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I would like to know if there is any algorithm which allows us to calulate precipation chance with following data: temperature, humidity, illuminance (in lux) and pressure. I've searched it in google, but no results satisfied me.

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    $\begingroup$ Could you maybe add results you did find, which did not satisfy you, or at least summarize what those results did tell, such that people trying to answer your question have an idea how much detail you are expecting. Your current description also only seems to ask about whether such an algorithm exists not about the physics/models involved. $\endgroup$ – fibonatic Nov 23 '14 at 23:45
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Simple climate models use the fact that the total amount of water is conserved, and assume that there are three processes that affect it: evaporation, precipitation, and moisture flux (i.e. how the moisture already in the atmosphere moves around). We can write the moisture flux vector simply as the product of the velocity vector and specific humidity: $\mathbf{F_M}=q\mathbf{v}$, and vertically integrated gives us total column water transport: $$ \mathbf{F}=\int_0^\infty \rho \mathbf{F_M}dz $$ where $\rho\equiv\rho(z)$ is the density of the air at a given height, $z$. So, the rate of change of precipitable water in the atmosphere can be written as the sum of the contributions from evaporation, precipitation, and column moisture flux convergence, hence: $$ \frac{\partial}{\partial t}\int^\infty_0 \rho qdz = \text{E}-\text{P} -\nabla\cdot \mathbf{F} $$ So an estimate for precipitation can be derived from surface evaporation rates (ignoring complicated transpiration terms from vegetation, this is essentially a function of sea surface temperature) and functions of specific humidity.

Insmaller scale models, things get significantly more complicated - depending on the scale, one must consider implicit convection terms and/or cloud microphysics as well.

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While Kieran Hunt's answer is very good, a more blunt answer is that no, there is not such an algorithm. The easiest way to see that there is not consider not physics but economics.

There are a large number of economically-important events which are crucially dependent on whether or not it rains -- as an example when wheat is harvested its moisture content makes a large difference to its price and this depends on whether it has rained in the immediately preceding period to the harvest among other factors. Insurance companies and financial markets are therefore very interested in the chance of rain in a given period and good estimates of this are very valuable. They would pay very well for such estimates.

Well, one of the things that weather-forecasting places try to do is produce such estimates. And the figures on how well they do and how well they have done in the past are available and they have two important characteristics: they don't do very well, but they are getting significantly better over time.

They are getting better for three reasons: better and much more detailed observational data, principally from satellites; better numerical models of weather; enormously more computing power to run these models.

And this really answers the question: the kind of model (algorithm) which can make a useful prediction of the chance of precipitation relies both on an enormous amount of observational data (far more than that you specified) and a really enormous amount of computer time. These are very expensive things: the Met Office in the UK recently spent £$10^8$ on a new computer for instance.

And finally we can put these together: if a simple algorithm was known which was any good at all then anyone using it could make a huge amount of money. Therefore no such algorithm is known.

Well I've been careful above and said that no such algorithm is known rather than making the claim that no such algorithm exists. However it is extremely unlikely that such an algorithm exists: the weather is a very complicated nonlinear system which has sensitive dependence on initial conditions: such systems are notoriously hard to predict in any useful way and there is no reason to think that the weather is any different.

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