The Bully Mnemonic

A mnemonic device (or memory device) is a technique for retaining information.

$$(2\times {0.40}\times {10330})\cdot ({\frac {10330}{0.40}})\cdot ({10330}\cdot {3055})=(8264)\cdot (25825)\cdot (31558150)$$

The Bully Mnemonic is a technique for remembering the exact number of seconds that occur in Earth's sidereal year, and the approximate relationship that exists between Earth's sidereal year, Earth's tropical Year, Earth's Great Year, and the Solar System's galactic year.

Sidereal Years & Galactic Years

Step 1

The first step is to write down the first five digits:

$${\begin{matrix}1&2&3&4&5\end{matrix}}$$

Step 2

The second step is to select odd digits:

$${\begin{matrix}{\color {Red}1}&\scriptstyle {\text{2}}&{\color {Red}3}&\scriptstyle {\text{4}}&{\color {Red}5}\end{matrix}}$$

Step 3

The third step is to intersperse the odd digits with zeros into the two (almost symmetric) integers shown below: (important to remember that the first integer ends with an extra 0, whereas the second integer ends with 5)

$${\begin{matrix}{\color {Red}1}&\scriptstyle {\text{2}}&{\color {Red}3}&\scriptstyle {\text{4}}&{\color {Red}5}\end{matrix}}$$

$${\color {Red}1}0{\color {Red}33}0$$

$${\color {Red}3}0{\color {Red}55}$$

Step 4

Multiply the first integer from Step 3 by itself to get the approximate number of sidereal years required for the Solar System to travel half way around the galactic center.

$${\color {Red}1}0{\color {Red}33}0\times {\color {Red}1}0{\color {Red}33}0=106708900\approx {\frac {{\tfrac {1}{2}}\,Galactic\,Year}{1\,Sidereal\,Year}}$$

Step 5

Multiply the two integers from Step 3 together to get the total number of seconds in a sidereal year.

$${\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}=31558150={\frac {1\,Sidereal\,Year}{1\,Second}}$$

Step 6

This step is an immediate consequence of steps 4 and 5 above:

$${{\color {Red}1}0{\color {Red}33}0}^{3}\times {\color {Red}3}0{\color {Red}55}\approx 3.3675355\times {10}^{15}\approx {\frac {{\tfrac {1}{2}}\,Galactic\,Year}{1\,Second}}$$

Tropical Years & Great Years

Step 7

The seventh step uses the fourth digit and zeros as shown below:

$${\begin{matrix}\scriptstyle {\text{1}}&\scriptstyle {\text{2}}&\scriptstyle {\text{3}}&{\color {Red}4}&\scriptstyle {\text{5}}\end{matrix}}$$

$$0.{\color {Red}4}0$$

Step 8

The tropical year has a slightly shorter duration than the sidereal year. Use the following variation of Step 5 to get the approximate number of seconds in a tropical year.

$$({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55}=31556928\approx {\frac {1\,Tropical\,Year}{1\,Second}}$$

Step 9

The Great Year is, by definition, a least common multiple of the sidereal year and the tropical year. It can be obtained by dividing 10330 by 0.40. This is an immediate consequence of steps 5 and 8 above:

$$1\,Great\,Year\approx 25824\,Sidereal\,Years\approx 25825\,Tropical\,Years$$

Proof:

$$25824\,Sidereal\,Years\approx 25825\,Tropical\,Years$$

$$25824({\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}\,sec)=25825(({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55}\,sec)$$

$$25824({\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}\,sec)=\,{\frac {10330}{0.40}}(({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55}\,sec)$$

$$25824({\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}\,sec)=10330(({\frac {{\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0}{0.40}})\times {\color {Red}3}0{\color {Red}55}\,sec)$$

$$25824({\color {Red}1}0{\color {Red}33}0\times {\color {Red}3}0{\color {Red}55}\,sec)=10330(({25825}-{1})\times {\color {Red}3}0{\color {Red}55}\,sec)$$

Step 10

This step is an immediate consequence of steps 4 and 9 above:

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx ({\color {Red}1}0{\color {Red}33}0\times {\color {Red}1}0{\color {Red}33}0)\,Sidereal\,Years$$

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx ({\color {Red}1}0{\color {Red}33}0\times 0.40)\times ({\frac {{\color {Red}1}0{\color {Red}33}0}{0.40}})\,Sidereal\,Years$$

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx {4132}\times {25825}\,Sidereal\,Years$$

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx {4132}\,Great\,Years$$

Summary and Binary Format

$$({10330}\cdot {0.40})\cdot ({\frac {10330}{0.40}})\cdot ({10330}\cdot {3055})=(4132)\cdot (25825)\cdot (31558150)$$

The following relationships are encoded in the above Bully Mnemonic:

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx {4132}\,Great\,Years$$

$$1\,Great\,Year\approx 25824\,Sidereal\,Years\approx 25825\,Tropical\,Years$$

$${1\,Tropical\,Year}\approx {31556928\,Seconds}$$

$${1\,Sidereal\,Year}={31558150\,Seconds}$$

Binary Format

Substituting the numbers 10330 and 0.40 with binary near equivalents, allows the Bully Mnemonic to be rewritten in a binary format

$$({10330}\cdot {0.40})\cdot ({\frac {10330}{0.40}})\cdot ({10330}\cdot {3055})=(4132)\cdot (25825)\cdot (31558150)$$

The following relationships are encoded in the above Bully Mnemonic:

$${{\tfrac {1}{2}}\,Galactic\,Year}\approx {4132}\,Great\,Years$$

$$1\,Great\,Year\approx 25824\,Sidereal\,Years\approx 25825\,Tropical\,Years$$

$${1\,Tropical\,Year}\approx {31556928\,Seconds}$$

$${1\,Sidereal\,Year}={31558150\,Seconds}$$