1

I need to find the constant α and β of the power law formular (d=α*D^(β)).

For that I have the folowing graph: enter image description here

I'm reading an article that says: Knowing that log10(d)= mlog10(D)+b. A power-law is then d=αD^(β) where α=10^(b) and β= m.

For this case, if I set α=10^(0.684), converting b from meters to km, and β= 0.054177, I get a strange result that do not match with the graph.

I don't know what am I doing wrong. Any insights?

The data for the blue line and points are (the data for diameter and depth are in meters):

crater Diameter Depth
    1   9703    1071
    2   11121   1668
    3   11198   1113
    4   12667   1221
    5   12786   1843
    6   13386   2055
    7   13462   903
    8   13821   961
    9   13870   963
    10  14104   1513
    11  14126   1813
    12  14830   1800
    13  16405   1983
    14  18915   1318
    15  19753   2184
    16  21074   1160
    17  22140   2162
    18  24052   1669
    19  24134   1716
    20  28690   1655
    21  29107   2591
    22  33187   3335
    23  33561   2468
    24  34412   2360
    25  34825   2585
    26  43102   2368
    27  49474   3855
Kajo
  • 113
  • If you have the raw data I think Excel (or any of its clones) can fit a power law. – Ethan Bolker Nov 30 '20 at 23:06
  • If they are not too many, could you provide the data points ? – Claude Leibovici Dec 01 '20 at 11:31
  • @ClaudeLeibovici I have added the data for the blue trend. Does this help? Sorry if I am doing something wrong but I am new to all of this. – Kajo Dec 01 '20 at 15:00
  • Is there any chance the software is working in natural logarithms rather than in base $10$? – J.G. Dec 01 '20 at 15:09
  • @ClaudeLeibovici Many thanks for the help. I have just slightly changed the data to match 100% with the graph. But it will be no problem to work with the old one. Many thanks again! – Kajo Dec 01 '20 at 19:49

1 Answers1

2

In order to be in the same conditions, I started defining $$x_i=\frac{\text{Diameter}_i} {1000} \qquad \text{and} \qquad y_i=\frac{\text{depth}_i } {1000}$$

Then, the log-log regression written as $$\log_{10}(y)=a + b \log_{10} x$$ leads to $$a=-1.29342\qquad \text{and} \qquad b=0.62046 \qquad (R^2=0.553897)$$ which for the model $$y =\alpha\, x^\beta$$ gives as estimates $$\alpha_0=10^{-1.29342}=0.0509\qquad \text{and} \qquad \beta_0=0.6205$$ Starting with these values, the nonlinear regression gives $$\alpha=0.255262\qquad \text{and} \qquad \beta=0.652586\qquad (R^2=0.953088)$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 0.255262 & 0.084156 & \{0.081572,0.428952\} \\ \beta & 0.652586 & 0.100473 & \{0.445219,0.859952\} \\ \end{array}$$

Back to the original data $$\left( \begin{array}{cccc} \text{crater} & \text{Diameter} & \text{Depth} & \text{predicted} \\ 1 & 9703 & 1071 & 1125 \\ 2 & 11121 & 1668 & 1229 \\ 3 & 11198 & 1113 & 1235 \\ 4 & 12667 & 1221 & 1338 \\ 5 & 12786 & 1843 & 1347 \\ 6 & 13386 & 2055 & 1387 \\ 7 & 13462 & 903 & 1393 \\ 8 & 13821 & 961 & 1417 \\ 9 & 13870 & 963 & 1420 \\ 10 & 14104 & 1513 & 1436 \\ 11 & 14126 & 1813 & 1437 \\ 12 & 14830 & 1800 & 1483 \\ 13 & 16405 & 1983 & 1584 \\ 14 & 18915 & 1318 & 1739 \\ 15 & 19753 & 2184 & 1789 \\ 16 & 21074 & 1160 & 1866 \\ 17 & 22140 & 2162 & 1927 \\ 18 & 24052 & 1669 & 2034 \\ 19 & 24134 & 1716 & 2038 \\ 20 & 28690 & 1655 & 2282 \\ 21 & 29107 & 2591 & 2303 \\ 22 & 33187 & 3335 & 2509 \\ 23 & 33561 & 2468 & 2528 \\ 24 & 34412 & 2360 & 2569 \\ 25 & 34825 & 2585 & 2589 \\ 26 & 43102 & 2368 & 2976 \\ 27 & 49474 & 3855 & 3256 \end{array} \right)$$

Claude Leibovici
  • 260,315