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I think you're confused about how models work in a scientific sense. They can be used for prediction, and if they couldn't they wouldn't have any value at all.

I think you're confused about how models work in a scientific sense. They can be used for prediction, and if they couldn't they wouldn't have any value at all. Of course they sometimes don't line up with the real world; why do you think there are a metric shit-ton of scientists in Antarctica? In terms of the article, the theory of climate is pretty well understood at this stage, and the models are sound at the macro scale. The hard parts are now in factoring in relatively small changes to make the numbers line up even better, and working through the implications of their predictions... which is what the article seems to be doing.

ars 14y ago
> They can be used for prediction, and if they couldn't they wouldn't have any value at all. Correct - they can't be used for prediction, and therefor they have no value at all (for prediction - they are useful for validation). > Of course they sometimes don't line up with the real world Then you improve them, but you don't try to predict from them. By improving them you increase your knowledge, but you don't predict from them because the only thing that comes out of a model is what you put into it. You learn nothing about a real world from a model. You learn about your understanding of the real world from the model. (They better you know the real world the better you model - the model teach you about your own knowledge. It has zero value in teaching you about the world since nothing can come out of it that you didn't put into it.) > why do you think there are a metric shit-ton of scientists in Antarctica? To do research about the real world? Is this supposed to be a big revelation? Not sure waht you were trying to imply here. > The hard parts are now in factoring in relatively small changes to make the numbers line up even better Well obviously. The details are always the hard part - the details is also where all the interesting stuff happens. You can make the best model in the world, and then miss the one tiny detail that only happens outside the domain of your model, and all the predictions are worthless.
- anthonyb 14y ago
  If what you are saying is true, then every scientific and mathematical formula would be worthless, since after all, you just get out of it what you put into it. e=mc^2? f=ma? v=ir? Yep, no bearing on the real world or any predictive power whatsoever.
  - ars 14y ago
    You are ktizo both have the same misconception. Perhaps this explains your belief in the value of models. Those formulas are not models! They are exact mathematical representations of a phenomena! A model by it's nature can not include everything, they include everything possible of course, but the world is too complex for them to include everything, so they must estimate. If you have a feedback loop with the real world you can tune your model to make it useful, but you can never get out of it anything you did not put in, since it's impossible to include everything. If a model did include everything then of course it would work perfectly. But it's not possible to do that in the real world.
    - anthonyb 14y ago
      Those formulas are models. They describe the world and can predict it's behaviour, but can also break down under certain conditions. For example, what happens when I put 240V across a regular 20Ohm resistor? According to V=IR you get 12A running through it - but I doubt you'll get that for very long. By your argument, V=IR must then be a "model" and is useless, with no predictive power, etc, etc. In practice, the only difference between those formulae and a more complex model is one of degree.
      - ars 14y ago
        No, those formulas are exact, you are just using them incorrectly. You think V=IR describes a resistor, it does not. It describes electricity. So it's not surprising that when applied to a resistor it doesn't work. You are trying to model a resistor, but you are not including everything in your model, so it fails. Which is exactly what I'm warning about. > By your argument, V=IR must then be a "model" and is useless, with no predictive power V=IR is not a model. Using it to describe a physical resistor is a model. Suppose I never touched a real resistor in my life, and the only information I had was an understanding based on some other law that V=IR. Now I want to use that information to predict what a resistor will do. I will fail - among other things it doesn't take into account inductance or capacitance. Which is why it is very very important to know what your equation describes. You thought that V=IR completely describes a physical resistor, when it does not. Now you know why it's incredibly dangerous to try to learn anything from a model. If even something as simple as Ohm's law can be confusing, imagine how many error a model with hundreds of equations has.

ars 14y ago

> They can be used for prediction, and if they couldn't they wouldn't have any value at all. Correct - they can't be used for prediction, and therefor they have no value at all (for prediction - they are useful for validation). > Of course they sometimes don't line up with the real world Then you improve them, but you don't try to predict from them. By improving them you increase your knowledge, but you don't predict from them because the only thing that comes out of a model is what you put into it. You learn nothing about a real world from a model. You learn about your understanding of the real world from the model. (They better you know the real world the better you model - the model teach you about your own knowledge. It has zero value in teaching you about the world since nothing can come out of it that you didn't put into it.) > why do you think there are a metric shit-ton of scientists in Antarctica? To do research about the real world? Is this supposed to be a big revelation? Not sure waht you were trying to imply here. > The hard parts are now in factoring in relatively small changes to make the numbers line up even better Well obviously. The details are always the hard part - the details is also where all the interesting stuff happens. You can make the best model in the world, and then miss the one tiny detail that only happens outside the domain of your model, and all the predictions are worthless.

anthonyb 14y ago
If what you are saying is true, then every scientific and mathematical formula would be worthless, since after all, you just get out of it what you put into it. e=mc^2? f=ma? v=ir? Yep, no bearing on the real world or any predictive power whatsoever.
- ars 14y ago
  You are ktizo both have the same misconception. Perhaps this explains your belief in the value of models. Those formulas are not models! They are exact mathematical representations of a phenomena! A model by it's nature can not include everything, they include everything possible of course, but the world is too complex for them to include everything, so they must estimate. If you have a feedback loop with the real world you can tune your model to make it useful, but you can never get out of it anything you did not put in, since it's impossible to include everything. If a model did include everything then of course it would work perfectly. But it's not possible to do that in the real world.
  - anthonyb 14y ago
    Those formulas are models. They describe the world and can predict it's behaviour, but can also break down under certain conditions. For example, what happens when I put 240V across a regular 20Ohm resistor? According to V=IR you get 12A running through it - but I doubt you'll get that for very long. By your argument, V=IR must then be a "model" and is useless, with no predictive power, etc, etc. In practice, the only difference between those formulae and a more complex model is one of degree.
    - ars 14y ago
      No, those formulas are exact, you are just using them incorrectly. You think V=IR describes a resistor, it does not. It describes electricity. So it's not surprising that when applied to a resistor it doesn't work. You are trying to model a resistor, but you are not including everything in your model, so it fails. Which is exactly what I'm warning about. > By your argument, V=IR must then be a "model" and is useless, with no predictive power V=IR is not a model. Using it to describe a physical resistor is a model. Suppose I never touched a real resistor in my life, and the only information I had was an understanding based on some other law that V=IR. Now I want to use that information to predict what a resistor will do. I will fail - among other things it doesn't take into account inductance or capacitance. Which is why it is very very important to know what your equation describes. You thought that V=IR completely describes a physical resistor, when it does not. Now you know why it's incredibly dangerous to try to learn anything from a model. If even something as simple as Ohm's law can be confusing, imagine how many error a model with hundreds of equations has.