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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