HP 12C Programming: Circles, Spheres, and Right Triangle
Links to other HP 12C
Programs:
HP 12C Programming Part I: Modulus, GCD, PITI: http://edspi31415.blogspot.com/2016/07/hp-12c-programming-part-i-modulus-gcd.html
HP 12C Programming Part II:
Weekday Number, Gross Up Calculation:
http://edspi31415.blogspot.com/2016/07/hp-12c-programming-part-ii-weekday.html
HP 12C Programming III:
Refinancing, Advance Payments in a Lease, NPV, NFV, NUS
HP 12C: Combination/Binomial
Distribution/Negative Binomial Distribution
If you want on
to calculate the date of Easter and you have the expanded HP 12C Platinum
Edition:
HP 12C Platinum: Finding the
Day of Easter
Approximating π
The HP-12C does
not have a π key. We can tackle this in
one of two ways:
* We can input
the full approximation of π until the display no longer accepts numbers, which
is up to 10 numbers. π typed to screen
capacity is 3.141592654. Since each
digit entered plus the decimal point takes a step, it will require 11 steps to
enter.
* We can use
the approximation π ≈ 355/113. 355/113 ≈
3.141592920. 355/113 is an accurate
approximation of π to 6 digits. It will
take a total of 8 steps to enter this approximation. Since most of the time the HP 12C is used at
Fix 2 mode (2 decimal places), this may be for most practical purposes an
adequate approximation. Just a
caution: make number of calculations low
and the factors should be relatively small.
The programs
represented on this blog will use the 355/113 to save space. If you require a better approximation of π
and have the space, feel free to replace 355/113 with the 3.141592654.
HP 12C: Circles – Circumference and Area
The program
calculates an approximate circumference and area of a circle given radius r.
C = 2*π*r
A = π*r^2
Here, we take
355/113 as an approximation for π.
|
STEP
|
CODE
|
KEY
|
|
01
|
44, 0
|
STO 0
|
|
02
|
3
|
3
|
|
03
|
5
|
5
|
|
04
|
5
|
5
|
|
05
|
36
|
ENTER
|
|
06
|
1
|
1
|
|
07
|
1
|
1
|
|
08
|
3
|
3
|
|
09
|
10
|
÷
|
|
10
|
44, 1
|
STO 1
|
|
11
|
20
|
*
|
|
12
|
2
|
2
|
|
13
|
20
|
*
|
|
14
|
31
|
R/S
|
|
15
|
45, 0
|
RCL 0
|
|
16
|
2
|
2
|
|
17
|
21
|
Y^X
|
|
18
|
45, 1
|
RCL 1
|
|
19
|
20
|
*
|
|
20
|
43, 33, 00
|
GTO 00
|
Registers used:
R0 = r, R1 =
335/113 ≈ π
Input:
Enter radius,
r, and press [R/S].
Output:
Obtain the
approximate circumference. Press [R/S]
for the area.
Examples (FIX
2):
Radius =
2.96. Results: Circumference ≈ 18.60, Area ≈ 27.53
Radius =
5.00 Results: Circumference ≈ 31.42, Area ≈ 78.54
Alternate: This uses the following shortcuts:
Number,
[ENTER], [ + ] doubles the number.
Number,
[ENTER], [ * ] squares the number.
That and the
use of LST X reduces the number of steps to 19 and only uses one register, R0.
|
STEP
|
CODE
|
KEY
|
|
01
|
44, 0
|
STO 0
|
|
02
|
36
|
ENTER
|
|
03
|
40
|
+
|
|
04
|
3
|
3
|
|
05
|
5
|
5
|
|
06
|
5
|
5
|
|
07
|
36
|
ENTER
|
|
08
|
1
|
1
|
|
09
|
1
|
1
|
|
10
|
3
|
3
|
|
11
|
10
|
÷
|
|
12
|
20
|
*
|
|
13
|
31
|
R/S
|
|
14
|
43, 36
|
LST X
|
|
15
|
45, 0
|
RCL 0
|
|
16
|
36
|
ENTER
|
|
17
|
20
|
*
|
|
18
|
20
|
*
|
|
19
|
43, 33, 00
|
GTO 00
|
Fun fact: A circle of radius 2 will have the same circumference
and area, approximately 12.56637.
HP 12C: Sphere – Surface Area and Volume
This program
calculates the surface area and volume of a sphere give the radius r. Again we take 355/113 as an approximation for
π. The well-known formulas:
S = 4*π*r^2
V = 4/3*π*r^3 =
S * r/3
|
STEP
|
CODE
|
KEY
|
|
01
|
44, 0
|
STO 0
|
|
02
|
2
|
2
|
|
03
|
21
|
Y^X
|
|
04
|
4
|
4
|
|
05
|
20
|
*
|
|
06
|
3
|
3
|
|
07
|
5
|
5
|
|
08
|
5
|
5
|
|
09
|
36
|
ENTER
|
|
10
|
1
|
1
|
|
11
|
1
|
1
|
|
12
|
3
|
3
|
|
13
|
10
|
÷
|
|
14
|
20
|
*
|
|
15
|
31
|
R/S
|
|
16
|
3
|
3
|
|
17
|
10
|
÷
|
|
18
|
45, 0
|
RCL 0
|
|
19
|
20
|
*
|
|
20
|
43, 33, 00
|
GTO 00
|
Registers used:
R0 = r
Input:
Enter radius,
r, and press [R/S].
Output:
Obtain the
approximate surface area. Press [R/S]
for the volume.
Examples:
Radius =
2. Surface area ≈ 50.27, Volume ≈ 33.51
Radius =
8.64. Surface area ≈ 938.07, Volume ≈
2701.65
Fun fact: A sphere of radius 3 will have the same
surface area and volume, at approximately 113.09734.
HP 12C: Right Triangles – Area, Hypotenuse, and Grade
given Rise and Run
Let y be the
rise (height) and x be the run (length) of a right triangle. Then:
Area = 1/2 * x
* y
Hypotenuse = √(x^2
+ y^2)
Grade = y/x *
100% (like slope)
|
STEP
|
CODE
|
KEY
|
|
01
|
44, 1
|
STO 1
|
|
02
|
34
|
X<>Y
|
|
03
|
44, 0
|
STO 0
|
|
04
|
20
|
*
|
|
05
|
2
|
2
|
|
06
|
10
|
÷
|
|
07
|
31
|
R/S
|
|
08
|
45, 1
|
RCL 1
|
|
09
|
2
|
2
|
|
10
|
21
|
Y^X
|
|
11
|
45, 0
|
RCL 0
|
|
12
|
2
|
2
|
|
13
|
21
|
Y^X
|
|
14
|
40
|
+
|
|
15
|
43, 21
|
√
|
|
16
|
31
|
R/S
|
|
17
|
45, 0
|
RCL 0
|
|
18
|
45, 1
|
RCL 1
|
|
19
|
10
|
÷
|
|
20
|
1
|
1
|
|
21
|
26
|
EEX
|
|
22
|
2
|
2
|
|
23
|
20
|
*
|
|
24
|
43, 33, 00
|
GTO 00
|
Registers Used:
R0 = rise (y),
R1 = run (x)
Input: rise [ENTER] run [R/S], height [ENTER] length [R/S]
Output: area of a triangle [R/S], hypotenuse [R/S], grade
Example: rise = 430, run = 1600
Input: 430 [ENTER] 1600 [R/S]
Results: Area: 344000, Hypotenuse: 1656.77, Grade:
26.88 (%)
I hope you find
this helpful. Can you believe it is
already October? How fast time flies,
Eddie
This blog is
property of Edward Shore, 2016.
