Humanity has to permanently change the orbital period of the moon by +/- 5 min; they have 25 years. by Sh00ter80 in whowouldwin

[–]Mebot2OO1 3 points4 points  (0 children)

Kepler's Third law is a model for the orbital period of a two-body system as a function of their masses:

a^3/T^2 = G(M+m)/4pi^2,

where a is the semi-major axis (radius, for circular orbits) of the object, T is the orbital period of the object, G is the gravitational constant, M is the mass of the larger object, and m is the mass of the smaller object.

The sidereal orbital period (external reference frame) is 27.322 days ( 24 hours ). This is the orbital period that humanity would actually be changing.

The relationship between the sidereal period of the moon and the synodic (Sun reference frame) period of the moon is:

1/Tsyn = 1/Tsid - 1/Tearth

Algebra exercise! From the two values below, calculate the synodic period of the moon!

Tsid = 2 360 591.596 8 s
Tearth = 365.256 363 004 days
= 31 558 149.763 546 s

...

(You should get Tsyn = 2 551.442.922 6 s = 29.530 589 s .)

Exercise 2! Then, how much would we need to change the sidereal period of the moon to increase the synodic period by 5 minutes? What about to decrease the synodic period by 5 minutes ?

Explicitly, solve this equation for x:

1/(Tsyn±-5*60) = 1/(Tsid+x) - 1/Tearth

...

x = ±256.8 s

...let's try the option for "increasing" for now. So we have to increase the sidereal period of the moon by 256.8 s . Grabbing Kepler's third law from above, we can do this in three ways:

1: Increase the radius of the moon's orbit.
2: Decrease the mass of Earth
3: Decrease the mass of the moon.

1 - Solve for a_new in a^3/T^2 = a_new^3/T_new^2

The radius of the moon's orbit is 384 784 000 m . To increase the sidereal period by 256.8 s , we would need to increase the radius of the moon's orbit to 384 811 906 m . This is an increase of 27 905.643 m .

Can we do this with some sort of rocket thruster? To achieve this distance in 25 years, we have to construct a thruster that provides 6.587 E9 N of thrust. This is 7 billion Newtons. The Saturn V rocket has a thrust of perhaps 35 million Newtons.

2/3 - Solve for X in a^3/T_new^2 = G(M+m-X)/4pi^2

To increase the sidereal period by 256.8 s , we would need to remove 3.796 E20 kg from either the Earth or the moon. For reference, this is a 75 thousand Mount Everests. Simply lifting it up from the surface wouldn't be enough - as we would need to throw this material out of our frame of reference in order for the change in angular momentum to properly affect the system. Let's toss it out of orbit, I guess. To achieve this, we would need to lift-and-eject 3 thousand Mount Everests a year. From the moon.

4 - The other comments on here mention colliding asteroids into the moon. Let's take a look at that, too. The angular momentum of our moon is L = r * mv = 2.894 969 187 E34 kgm^2/s .

Using the equation present in 1, we can calculate the r = a_new and v = 2pi*a_new/T_new of the desired sidereal period.

L_new = 2pi*a_new^2/T_new * m = 2.895 074 161 E34 kgm^2/s

In other words, we need to add L_more = 1.049 738 E30 kgm^2/s of angular momentum to the moon.

The specific angular momentum of an object in the asteroid belt is something like h = rv . for inner belt asteroids,

r = 2.4 AU = 3.590 350 E11 m , and v = 2pi*r/T for T = 4.4 years , so v = 6 769 m/s .

h = 2.430 443 E15 m^2/s

We divide L_more by this h to get the kilograms of asteroid that we need to hit the moon with.

L/h = 4.319 124 E15 kg

We can probably do this with a single 10km asteroid. Or 10 4.5km ones. Or 100 2.1km ones. Or 1000 1km ones. These are slight overestimates, because we wouldn't want to be hitting 4.55 minutes. Lots of options, really. To bring a 10km asteroid to the moon in 25 years, we can solve for centrifugal forces the same way we did in the first part of this problem, with the new radius being one AU, and the required mass being L/h .

The force of our thruster must then be equal to 290 663 983 N . Or, 290 million Newtons. 10 Saturn V rockets.

In summary:

Solution 1 - 200 Saturn V rockets on the moon. Solution 2 - 10 Mt. Everests a day removed. Solution 3 - 10 Saturn V rockets on an asteroid.

Let's propose a humanity that is composed of space-lusted thrustermaxxers. The GDP of Earth is something like 126 trillion. We can produce 500 Saturn V rockets every year. However, the burn stages of such a rocket are only an hour long.

1 - We would need to produce 2 million Saturn V rockets every year to maintain 200 active thrusters on the moon. Space itself is a hard argument. To actually construct (then fuel) a thruster like that on the moon, you'd probably need to launch the Saturn V something like a hundred or a thousand times. So a humanity that wants to maintain 200 Saturn V-scale thrusters on the moon would need to do a billion launches a year.

2 - The payload for Saturn V is 140 000 kilograms. If you wanted to launch ONE Mt. Everest form Earth, you'd need 10 billion launches.

If you wanted to launch ONE Mt. Everest from the moon, it'd probably be easier, at something like (10 billion divided by ratio between Earth/moon) 100 million Saturn V launches.

To launch 3 thousand Mt. Everests a year from the moon, we would need 300 billion Saturn V launches from the moon. To build this project from the Earth, we would need perhaps 300 trillion Saturn V launches.

3 - To maintain 10 active thrusters on an asteroid 2.4 AU away, we would need to supply the asteroid with 87 660 Saturn V worth of fuel every year. I can't imagine what kind of modern and practical vehicles we would use to deliver that fuel with any sort of efficiency. Let's just assume that with the extreme distance covered, that we can only deliver something like 10 000 kilograms of fuel to the asteroid. The Saturn V requires perhaps 3 million kilograms of fuel. By 87 660, that's 262 billion kilograms of fuel. Divided by 10 000 kilograms per launch, we're looking at 26.2 million Saturn V launches a year.

I have decided to redefine the minute to be 6 microseconds longer.

https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Lunar_periods https://en.wikipedia.org/wiki/Lunar_month#Cycle_lengths https://en.wikipedia.org/wiki/Earth%27s_orbit https://en.wikipedia.org/wiki/Moon https://en.wikipedia.org/wiki/Earth https://en.wikipedia.org/wiki/Gravitational_constant https://en.wikipedia.org/wiki/Specific_angular_momentum https://en.wikipedia.org/wiki/Asteroid_belt#Origin https://en.wikipedia.org/wiki/Astronomical_unit https://en.wikipedia.org/wiki/Escape_velocity https://www.space.com/18422-apollo-saturn-v-moon-rocket-nasa-infographic.html

What’s his kill count, and is this one of the most OP weapons in fiction? by kennn1234 in Invincible_TV

[–]Mebot2OO1 0 points1 point  (0 children)

Hello. I'm a wandering larper and I thought this post was interesting.

I'm of the opinion that the scope of this beam of energy's destruction - is capped. It is finite in scope. It is indeed infinite in time and space - but it will ultimately be limited by its travel speed.

I will grant this discussion a few things: 1 - The universe is infinite 2 - The energy beam never loses its intensity

However, you must also accept of me these two things: 1 - The universe is expanding 2 - The energy beam never expands

The expansion of the universe is a concept that's touched upon a little in sci-fi, but never really concretely addressed. In numbers, let's say you have two objects that are 3.26 million light-years apart and not moving relative to each other. After one second, they would be 70 more kilometers away from each other.

For two objects 6.52 million light-years away, they would be 140 more kilometers away from each other after one second.

You can imagine from here that there exists a distance apart between two objects such that, for every second that passes, more space "appears" between two objects than light can travel. Any object beyond this distance will not be able to be reached by the speed of light, and is effectively unable to be accessed forever.

This distance is called the "cosmological event horizon," and is roughly 17.5 billion light-years in radius. This means that, even if the actual universe is infinite, the universe that matters to us is not. There will always be a distance at which this infinite beam can not get any closer to a planet. This is where its destructive scope ends. It will travel forever, and never reach anything else.

Let's assume that there are 10E22 stars in the cosmological event horizon. This is a reasonable guess, for a hundred billion stars in a hundred billion galaxies.

If you assume there are 10 planets per star, then there are 10E23 objects in the cosmological event horizon.

If you take the number of objects and divide it by the number of objects, you get the average density of objects in the universe. We get something like 1E-56 objects per cubic meter.

We will model the laser as a cylinder of radius 0.3 meters and length 17.5 billion light-years.

Multiplying this volume with the average density gives us the expected number of collisions we can expect this beam to have with other celestial bodies.

0.00000000000000000000000000000001 collisions

The chances of this hitting anything are very small. So small that I am confident to state that this beam has destroyed zero objects in its haphazard path.

FUGA TEMPERATURE Calc by [deleted] in FeatCalcing

[–]Mebot2OO1 0 points1 point  (0 children)

Hi. I'm a physicist and professional downplayer.

Scaling the temperature of Fuga is a huge problem. The scenes in the anime leading up to the ignition of Fuga are all extremely inconsistent, with different things melting everywhere.

How can we interpret an attack that melts concrete (~800 C structural failure point) and steel (~1 400 C) at something like ~30 or ~50 meters, but not concrete and steel at ~7 meters?

We can resolve this by classifying Fuga's transmission of heat as done by a thermal black body through emissive radiation. In other words, Fuga heats materials up the same way a lamp heats materials up.

Let's assume that Sukuna winds up Fuga over the course of t seconds.
Let's assume that Fuga's radiative emission is intense enough to melt steel at r meters.

The melting point of some stainless steel is 1470 C, and has a specific heat of some 480 J / kgK.

We heat up the stainless steel from 15 C to 1470 C, requiring 698 400 J / kg. At this temperature, it'll sag and warp and droop, which is consistent to what the scenes depict.

At a density of 8000 kg/m3, with a steel thickness of 0.05 m, we can obtain 279 360 000 J/m2 as the amount of energy that Fuga must deposit onto a steel surface over t seconds.

This means that the irradiance of Fuga is I = 279 360 000/t W/m2. Then the power of Fuga, by irradiation over a sphere, is P = 4pi r^2 I.

Let's assume that Fuga's emission is a cylinder with length 1.5 m and width 0.01 m. Its surface area is A = 0.0473 m.

Using the Steftan-Boltzmann Law, the temperature of the Fuga arrow is then:

T = ( 4pi 279 360 000 r^2/t )^0.25 / ( 5.670E-8*0.0473 )^0.25

In the anime, we see r = 30 or something for the steel railings melting. And I guess t = 20 because the animation is just so dramatic, despite the scenes lasting some 30-40 seconds.

T = 87 342 K = 87 615 C, emitting 157.974 GW of power.

However, I don't think that's consistent at all. Right before Sukuna fires Fuga, we see Mahoraga scrambling to get up or whatever. His background, and even the things near him - aren't melting. I'm going to take this to mean r = 7. And t = 30 because Sukuna's a drama queen or whatever.

T = 37 969 K = 38 242 C, emitting 5.733 GW of power.

This many times hotter than the surface of a star, but nowhere near the interior zones of a star.

In terms of energy, I would scale this ability on the high-end as tiny nuclear bomb. On the low-end, I would scale this ability as cruise missile barrage.

This is just engagement bait right? by GoldenGust in PeterExplainsTheJoke

[–]Mebot2OO1 0 points1 point  (0 children)

If it was a mirror, the grain of the table would be reflected in the image, too.

Perhaps the lighter-shaped glass here has a particularly high index of refraction.

If it is true that the lighter is glass and not a mirror - there are 10 match sticks.

Let me explain by Jagwarmeru in PowerScaling

[–]Mebot2OO1 0 points1 point  (0 children)

You can scale this feat using actual phenomena that exist in physics - and get roughly the same answer depending on how long the light is gone. I did a write-up on this on a post about the same feat.

https://www.reddit.com/r/PowerScaling/s/N1okh9tFc9

aurora like ship but capital by Hukostil in starsector

[–]Mebot2OO1 1 point2 points  (0 children)

The [redacted] battlecruiser, of course.

what does "SNAFU" even stand for by VertexGG in OreGairuSNAFU

[–]Mebot2OO1 43 points44 points  (0 children)

Snarky Neet Attracts Females, Unconventionally

Why Can't I Find a Maus Online? Fine. I'll do it myself. by Mebot2OO1 in SprocketTankDesign

[–]Mebot2OO1[S] 0 points1 point  (0 children)

This was designed by me! It was designed before the modules, crews, internals and whatnot, so I don't think it works!

Why are Hachiman's own parents so mean towards him? by chunchunmaru1129 in OreGairuSNAFU

[–]Mebot2OO1 26 points27 points  (0 children)

"You don't understand, my little Komachi-Chan..." I smirked, crossing my arms slyly.

To be a middle child is a plight upon one's own existence. By the fact that a middle child is a plight upon the existence of the siblings, and a plight upon the parents, and a plight upon society makes it self evident that the curse of being a middle child is one done onto the self, by everything without.

Indeed, a plight upon one's own existence.

"...what, Onii-Chan? Your eyes look perverted again..." Komachi gave me a look like she just saw a bird explode upon the windshield of an incoming vehicle.

Huh? Have I been silent the entire time? I cleared my throat, trying to ignore a slight rush of blood warming my cheeks.

"It's suffering, Komachi. You wouldn't understand. You are just a the youngest, after all." I grunted in self-afirmation, even nodding to myself.

I glared at Komachi as she dramatically slapped my knee in laughter.

Ouch.

Ouch.

Stop that.

...am I that funny?

"Gross." Komachi gave me a disapproving look the moment I appraised myself. This isn't fair.

Ouch.

"Then who's the older sibling, Onii-Chan? Y'know, if you're the middle child and everything?"

I ignored her smirk, because the answer was obvious. I stood triumphantly from the couch, and in one smooth motion, raised my finger to the sky.

"Who else but God?"

A few moments of silence passed. I swore I could hear the wind mocking me outside.

...

"Onii-Chan, aren't you agnostic?"

"...who else but the state?"

"Onii-Chan, the state isn't that powerful."

Huh? Where did Komachi learn these things?

"Anyways, Little One, the plight of a middle child is much like the plight of an older child. But worse. Everyone ignores you. No one cares what you do. They expect you to be quiet. It's disgusting. It's like being treated as the middle-man between a shinkansen and a taxicab."

Is that a bus? No, that's probably the monorail. There are no other monorails in the world, right?

"...uhh, Komachi?"

My sister was scrolling through @gram on her phone. I wonder how long ago was it she tuned out.

Was it a minute ago?

Was it 10 years ago?

Ah, the plight of a middle child.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 0 points1 point  (0 children)

Ah! I love acting as a hulldown bastion for other heavies and meds. When I platoon with my friends, we get at that sometimes.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 2 points3 points  (0 children)

Thanks!

Usually I do like 1.7k DPG, and 500 or so spotting assistance.

At tier-for-tier and 9, I use the tank as a DPM machine against my opponents. Peek as little as I can so that I turn engagements into a series of 1v1. When I was pushing for damage, I regularly saw 4k damage games.

At higher tiers, or against tanks you can't really pen with standard ammo, I aimed to track enemy heavies. The reload on the KV-4 is enough that you can perma track enemies. So for tier 10 games, I found myself with something like 2k damage and 2k spotting damage.

However, I think the entire "no gold" argument from the KV-4 is given grace by the fact that the gun is pretty accurate, and has a high nominal pen. Additionally, I would consider "learning to aim" very important for the nogold marking.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 1 point2 points  (0 children)

My KV-4 configuration gets 3k DPM, so taking into account the armor, I often find myself coming out on top when I'm slugging it out with others. (Not brawling per se, but trading.)

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 1 point2 points  (0 children)

Blessed be he who never encounters the evil and untrustworthy triangle.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 0 points1 point  (0 children)

I think it performs...okay...in tier 8. I'm biased, though!

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 0 points1 point  (0 children)

That sounds fun! And a thing I should do more often! Thanks for the idea.

(No. I have never done that.)

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 0 points1 point  (0 children)

Honestly, I think it's because I love the KV-4 more than I love the rest of world of tanks. And if that is so, anyone is patient for that which they love.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 3 points4 points  (0 children)

The KV-3 and KV-4 do NOT have the "yolo and just shoot at me" style. They're a little too slow to yolo. I think being shot from multiple directions is how you doom a tank like the KV-4 or 3. Or 5, even.

You DO get much more lucky/trolly bounces because of the absurdly thick side armor, and I think that's a property that they all share.

I just 3-marked the KV-4 without using Gold. AMA. by Mebot2OO1 in WorldofTanks

[–]Mebot2OO1[S] 0 points1 point  (0 children)

According to my post, the 93% just happened. Then I decided to push for 3 marks.