The h Bar

1:12 PM

Idk why you're complaining though, all this is just extra conversation. I already told you how to compute it

I told you literally every step

phase

you told but

I got nothing

I told you take v = rw, sub it into the expression for a and compute a using 2pi / T

@Blue I guess he's using . as a comma

@Blue is it my imagination though or is . as a decimal point sort of accepted as standard?

Anonymous

@Phase May god halp him

my god

my brain is firing atm

Anonymous

@Phase Never in my 18 years of existence I saw . being used as a comma :P

1:13 PM

@Lagranian $v = r\omega$, yes?

Anonymous

I dunno if some country uses that

I'm gonna use this $v = r\omega$ now

gimme a min

Yeah but you shouldn't compute v using it

It's a waste of time imo

you can if you want tho

what is the easiest way then

Well sub it into $a$

$a = \frac{v^2}{r} = \frac{(r\omega)^2}{r} = r\omega ^2$

Then just compute $\omega$ using $\omega = \frac{2\pi}{T}$ where T is the period in seconds and you're done

Although

Idk where Ecuador is

Is it on the Equator?

I hope for your sake it is

If it's not, you'll have to use $r = r\sin (\phi)$.

1:18 PM

Yes That's in equator

I mean the line

the omega is $\omega = 530.9166666666667$ right?

nooooooooooooooooooooooooooooooooooooooooooooooo

noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

my god

@Lagranian WHAT NOTATION DO YOU EVEN USE

:(

anyway, regardless of notation, that value for omega is wrong

You're dividing 2pi by 24*60*60

I swear

my brain is firing

Won't we use $\omega = \frac{2\pi}{T}$ to compute $\omega$

You do

It's just you made a mistake somewhere

1:22 PM

Where did I do mistake

tell me let me fix up that

your value of omega would imply that it takes a fraction of a second for the earth to revolve

Idk where your error is but I told you what to do

divide 2pi by 24*60*60

Anonymous

Can you divide 6.28 by 86400 ? @Lagranian

Wait I've seen where I did mistake

Anonymous

What is $$\frac{6.28}{86400}$$ ?

I thought that $2\pi$ is radius

1:24 PM

No, $2 \pi r =$ circumference

@Phase Exactly, that's why I did mistake.

by the way Lagranian, if I may Pique your interest for a second

@Phase How to find $T$?

Is that 24*60*60

Have you ever considered the centripetal acceleration and the fact that it takes away from your weight?

Yes, T is the period in seconds

$T$ is $84.600$ then

1:27 PM

$\omega$ is defined as the 'angular velocity'. It's a measure of angle per unit time, so if you take any angle travelled [as long as its travelled at a constant rate] and divide it by the time taken for that, you'll get $\omega$.

Idk, probably. Sounds right

@Lagranian No, it's $86400$ (you swapped the digits)

I'd really recommend you stop using the decimal though

@Mr.Xcoder he's using the decimal as a comma

Okay, right.

$\omega = \frac{2\pi}{T}$

O

I didn't even see that

from this

1:27 PM

@Phase Not talking about that, the order of digits in wrong

yeah I didn't notice

ok then

Anonymous

@Lagranian Please stop using a decimal point as a comma

eyes were drawn by the purrty zeroes on the end.

And Hi! btw

1:28 PM

$\omega = 43200$ right?

Who moi? @Mr.Xcoder

from $\omega = \frac{2\pi}{86400}$

if so hi back!

Anyone.

....

How

Do you get a larger number than 2pi

by dividing 2 pi by a number greater than 1

Anonymous

1:29 PM

@Mr.Xcoder Hi. Seeing you after long...:)

@Lagranian How come you divide something roughly 6 by 86400 and get 43200?

Did you half the period...?

He divided the period by two for some reason

@Blue Long time no see indeed. Code Golf takes away most of my (other) free time :D

Wait I know that $T$ is $86400$

then I plugged it into the equation

Yes.

1:30 PM

$\omega = \frac{2\pi}{86400}$

Correct...

which will give you a very small number

on the order of 10^-5 prolly

Namely $0.00007272205217$

Anonymous

@Mr.Xcoder So, learning any new stuff lately ?

Thermodynamics

Anonymous

1:30 PM

Nice!

And fluid mechanics. How are you?

Anonymous

I'm good :)

I see

Anonymous

Thermo is one of my favourite topics :D

Anonymous

We can discuss someday if you find some interesting problem

1:31 PM

Oh really? I like it too

Anonymous

Yeah, especially statistical mechanics

by the way

I do however tend to like classical mechanics more

$\omega$ is $0.00007272205217$

right?

Correct

Anonymous

1:32 PM

You should look up the Monte Carlo algorithm and Wolff algorithm sometime (if you didn't already)...it's related to thermodynamics and statistical mech. Also it's "algorithm", so I hope you'll like it :P

Great now we know what $\omega$ is.

from $\omega^2r$

@Blue Will keep in mind

we get $0.92662438875014$

right?

that seems a little large

just a factor of 3 or so tho

Is that correct?

1:35 PM

Our school curriculum tends to be rather odd, since we study kinematics in the 6th grade, then mechanics in the 7th grade, then thermodynamics, fluid mechanics and electricity in the 8th, then in the 9th we go back to mechanics (because we couldn't learn more about acceleration and stuff because the mathematics curriculum is even weirder)

I wouldn't say so

it's too large

It's the right power of 10 though.

Where did we go wrong?

I did all correct.

Anonymous

@Mr.Xcoder It was similar in our school. It's good to gain a basic knowledge of all fields in physics and then build upon previous knowledge. It's not necessarily bad

It's not bad, but it's a bit odd

The thing is that we have to learn acceleration from the 7th grade rather than the ninth grade if we want to take part in competitions, and it's boring afterwards

@Blue Huh? That’s standard European.

1:41 PM

Question: Is the formula of the centrifugal force $\frac{mv^2}{r}$, whereas $r$ is mass, $v$ is velocity and $r$ radius? FWIW I cannot completely remember it

Yeah

Anonymous

@Mr.Xcoder Yeah, that's true. I think you're too young to worry about school grades anyway. Go ahead and learn whatever you want to learn by yourself. It's difficult to find time to learn extra stuff once you're in college.

It's just centripetal acceleration multiplied by mass

Anonymous

@0celo7 That's such bad notation....

Since the centripetal force and centrifugal force must cancel out in a non inertial frame like on the surface of the Earth

@0celo7 its trash

Anonymous

1:42 PM

@Mr.Xcoder Yes. It's a pseudo force

@Blue FWIW I don't need to learn all by myself, since I have a great professor and we have additional preparation (3 hours) on Sunday, as well as other 3 hours during the week, and we work at a very high level compared to the rest of the class (of course there are more children passionate about Physics other than me, and we form a great team :P)

Anonymous

@Mr.Xcoder Awesome!

@Blue It's their cultural heritage

have some respect

Anonymous

@0celo7 What do they use for decimal, then?

Anonymous

@0celo7 Okay, if you say so :P

1:45 PM

@Blue comma ,

Anonymous

Sorry, if that's true

Anonymous

@Mr.Xcoder ......My whole life was a lie :P

@Blue 1 million: 1.000.000

@0celo7 Don't worry about cultural heritage, I have to use it too (since I live in Europe), and still think it's trashy :D

It's why the UK went through Brexit

The disagreement over commas and decimals

1:46 PM

@Phase ACO got the series back on track.

Really?

It's not nearly as good as AC2, of course

I would get it but..

The performance

I recall a meme regarding the UK leaving Europe twice in the Same week (Brexit and losing a match at the European Football (?) Championship) :D

Yeah I had a few hard crashes last night

1:47 PM

I can't support the practices of a company that willfully destroy people's experience of the game just to try and delay a cracked release

It's because of the two DRM's they have

I think it was because I was running my CPU overclocked for hours

at 100%

@Mr.Xcoder no-one stopped to think about the consequences of Brexit really.

I mean just look at this

@Phase I bet if I turn down AA my fps will go way up

@Phase ಠ_ಠ UK

You should get it

1:49 PM

Yeah but you shouldn't have to that's what pisses me off

What chokes first, cpu or gpu?

GPU in the country, CPU in the city.

My CPU is weaker than my GPU

I'm thinking about upgrading

What OS do you use?

...Windows

What else would one use

What CPU do you have?

@Phase R5 1600X

1:51 PM

R5...? Radeon?

Anonymous

@0celo7 I thought you use Mac?

of course... Get rid of windows and enjoy life!

Ryzen

@Blue I have two computers.

Oh, is't that the meme that came out a while ago?

Or was that the Zen

And I use Windows, Mac, and Linux

I am multicultural

Anonymous

1:51 PM

@0celo7 Means you aren't a whole-hearted apple fanboy

Anonymous

Noice

Never said I was

@Blue I am instead :-)

You [redacted] [redacted] keep attacking me as though I was

huh

cant find any benchmark tests of mine against yours

Anonymous

1:53 PM

@Mr.Xcoder I'm not sure we can be friends anymore :P

Anonymous

Okay, I'm joking XD

@Phase Problem with AMD is the single-core performance, which is important for some games

Lol did I find it wrong

not sure about this one

it seems to be using every core

$\omega$ = $092662438875014$

1:53 PM

so it's just awfully programmed

from $\omega^2r$

wonder if they'll fix it

would be nice to get a steady 100

@Blue xP... come on we still both like thermodynamics :P

Nvm found it

Im on a laptop so mines as-expected weaker

@Phase Are you sure?

I've used all equations you showed.

1:55 PM

@Phase Oh I also have an 11% overclock.

@Lagranian im pretty sure.

@0celo7 dang it

I can run witcher on virtually full ultra

Is ACO that much a fucking tanker

Your GPU doesn’t crap out?

I can only get 100fps on ultra in Witcher

Nah, my GPU isn't too bad, again it's limited by laptops

Full ultra

I set a limit to 60fps anyway

1:56 PM

Hairworks, etc.

I dont wanna spoil myself

Oh I dont use Hairworks because I dont think its worth it

That’s like 10 FPS lol

Exactly, its fine as long as you dont get accustomed to like 100+ fps

which is why I dont

Ok playing ACO again, i hate you for making me do this

I spend a lot of time playing fighting games

1:57 PM

@Phase it IS an issue dude

so 60fps is what I'm naturally acclimatised to

I played old splinter cells last weekend at 185

I really cant decide whether to try ACO or not

Now ACO at 85

Its not as if I can even be guaranteed I can get a refund if it tanks

1:57 PM

My eyes are bleeding

Man up boy

Damn, that could be flagged

goes in time using my [redacted] to flag

@0celo7 what are you actual benchmarks?

I’ve done a passmark test but I don’t remember the score

13.something for the CPU

Eh

I am too lazy to search, so is the period of a pendulum equal to $2\pi\sqrt{\frac{l}{g}}$ or am I mistaken?

2:02 PM

@Phase GPU is 14,121

hang on gonna close chrome for benchmark

cus that thing eats cpu like nothing else

Now I don’t remember if I overclocked for that number

Looool they bury you with the hidden blade

Poll: What is your favourite subject after Physics (or if Physics isn't your favourite, which is)?

Q: Can't logout from stack exchange

wait

crap used wrong tool

RIP

yeah I dont think I can commit to ACO

Physics Meta

0

I can't log out from stack exchange in browser .in pc and mobile. Afted loged out when i open the site again it appears as i am already loggedin. I have not enabled to remember my password

bug

2:12 PM

Maybe if it wasn't a piece of shit because of DRM

Slereah

hello

hey

The gameplay is really

really

good

my god phone

I need a new phone

I just shot a guard in the head with a bow and hid him in a bush! Incredible!

Such tactics

Surely there must be some coding/programming project you can do that is related to your field of research. — Peter Shor 1 hour ago

my god

the irony

Yeah but the performance is inexcusible. The best bit is that I keep finding articles online praising ACO for it's port...

2:25 PM

@EmilioPisanty why?

@0celo7 not sure

but being told to "c'mon, just program something" by Peter Shor has to count as a minor life event, no?

Oh, for sure

I was doubting the irony part

Slereah

Program Shor's algorithm

Anonymous

Oh wow, the MIT guy. Didn't expect him to be on SE

Q: Was the reduction in Shor's algorithm originally discovered by Shor?

though I guess it doesn't quite beat this one

71

This is a "historical question" more than it is a research question, but was the classical reduction to order-finding in Shor's algorithm for factorization initially discovered by Peter Shor, or was it previously known? Is there a paper that describes the reduction that pre-dates Shor, or is it ...

quantum-computing ho.history-overview factoring

one of the all-time highs of SE as a format in science if you ask me

2:31 PM

I dont get it

what do you mean @EmilioPisanty ?

I'm an SE plebian

@Phase you've seen who wrote the accepted answer, right?

Yeah

I just don't get the whole "just program something" comment or the one about it being a "high of SE..."

maybe im just slow today

@Phase Shor's algorithm is the foundational result of quantum computing, at least in the sense of "it made people realize this had any potential at all" if nothing else. Getting an answer from Peter Shor about his frame of mind when he wrote is about as straight-from-the-horse's-mouth as it gets.

oh I see

Anonymous

2:41 PM

Is choosing this kind of a contour wrong for integrating $\int_{0}^{\infty} \frac{\sqrt x}{(x^2+1)^2} dx$ ?

Anonymous

They used that for finding $\int_{0}^{\infty} \frac{\ln x}{(x^2+1)^2} dx$

Anonymous

But used a keyhole contour for this one

@Blue there's nothing immediately wrong with it

so long as the circle at infinity converges

which looks dubious

no, wait

looks OK for both

Anonymous

@EmilioPisanty I mean the contour chosen in the picture I uploaded is used by the author to find $\int_{0}^{\infty} \frac{\ln x}{(x^2+1)^2} dx$. It works fine for that. But when he is integrating $\int_{0}^{\infty} \frac{\sqrt x}{(x^2+1)^2} dx$, he uses a keyhole contour. I don't know what was the need to switch

Anonymous

@EmilioPisanty Yeah, same here

Anonymous

2:45 PM

I think this:

Anonymous

@Blue well, you can do the integral with the half-circle contour and see whether you get the same

Anonymous

Should work for both

Anonymous

@EmilioPisanty Yeah, checking that...