Bully Mnemonic Extension

The Bully Mnemonic Extension is a technique for remembering a the exact number of meters that light travels in one second, and the approximate range of gravitational accelerations that occur on the surface of the Earth due to Newton's law of universal gravitation.

The following relationships are encoded in the Bully Mnemonic Extension:

$$Speed\,of\,light\,in\,vacuum={299,792,458\,meters\,per\,second}$$

$${High\,end\,of\,g_{Earth}}={9.8624\,meters\,per\,second^{2}}$$

$${Low\,end\,of\,g_{Earth}}={9.7644\,meters\,per\,second^{2}}$$

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

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

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

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

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
————————————
       0000
      9165
     9165
    0000
   3055
————————————
   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 = (10330 × 3055) - (0.40 × 3055)
                      =    31558150    -     1222
                      =    31556928

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:

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

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
————————————
       2.00
      08.0
     320
    200
   080
————————————
   10330.00

Galactic Years

Step 7

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).

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

Using Long Multiplication:

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

And finally:

106708900 × 2 = 213417800