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. π/2 radians in degrees is 90° and π/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.)