The Bully Mnemonic

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

What is the Bully Mnemonic?

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

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:

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

Step 2

The second step is to select odd digits:

$$\begin{array}{ccccc}1& \text{2}& 3& \text{4}& 5\end{array}$$

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)

$$\begin{array}{ccccc}1& \text{2}& 3& \text{4}& 5\end{array}$$

$$10330$$ $$3055$$

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.

$$10330\times 10330=106708900\approx \frac{\frac{1}{2}GalacticYear}{1SiderealYear}$$

Step 5

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

$$10330\times 3055=31558150=\frac{1SiderealYear}{1Second}$$

Step 6

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

$${10330}^3\times 3055\approx 3.3675355\times {10}^{15}\approx \frac{\frac{1}{2}GalacticYear}{1Second}$$

Tropical Years & Great Years

Step 7

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

$$\begin{array}{ccccc}\text{1}& \text{2}& \text{3}& 4& \text{5}\end{array}$$

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.

$$(10330-0.40)\times 3055=31556928\approx \frac{1TropicalYear}{1Second}$$

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 an immediate consequence of steps 5 and 8 above:

$$1GreatYear\approx 25824SiderealYears\approx 25825TropicalYears$$

Proof:

$$25824SiderealYears\approx 25825TropicalYears$$

$$25824(10330\times 3055sec)=25825((10330-0.40)\times 3055sec)$$

$$25824(10330\times 3055sec)=\frac{10330}{0.40}((10330-0.40)\times 3055sec)$$

$$25824(10330\times 3055sec)=10330((\frac{10330-0.40}{0.40})\times 3055sec)$$

$$25824(10330\times 3055sec)=10330((25825-1)\times 3055sec)$$

Step 10

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

$$\frac{1}{2}GalacticYear\approx (10330\times 10330)SiderealYears$$

$$\frac{1}{2}GalacticYear\approx (10330\times 0.40)\times (\frac{10330}{0.40})SiderealYears$$

$$\frac{1}{2}GalacticYear\approx 4132\times 25825SiderealYears$$

$$\frac{1}{2}GalacticYear\approx 4132GreatYears$$

Tropical Years & Great Years

Step 7

The seventh step is to select even digits:

$$\begin{array}{ccccc}\text{1}& 2& \text{3}& 4& \text{5}\end{array}$$

Step 8

The eighth step is to organize even digits and zeros into the two numbers shown below:

$$\begin{array}{ccccc}\text{1}& 2& \text{3}& 4& \text{5}\end{array}$$

$$2^{40}$$ $$0.40$$