Talk:3179: Fishing

The tug on a fishing line would be measured in newtons, not kilograms. 76.187.17.7 04:30, 11 December 2025 (UTC)

It's at least a C+ 65.35.15.18 05:12, 11 December 2025 (UTC)

Yes, but he is not measuring the force, he is measuring the "weight" (mass) of the thing he thinks he hooked. (e.g. a 5 lb fish)2603:8000:5E00:2913:EE02:2D56:E960:2CDE 05:21, 11 December 2025 (UTC)

YAY BERET GUY!Mathmaster (talk)

Beret Guy's estimate is, of course, absurd (or it would be for anyone else). His lifting capacity, the breaking points of his line and rod, the buoyancy of his boat and the force to break loose an individual piece of rock (the lowest of which would mark the upper bounds for his estimate) are (many) orders of magnitude lower than the force required to haul a 10^24 kg catch into the boat [citation needed, I guess] 627235 (talk) 11:42, 11 December 2025 (UTC)

He's not saying that he can reel it in, just estimating the size of the "fish" he's hooked. Barmar (talk) 14:57, 11 December 2025 (UTC)

When he pulls, there is a little bit of give (infinitesimal) as the planet moves - he might be sensing that and estimating based on it. Assuming you know the properties of the fishing line, like its stretch, and of the boat's surface area, buoyancy. I don't see why it wouldn't be possible to estimate given perfect knowledge of the water, boat, line, forces, etc, even if the Earth is much more massive than the boat being pulled downwards R128 (talk) 15:22, 11 December 2025 (UTC)

At that massive scale, wouldn’t the upper bound of weight he could detect the buoyancy of the boat - beyond that he is no longer pulling anything up, but pulling himself down - so that the resistance he feels is the buoyancy keeping the boat up? 71.17.36.59 16:18, 11 December 2025 (UTC)

You have to have a 'feel' for the potential elongation of the line (which depends a bit upon knowing how much you've deployed), the properties of the rod and (at least for fish, which weigh 'nothing', when they wish to be neutrally buoyant, but have a muscle-related resistive power and a degree of swim-bladder 'weight adjustment) the pseudo-weight that a given size of likely catch (of a likely species) fights against your line-tug with. You can get fooled that a static-snag is actively fighting you if you think you're only dealing with a fairly shallow fish when you've managed to snag onto something somewhat deeper so that the resonance of your attempts to pull give you the feeling of a reaction that's more just some kind of Young's Modulus/Hooke's Law artefact. It all feels completely different from a fish finally wrested entirely out of the water, and even allows a catch that's heavier than the eventual breaking-stress of your line (or weight+its effort to be greater than that) to be gradually played towards you while in the water, tiring it and bringing it into reach of your grasp/catch-net that will do the actual work of landing it.

Though, in this case, I think that Beret Guy knows how much line he is using/etc, and that he's hooked a planet (however that feels) rather than a more actively resisting (and tiring) seacreature. But by the very subtle reaction of the body to 'test tugs' (accounting for how this also moves the boat) he has correctly (within an order of magnitude!) assessed its 'dry mass'.

(A cynic might ask what other masses he might have caught, between Earth-sized and the next most massive mass that's hookable in this manner... a large part of the Titanic..? Having established that it 'tugs back' more than the latter, then Earth would be the only logical target to lead to such a 'logical guess' of mass. Though even establishing that it isn't a massive shipwreck probably needs extremely well tuned supersenses to rule out. So we can probably still consider it a preternatural Skill Of Beret Guy, even to do that.) 78.144.255.82 19:56, 11 December 2025 (UTC)

Planets don't have a catch size limit. By definition, a planet has cleared its neighborhood, meaning there's no need to maintain a breeding population. Now, dwarf planets and small solar system bodies are a different story, and the rules are rather strict. 209.188.63.98 19:08, 11 December 2025 (UTC)

Beret Guy is surprisingly inaccurate on this one. You'd need to divide the actual weight of Earth by 6 to his estimated weight. --DollarStoreBa'al^Converse 19:23, 11 December 2025 (UTC)

Any reason to believe this is an ocean instead of a lake? I changed the transcript to make it more generic, but like "seafloor" better than "bottom," so would change it back if there is some reason to know it is an ocean. Bobthegoat123 (talk) 19:48, 11 December 2025 (UTC)