Just put two π ohm resistors in series duh
this post was submitted on 30 Mar 2025
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Whats wrong with your 3!
Ugh, 3 factorial is most definitely not equal to π. It's something more like, idk, 9? Honestly I don't even know how I got here; I majored in Latin and barely past
Barely passed your English classes as well I assume. /s
They barely passed me.
In case anyone wondering factorial is
n! = n * n-1 * n-2 * ... * 3 * 2 * 1
Erm. In what world do you live that the precedent in your expression is right?
In all languages and countries I know multiplication binds more strongly than addition. So what you wrote would be
n^2 - n - 2n - 3n....
I wrote it correctly. It is the definition of a factorial.
Well....
g^1/2^ = e = 3 = pi
e = π = σ = ε = µ = Avogadro's Number = k = g = G = α = i = j = 3
(at least that's how they all look when viewed from ∞)
I'm a physicist. If you are an engineer that sounds like a "you" problem.
Numbers like that are why I quit majoring in mechanical engineering. Physics took the beauty of math and made it ugly.
You knew something was wrong in calculus when you got a fucked up coefficient that wasn’t a nice number.
I actually really like physics, and it's 100% because I'm fucked up and evil
Respect. Physics is way up there in terms of hard science nerd cred.
the philosopher floating on a cloud: So how do you guys really know what's real?
Without using fancy components: Just simply adding a 6.2 and a 2400 Ohm resistor in parallel already gives you 6.18402 Ohm ⚡️
Real world resistors usually have a tolerance of ±5%, so you'll never get anything that precise.
That's why I keep a roll of 20 AWG nichrome on hand. Spool off 9.7195853528209 feet and it'll be bang on.
So 1 inch of your wire would weigh ~0.0987 grams, so to measure down to 8.6350242338508 inches of wire your scale would need to weigh down to ~0.00000000000007 grams. Which is the weight of about a dozen atoms or so.
Yeah which is why you use a Kibble balance. Are you sure you're cut out for this kind of work?
couldnt you technically fine tune a potentiometer to be this resistance if you were precise enough?
Mathematically yes. Practically, right now? No.
So you need a resistor of this value for your widget.
For that many places of precision you're looking at a potentiometer with a 10 nano-ohm precision.
I am not aware of any commercially available resistor that can do that but you could create one using microelectronic structures used for ICs and derive a 10 nano-ohm resistor by design and then chain enough of these elements into a resistor network or potentiometer to create the super precise resistance value you want.
Cool, congratulations.
Now how are you going to use this 10 nano-ohm resistor? What voltage will you be applying across it? What current do you expect it to handle? And therefore what are your power requirements? What are your tolerances, how much can the true value deviate from the designed ideal?
Because power generates heat through losses, and that will affect the resistance value so how tightly do you need to manage the power dissipation?
How will you connect to this resistor to other circuit components? Because a super precise resistor on it's own is nothing but an over-engineered heating element.
If you tried connecting other surface mount devices (SMDs) from the E24 or even E96 series to this super precise resistor then the several orders of magnitude wider tolerances of these other components alone will swallow any of the precision from your super accurate resistor.
So now your entire circuit has to be made to the same precision else all of your design work has been wasted.
Speaking of which, now your heat management solution now needs to be super precise as well and before you know it you've built the world's most accurate widget that probably took billions of dollars/euros/schmeckles and collaboration from the worlds leading engineers and scientists that probably cost more time and money than the Large Hadron Collider.
9 significant figures? good luck!
This guy looks like the dude from Programmers Are Human Too