We use quite a bit of energy. In an attempt to reduce our impact on the environment, we are searching for so called renewable power sources as alternatives to nuclear power and fossil fuels. Often we hear about solar power and wind power as serious candidates.
Here in the Netherlands we used to have a commercial in which a child asks her father: “Will the sun ever stop shining?” and “will the winds ever tire?” Of course the message is that sun and wind can be used as limitless sources of energy.
We want to find ways to limit our impact on the climate, but is it sensible then to do that by extracting energy from the dynamics of the atmosphere and even from sunshine, the great engine behind the climate? There are also plans of extracting energy from the sea: there is a lot of energy stored in currents, the tidal movements and even in ordinary waves.
I do not know much (if anything) about the subject, it’s just that I am so curious and always I am scrutinising. The problem I see here is that the climate is a highly chaotic system. Changing the trajectory of such a system through its state space may result in a trajectory towards just about any unforeseen equilibrium. Extracting energy from for example the tides may have little effect at first, but no one knows what will happen after a while.
Either I am not looking in the right places, or there is not much known on this subject, in any case I cannot find much on the Internet. One of the few articles on wind power addressing the question of the impact on the climate (David Keith, (2004) Wind Power and Climate Change) states that large parks of wind mills may have an effect on the local climate. Keith does not seem to be worried by the effect of wind farms on the global climate, however. Another article states that the turbulence generated by the rotors causes different layers of air to mixing (S.B. Roy, et al. (2004) Can large wind farms affect local meteorology?), also P. Rooijmans (2003, 4) did some studies that "show different effects of the wind farm [a large-scale (9000 km2) offshore wind farm in the North Sea] on the meteorological variables, especially for cloud formation."
Also, I am sceptical about parks of solar power plants. These panels have quite different properties (how they warm up and reflect light) than the ground that they are standing on (more here:albedo). For example, when one thinks about the ground of the Sahara desert, typically, there will be quite a lot of reflection of light. Solar panels do not reflect much light; that is the whole idea behind them, so maybe the presence of solar panels in desert areas may cause a local increase of temperature. The planting of crops in these areas (e.g. in Saudi Arabia and in Nevada) may have similar effects: vegetation absorbs sunlight much better than rock and sand. This will cause an increase in regional temperature (and it's quite hot there allready! Furthermore, these crops need water, so there will be displacement of water and an excess of evaporation in normally arid regions.
Now here is my next quest: what kind of power source is not itself part of the climate system AND has no known nonlinear properties (i.e. does not have chaotic behaviour)?
Monday, 30 March 2009
Tuesday, 17 March 2009
Kendo match schedules fair? (rev.)
I have not yet had time to think things over, but I realised that a generalisation of the problem might be more interesting:
What is the probability distribution of X, the number of groups with more than one fellow dojo member, when there are d dojo competing with each md members and all competitors are randomly assigned to any one of g groups, each consisting of n group competitors?
Labels:
probability
Sunday, 15 March 2009
Kendo match schedules fair?
So once in a while I go to a kendo match in a delegation of our dojo. Today, I went to the Ijima-cup tournament; named after Ijima sensei, who has meant a great deal for the development of kendo in the Netherlands. Our dojo sent 8 members to the competition.
There were no less than 29 groups in the first stage of the tournament. All but one group contained three kendoka. To our surprise two of our dojo members were in group number 20. All the rest were divided over different groups, so the rest of us didn't have to fight friends from the dojo.
The reason why it surprised us was that until today, we were much less fortunate. Some began even to suspect that there was some evil scheming at work. I told them that it was probably just a matter of probability. However, I have not yet come to make a model.
So this will be my next 'challenge':
At the moment, I am tired. I still have some preparations to do for tomorrow, so I will not try to figure out the answer just now.
P.S. The tournament went well. I stranded in third round. After a period in which my kendo seemed to get less flexible, it slowed down, cramped up and only felt frustrating, today I did get "in the flow" again and I could do some fluid, strong and fast kendo. Most of our dojo members did well.
There were no less than 29 groups in the first stage of the tournament. All but one group contained three kendoka. To our surprise two of our dojo members were in group number 20. All the rest were divided over different groups, so the rest of us didn't have to fight friends from the dojo.
The reason why it surprised us was that until today, we were much less fortunate. Some began even to suspect that there was some evil scheming at work. I told them that it was probably just a matter of probability. However, I have not yet come to make a model.
So this will be my next 'challenge':
What is the probability distribution of X, the number of groups with more than one fellow dojo member, when m members of our dojo are randomly assigned to any one of g groups, each consisting of n group competitors?
At the moment, I am tired. I still have some preparations to do for tomorrow, so I will not try to figure out the answer just now.
P.S. The tournament went well. I stranded in third round. After a period in which my kendo seemed to get less flexible, it slowed down, cramped up and only felt frustrating, today I did get "in the flow" again and I could do some fluid, strong and fast kendo. Most of our dojo members did well.
Labels:
probability
Thursday, 12 March 2009
The human footprint solved
In this post I will give you what I think is the answer to the question in my previous post (The human footprint).
We wondered how to calculate the average frequency of human sexual behavior (in sexually active adults) given a set of answers to the question "how long has it been since the last time you had sex?" Here is an attempt to find the solution.
To simplify things a bit, we assume that the event (sexual intercourse) occurs perfectly periodic for each individual, with period I (interval). Another simplification is that the period is an integer (discrete stochastic variable) and that the minimal period equals one day. The only data we have are the number of days (D) since the event occurred last. Obviously we try to estimate the expected value of the interval period E (I). If N is the size of the data set (the number of respondents), then the estimated value of the average interval equals:
Bayes states that:
We are able to estimate P(D=d | I = i). Here is how. What we do know is that for a person who has sex every day (I = 1) there will be a 'fifty-fifty' distribution between the answers "just today" (D = 0) and "that was yesterday" (D = 1). Even though we cannot know the value of I for any individual, we do know that for all levels of I the chance of D = d equals P(D = d | I = i) = 1 : (i+1).
This yields the following table:
Distribution of P(D = d | I = i) | |||||||
---|---|---|---|---|---|---|---|
D number of days since last event | |||||||
0 | 1 | 2 | 3 | 4 | ... | ||
I (interval between events) | 1 | 1/2 | 1/2 | 0 | 0 | 0 | ... |
2 | 1/3 | 1/3 | 1/3 | 0 | 0 | ... | |
3 | 1/4 | 1/4 | 1/4 | 1/4 | 0 | ... | |
... |
To estimate P (I = i) we have to know P (D = d ∧ I = i). At first I had hoped to use the above table for this, but we cannot use it directly. We would like to estimate P (D = d ∧ I = i) with our given set of nd and P (D = d | I = i), but we cannot use the table here because if we allow I to be unbound then:
P (D = d ∧ I = i) remains unknown. Note, however that for all levels of i (with d < i) all P(D = d ∧ I = i) are equal, so P(D=0∧I=i)=P(D=1∧I=i)=P(D=2∧I=i). This means that the actual number in cell (d, i) should be about equal to the numbers is the non empty cells in the same row (with the same level of I). I will call the estimated number in the non empty cells of the first row .
Now we are getting somewhere; we do know is nd, the number of people in the sample who answered D = d. We would expect n0 ≈ n1. The important thing is that , et cetera.
We can now estimate .
In general a (crude) estimation of n(d,i) would be
.
Now we know enough:
thus
This estimation is very crude indeed. At this stage I have not tried to integrate the information of ni+2, ni+3, and so forth in the estimation of n(d,i).
Labels:
answers,
human behaviour,
mathematics,
probability
Sunday, 8 March 2009
The human footprint
Some time ago I saw the British documentary The Human Footprint. This documentary shows u how much milk an average British person drinks, how many chickens they consume, the total length of fingernails such a person grows during their lifespan and how often they have sex.
About this last topic an interesting remark was made: researchers don't just ask “how often do you have sex?” to a large group of individuals, that would give very unreliable results. To estimate the average frequency of sexual activity in British adults, the researchers asked: “how long ago did you last have sex?”
How does one calculate the average frequency of sexual activity on the basis of data on time since last activity? For example, if a man answers “the day before yesterday”, we know he doesn’t have sex every day. But we do not know whether he has sex about every two days, once a week, or once a month. It may even be a once-a-lifetime experience for the man.
Because of this asymmetry, one cannot simply assume that the given length of time is half way the average inter-intercourse period. If I want to learn how to calculate average frequency of sexual activity (the average occurrence of volcanic eruptions, or other semi-periodic events), I need to find out about waiting time theory. It is a topic I have not had to deal with since 1994; luckily I still have most of the university books I used.
About this last topic an interesting remark was made: researchers don't just ask “how often do you have sex?” to a large group of individuals, that would give very unreliable results. To estimate the average frequency of sexual activity in British adults, the researchers asked: “how long ago did you last have sex?”
How does one calculate the average frequency of sexual activity on the basis of data on time since last activity? For example, if a man answers “the day before yesterday”, we know he doesn’t have sex every day. But we do not know whether he has sex about every two days, once a week, or once a month. It may even be a once-a-lifetime experience for the man.
Because of this asymmetry, one cannot simply assume that the given length of time is half way the average inter-intercourse period. If I want to learn how to calculate average frequency of sexual activity (the average occurrence of volcanic eruptions, or other semi-periodic events), I need to find out about waiting time theory. It is a topic I have not had to deal with since 1994; luckily I still have most of the university books I used.
Labels:
human behaviour,
mathematics
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