**fx-260 Solar Algorithms Part II**

**Decimal to Binary Conversions**

This is probably best demonstrated by example.

Algorithm:

To the decimal integer D to binary integer B:

1. Determine the number of digits (zeroes or ones) that the binary integer is going to have. Also, we’ll store D in memory.

n = int(log D/log 2)

Each digit will represent the powers 2^(n) to 2^0.

Keystrokes: D [SHIFT] (Min) [ log ] [ ÷ ] 2 [ log ] [ = ] // ignore the decimal part

2. Starting with n and going to 0, calculate 2^n. Compare 2^n to the number in memory.

If 2^n ≤ Memory, then write a 1. Subtract 2^n from memory: 2^n [ +/- ] [M+]. Decrease n by 1 and continue.

If 2^n > Memory, then write a 0. Decrease n by 1 and continue.

Each digit will be written to the right of the preceding digit.

Example: Convert 462 to binary.

Determine n:

462 [SHIFT] (Min) [ log ] [ ÷ ] 2 [ log ] [ = ]

Result: 8.851749041

Start with n = 8. 462 is stored in Memory.

In M: 462 (n = 8)

2 [ x^y ] 8 [ = ] 256, 256 ≤ 462, [+/-] [ M+ ] // first digit is 1

Binary: 1________

In M: 206 (n = 7)

2 [ x^y ] 7 [ = ] 128, 128 ≤ 206, [+/-] [ M+ ] // next digit is 1

Binary: 11_______

In M: 78 (n = 6)

2 [ x^y ] 6 [ = ] 64, 64 ≤ 78, [+/-] [ M+ ] // next digit is 1

Binary: 111______

In M: 14 (n = 5)

2 [ x^y ] 5 [ = ] 32, 32 > 14 // next digit is 0

Binary: 1110_____

In M: 14 (n = 4)

2 [ x^y ] 4 [ = ] 16, 16 > 14 // next digit is 0

Binary: 11100____

In M: 14 (n = 3)

2 [ x^y ] 3 [ = ] 8, 8 ≤ 14, [+/-] [ M+ ] // next digit is 1

Binary: 1110001___

In M: 6 (n = 2)

2 [ x^y ] 2 [ = ] 4, 4 ≤ 6, [+/-] [ M+ ] // next digit is 1

Binary: 11100011__

In M: 2 (n = 1)

2 [ x^y ] 1 [ = ] 2, 2 ≤ 2, [+/-] [ M+ ] // next digit is 1

Binary: 111000111_

In M: 2 (n = 0)

2 [ x^y ] 01 [ = ] 1, 1 > 0 // last digit is 0

Binary: 1110001110

Result: 462_10 = 1110001110_2

**Combinations that Allow for Repeated Picks**

Sometimes when we are choosing r objects out of a group of n objects, repeated picks are allowed. That is, any object that is picked is put back in the pool and has a chance to be picked again. The formula to calculate such calculations is:

nHr = (n + r – 1)! / (r! * (n – 1)!)

We can state nHr in terms of nCr (number of combinations where no repeats are allowed).

aCb = a! / (b! * (a – b)!)

Let a = n + r – 1 and b = n – 1.

Then a – b = n + r – 1 – (r – 1) = r

Then:

nHr = (n + r -1)C(n – 1)

Algorithm:

[ ( ] n [ + ] r [ – ] 1 [ ) ] [SHIFT] (nCr) [ ( ] n [ – ] 1 [ ) ] [ = ]

Example:

Find the number of combinations of picking 10 objects out of the pool of 38, where repeats are allowed.

n = 38, r = 10

[ ( ] 38 [ + ] 10 [ – ] 1 [ ) ] [SHIFT] (nCr) [ ( ] 38 [ – ] 1 [ ) ] [ = ]

Result: 5,178,066,751

**Harmonic Mean of Numbers**

The harmonic mean of a set of n numbers is calculated by:

HM = n / Σ(1 / x_i)

We can use the Statistics mode to calculate the harmonic mean.

Algorithm:

[ON] // clear everything and reset calculator to COMP mode

[MODE] 0 // Mode 0 is SD mode (single data, standard deviation)

x_i [SHIFT] (1/x) [M+](DATA)

….

[SHIFT] (n) [ × ] [SHIFT] (Σx) [ = ]

Example:

Data: 3.8, 4.6, 5.9, 7.1, 7.6, 9.0 (n = 6)

[ON]

[MODE] 0

3.8 [SHIFT] (1/x) [M+](DATA)

4.6 [SHIFT] (1/x) [M+](DATA)

5.9 [SHIFT] (1/x) [M+](DATA)

7.1 [SHIFT] (1/x) [M+](DATA)

7.6 [SHIFT] (1/x) [M+](DATA)

9.0 [SHIFT] (1/x) [M+](DATA)

[SHIFT] (n) [ × ] [SHIFT] (Σx) [ = ]

Result: 6.20145512

**Atwood Machine**

Given the masses of two weights (in kg) on an Atwood Machine, the following system describes the relationship between the masses, tension, and acceleration of the system:

T + M1 * a = M1 * g

T – M2 * a = M2 * g

where:

T = tension of the system (N)

a = acceleration, positive means the pulley rotates counter-clockwise; negative means the pulley rotates clockwise (m/s^2)

g = Earth’s gravity constant, 9.80665 m/s^2

Assumptions:

1. The mass of both the pulley and the string are negligent

2. Mass 1 is on the left side of the pulley while Mass 2 is on the right.

Solving for T and a give:

a = (M1 – M2) * g / (M1 + M2)

T = M1 * (g – a) = M2 (g + a)

Algorithm:

[ ( ] M1 [ – ] M2 [ ) ] [ × ] 9.80665 [ ÷ ] [ ( ] M1 [ + ] M2 [ ) ] [ = ]

// acceleration is displayed

[ +/- ] [ + ] 9.80665 [ = ] [ × ] M1 [ = ]

// tension is displayed

Example:

M1 = 18 kg, M2 = 12 kg

[ ( ] 18 [ – ] 12 [ ) ] [ × ] 9.80665 [ ÷ ] [ ( ] 18 [ + ] 12 [ ) ] [ = ]

Acceleration: 1.96133 m/s^2 (pulley is rotating counter-clockwise)

[ +/- ] [ + ] 9.80665 [ = ] [ × ] 18 [ = ]

Tension: 141.21576 N

Eddie

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