no measures they could have taken to prevent this crime (other than not shopping online at all). If OP gets a police report, OP is taking up the task of being the victim, and then BestBuy has no legal obligation to refund them at all, other than out of the kindness of their heart. Rather, BestBuy is the victim in this crime, same as if the item was stolen off the shelf at their warehouse and scanner records forged. It is their responsibility to file a police report, if they want the numbers in their system to add up. Only then could they ask OP to kindly provide the video evidence to help them out, and they’d be lucky if OP would give it to them, having no obligation to do so.

Yes! It’s an olympics game of mental gymnastics where everyone - BestBuy, DoorDash, OP, the police - try to offload responsibility onto someone else. However, a crime WAS committed. Someone is the victim. The victim is the one who was deprived of property/money and will not have access to it until/unless the thief is caught and property recovered. BestBuy thinks OP is the victim, since the item was stolen off (not)their porch. OP thinks BestBuy should be the victim, since OP had no involvement in organizing the delivery. DoorDash could also take up responsibility of being the victim, since it was their (not)employee that stole from them.

If OP goes to the police now, they would be losing the mental gymnastics by accepting the status of the victim. BestBuy would never refund them in this case. It is in OP’s best interest to pursue the chargeback first. If OP succeeds in the refund or the chargeback, then BestBuy will have no package and no money, so BestBuy would be the victim. Then it will be BestBuy’s responsibility to report the crime.

The “libertarian paradise” idea is that as far as Best Buy is concerned, the item was delivered. If the DoorDash delivery driver happened to turn right around and steal the package, that’s a separate crime and a matter for the police to deal with, same as if anyone else had stolen it. And it’s OP’s fault for not picking the box up sooner, during the 3 seconds it was sitting on the porch. The porch that wasn’t even theirs. So anyway, the libertarian solution is for OP to contact police to track down the thief and either recover the stolen item or sue the thief for monetary compensation. Best Buy is innocent and no refund is coming. DoorDash is innocent too because they contracted with an independent contractor to deliver the item, and what the contractor does after the item has been delivered is not their responsibility.

I run a Minecraft server. Couldn’t have done it without port forwarding.

New York City reporting in, can confirm the subways have been plastered in Mullvad ads for months. I’m so glad I switched away last year, so that at least I know none of my money is used to spam my own eyeballs. Though in some sense it is nice that VPNs are common in public discourse now.

Ah, I can see OP’s line of thought now:

• you have a point A’ on a plane and a random point A
• you find a midpoint B and draw a sphere around it. A and A’ are now a diameter of the sphere
• pick two random points D and C at the intersection of the plane and the sphere
• by the “triangle inscribed in a circle/sphere where one side is a diameter” rule, such a triangle must be a right triangle
• therefore both angles ACA’ and ADA’ are right angles
• thus C and D both satisfy the conditions of the initial question (with all points renamed: A=P, (C or D)=H, A’=A)
• OP never defined what a projection is, it being “4th grade math”, but one of the requirements is being unique
• C and D cannot both be the projection, therefore the initial question must be answered “false”: just because AH is perpendicular to PH doesn’t make H a projection.

I like treating posts as puzzles, figuring out thread by thread WTF they are talking about. But dear OP, let me let you know, your picture and explanation of it are completely incomprehensible to everyone else xD. The picture is not an illustration to the question but a sketch of your search for a counterexample, with all points renamed of course, but also a sphere appearing out of nowhere (for you to invoke the inscribed-triangle-rule, also mentioned nowhere). Your headline question is a non-sequitur, jumping from talking about 4D (never to be mentioned again) into a ChatGPT experiment, into demanding more education in schools. You complain about geometry being hard but also simple. The math problem itself was not even your question, yet it distracted everyone else from whatever it is you were trying to ask. If you ever want to get useful answers from people other than crazed puzzleseekers like me, you’ll need to use better communication!

PH would be parallel to n

The question doesn’t posit that.

Spoiler alert: the image has no relation to the question, it’s just something OP picked to elicit that “3D geometry” feeling. There is a point A’, a point B, point C, and point D, but no point P or H or n.

Then you’d be surprised when you calculate the numbers!

A Falcon 9 delivers 13100kg to LEO and has 395,700kg propellant in 1st stage and 92,670kg in 2nd stage. Propellant in both is LOX/RP-1. RP-1 is basically long chains of CH2, so together they burn as:

3 O2 (3x32) + 2 CH2 (2x14) -> 2 CO2 (2x44) + 2 H2O (2x18)

Which is 2*44/(2*44+2*18) = 71% CO2. Meaning each launch makes (395700+92670)*.71 = 347 tons CO2 or 347/13.1 = 26.5 tons of CO2 per ton to orbit. A lot of it is burned in space, but I’m guessing the exhaust gases don’t reach escape velocity so they all end up in the atmosphere anyway.

As for how much a compute satellite weighs, there is a wider range of possibilities, since they don’t exist yet. This is China launching a test version of one, but it’s not yet an artifact optimized for compute per watt per kilogram that we’d imagine a supercomputer to be.

I like to imagine something like a gaming PC strapped to a portable solar panel, a true cubesat :). On online shopping I currently see a fancy gaming PC at 12.7kg with 650W, and a 600W solar panel at 12.5kg. Strap them together with duct tape, and it’s 1000/(12.7+12.5)*600 = 24kW of compute power per ton to orbit.

Something more real life is the ISS support truss. STS-119 delivered and installed S6 truss on the ISS. The 14,088kg payload included solar panels, batteries, and truss superstructure, supplying last 25% of station’s power, or 30kW. Say, double that to strap server-grade hardware and cooling on it. That’s 1000*30/(2*14088) = 1.1kW of compute per ton to orbit. A 500kg 1kW server is overkilling it, but we are being conservative here.

In my past post I’ve calculated that fossil fuel electricity on Earth makes 296g CO2 per 1 kilowatthour (using gas turbine at 60% efficiency burning 891kJ/mol methane into 1 mol CO2: 1kJ/s * 3600s / 0.6 eff / (891kJ/mol) * 44g/mol = 296g, as is the case where I live).

The CO2 payback time for a ton of duct taped gamer PC is 1000kg * 26.5kg CO2/kg / ( 24kW * 0.296kg/kW/hour) / (24*365) = 0.43 years. The CO2 payback time for a steel truss monstrosity is `1000kg * 26.5kg/kg / (1.1kW * 0.296kg/kW/hour) / (24*365) = 9.3 years.

Hey, I was pretty close!

A solar-powered computer in space could recoup the CO2 cost of its launch fuel over its lifecycle (say 10 years?) when compared to coal-fueled electricity on the ground. After that it’s free. Of course, you’d benefit more by filling up every available spot on the ground with solar arrays first! But you will eventually run out, or you might not want to do that.

If you have a megawatt solar array, you can also afford a megawatt cooling array. The size is comparable.

quantum computing is revolutionary because it allows instant transmission of information across unlimited space

False and common misconception! In quantum mechanics there is even a theorem called the no-communication theorem that mathematically proves that no information can ever be transmitted by quantum entanglement. Sorry!

Gravity—the curvature of space-time—can’t stop changes in the curvature of spacetime from propagating outward.

This is false. If it were true, you could build a device to communicate from inside a black hole event horizon. By waving around a heavy ball you would create gravitational waves that a sensitive enough LIGO outside the black hole could detect. But this is impossible. You would create gravitational waves, yes, but they would fall towards the center singularity same as you, and will never penetrate and escape the event horizon.

Use wireshark and listen on your ethernet interface. When you use tailscale, are the packets coming in/out from the tailscale server IP or the VPN ip? Check through the ip route routing table and figure out which pathway a packet will take in each use case. Might need to add a route exception specifically for the tailscale server IP to go out on the ethernet device.

PostUp = ip route add 100.64.0.0/10 dev tailscale0

Looks like you need to stick this line in the tailscale service file, since it’s the only time that the existence of the tailscale0 device is guaranteed. If you don’t want to modify the service file inside the package, could you write your own systemd service file and include the tailscale service as a prerequisite?

Also make sure that when you start the VPN first and then tailscale, you don’t get a double tunnel situation where tailscale goes out through the VPN (unless that’s what you wanted).

That allows sending packets inside the VPN tunnel, but the outer envelope packets still need to be able to reach the VPN server.

sudo ufw default deny outgoing

I’m guessing this would block the VPN packets themselves as well.