New Government Report spells disaster for on-shore Wind Energy

Paul Thomas Rowland

LiamMayo wrote: »

very good. even some of these experts must be starting to realise something is very very rotten in dept of energy. not that they will admit a thing. professional posters some of them.

Experts? Yeah, right.

Kenny, like Putin, has his internet trolls paid to attack anyone who might criticise his regime and point out criminal behaviour by the State.

Realise this, it would be difficult, if not impossible, to show mathematically that Oweninny Wind Farm is a well planned €10billion natural resources fraud by the same people who brought you the Corrib Gas disaster, if institutionalised crime had not been so bloody greedy. They could flog off OPL tomorrow and the first we would hear of it would be when the annual accounts for the relevant reporting period appeared 18 months or 2 years time.

Those who posted on this thread shouting Conspiracy Theory have fled the field with their tails between their legs. Not a whit of data has appeared in support of the fraudsters.

Dermot, what did the guards do? nothing I suppose.

Come on man tell us, did the cops do anything or not. Board Pleanala are supposed to make a decision today.

djpbarry

Paul Thomas Rowland wrote: »

If 1.4m/s is their 90% confidence level, what's the 2.0m/s confidence level? 99% or 99.9%??

You do understand what a confidence interval is, don’t you? Because the above suggests you do not.

Paul Thomas Rowland wrote: »

2.0m/s is, as I wrote, is almost 50% higher than 1.4m/s, is it not? When the state publishes an estimate, with 90% confidence, I expect them to hit the barn door not miss it by a wide margin.

Fine, but 90% confidence is not considered especially high in the scientific community.

Paul Thomas Rowland

djpbarry wrote: »

You do understand what a confidence interval is, don’t you? Because the above suggests you do not.

SEAI, aka the State, are 90% confindent that at 100m a.g. at Oweninny the mean wind speed will be 7.7m/s plus or minus 1.4m/s.

Therefore, the confidence level at plus or minus 2.0m/s must be greater than 90%, significantly greater imo because 2.0 is 43% larger than 1.4. Is there any part of that you do not understand?

Fine, but 90% confidence is not considered especially high in the scientific community.

We are interested in what is going is likely to be produced at Oweninny on average, best guess, not some artful fiction that predicts significantly less production than the old wind atlas.

djpbarry

Paul Thomas Rowland wrote: »

SEAI, aka the State, are 90% confindent that at 100m a.g. at Oweninny the mean wind speed will be 7.7m/s plus or minus 1.4m/s.

Therefore, the confidence level at plus or minus 2.0m/s must be greater than 90%, significantly greater imo because 2.0 is 43% larger than 1.4.

Ok, I think I see what you’re getting at.

What I don’t understand is, what’s your point exactly?

Paul Thomas Rowland

djpbarry wrote: »

Ok, I think I see what you’re getting at.

What I don’t understand is, what’s your point exactly?

The equilibrium of the fluid has shifted considerably.

Based on the new wind atlas,

1. what is the probability of the mean wind speed at 100m a.g. at Oweninny being 9.1m/s in a given year? (Hint: 9.1 = 7.7 +1.4)

2. what is the probability of the mean wind speed being 9.1m/s over a 10 year period at that location? (approximately)

djpbarry

Paul Thomas Rowland wrote: »

Based on the new wind atlas,

1. what is the probability of the mean wind speed at 100m a.g. at Oweninny being 9.1m/s in a given year? (Hint: 9.1 = 7.7 +1.4)

2. what is the probability of the mean wind speed being 9.1m/s over a 10 year period at that location? (approximately)

Without having access to the annual data, it's impossible to say.

Dermot McDonnell

LiamMayo wrote: »

...did the guards do? nothing I suppose.

I will post the letters sent to me by the Gardai as soon as I lay hands on them.

Dermot McDonnell

djpbarry wrote: »

Without having access to the annual data, it's impossible to say.

It is perfectly possible.

Based on the new wind atlas,

what is the probability of the mean wind speed at 100m a.g. at Oweninny being 9.1m/s in a given year? (Hint: 9.1 = 7.7 + 1.4)

Wind speed data may be treated as a gaussian, or normal, probability distribution, graphed as a bell curve with wind speed on the X-axis and probability on the Y-axis. The 90% confidence level represents 90% of the area under the curve, containing 90% of outcomes. The lower bound of that 90% area is at 6.3m/s on the X-axis, 7.7 - 1.4 = 6.3, and the upper bound is at 9.1m/s, 7.7 + 1.4 = 9.1. Thus the areas under the curve, to the left of the lower bound and to the right of the upper bound, each contain 5% of the area for a total area of 100% = 90% + 5% + 5%.

The 2013 SEAI wind atlas predicts that the probability of a mean wind speed of 9.1m/s for any one year is exactly 5%, or .05, or 1 in 20.

what is the probability of the mean wind speed being 9.1m/s over a 10 year period at that location? (approximately)

To achieve a mean wind speed of 9.1m/s over a 10 year period implies that half of those years will exceed 9.1m/s, each at a probability of less than .05, the other years will be less than 9.1m/s. To approximate an answer, we can treat the problem as rolling a fair 20 sided dice, each side having a probability of .05, 10 times and determining the probabilty of any 5 throws comming up 20.

The solution can be written

(10! / (5! x 5!)) x (1/20)^5 x (19/20)^5 = 6.09 x 10^-5% or 0.0000609%

Now consider the probability of a 36 year period in which the mean wind speed is 9.1m/s. It would not be expected to happen in the lifetime of the universe.

The same order of probabilities apply if you turn the question around and ask

Based on the old wind atlas,

What is the probability of the mean wind speed at 100m a.g. at Oweninny being 7.7m/s in a given year? (7.7m/s is the new mean)

What is the probability of the mean wind speed being 7.7m/s over a 10 year period at that location?

What is the probability of the mean wind speed being 7.7m/s over a 36 year period at that location?

The problem for the State is that it absolutely needs a mean wind speed of 7.7m/s at 100m a.g. over the entire 36 year lifetime of Oweninny to achieve a capacity factor as low as 33% at 90m a.g. for the smallest turbine considered in their planning application.

An Bórd Pleanála have yet again decided to postpone their decision on Oweninny until the end of May. They can sit until doomsday, it is quite impossible to escape the numbers nor the sheer scale and sophistication of this fraud. If the Bord had done any due diligence in the first instance, before the 2013 wind atlas arrived, we would not be here having this discussion.

Thank you for your excellent contribution, it is greatly appreciated.

LiamMayo

Dermot McDonnell wrote: »

I will post the letters sent to me by the Gardai as soon as I lay hands on them.

sound. thx

Dermot McDonnell wrote: »

...
The solution can be written

(10! / (5! x 5!)) x (1/20)^5 x (19/20)^5 = 6.09 x 10^-5% or 0.0000609%

Now consider the probability of a 36 year period in which the mean wind speed is 9.1m/s. It would not be expected to happen in the lifetime of the universe....

would you do the calculation for 36 years in case they ask it in statistics and probability next month. haha. i think i got it right but it would be good to know for sure. thx

djpbarry

Dermot McDonnell wrote: »

Wind speed data may be treated as a gaussian, or normal, probability distribution...

That would allow the possibility of negative wind speeds.

Dermot McDonnell wrote: »

The 90% confidence level represents 90% of the area under the curve, containing 90% of outcomes.

Nope – you fundamentally misunderstand what a confidence interval is.

Dermot McDonnell wrote: »

To achieve a mean wind speed of 9.1m/s over a 10 year period implies that half of those years will exceed 9.1m/s...

No it doesn’t.

beeno67

Dermot McDonnell wrote: »

To achieve a mean wind speed of 9.1m/s over a 10 year period implies that half of those years will exceed 9.1m/s, each at a probability of less than .05, the other years will be less than 9.1m/s. To approximate an answer, we can treat the problem as rolling a fair 20 sided dice, each side having a probability of .05, 10 times and determining the probabilty of any 5 throws comming up 20.


.

To be honest I cannot comment on the accuracy of your figures but your extrapolation of them is not quite right. You say:
To achieve a mean wind speed of 9.1m/s over a 10 year period implies that half of those years will exceed 9.1m/s, each at a probability of less than .05, the other years will be less than 9.1m/s.

This is not quite true. You are mistaking median for mean. So to achieve a mean wind speed of 9.1m/s over a 10 year period only implies that at least 1 of those years be above 9.1m/s. All the rest can be below. Therefore your dice analogy doesn't work at all.

fclauson

beeno67 wrote: »

This is not quite true. You are mistaking median for mean. So to achieve a mean wind speed of 9.1m/s over a 10 year period only implies that at least 1 of those years be above 9.1m/s. All the rest can be below. Therefore your dice analogy doesn't work at all.

So median is if you line them up in a row and pick the middle one and mean can also be the average

So to achieve of 9.1 over 10 years you could have some - or many years below 9.1 if you have a stonking high mean one year to offset and bring up the mean to 9.1 but I think if you only have 1 year above 9.1 you would find it hard to have a mean of 9.1

fclauson

djpbarry wrote: »

That would allow the possibility of negative wind speeds.

How come if the distribution is around 9.1 as the mean ?

Can you explain for the less maths literate of us

fclauson

For discussion purposes I have created this graph

7.7 m/s mean
Std 1.4

Paul Thomas Rowland

Dermot McDonnell wrote: »

Wind speed data may be treated as a gaussian, or normal, probability distribution..

djpbarry wrote: »

That would allow the possibility of negative wind speeds.

You manage in a single sentence to expose your complete ignorance of a Gaussian probability distribution. Nobody with any clue at all about the topic would ever make such a statement. As Dirac might have said, it's not even wrong. It is a truly mind boggling assertion conceived at the very heart of the pit of ignorance.

There is a son of Mayo that can help you, Prof Adam Hugh Monaghan. See The Gaussian predictability of wind speeds published in the Journal of Climate or any of the many scientific papers that use this approach. Satisfy yourself that the area under the bell curve is 1, and will contain all of the wind atlas predicitive space. The wind speeds will be positive and each will be assigned a probability.

djpbarry wrote: »

.Nope – you fundamentally misunderstand what a confidence interval is.

What has confidence interval got to do with it? The analysis given in reply to the problem I set makes no mention of it at all. He used the confidence level, 90% as per my post.

To date you have not done a single calculation in support of your position, all you have managed to do is convince readers that you know nothing about the subject matter at all.

Paul Thomas Rowland

beeno67 wrote: »

To be honest I cannot comment on the accuracy of your figures but your extrapolation of them is not quite right. You say:
To achieve a mean wind speed of 9.1m/s over a 10 year period implies that half of those years will exceed 9.1m/s, each at a probability of less than .05, the other years will be less than 9.1m/s.

This is not quite true. You are mistaking median for mean. So to achieve a mean wind speed of 9.1m/s over a 10 year period only implies that at least 1 of those years be above 9.1m/s. All the rest can be below. Therefore your dice analogy doesn't work at all.

Wind speeds are stochastic "having a random probability distribution or pattern that may be analyzed statistically but may not be predicted precisely". When SEAI say a wind farm can expect a mean speed of 7.7m/s, or 9.1m/s, each year that passes may be described as a good, or bad, year relative to the value predicted. As time tends toward infinity, 50% of years will be good and 50% will be bad if the SEAI statement is valid. It is after all how you test the prediction. It's a coin flip whether next year is good or bad, it's 50/50. A coin is a two sided die and SEAI provide a means of making it a 20 sided dice if you can estimate the wind speeds with a probability of .05. So the probability calculation given is a good approximation.

Although is might appear at first glance that you could achieve a 10 year mean with one figure above 9.1m/s and the rest below, as you say, it is not in fact true. You cannot just use any numbers, you must use those under the area bounded by the bell curve in the gaussian distribution because only those wind speeds have a probability greater than zero - the wind atlas will never ever predict wind speed values outside of that area and to test the wind atlas you can only use its predictions.

djpbarry

Paul Thomas Rowland wrote: »

You manage in a single sentence to expose your complete ignorance of a Gaussian probability distribution. Nobody with any clue at all about the topic would ever make such a statement. As Dirac might have said, it's not even wrong. It is a truly mind boggling assertion conceived at the very heart of the pit of ignorance.

So a Gaussian is not defined over the interval -∞ to +∞?

Paul Thomas Rowland wrote: »

There is a son of Mayo that can help you, Prof Adam Hugh Monaghan. See The Gaussian predictability of wind speeds published in the Journal of Climate or any of the many scientific papers that use this approach.

Yeah, that’s the first result I found when I googled “Gaussian Wind Speed” too. Of course, there’s nothing in there that contradicts my point.

Paul Thomas Rowland wrote: »

Satisfy yourself that the area under the bell curve is 1…

…if the distribution is infinitely wide.

Paul Thomas Rowland wrote: »

What has confidence interval got to do with it?

Eh, +/- 1.4 m/s is the confidence interval?

Paul Thomas Rowland wrote: »

To date you have not done a single calculation in support of your position...

In support of what position? That you don’t seem to understand what a Gaussian distribution is? That you’re conflating confidence levels, confidence intervals and predictive probabilities?

Paul Thomas Rowland wrote: »

Although is might appear at first glance that you could achieve a 10 year mean with one figure above 9.1m/s and the rest below, as you say, it is not in fact true. You cannot just use any numbers, you must use those under the area bounded by the bell curve in the gaussian distribution because only those wind speeds have a probability greater than zero...

All wind speeds have a probability greater than zero if you’re describing them with a Gaussian distribution. That probability might be very, very low, but it will be non-zero.

Paul Thomas Rowland

djpbarry wrote: »

So a Gaussian is not defined over the interval -∞ to +∞?

What you said was

djpbarry wrote: »

That would allow the possibility of negative wind speeds.

Take a look at the graph posted by fclauson, there are no negative numbers under the bell curve, are there? There never has been, nor never will be, a gaussian distribution of wind speed probabilites with negative wind speeds. Your statement is just nonesense and your explanation bluster to save face.

djpbarry

Paul Thomas Rowland wrote: »

Take a look at the graph posted by fclauson, there are no negative numbers under the bell curve, are there?

The equation of that curve is...

[latex]\displaystyle g(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}[/latex]

where [latex]\mu = 7.7[/latex] and [latex]\sigma = 1.4[/latex]. If we evaluate that equation at, say, [latex]x = -1.0[/latex], we get...

[latex]\displaystyle g(-1.0)=\frac{1}{1.4\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{-1.0-7.7}{1.4}\right)^2}[/latex]

which gives us...

[latex]\displaystyle g(-1.0)=1.1726 \times 10^{-9}[/latex]

Is that, or is that not, a value greater than zero?

fclauson

What is the point of arguing that there could be negative wind speeds ?

a normal distribution is just that - the probability of -ve wind speed is as you have calculated very low. The data set which bound the distribution will always be positive so although there is a probability of a -ve wind speed it can be discounted - its just a probability.

djpbarry

fclauson wrote: »

What is the point of arguing that there could be negative wind speeds ?

a normal distribution is just that - the probability of -ve wind speed is as you have calculated very low. The data set which bound the distribution will always be positive so although there is a probability of a -ve wind speed it can be discounted - its just a probability.

You're absolutely right. It's a relatively minor point I was making that another poster seems to have taken issue with.

rosie34

Sorry to intrude. I need hard information, please.

My partner and I recently returned to Ireland. We have only heard about this wind farm and have a property nearby.

A friend said the government want one hundred and sixty very very large turbines. Is that correct? What does very very large mean? More than an acre? We want to know the worse case not some watered down story. We are going to have to look at them for a long time or sell the property.

How much electricity will one of these monsters make?

Is it too late to object?

When will we be told the decision?

LiamMayo

rosie34 wrote: »

..A friend said the government want one hundred and sixty very very large turbines. Is that correct?

yes 160 turbines

What does very very large mean?

vesta v112 3.075mw with 56m long blades and hub hight 120m above ground

More than an acre?

pi by 56 by 56 is nearly 10,000 square m about 2 and a half acres each

How much electricity will one of these monsters make?

esb and bord na mona expect 9973mwhours from that vestas but coillte will get about 53% more at 15769mwhours per year

Is it too late to object?

yes

When will we be told the decision?

i do not know. end of may or maybe in a year

beeno67

Paul Thomas Rowland wrote: »

Although is might appear at first glance that you could achieve a 10 year mean with one figure above 9.1m/s and the rest below, as you say, it is not in fact true. You cannot just use any numbers, you must use those under the area bounded by the bell curve in the gaussian distribution because only those wind speeds have a probability greater than zero - the wind atlas will never ever predict wind speed values outside of that area and to test the wind atlas you can only use its predictions.

So you cannot have for example 1 year with wind speed 9.2m/s and 9 years of 9.08? Why not

rosie34

LiamMayo wrote: »

yes 160 turbines

vesta v112 3.075mw with 56m long blades and hub hight 120m above ground

pi by 56 by 56 is nearly 10,000 square m about 2 and a half acres each

Oh My God. 400 acres of turbine blades. omg. It's complete mad.

esb and bord na mona expect 9973mwhours from that vestas but coillte will get about 53% more at 15769mwhours per year

Who do you believe? That a big difference.

It is madness. We will sell if they get planning.

Thank you. I do not mean to shoot the messenger.

LiamMayo

rosie34 wrote: »

Oh My God. 400 acres of turbine blades. omg. It's complete mad.

Who do you believe? That a big difference.

I believe coillte. they have no reason to lie. nobody would believe d of energy around here

It is madness. We will sell if they get planning.

it will go to court anyway

Thank you. I do not mean to shoot the messenger.

sound

Paul Thomas Rowland

beeno67 wrote: »

So you cannot have for example 1 year with wind speed 9.2m/s and 9 years of 9.08? Why not

You can have any scenario you wish so long as you use the numbers under the bell curve. Pick any, it has a probability associated with it. The probability of your scenario is the product of the probabilities. Scenarios populated with values far from the mean are many orders of magnitude less likely than scenarios populated with values close to the mean.

beeno67

Paul Thomas Rowland wrote: »

You can have any scenario you wish so long as you use the numbers under the bell curve. Pick any, it has a probability associated with it. The probability of your scenario is the product of the probabilities. Scenarios populated with values far from the mean are many orders of magnitude less likely than scenarios populated with values close to the mean.

In other words
I was correct in my previous post.
You were incorrect saying 5 had to be greater than 9.1.
You were also wrong when you said I was wrong.

Paul Thomas Rowland

beeno67 wrote: »

...dice analogy doesn't work at all.

You were wrong too. Probability of your scenario is the product of the individual probabilities. Dice analogy is an excellent method.

Paul Thomas Rowland

LiamMayo wrote: »

I believe coillte. they have no reason to lie. nobody would believe d of energy around here

Not to mention the fact that Coillte's production estimates are backed by wind speed measurments on site and all of the scientific data. Extraordinarily, Dept of Energy co-incidently released new guidance for Planning Authorities last year, in secret, that precisely agree with their own planning estimates, just in time to provide An Bord Pleanala with "evidence" in support of their Oweninny planning application.