Okay… So quick math.
A 1TB Nvme weighs about eight grams. A typical takes up 13KB of disk space.

So (\frac {1 TB}{13 KB}) = (7.692×10^7) times. That means that the weight of 8 grams divided by the number of times a fits on a 1 TB Nvme (\frac {8 grams} {7.692×10^7}) = (104ng)

So one typical weighs about 104 nano grams!

cc @volpeon

@Erpel @volpeon > math

👀

@schappi@brotka.st @volpeon@icy.wyvern.rip Did.. did I miscalculated my ​​?

@Erpel
We love math nerds.
@volpeon @darkphoenix

@VanessaNightshade@beige.party I am more of a neomoji nerd ​​

@Erpel
🥔🥔

@VanessaNightshade@beige.party ​​

@Erpel @volpeon But if you use a bigger drive.... Does it make them lighter?

@Erpel @volpeon Yes but how much does weight ?

@garfieldairlines @volpeon uses 2KB of disk space.

So we do: (\frac {8 grams} {\frac {1 TB}{20 KB}} = 160ng)

So weighs 58 nano grams more than a regular .

That fur is chonky it seems!

@Erpel what it you put on a 2TB micro SD card?

@ieatbeees@tech.lgbt then they get light weight and weigh only 52 nano grams. Thats way below the recomended minimum mass of a ​​

Don't starve your ​​!

@stefan@akko.lightnovel-dungeon.de @volpeon@icy.wyvern.rip yes. You starve them this way!

@Erpel @volpeon This is a bit confusing.

Asuming this is in true bytes, its fine because everyything is in powers of 10. But if they are Microsoft units, the units are in a power of 2. And grams are in standard units which means its a power of 10. These are not equivalent and I dont think that would be correct.

If so,you need to convert 1TB and 13KB to real TB and KB by converting them to its equivalent as a base 10 unit, then divide, then divide the grams by the result.

@volpeon@icy.wyvern.rip @enigmatico@mk.absturztau.be you are putting to much thought into a shitposting​​

@Erpel @volpeon Yes.

I’d rather put that much thought into something that doesn’t even make sense to begin with, than in whatever is going on around me rn

@enigmatico@mk.absturztau.be basically I just throw a few numbers into wolfram alpha.
1TB/13KB gave me a factor and I just divided the 8 grams by that factor. (Also in Wolfram Alpha).

This is all wrong but makes for a funny shitpost

@Erpel Well, I know this is a shitpost but let’s suppose we were trying to do something serious instead, and ignoring the fact that the weight of your SSD is not given by the amount of storage in it but the density and amount of all the materials in it.

Let’s assume the following data:

• The link between the weight of the SSD and it’s capacity is given by some linear function (that is, it’s weight is directly proportional to the amount of storage in it).
• Your SSD contains 1TB (true bytes, that is (1\times10^{12} ) bytes).
• The weight of your SSD is 8g (SI units, also a power of 10)
• A takes 13KB of space (also true KB).

Since we are hypothetically linking the mass to the amount of storage, we can simply divide (\frac{8g}{1\times10^{12} B} = 8 \times 10^{-12} g) per byte (B).

What’s the weight of a then?

(13KB = 1.3 \times 10^{4} B) or (13,000 B)

So ((13\times10^{3})\cdot\frac{8}{10^{12}} = \frac{104}{10^{9}}g = 1.04 \times 10^{-7}g) or 0.000000104g

So in conclusion, yes. This is correct, assuming the true definition of a Kilobyte and it’s subsequent multiples. However, what Microsoft calls a Kilobyte is a power of 1024 or a power of 2 (1024² = 2²⁰). So in this case, it would be required to convert it to true Kilobytes.

(1 \times 10^{12} \times \frac{1}{1024^{4}} = 0.9094947017729282TB)

Or in GB

(1 \times 10^{12} \times \frac{1}{1024^{3}} = 931.3225746154785GB)

And from there, repeat the same math as before.