Announcement: Calculator Reviews
In the coming weeks, I have acquired a lot of calculators and plan to give a short review of each. They include, the original Hewlett Packard HP 10B, Casio EL-5500 III, and the Calculated Industries Construction Pro/Trig App.
Fun with the HP 12C (I lost count on how segments I done so far)
More fun with the HP 12C! (The HP 12C is on my list of top ten calculators of all time - the other nine I have to think about... subject of a future post?). I like using the HP 12C for a variety of applications and not just strictly finance. Here are four more programs, enjoy!
HP 12C: Rule of 78
When a mortgage, a loan, a lease, or other annuity is paid early, we can determine how much interest rebate is due by the Rule of 78:
Rebate = ( (n – k + 1) * FC ) / ( (n + 1) * n)
Where:
n = the length of the annuity (number of periods)
k = the period where the loan is paid off
FC = total interest, finance charge = PMT * n – PV
The program will require the user to input and compute the annuity variables [ n ], [ i ], [ PV ], and [PMT] ([FV] if a balloon payment is required). Then enter the period # where the loan is paid off (k), and press [R/S].
Program:
Keep in mind: this is done on the HP 12C (regular). For the HP 12C Platinum, the code for Last X is 43, 40 ([ g ] [ + ])
| STEP | KEY | CODE NUMBER |
| 01 | STO 1 | 44, 1 |
| 02 | RCL PMT | 45, 14 |
| 03 | RCL n | 45, 11 |
| 04 | * | 20 |
| 05 | RCL PV | 45, 13 |
| 06 | + | 40 |
| 07 | CHS | 16 |
| 08 | STO 0 | 44, 0 |
| 09 | RCL n | 45, 11 |
| 10 | RCL 1 | 45, 1 |
| 11 | - | 30 |
| 12 | 1 | 1 |
| 13 | + | 40 |
| 14 | * | 20 |
| 15 | RCL 1 | 45, 1 |
| 16 | ÷ | 10 |
| 17 | LST x | 43, 36 |
| 18 | 1 | 1 |
| 19 | + | 40 |
| 20 | ÷ | 10 |
| 21 | GTO 00 | 43, 33, 00 |
Example: On a 48 month purchase of a $20,000 car, financed at 5%, the purchaser pays the car off early after 24 payments (k = 24). What is the rebate?
Output:
Find the payment:
[ f ] [X<>Y] (CLEAR FIN) (if necessary)
48 [ n ]
5 [ g ] [ i ] (12÷)
20000 [ PV ]
[ PMT ] (payment = -460.59)
24 [R/S]
Rebate: $87.84
Source: Rosenstein, Morton. Computing With the Scientific Calculator Casio: Tokyo, Japan. 1986. ISBN-10: 1124161430
HP 12C: Slicing a Right Triangle
The program finds slices a right triangle into equal parts. Using similar triangles, the bases and heights of similar triangles are found.
| STEP | KEY | CODE NUMBER |
| 01 | RCL 0 | 45, 0 |
| 02 | RCL 2 | 45, 2 |
| 03 | ÷ | 10 |
| 04 | INTG | 43, 25 |
| 05 | STO 3 | 44, 3 |
| 06 | 1 | 1 |
| 07 | STO 4 | 44, 4 |
| 08 | RCL 0 | 45, 0 |
| 09 | RCL 3 | 45, 3 |
| 10 | RCL 4 | 45, 4 |
| 11 | * | 20 |
| 12 | - | 30 |
| 13 | R/S | 31 |
| 14 | RCL 1 | 45, 1 |
| 15 | * | 20 |
| 16 | RCL 0 | 45, 0 |
| 17 | ÷ | 10 |
| 18 | R/S | 31 |
| 19 | 1 | 1 |
| 20 | STO+ 4 | 44, 40, 4 |
| 21 | RCL 2 | 45, 2 |
| 22 | RCL 4 | 45, 4 |
| 23 | X≤Y | 43, 34 |
| 24 | GTO 08 | 43, 33, 08 |
| 25 | GTO 00 | 43, 33, 00 |
Input: Pre-store the following values:
Run in Register 0 (R0)
Rise in Register 1 (R1)
Number of partitions in Register (R2)
Output: Loop:
Base of the smaller triangle (x), press [ R/S ]
Height of the smaller triangle (y), press [ R/S ]
Loop ends after n pairs
Example: Run = 5 (R0), Height = 3 (R1), Number of Partitions = 5 (R2)
Output:
| X | 4.00 | 3.00 | 2.00 | 1.00 | 0.00 |
| Y | 2.40 | 1.80 | 1.20 | 0.60 | 0.00 |
HP 12C: Sums of Σx, Σx^2, Σx^3
This program takes two arguments:
Y: x
X: n (where n=1, n=2, n=3)
If n = 1, the sum Σ x from 1 to n is calculated
If n = 2, the sum Σ x^2 from 1 to n is calculated
If n = 3, the sum Σ x^3 from 1 to n is calculated
If n is not 1, 2, or 3, an error occurs.
Program:
Keep in mind: this is done on the HP 12C (regular). For the HP 12C Platinum, the code for Last X is 43, 40 ([ g ] [ + ]) and the step numbers are three digits (000 instead of 00).
| STEP | KEY | CODE NUMBER |
| 01 | X<>Y | 34 |
| 02 | STO 1 | 44, 1 |
| 03 | X<>Y | 34 |
| 04 | STO 0 | 44, 0 |
| 05 | 1 | 1 |
| 06 | - | 30 |
| 07 | X=0 | 43,35 |
| 08 | GTO 21 | 44, 33, 21 |
| 09 | RCL 0 | 45, 0 |
| 10 | 2 | 2 |
| 11 | - | 30 |
| 12 | X=0 | 43, 35 |
| 13 | GTO 29 | 44, 33, 29 |
| 14 | RCL 0 | 45, 0 |
| 15 | 3 | 3 |
| 16 | - | 30 |
| 17 | X=0 | 43, 35 |
| 18 | GTO 46 | 43, 33, 46 |
| 19 | 0 | 0 |
| 20 | ÷ | 10 |
| 21 | RCL 1 | 45, 1 |
| 22 | ENTER | 36 |
| 23 | * | 20 |
| 24 | LST X | 43, 36 |
| 25 | + | 40 |
| 26 | 2 | 2 |
| 27 | ÷ | 10 |
| 28 | GTO 00 | 43, 33, 00 |
| 29 | RCL 1 | 45, 1 |
| 30 | ENTER | 36 |
| 31 | * | 20 |
| 32 | LST X | 43, 36 |
| 33 | X<>Y | 34 |
| 34 | 3 | 3 |
| 35 | * | 20 |
| 36 | + | 40 |
| 37 | RCL 1 | 45, 1 |
| 38 | 3 | 3 |
| 39 | Y^X | 21 |
| 40 | 2 | 2 |
| 41 | * | 20 |
| 42 | + | 40 |
| 43 | 6 | 6 |
| 44 | ÷ | 10 |
| 45 | GTO 00 | 43, 33, 00 |
| 46 | RCL 1 | 45, 1 |
| 47 | ENTER | 36 |
| 48 | ENTER | 36 |
| 49 | 1 | 1 |
| 50 | + | 40 |
| 51 | * | 20 |
| 52 | ENTER | 36 |
| 53 | * | 20 |
| 54 | 4 | 4 |
| 55 | ÷ | 10 |
| 56 | GTO 00 | 43, 33, 00 |
Example: n = 5
Y: 5, X: 1. Result: 15
Y: 5, X: 2. Result: 55
Y: 5, X: 3. Result 225
For an object that travels in a projectile motion, we can track its range (distance traveled from the beginning) and height by:
R = v^2 * sin (2 * θ)/g
H = (v^2 * (sin θ)^2) / (2 * g)
Where:
v = initial velocity
θ = initial angle
g = Earth’s gravity. For in US units, g = 32.1740468 ft/s^2.
This program uses the approximation g ≈ 32.174 ft/s^2
The projectile will have maximum range (distance) if we aim the object at 45°.
--------------------------
Aside: Why?
Let’s let range (R) be a function of angle (θ):
R = v^2/g * sin(2 * θ)
Find the critical points by finding the zero of the first derivative:
dR/dθ = 2 * v^2/g * cos (2 * θ)
0 = 2 * v^2/g * cos (2 * θ)
0 = cos (2 * θ)
arccos 0 = 2 * θ
π/2 = 2 * θ
θ = π/4
Now we can use the second derivative to test whether the function is at a maximum (less than 0) and minimum (more than 0) at the crucial point.
d^2 R/dθ^2 = -4 * v^2/g * sin(2 * θ)
Let θ = π/4
-4 * v^2/g * sin(2 * π/4) = -4 * v^2/g * sin(π/2) = -4 * v^2/g < 0
(We are assuming the initial velocity is positive, and g ≈ 32.174 >0)
Since the second derivative at θ = π/4 is negative, the range is at its maximum.
Note that in calculus, angles are measured in radians. Ï€/2 radians in degrees is 90° and Ï€/4 radians in degrees is 45°. (We are only concentrating on angles between 0° and 90°)
--------------------------
To find the maximum range and height, substitute at θ = 45° and range and height are:
R = v^2 /g
H = v^2 / (4 * g)
The time this certain projectile lasts is:
T = (v * √2) / (2 * g)
Program:
Keep in mind: this is done on the HP 12C (regular). For the HP 12C Platinum, the code for Last X is 43, 40 ([ g ] [ + ])
| STEP | KEY | CODE NUMBER |
| 01 | STO 1 | 44, 1 |
| 02 | 2 | 2 |
| 03 | ÷ | 10 |
| 04 | LST x | 43, 36 |
| 05 | √x | 43, 21 |
| 06 | * | 20 |
| 07 | 3 | 3 |
| 08 | 2 | 2 |
| 09 | . | 48 |
| 10 | 1 | 1 |
| 11 | 7 | 7 |
| 12 | 4 | 4 |
| 13 | STO 0 | 44, 0 |
| 14 | ÷ | 10 |
| 15 | R/S | 31 |
| 16 | RCL 1 | 45, 1 |
| 17 | ENTER | 36 |
| 18 | * | 20 |
| 19 | RCL 0 | 45, 0 |
| 20 | ÷ | 10 |
| 21 | R/S | 31 |
| 22 | 4 | 4 |
| 23 | ÷ | 10 |
| 24 | GTO 00 | 43, 33, 00 |
Input: velocity in ft/s (convert from mph to ft/s by multiplying it by 22/15)
Output:
time of projectile in seconds, [R/S]
range of projectile in feet, [R/S]
height of projectile in feet
Example:
V = 25 mph = 36.6666667 ft/s (110/3)
Output:
Time: 0.81 sec, Range: 41.79 ft, Height: 10.45 ft
Source: Rosenstein, Morton. Computing With the Scientific Calculator Casio: Tokyo, Japan. 1986. ISBN-10: 1124161430
Eddie
This blog is property of Edward Shore, 2017. (2017, wow! 7 days already have passed.)
HP 12C: Rule of 78, Slicing a Right Triangle, Sums, Projectile Motion
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