Lines, Sines, and Curve Fitting 9 – Girma
y = (m*(x-1880) + b) + (A * cos(((x-1880)/T)*(2*pi)))
intercept (1880) b = -0.53
slope m = 0.0059 C / yr
amplitude A = 0.3 C
period T = 60 years
Which he displays as such:
I’ve recharted Dr Orssengo’s graph as follows:
Dr Orssengo has commented here that his correlation is 0.88. However, using annualized HadCRUv3 data, the correlation I calculate from 1880-2009 is a bit lower at 0.87. However, just reducing the amplitude of the cosine to 0.25C increases the correlation to 0.89. I suspect that Girma fit his model by hand and did not seek to optimize the values used.
Dr Orssengo has put a lot of emphasis on the correlation of his model, and has allowed that if an equally simple model had a higher correlation that it would be a better model. I now present a simple model based on the same data with a higher correlation.
First a simple exponential trend.
y1 = -a/100 + b/100 * exp((k/10000)*(x-1880))
a <- 44.40
b <- 6.02
k <- 211
This model has a correlation of 0.89 for HadCRU, but 0.90 for NCDC, and 0.91 for GISTEMP all through 2009. Since HadCRU is the data set that Dr Orssengo used, we will judge on the correlation against it. And 0.89 is higher than the 0.88 that Girma has calculated for his model or the 0.87 that I calculated for it.
Based on the criteria that the best correlation makes the best model, my exponential warming model beats Dr Orssengo’s line+sine model.
Will the model hold … no, it will probably not. That will be a whole bunch of warming if it does!
To be fair, Dr Orssengo points out that the future is a better filter for selecting models than correlation. I’ll address that – with a slightly better model-of-fit – in a couple of days.
And one more note. While curve fitting is fun, divorced of a physical explanation, they aren’t much more than fun. What drives the linear (or exponential) trend? What drives the 60 year sine? Without knowing these, the mathematical model is a sterile construct.