The Bully Mnemonic

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

Bully Mnemonic Steps

Initial Definitions

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 and intersperse them with zeros into integers a) and b) as shown below: (important to remember that the first integer ends with an extra 0, whereas the second integer ends with 5)

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

$$a)\,{\color {Red}1}0{\color {Red}33}0$$

$$b)\,{\color {Red}3}0{\color {Red}55}$$

Step 3

The third step is to select even digits and define numbers c) and d) as shown below:

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

$$c)\,{\color {Red}2}$$

$$d)\,0.{\color {Red}4}0$$

Sidereal Years & Galactic Years

Step 3

Multiply integers a) and b) from Step 2 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 4

Multiply integer c) by the square of integer a) to get the approximate number of sidereal years required for the Solar System to orbit around the galactic center.

$${\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}^{2}=213417800\approx {\frac {1\,Galactic\,Year}{1\,Sidereal\,Year}}$$

Step 5

This step is an immediate consequence of steps 3 and 4 above:

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

Tropical Years & Great Years

Step 6

The tropical year has a slightly shorter duration than the sidereal year. Use the following variation of Step 3 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 7

The Great Year is, by definition, a least common multiple of the sidereal year and the tropical year. It can be obtained by dividing a) by d). This is an immediate consequence of steps 3 and 6 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 8

This step is an immediate consequence of steps 4 and 7 above:

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

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

$${{1}\,Galactic\,Year}\approx {8264}\times {25825}\,Sidereal\,Years$$

$${{1}\,Galactic\,Year}\approx {8264}\,Great\,Years$$

Summary

The Bully Mnemonic can be represented as follows:

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

$$(2\times {0.40}\times {10330})\cdot ({\frac {10330}{0.40}})\cdot (({10330}-{0.40})\times {3055})=(8264)\cdot (25825)\cdot (31556928)$$

$$(2\times {0.40}\times {10330})\cdot ({\frac {10330}{0.40}}-1)\cdot ({10330}\times {3055})=(8264)\cdot (25824)\cdot (31558150)$$

The following relationships are encoded in the Bully Mnemonic:

$${{1}\,Galactic\,Year}\approx {8264}\,Great\,Years$$

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

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

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

Julian Years & Gregorian Years

There are exactly 86400 seconds in a day:

$${1\,Day}={24\times 60\times 60\,seconds}=86400\,seconds$$