Doubling Numbers – part 3

In  my post on February 10 I predicted that there would be a million cases of coronavirus by March 1. Today, March 6, I am looking at the number of cases outside China reported as 2400 on February 24 and 1200 on February 20. One can use these values to get a rough estimate of doubling of 4 days.

Using the value of M=21K, for today, March 6, yields the equation, n log 2 + log M = 6 log 10.
Or, n = 1/.69 ( 6 log 10 – log 21K) = 5.6 doubling. This rough calculation extrapolates the reported exponential cases.

(%i5) 1/.69*(6*log(10)-log(21000)),numer;
(%o5) 5.598888175737271

With 5.6 as the doubling number, and 4 days for doubling, predicts 5.6 x 4 = 22.4 days for 1 million cases.
The surprising result is that there will be 1 million cases worldwide as of March 28.

This calculation assumes that there are no actions now to halt, contain, or mitigate the spread of the disease.

The equation above intimidates those who might not have aced their ‘A’ levels. It is basic logarithmic identity. log(AB)=log(A)+log(B), so log(2^nM)=n log 2 + log M. This transforms complicated relationships into a simple linear straight line relation.

Nick Strauss.COM

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