226
Guess I'll just burn (sh.itjust.works)
submitted 5 months ago by RmDebArc_5@sh.itjust.works to c/memes@lemmy.ml
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[-] Funkytom467@lemmy.world 80 points 5 months ago* (last edited 5 months ago)

This is a fucking differential equation with unknown constants, so yeah, everyone will be burning...

[-] sharkfucker420@lemmy.ml 6 points 5 months ago* (last edited 5 months ago)

Maybe use c~1~ and c~2~?

[-] transientpunk@sh.itjust.works 42 points 5 months ago

The Navier Stokes equations represent the universal laws of physics that can model any fluid in the universe.

These equations have been around since almost two centuries now but we still understand very little about them. When we have a set of equations we expect the following to happen-

  1. Solution should exist- One should be able to solve the equations

  2. Solution should be unique- Given particular initial conditions, one should obtain an unique solution to the problem. For example if you and your friend pour water into a container in an identical way, keeping all parameters (pouring velocity, direction, geometry and dimensions of the container, etc) identical then you both should get the same flow pattern. Water in both the containers should behave in exactly the same way. If your friend gets air bubbles at a point then you should get them at the exact same point as well.

  3. Solution should be smooth- A finite change in the input should produce a finite change in the output. It should not be erratic and unpredictable.

Unfortunately, Navier Stokes equations do not satisfy any of the conditions mentioned above.

https://medium.com/@ases2409/navier-stokes-equations-the-million-dollar-problem-78c01ec05d75

[-] Zerush@lemmy.ml 7 points 5 months ago* (last edited 5 months ago)

There is still waiting a prize of $1 Millon solving it, fuck the Thermostat

[-] deegeese@sopuli.xyz 35 points 5 months ago

One thing I remember from physics is if you have to guess, the most common answer is zero.

[-] sharkfucker420@lemmy.ml 34 points 5 months ago

As a physics major this is very true. The answers are always 0, 1, or π

[-] captainlezbian@lemmy.world 9 points 5 months ago
[-] chemical_cutthroat@lemmy.world 17 points 5 months ago

Only if you are blowing shit up.

[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 5 points 5 months ago* (last edited 5 months ago)

If your signal looks like f(t) = K•u(t)e^at with u(t) = {1 if t≥0, 0 else}:

  • If Real(a) > 0, then your signal will eventually blow up.
  • If Real(a) < 0, then you signal will not blow up. In fact, your signal will have a maximum absolute value of |K|, and it will approach zero as time goes on.
  • If Real(a) = 0, it is either a complex sinusoid or a constant. In either case, it is bounded with maximum absolute value of |K|. It very much does not blow up.

So e pops up all the time in stable systems and bounded signals because the function e^at solves the common differential equation dx/dt = ax(t) with x(0)=1 regardless of the value of a, particularly regardless of whether or not the real part of a causes the solution to blow up.

[-] sharkfucker420@lemmy.ml 5 points 5 months ago* (last edited 5 months ago)

Usually it is e^t or something similar

[-] Mr_Dr_Oink@lemmy.world 23 points 5 months ago

Used a screen shot to Google this and it turns out to be some unsolved ancient equation regarding the laws of physics. Or something that I dont understand. And have probably misrepresented here.

[-] nikaaa@lemmy.world 18 points 5 months ago

Wait isn't this just some fluid physics equation?

Like, incompressible fluid flow, iirc.

[-] crapwittyname@lemm.ee 12 points 5 months ago

It looks like it given the symbols used. P for pressure, rho for density etc. u-arrow is definitely a vector field, so it could be fluid flow. Otherwise it could be equally anything described by a vector field, like electromagnetism or gravity but they usually have a lot more E and G involved I think. I used to solve these but then I got a certificate so now I don't have to.

[-] supercriticalcheese@lemmy.world 4 points 5 months ago

It's a fluid dynamics equation, cannot be analytically solved unless laminar flow assumption is valid.

[-] InputZero@lemmy.ml 3 points 5 months ago* (last edited 5 months ago)

It really reminds me of all those static and velocity pressure calculations I had to do in undergrad, until I got the degree.

[-] nikaaa@lemmy.world 2 points 5 months ago

u stands for velocity.

[-] supercriticalcheese@lemmy.world 4 points 5 months ago

Naviers stokes equation looks incompressibility fluid. Only possible to solve it for strictly laminar flow.

https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations#Incompressible_flow

[-] plz1@lemmy.world 15 points 5 months ago

Irony is, a lot of larger office building thermostats are really only there for display purposes (thermometer), not for control purposes (actually functional).

[-] rbesfe@lemmy.ca 7 points 5 months ago

Often tenants can change the thermostat to whatever they want visually, but in the background it caps at a certain value or doesn't change the set point at all

[-] Feathercrown@lemmy.world 11 points 5 months ago* (last edited 5 months ago)

Am I missing something or does the top equation knock out half the terms? It simplifies to just F = delta + deriv. of u wrt t, right?

(Assuming p =/= 0)

Wait nope some of those ps and us are different. Oh no

[-] mexicancartel@lemmy.dbzer0.com 4 points 5 months ago

Also its not delta, its upside down delta called "Del" (or nabla) which is an operator like derivative but in vector feilds

[-] Feathercrown@lemmy.world 6 points 5 months ago

This deal keeps getting worse every day 😭

[-] Wizard_Pope@lemmy.world 3 points 5 months ago

Because some of those are rho and mu

[-] brokenlcd@feddit.it 8 points 5 months ago

Or if you know what you are doing electrically speaking pull the thermostat off the wall and connect the cooling/heating line to common for a bit; I think it would actually be less effort

[-] unmagical@lemmy.ml 5 points 5 months ago

My work set the password to the facilities manager's phone extension which could just be looked up in the internal directory.

[-] zonsopkomst@lemmy.ml 4 points 5 months ago

Easy, I just need a wire, toothpaste, bubblegum, and and a watch crystal. Problem solved!

[-] getoffthedrugsdude@lemmy.ml 4 points 5 months ago

Aren't those locked with actual keys?

[-] ftbd@feddit.de 3 points 5 months ago

No boundary and initial conditions?

[-] vk6flab@lemmy.radio 3 points 5 months ago

Or you could point a flamethrower at the sensor..

For less drastic response, put your hand on the sensor.

this post was submitted on 22 May 2024
226 points (100.0% liked)

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