[HN Gopher] LLMs Predict My Coffee
___________________________________________________________________
LLMs Predict My Coffee
Author : surprisetalk
Score : 37 points
Date : 2026-03-18 13:56 UTC (4 days ago)
(HTM) web link (dynomight.net)
(TXT) w3m dump (dynomight.net)
| amha wrote:
| There's a simple differential equation often taught in intro calc
| courses, "Newton's Law of Cooling/Heating," which basically says
| that the rate of heat loss is proportional to the difference in
| temperature between a substance and its environment. I'm curious
| what that'd look like here. It's a very simple model, of course,
| not taking into account all the variables that Dynomight points
| out, but if a simple model can be nearly as predictive as more
| complex models...
|
| I'm also curious to see the details of the models that
| Dynomight's LLMs produced!
| 3eb7988a1663 wrote:
| The appendix lists the equations transcribed from the raw
| answers. LLM T(t) Cost Kimi K2.5
| (reasoning) 20 + 52.9 exp(-t/3600)+ 27.1 exp(-t/80) $0.01
| Gemini 3.1 Pro 20 + 53 exp(-t/2500) + 27 exp(-t/149.25) $0.09
| GPT 5.4 20 + 54.6 exp(-t/2920) + 25.4 exp(-t/68.1) $0.11
| Claude 4.6 Opus (reasoning) 20 + 55 exp(-t/1700) + 25
| exp(-t/43) $0.61 (eeek) Qwen3-235B 20 + 53.17
| exp(-t/1414.43) $0.009 GLM-4.7 (reasoning) 20 + 53.2
| exp(-t/2500) $0.03
| kurthr wrote:
| It looks like a lot of them are missing something big. I'd
| think the two big ones are the evaporative cooling as you
| pour into the cup, and heating up the cup (by convection)
| itself. The convective cooling to the air is tertiary, but
| important (and conduction of the mug to the table probably
| isn't completely negligible). If there's only one
| exponential, they're definitely doing something wrong.
|
| I'd like to see a sensitivity study to see how much those
| terms would need to be changed to match within a few %.
| Exponentials are really tweaky!
| amelius wrote:
| That model doesn't explain the relatively sharp drop in the
| beginning.
| coder68 wrote:
| It does? There is a fast drop followed by a long decay,
| exponential in fact. The cooling rate is proportional to the
| temperature difference, so the drop is sharpest at the very
| beginning when the object is hottest.
| amelius wrote:
| I mean that initial drop doesn't look like it is part of
| the same exponential decay.
| bryan0 wrote:
| Are you sure? I believe Newtown's law of cooling says the
| temperature will drop sharply at the beginning:
|
| dT/dt = -k _(T_0 - T_room)
|
| so T(t) = T_room + (T_0 - T_room) _ exp(-kt)
|
| exp(-x) has a fast drop off then levels off.
| andy99 wrote:
| Does that seem hard? I think it's hard. The relevant physical
| phenomena include at least..,
|
| In most engineering problems, the starting point is recognizing
| that usually one or two key things will dominate and the rest
| won't matter.
| kaelandt wrote:
| It isn't that surprising that it works well, this problem is
| fairly well known and some simple heat equations would lead to
| the result, about which there is a lot of training data online.
| leecommamichael wrote:
| ... and so another benchmark is born.
| IncreasePosts wrote:
| The water temperature drops quickly because the room temperature
| ceramic mug is getting heated to near equilibrium with the water.
| If you used a vacuum sealed mug(thermos) then the water temp
| would drop a bit but not much at all initially.
___________________________________________________________________
(page generated 2026-03-22 23:00 UTC)