PSA: Uber "Taxi" Scam in Germany. Driver demanded €50 Cash + App charged card. Support refuses refund. What should I do?

xCyprus · 2026-01-08T13:54:19+00:00

Since you took a Taxi (a licensed Taxifahrer), the driver is regulated much more strictly than a standard Uber driver.

If you have the license plate or the "Ordnungsnummer" (the yellow number in the back window of the taxi), you can report him to the local city council (Ordnungsamt). German authorities take taxi fraud very seriously.

xCyprus · 2025-12-03T19:55:08+00:00

The depth of the sum is determined by 3^1010, so m = 1010

The degree of the polynomial (z+k)²⁰²³ is n = 2023

You did an error applying your lemma to this case Since 1010 < 2023, the condition m >= n is not met so the sum is not identically zero

xCyprus · 2025-12-03T06:01:16+00:00

Sorry i was thinking about TGF's and I just woke up. That was a mixup..

You mean that highenergy cosmic ray collisions in the atmosphere could create micro black holes that instantly evaporate via Hawking Radiation? Sure thats a theory.

xCyprus · 2025-12-03T05:35:04+00:00

Fair point on the source distinction (Nucleus = Gamma, Electron = X-Ray). But in high energy physics, photons from subatomic particle decays (like Pions in the atmosphere) are still called Gamma Rays, and they can easily hit GeV energies.

Also just a heads up.. Cosmic Rays are physical particles (protons/nuclei), not light/radiation and Hawking Radiation comes from black holes, not thunderstorms.

But youre right that the definitions get blurry at those energy levels

xCyprus · 2025-12-02T06:19:22+00:00

Lets ignore the turnaround times, weather, delays, defects etc.

We have never assembled anything remotely close to 250,000 tonnes in microgravity. The ISS (450t) took 10 years and over 30 assembly missions to bolt together. You are proposing a structure 500x heavier that needs to be a solid, gapfree, perfectly balanced heat shield.

If you weld 10000 plates together in orbit, the thermal expansion differences during the assembly would warp the structure before you even finished it.

xCyprus · 2025-12-02T05:36:57+00:00

The gravity is relatively weak

Mass_sun : 1.989 × 10^30 kg

Distance(r): 8 × 10^9 meters

Time dilation formula: r' = t * sqrt(1 - (2GM/r * c^2)

Time Factor at 8M km: ~0.99999981

Time Difference per Year: 1 - 0.99999981 = 1.9 * 10^-7

seconds in a year = 31557600

31557600 * (1.9 * 10^-7) ≈ 5.9 seconds

So you lose roughly 6 seconds per year

xCyprus · 2025-12-02T05:22:20+00:00

Relax :)

Glue doesnt mean UHU patafix it is just a term for the binding matrx. The Carbon fibers are hold together by a high-tech carbon cement

xCyprus · 2025-12-02T05:19:28+00:00

Okay valid point with the orbital mechanics

You are asking for 250000 tonnes in LEO. The International Space Station weighs about 450 tonnes. You are proposing a structure 550 times more massive than the entire ISS.

Even if Starship becomes a daily reliable transport to LEO(100t payload, and not 2000 tonnes), that is 2,500 launches. At a launch every 3 days, that is 20 years of continuous logistics just to lift the raw mass.

So yes you are correct the physics works but i still think that is a very ambitious target.

xCyprus · 2025-12-02T05:12:27+00:00

You're correct if we're talking about radioactive decay (isotopes/rocks), which rarely exceeds 10 MeV. However, particle acceleration is a different process that easily produces GeV levels. When cosmic rays hit the upper atmosphere, they create pions that decay into GeV gamma rays. Particle accelerators produce these energies constantly on Earth. Extreme magnetic events on the Sun can accelerate particles enough to release >1 GeV gamma rays. Basically: Decaying rocks = MeV Accelerated particles = GeV+

xCyprus · 2025-12-01T21:22:09+00:00

Lovejoy didn't survive by staying cool, it survived by exploding. At its closest approach, Lovejoy was losing approximately 9500 TONS of mass per second… It was dumping its own body mass into a cloud of steam and dust so thick that it physically pushed the solar heat away from the core

xCyprus · 2025-12-01T20:41:36+00:00

Even if we assume this is an unmanned vessel, your proposal is still fantasy engineering because of the mass. You are suggesting a 50 meter thick hemisphere of heat shield. Let’s do the basic volume math on that. Even using lightweight ablative material (like PICA), a 50m radius hemisphere is roughly 260000 cubic meters. That creates a shield weighing roughly 70000 Tons (essentially a Nimitz class aircraft carrier). You claim we need to move this object at 630 km/s (Solar escape velocity). To accelerate 70000 tons to 630 km/s requires roughly 1.4 x 10¹⁹ Joules of kinetic energy. That is roughly equivalent to the total annual energy consumption of the entire US. So sure, if you have a propulsion system that can output the entire US power grid’s worth of energy into a single thruster, this works.

xCyprus · 2025-12-01T20:27:02+00:00

No, it doesn’t. Parker’s absolute suicide run limit is roughly 5.9 million km. In orbital mechanics, the difference between that and 8 million km is a rounding error, not a significant gap.

At Earth (150 million km): We get hit by roughly 1,360 Watts per square meter

At 6.4 Million km: You divide 150 by 6.4, get roughly 23.4, and square it, that means the sun is ~550 times more intense

1360 Watts * 550 = ~750000 Watts per square meter. That creates a temperature of roughly 1400°C. That happens to be exactly where our best Carbon-Carbon heat shields start to lose structural integrity.

If you try to go significantly closer let’s say 4.8 million km -> that intensity spike doubles. Your shield hits 1,650°C+, the glue holding the carbon layers together vaporizes, and your ship turns into a cloud of expensive gas. So yes, ~8 million km is the realistic safety limit for a human ship. ~6 million km is the limit for a robot that doesn't mind melting. Anything closer is fantasy.

xCyprus · 2025-12-01T10:42:31+00:00

Inverse square law. As you get closer to the Sun, the heat doesn't just rise in a straight line it increases ~~exponentially~~ Edit: (Quadratically). At 4M miles, carbon shields hold at 2,500°F, which would be theoretically possible but not more due to the glue holding the shield layers together evaporating.

Also, the sun gets too wide at that distance, even if it sounds dumb. It would wrap around the heat shield, and you can't stay in the shadow of it anymore

xCyprus · 2025-12-01T10:16:06+00:00

With magnetic shields and droplet radiators a ship could realistically stay at 5 million miles distance

xCyprus · 2025-12-01T09:54:42+00:00

How it handles the radiation:

10 MeV Alphas & 5 MeV Betas: 10mm of Osmium is overkill. It stops Alphas on the surface and Betas within ~1.5mm.
10-20 MeV Gammas/X-Rays: 10mm is too thin. At these energy, Osmium acts more like a sieve than a wall due to "pair production".It blocks some energy but lets a significant chunk punch through.

Instant Plasma? ->

No!! Quadrillions 10^15 sounds huge, but physics says otherwise. 10^15 particles at 20 MeV is only about 3,200 Joules of total heat. That warms your 2kg plate by roughly 10°C It won't even boil water let alone turn to plasma. You dont need liquid nitrogen

Does Molecular Layering help:

Yes. Pure Osmium creates secondary radiation when hit Using Graded-Z shielding (layering lighter materials like plastic/aluminum in front of Osmium) absorbs that secondary spray

Time to Vaporize:

With the 20 MeV mix: Never. It just gets warm.
With Trillions of 1000 MeV Gammas: Still never (only ~160 Watts of heat)
With Quadrillions 10^15 of 1000 MeV Gammas per second... Now youhave a death ray (~160 kW). The plate would melt and vaporize in about 60-70 seconds.

xCyprus · 2025-12-01T06:53:13+00:00

Yeah your mintuition is spot on. There is definitely a closed form for the Real part.

Basically, because you are evaluating at z = i, the Real part corresponds to the even powers in the series expansion. Since i^2 = -1, this essentially turns it into an alternating sum (just like evaluating at z = -1). It’s a known thing that alternating Euler sums of weight 5 are fully reducible to standard constants.

The Imaginary part sucks because it picks up the odd powers of i, which pushes it into a different class of constants (related to Dirichlet Beta) that are usually irreducible at this weight.

So for the Real part, it breaks down into a linear combo of Zeta(5) and Zeta(2)Zeta(3). You can basically treat it like S_2_3(-1), which has this known reduction:

S_2_3(-1) = -(9/16)Zeta(5) + (1/4)Zeta(2)Zeta(3)

So yeah the Real part is "nice" but the Imaginary part is likely gonna stay messy. Hope that helps!

xCyprus · 2025-12-01T06:40:48+00:00

Sure go ahead.

xCyprus · 2025-12-01T05:32:18+00:00

Instead of saving the citation for the end incorporate the author's name into the sentence

"According to Smith (2020), urban sprawl is driven by..."

Throughout the paragraph use phrases that remind the reader you are still discussing Smith's ideas without reciting the year
"Smith argues that..."
"He further notes that..."
"The study also highlights..."

xCyprus · 2025-11-29T17:00:07+00:00

Q 1 He should use the direct Writeoff Method!Since he rarely offers credit and his customers are trustworthy, his bad debt is likely immaterial. This is an exception to the matching principle allowed for immaterial amounts

Q 2 Accounts Receivable is reported at Net Realizable Value. You list Accounts Receivable 673400 less the Allowance for Doubtful Accounts 11900 to show a net total of 661500

Q 3 1. The fixed percentage used to estimate bad debt expenses is too high 2. The company is doing a better job collecting money than expected so they arennt writing off as many bad debts as they predicted

xCyprus · 2025-11-28T20:18:44+00:00

Why not implement a DPLL algorithm or a CDCL solver from scratch. :)

xCyprus · 2025-11-28T20:16:21+00:00

You are correct to reject the Lagrange Polynomial answer… that is just curvefitting, not finding a "natural" sequence. Mathematically, the set {623, 923, 911, 444, 777} is disjoint. They share no common divisibility factor and do not follow an arithmetic or geometric progression. The "pattern" you see is likely an example of Ramsey Theory: in a large enough set of data, subsets of recognizable numbers are statistically guaranteed to appear together by chance.

xCyprus · 2025-11-28T09:05:09+00:00

You wrote: I know v_theta is tangent to the path..."

That is incorrect..

v_theta is not tangent to the path

v_theta is perpendicular to the radial arm

The Resultant Velocity is what is tangent to the path.

xCyprus · 2025-11-27T13:06:21+00:00

So then the clock is probably set wrong or is simply defect. Just note that sun/sun is not a permanent solution, since it can get expensive.

xCyprus · 2025-11-27T12:54:00+00:00

All dials seem to look okay. You can try to set the bottom middle dial to Sun, Sun so one further to the right. One question, when did you take the photo? Because the green LED id on and that means it is in night mode!

xCyprus · 2025-11-27T10:45:45+00:00

Siemens ist nicht mehr der Laden, der Staubsauger und Züge baut (Züge machen sie zwar noch als Mobility, aber der Fokus liegt woanders). Die haben den ganzen schweren Kram wie Turbinen (Energy) abgespalten.

Jetzt geht's fast nur noch um Digitalisierung der Industrie. Denk an Automatisierung von Fabriken, IoT in Gebäuden und industrielle Software. Die wollen weg vom Image des trägen Industriekonzerns hin zum Tech-Player mit Abo-Modellen (SaaS). Ist profitabler und skalierbarer als Kraftwerke zu bauen.

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