this post was submitted on 28 Oct 2025
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I tried to calculate how much this kitty would have to weigh to make a blackhole and... I found 3.13*(e)^(-27). Society is so fucking lucky I didn't decide to study math
Edit: my painkillers finally kicked in and I redid my calculation: Assuming the cat as a sphere with 15 cm radius, the cat would have to weigh 1.00994318 e26 kg which isn't as glaringly wrong as my previous calculation. (omg I wrote e^-27 whats wrong with me)
I don't even know what that equation means. I am in awe of you
(no one does ... he doesn't say if that's a number of ounces or a number of metric tons ...)
uh it should be in kgs sorry, I was just bewildered how I managed to fuck up an equation with it literally being in front of me (cus I googled it). The number comes up to a little less than 2 protons... which means (according to my food poisoned brain's calculation) that if there were 2 protons in a sphere with a radius of 15 cm, it would collapse into a black hole.
So you're saying that 2 protons closer than 15 cm would collapse into a black hole? That's got me pretty worried, because I have a lot more than 2 protons within that proximity in my body, by at least an order of magnitude or so
yeah sorry, its all blackholes now
Well anyway, that's wrong too. Density makes a black hole, not just weight. And a tiny tiny tiny weight, for a normal cat size, is on the wrong end of density.
is density not just weight divided by volume? I had a set volume (assumed the cat was a sphere with a radius of 15 cm), where am I wrong....
And yes the calculation is wrong, that was the point, I WAS TRYING TO BE FUNNY.
I know this is only for fun, but what you have to do is: set cat weight (1 kg) and calculate cat radius (instead of setting cat radius and calculating cat weight). This will give a crazy small size and crazy high density. 🙂 (poor cat tho!)
Alright mate, you and I seem to be having a major problem with communication. I am not trying to see how much I would have to squeeze a normal cat to form a blackhole, I am trying to see how much a cat would have to weigh to form a blackhole as it curled up. And an equation can be used to find any component of said equation as long it is the only one missing, so there is no one thing I "have to do". I am really feeling like you're disrespecting me here.
A cat of any weight can become a black hole, so I don't understand what you're trying to find. What I showed you is a more common problem with a clear solution. If you don't want communicate then just don't.
I always thought that leaving a conversation unfinished was a rude thing. What I was trying to find (as I have previously said) is how much a cat would have to weigh to collapse into a blackhole (assuming the cat when curled up would resemble a sphere with a radius of 15 cm). Last thing, how is "a more common problem" any helpful? I appreciate you trying to help but it feels like telling somebody, who is drowning, that breathing air instead of water would prevent drowning. Again, thank you for trying to be helpful.
“How is a more common problem any helpful?" - I never saw anyone trying to solve for weight so I brought you where most people discussing black holes operate: squeezing mass until it's a black hole. But you actually wanted weight: feed the cat until it weighs as much as a planet or sun. No problem man, you do you. And smiles don't mean disrespect.
The number in the exponent is negative, so it's a very tiny value, whatever the unit.
if the problem was with the e, that just means how many times you multiply it with 10, 1.87e4 = 1.87 * 10^4 = 18700. (ignore me writing it with an exponent in my original "calculation")