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# fx-260 Solar Algorithms Part II

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

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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