On a related note, I have been working on an app that helps determine the correct grinder setting when dialing in espresso. After logging two shots with the same setup (grinder, coffeee machine, basket etc), it then uses machine learning (and some other stuff that I am still improving) to predict the correct setting for your grinder based on the machine temperature, the weight of the shot etc.
Its far from perfect when it comes to predictions right now but I expect to have massive improvements over the coming weeks. For now it works ok as an espresso log at least.
I'm hoping after a few tweaks I can save people a lot of wasted coffee!
Funnily enough I have built essentially the exact same thing in HomeAssistant. Shot collection is completely automated as I have a LM Linea Micra and Acaia Lunar scales (Both have integrations that use Bluetooth). You should consider support for bluetooth scales etc!
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!
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!
It's a mix of course, but I think it should be mainly that and evaporative cooling. Evap is _very_ effective but will fall off rapidly as you get away from boiling. The conduction into the mug will depend a lot on the mug material but will slow down a lot as the mug approaches the water temperature.
I'd be very interested in seeing separate graphs for each major component and how they add up to the total. Even asking the LLMs to separate it out might improve some of their results, would be interesting to try that too.
probably dominated by the cup as the ambient temperature initially and then as air/the counter top as the ambient temperature on the longer time scale, once the cup and the liquid near equilibrium
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.
The problem is both highly complex, but fairly easy to model. Engineers have been doing this for over a century.
Of all the cooling modes identified by the author, one will dominate. And it is almost certainly going to have an exponential relationship with time.
Once this mode decays below the next fastest will this new fastest mode will dominate.
All the LLM has to do, then, is give a reasonable estimate for the Q for:
$T = To exp(-Qt)$
This is not too hard to fit if your training set has the internet within itself.
I would have been more interested to see the equations than the plots, but I would have been most interested to see the plots in log space. There, each cooling mode is a straight line.
The data collected, btw, appears to have at least two exponential modes within it.
[The author did not list the temperature dependance of heat capacity, which for pure water is fairly constant]
You don't need a full model of every atomic interaction because all of those chaotic interactions end up averaging out. Given enough coin flips you will end up on a 50/50 split even if the individual flips are unpredictable. Given enough atomic interactions the heat will transfer in the same way every time.
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.
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.
On a related note, I have been working on an app that helps determine the correct grinder setting when dialing in espresso. After logging two shots with the same setup (grinder, coffeee machine, basket etc), it then uses machine learning (and some other stuff that I am still improving) to predict the correct setting for your grinder based on the machine temperature, the weight of the shot etc.
https://apps.apple.com/ph/app/grind-finer-app/id6760079211
Its far from perfect when it comes to predictions right now but I expect to have massive improvements over the coming weeks. For now it works ok as an espresso log at least.
I'm hoping after a few tweaks I can save people a lot of wasted coffee!
Funnily enough I have built essentially the exact same thing in HomeAssistant. Shot collection is completely automated as I have a LM Linea Micra and Acaia Lunar scales (Both have integrations that use Bluetooth). You should consider support for bluetooth scales etc!
https://i.imgur.com/a5ztsco.jpeg
Missed chance to call it Grind Finr
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!
The appendix lists the equations transcribed from the raw answers.
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!
Is that what that first drop is? The cold cup stealing heat from the coffee?
It's a mix of course, but I think it should be mainly that and evaporative cooling. Evap is _very_ effective but will fall off rapidly as you get away from boiling. The conduction into the mug will depend a lot on the mug material but will slow down a lot as the mug approaches the water temperature.
I'd be very interested in seeing separate graphs for each major component and how they add up to the total. Even asking the LLMs to separate it out might improve some of their results, would be interesting to try that too.
That model doesn't explain the relatively sharp drop in the beginning.
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.
https://www.electronics-tutorials.ws/rc/time-constant.html
scroll down, these graphs just don't look similar.
probably dominated by the cup as the ambient temperature initially and then as air/the counter top as the ambient temperature on the longer time scale, once the cup and the liquid near equilibrium
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.
I mean that initial drop doesn't look like it is part of the same exponential decay.
The problem is both highly complex, but fairly easy to model. Engineers have been doing this for over a century.
Of all the cooling modes identified by the author, one will dominate. And it is almost certainly going to have an exponential relationship with time.
Once this mode decays below the next fastest will this new fastest mode will dominate.
All the LLM has to do, then, is give a reasonable estimate for the Q for:
$T = To exp(-Qt)$
This is not too hard to fit if your training set has the internet within itself.
I would have been more interested to see the equations than the plots, but I would have been most interested to see the plots in log space. There, each cooling mode is a straight line.
The data collected, btw, appears to have at least two exponential modes within it.
[The author did not list the temperature dependance of heat capacity, which for pure water is fairly constant]
" Does that seem hard? I think it’s hard. The relevant physical phenomena include at least"
Imo no, this seems like something that would be in multiple scientific papers so a LLM would be able to generate the answer based on predictive text.
A full model of a cup of water cooling is, in fact, incredibly difficult.
Impossible, since it is chaotic.
But a T(t) model should not be too hard for an LLM with a basic heat transfer book in its training set.
You don't need a full model of every atomic interaction because all of those chaotic interactions end up averaging out. Given enough coin flips you will end up on a 50/50 split even if the individual flips are unpredictable. Given enough atomic interactions the heat will transfer in the same way every time.
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.
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.
Or preheated the mug
... and so another benchmark is born.