Bully Mnemonic

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Tropical Year

The Bully Mnemonic is a technique for remembering the exact (eight digits) number of seconds that occur in Earth's sidereal year and tropical year; a good approximation (four digits) 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 S i d e r e a l Y e a r = 31 , 558 , 150 S e c o n d s {\displaystyle {1\,Sidereal\,Year}={31,558,150\,Seconds}}

1 T r o p i c a l Y e a r = 31 , 556 , 926 S e c o n d s {\displaystyle {1\,Tropical\,Year}={31,556,926\,Seconds}}

1 G r e a t Y e a r ≈ 25 , 824 S i d e r e a l Y e a r s ≈ 25 , 825 T r o p i c a l Y e a r s {\displaystyle 1\,Great\,Year\approx 25,824\,Sidereal\,Years\approx 25,825\,Tropical\,Years}

1 G a l a c t i c Y e a r ≈ 213 , 417 , 800 T r o p i c a l Y e a r s {\displaystyle {1\,Galactic\,Year}\approx 213,417,800\,Tropical\,Years}

# Bully Mnemonic Steps

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## Initial Definitions

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### Step 1

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

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

### Step 2

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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 33 followed by a 0, whereas the second integer ends with 55 with no trailing 0)

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

a ) 1 0 33 0 {\displaystyle a)\,{\color {Red}1}0{\color {Red}33}0} b ) 3 0 55 {\displaystyle b)\,{\color {Red}3}0{\color {Red}55}}

### Step 3

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

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

c ) 2 {\displaystyle c)\,{\color {Red}2}} d ) 0. 4 0 {\displaystyle d)\,0.{\color {Red}4}0}

## Sidereal & Tropical Years

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### Step 4

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

1 0 33 0 × 3 0 55 = 31558150 = 1 S i d e r e a l Y e a r 1 S e c o n d {\displaystyle {\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
————————————
0000
9165
9165
0000
3055
————————————
31558150

### Step 5

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The tropical year has a slightly shorter duration than the sidereal year. 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).

( ( 1 0 33 0 − 0. 4 0 ) × 3 0 55 ) − 2 = 31556926 = 1 T r o p i c a l Y e a r 1 S e c o n d {\displaystyle (({\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 = (10330 × 3055) - (0.40 × 3055)
=    31558150    -     1222
=    31556928

## Great Years

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### Step 6

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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:

1 T r o p i c a l Y e a r 1 S i d e r e a l Y e a r ≈ ( 10330 − 0.40 ) × 3055 s e c 10330 × 3055 s e c {\displaystyle {\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:

1 T r o p i c a l Y e a r 1 S i d e r e a l Y e a r ≈ ( 10330 0.40 − 0.40 0.40 ) × 3055 s e c ( 10330 0.40 ) × 3055 s e c {\displaystyle {\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:

1 T r o p i c a l Y e a r 1 S i d e r e a l Y e a r ≈ ( 25825 − 1 ) × 3055 s e c ( 25825 ) × 3055 s e c {\displaystyle {\frac {1\,Tropical\,Year}{1\,Sidereal\,Year}}\approx {\frac {(25825-1)\times 3055\,sec}{(25825)\times 3055\,sec}}}

Consequently:

25825 T r o p i c a l Y e a r 25824 S i d e r e a l Y e a r ≈ 25825 × ( 25824 ) × 3055 s e c 25824 × ( 25825 ) × 3055 s e c = 1 {\displaystyle {\frac {25825\,Tropical\,Year}{25824\,Sidereal\,Year}}\approx {\frac {25825\times (25824)\times 3055\,sec}{25824\times (25825)\times 3055\,sec}}=1}

Finally:

1 G r e a t Y e a r ≈ 25825 T r o p i c a l Y e a r s ≈ 25824 S i d e r e a l Y e a r s {\displaystyle 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
————————————
2.00
08.0
320
200
080
————————————
10330.00

## Galactic Years

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### Step 7

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

2 × 1 0 33 0 2 = 213417800 ≈ 1 G a l a c t i c Y e a r 1 T r o p i c a l Y e a r {\displaystyle {\color {Red}2}\times {{\color {Red}1}0{\color {Red}33}0}^{2}=213417800\approx {\frac {1\,Galactic\,Year}{1\,Tropical\,Year}}}

Using Long Multiplication:

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

And finally:

106708900 × 2 = 213417800