The Bully Mnemonic

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Bully Mnemonic Topics:

• The Bully Mnemonic
• The Bully Mnemonic & Calendar Variations

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The Bully Mnemonic is a technique for remembering the exact number of seconds that occur in Earth's sidereal and tropical year; a good approximation of the Earth's Great Year; and a rough approximation of the Solar System's galactic year.

The following relationships are encoded in the Bully Mnemonic:

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

$${1\,Tropical\,Year}={31556926\,Seconds}$$

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

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

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 to form 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 & Tropical Years

Step 4

Multiply integers a) and b) from Step 2 to get the total number of seconds in a sidereal year.

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

Using Long Multiplication:

       3055
×     10330
————————————
   3055
    0000
     9165
      9165
+      0000
————————————
   31558150

Step 5

The tropical year has a slightly shorter duration than the sidereal year. The approximate number of seconds in a tropical year is obtained by reducing integer a) by amount d), and then multiplying by b).

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

The exact number of seconds in a tropical year is obtained by reducing integer a) by amount d), multiplying by b), and then reducing by c).

$$(({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55})-{\color {Red}2}=31556926={\frac {1\,Tropical\,Year}{1\,Second}}$$

Using the Distributive Property of Multiplication:

((10330 - 0.40) × 3055) - 2 = (10330 × 3055) - (0.40 × 3055) - 2
                            =    31558150    -     1222      - 2
                            =    31556926

Great Years

Step 6

The Great Year is, by definition, a least common multiple of the sidereal year and the tropical year. From steps 4 and 5 above, we have that the ratio of tropical years to sidereal years is:

$${\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {(10330-0.40)\times 3055\,sec}{10330\times 3055\,sec}}$$

Divide top and bottom by amount d) and use the Distributive Property of Multiplication to obtain:

$${\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {({\frac {10330}{0.40}}-{\frac {0.40}{0.40}})\times 3055\,sec}{({\frac {10330}{0.40}})\times 3055\,sec}}$$

From whence:

$${\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {(25825-1)\times 3055\,sec}{(25825)\times 3055\,sec}}$$

Consequently:

$${\frac {25825\,Tropical\,Year}{25824\,Sidereal\,Year}}\approx {\frac {25825(25824)\times 3055\,sec}{25824(25825)\times 3055\,sec}}=1$$

Finally:

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

In terms of Long Multiplication; 0.40, 25825, and 10330 are related as follows:

       0.40
×  25825
————————————
   080
    200
     320
      08.0
+      2.00
————————————
   10330.00

Galactic Years

Step 7

Multiply integer c) by the square of integer a) to get a galactic year (the rough approximate number of tropical years required for the Solar System to orbit once around the galactic center).

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

$${\color {Red}2}\times {\color {Red}1}0{\color {Red}33}0\times ({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)=213409536\approx {\frac {1\,Galactic\,Year}{1\,Sidereal\,Year}}$$

Using Long Multiplication:

       10330
×      10330
——————————————
   10330
    00000
     30990
      30990
+      00000
——————————————
   106708900

Using the Distributive Property of Multiplication:

(10330 - 0.40) × 10330 = (10330 × 10330) - (0.40 × 10330)
                       =    106708900    -      4132
                       =    106704768

And finally:

106708900 × 2 = 213417800
106704768 × 2 = 213409536

Step 8

This step is an immediate consequence of steps 6 and 7 above:

$${\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}^{2}\times ({\color {Red}1}0{\color {Red}33}0-0.{\color {Red}4}0)\times {\color {Red}3}0{\color {Red}55}=6.7348101\times {10}^{15}\approx {\frac {1\,Galactic\,Year}{1\,Second}}$$

Summary

The Bully Mnemonic can be represented as follows:

$${\frac {1\,Galactic\,Year}{1\,Second}}\approx (2\times {0.40}\times {10330})\cdot ({\frac {10330}{0.40}})\cdot (({10330}-{0.40})\times {3055})=(8264)\cdot (25825)\cdot (31556928)$$

$${\frac {1\,Galactic\,Year}{1\,Second}}\approx (2\times {0.40}\times {10330})\cdot ({\frac {10330}{0.40}}-1)\cdot ({10330}\times {3055})=(8264)\cdot (25824)\cdot (31558150)$$