HP 20S: The July 2026 Program Collection

HP 20S: The July 2026 Program Collection

Triangulation

d = l * sin(α) * sin(ß) ÷ sin(α + ß)

Store in the following registers before calculation:

R1 = measure of angle A

R2 = measure of angle B

R3 = length l from point A to B

Solve:

R4: distance

Code:

01: LBL A; 61, 41 ,A

02: DEG; 61, 23

03: RCL 1; 22, 1

04: SIN; 23

05: ×; 55

06: RCL 2; 22, 2

07: SIN; 23

08: ÷; 45

09: (; 33

10: RCL 1; 22, 1

11: +; 75

12: RCL 2; 22, 2

13: ); 34

14: SIN; 23

15: ×; 55

16: RCL 3; 22, 3

17: =; 74

18: STO 4; 21, 4

19: RTN; 61, 26

Examples

Example 1:

Inputs: R1 = 60°, R2 = 50°, R3 = 10

Output: R4: 7.05990377592

Example 2:

Input: R1 = 30°, R2 = 80°, R3 = 27.5

Output: R4: 14.4101446626

Source:

“Triangulation (surveying)” Wikipedia. https://en.wikipedia.org/wiki/Triangulation_(surveying) (last edited October 24, 2025). Retried March 17, 2026

Payment: Continuous Compounding

PMT = PV * (e^r – 1) ÷ (1 – e^(-r * t))

Store in the following registers before calculation:

R1 = t: number of payments

R2 = r: periodic interest rate (as a decimal)

R3 = PV: present value

Solve:

R4: PMT: present

Code:

01: GTO B; 61, 41, b

02: RCL 3; 22, 3

03: ×; 55

04: (; 33

05: RCL 2; 22, 2

06: e^x; 12

07: -; 65

08: 1; 1

09: ); 34

10: ÷; 45

11: (; 33

12: 1; 1

13: -; 65

14: (; 33

15: RCL 1; 22, 1

16: ×; 55

17: RCL 2; 22, 2

18: ); 34

19: +/-; 32

20: e^x; 12

21: ); 34

22: =; 74

23: STO 4; 21, 4

24: RTN; 61, 26

Examples

Example 1:

Input: t: 36, r: 0.10 ÷ 12, PV: 5,000.00

Output: PMT: 161.434037378

Example 2:

Input: t: 60, r: 0.05 ÷ 12; PV: 26,349.56

Output: PMT: 497.374636585

Electrical Engineering: System Temperature to Noise Figure

When an amplifier is activated, two ways to express noise are the noise temperature (in Kelvin) and the noise figure (in decibels, dB). Using a reference temperature of 290 K, when the noise temperature (T) is known, the noise figure (F) can be calculated by:

F = 10 * log((T + 290) ÷ 290)

If temperature is given in degrees Celsius (°C), then the formula for noise figure becomes:

F = 10 * log((T°C + 563.15) ÷ 290)

since T = T°C + 273.15.

The following program assumes that temperature is given in degrees Celsius.

Store in the following registers before calculation:

R1 = T°C; temperature in degrees Celsius

Solve:

R2: F: noise figure

Code:

01: LBL C; 61, 41, C

02: RCL 1; 21, 1

03: +; 75

04: 5; 5

05: 6; 6

06: 3; 3

07 . ; 73

08: 1; 1

09: 5; 5

10: =; 74

11: ÷; 45

12: 2; 2

13: 9; 9

14: 0; 0

15: =; 74

16: LOG; 51, 13

17: ×; 55

18: 1; 1

19: 0; 0

20: =; 74

21: STO 2; 21, 2

22: RTN; 61, 26

Examples

Example 1:

Input: T = -160 °C (store in R1)

Output: F: 1.43068666235 dB

Example 2:

Input: T = 15°C

Output: F: 2.99642532081 dB

Source:

Ball, John A. Algorithms for RPN Calculators John Wiley & Sons: New York. 1978. ISBN 0-47-03070-8. pp. 266-267

Moderate Exercise: Target Heart Rate Range

The following equations calculate the target heart range for a typical person engaging in moderate exercise. According to the particle by Jenna Fletcher (see Source), the American Heart Society (AHA) states the person is engaged in moderate exercise when their heart rate is 50% to 70% of their maximum heart rate. The maximum heart rate is 220 minus the person’s age.

Disclaimer: This is not medical advice, any questions should be discussed with a doctor or health professional.

The moderate range is determined by:

high = (220 – age) * 0.7 = low * 1.4

low = (220 – age) * 0.5 = high * 5/7

For high intensity, use the range 70% to 85%.

Code:

01: LBL D; 61, 41, d

02: +/-; 32

03: +; 75

04: 2; 2

05: 2; 2

06: 0; 0

07: =; 74

08: ÷; 2

09: 2; 2

10: =; 74

11: R/S; 26

12: ×; 55

13: 1; 1

14: . ; 73

15: 4 ; 4

16: =; 74

17: RTN; 61, 26

Examples

Example 1:

Input: Age 49 (my age at the time of this blog)

Output: low: 85.5, high: 119.7

Example 2:

Input: Age 25

Output: low: 97.5, high: 136.5

Source:

Fletcher, Jenna. “What a target heart rate is and how to calculate it”. Medically Reviewed by Debra Sullivan Ph. D. Medical News Today. January 22, 2024. https://www.medicalnewstoday.com/articles/target-heart-rate-calculator March 16, 2026.

Percentile in a Range

The program calculates the percentile of x in the range [a, b].

percentile = (x – a) ÷ (b – a) * 100%

Store before calculating:

R1 = a, R2 = b, R3 = x

Code:

01: LBL E; 61, 42, E

02: ( ; 33

03: RCL 3; 22, 3

04: - ; 65

05 RCL 1; 22, 1

06: ) ; 34

07: ÷; 45

08: ( ; 33

09: RCL 2; 22, 2

10: - ; 65

11: RCL 1; 22, 1

12: ) ; 34

13: ×; 55

14: 2; 2

15: 10^x; 51, 12

16: =; 74

17: RTN; 61, 26

Examples

Example 1:

Input: R1 = 10, R2 = 50, R3 = 30 (30 in [10, 50])

Output: 50 (50%)

Example 2:

Input: R1 = 85.5, R2 = 117.5, R3 = 110 (110 in [85.5, 117.5])

Output: 76.5625 (76.5625%)

Programming all five of these programs will fill the program space of the HP 20S entirely (99 steps)!

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

HP 20S: Acoustics Programs

HP 20S: Acoustics Programs

Program A: Speed of Sound in Dry Air

cs = 20.05 × √(273.15 + T°C)

Code:

01: 61, 41, A: LBL A

02: 75: +

03: 2: 2

04: 7: 7

05: 3: 3

06: 73: .

07: 1: 1

08: 5: 5

09: 74: =

10: 11: √

11: 55: ×

12: 2: 2

13: 0: 0

14: 73: .

15: 0: 0

16: 5: 5

17: 74: =

18: 61, 25: RTN

Program B: Mersenne's Law: Fundamental Frequency

f0 = √(F ÷ µ) ÷ (2 × L)

F: force (N)

µ: mass per unit length (kg/m)

L: string length (m)

f0: fundamental frequency (Hz)

(Marshall, pg. 19)

Store before running:

R1: force

R2: µ

R3: L

Code (continuing from the previous section):

19: 61, 41, b: LBL B

20: 33: (

21: 22, 1: RCL 1

22: 45: ÷

23: 22, 2: RCL 2

24: 34: )

25: 11: √

26: 45: ÷

27: 33: (

28: 2: 2

29: 55: ×

30: 22, 3: RCL 3

31: 34: )

32: 74: =

33: 61, 26: RTN

Program C: Standing Wavelength in an Open Pipe for the 1st through 5th Harmonic

λ = 2 × L ÷ n

L: length of the pipe

n: nth harmonic (positive integer)

(Marshall, pg. 21)

Code (continuing from the previous section):

34: 61, 41, C: LBL C

35: 21, 1: STO 1

36: 1: 1

37: 21, 2: STO 2

38: 61, 41, 3: LBL 3

39: 2: 2

40: 55: ×

41: 22, 1: RCL 1

42: 45: ÷

43: 22, 2: RCL 2

44: 74: =

45: 26: R/S

46: 1: 1

47: 21, 75, 2: STO+ 2

48: 22, 2: RCL 2

49: 31: INPUT

50: 5: 5

51: 61, 42: x≤y?

52: 51, 41, 3: GTO 3

53: 61, 26: RTN

Program D: Finding the perfect length and height for recording studios given the width.

Length = W × φ (stored in R1)

Width = W (stored in R2)

Height = W ÷ φ (stored in R3)

φ = (1 + √5) ÷ 2

(Marshall, pg. 34)

Code (continuing from the previous section):

54: 61, 41, d: LBL D

55: 21, 1: STO 1

56: 22, 2: STO 2

57: STO 3: STO 3

58: 33: (

59: 1: 1

60: 75: +

61: 5: 5

62: 11: √

63: 34: )

64: 45: ÷

65: 2: 2

66: 74: =

67: 21, 55, 1: STO× 1

68: 21, 45, 3: STO÷ 3

69: 22, 1: RCL 1

70: 26: R/S

71: 22, 2: RCL 2

72: 26: R/S

73: 22, 3: RCL 3

74: 61, 26: RTN

HP 20S - The Complete Code:

01: 61, 41, A: LBL A

02: 75: +

03: 2: 2

04: 7: 7

05: 3: 3

06: 73: .

07: 1: 1

08: 5: 5

09: 74: =

10: 11: √

11: 55: ×

12: 2: 2

13: 0: 0

14: 73: .

15: 0: 0

16: 5: 5

17: 74: =

18: 61, 25: RTN

19: 61, 41, b: LBL B

20: 33: (

21: 22, 1: RCL 1

22: 45: ÷

23: 22, 2: RCL 2

24: 34: )

25: 11: √

26: 45: ÷

27: 33: (

28: 2: 2

29: 55: ×

30: 22, 3: RCL 3

31: 34: )

32: 74: =

33: 61, 26: RTN

34: 61, 41, C: LBL C

35: 21, 1: STO 1

36: 1: 1

37: 21, 2: STO 2

38: 61, 41, 3: LBL 3

39: 2: 2

40: 55: ×

41: 22, 1: RCL 1

42: 45: ÷

43: 22, 2: RCL 2

44: 74: =

45: 26: R/S

46: 1: 1

47: 21, 75, 2: STO+ 2

48: 22, 2: RCL 2

49: 31: INPUT

50: 5: 5

51: 61, 42: x≤y?

52: 51, 41, 3: GTO 3

53: 61, 26: RTN

54: 61, 41, d: LBL D

55: 21, 1: STO 1

56: 22, 2: STO 2

57: STO 3: STO 3

58: 33: (

59: 1: 1

60: 75: +

61: 5: 5

62: 11: √

63: 34: )

64: 45: ÷

65: 2: 2

66: 74: =

67: 21, 55, 1: STO× 1

68: 21, 45, 3: STO÷ 3

69: 22, 1: RCL 1

70: 26: R/S

71: 22, 2: RCL 2

72: 26: R/S

73: 22, 3: RCL 3

74: 61, 26: RTN

Examples

LBL A:

Input: T = 0°C: Speed of sound: 331.371367011 m/s

Input: T = 10°C (50°F): Speed of sound: 337.82583835 m/s

Input: T = 26.66666667 (80°F) C: Speed of sound: 347.170058532 m/s

LBL B:

R1: 45 N, R2: 0.008 kg, R3: 0.06 m; f: 625 Hz

R1: 3 N, R2: 0.006 kg, R3: 0.058 m; f: 192.764480819 Hz

LBL C:

Open ended pipe length: 1.2 m

The wavelengths for the 1st, 2nd, 3rd, 4th, and 5th harmonics respectively:

2.4, 1.2, 0.8, 0.6, 0.48

Open ended pipe length: 0.059 m

The wavelengths for the 1st, 2nd, 3rd, 4th, and 5th harmonics respectively:

0.118, 0.059, 3.933333333E-2, 0.0295, 0.0236

LBL D:

Width: 12 ft; Length: 19.416407865 ft, Height: 7.416407865 ft

Width: 18 ft 6 in = 18.5 ft; Length: 29.9336287919 ft, 11.4336287919 ft

Source

Marshall, Steve. Acoustics: The Art of Sound. Wooden Books, LLC. San Rafael, California. 2023. ISBN 978-1-952178-33-7

Eddie

All original content copyright, © 2011-2025. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

The author does not use AI engines and never will.

HP 20S - Normal Distribution, Direction Cosine, Fire Friction Loss

HP 20S - Normal Distribution, Direction Cosine, Fire Friction Loss

HP 20S Normal Distribution

Goal: To estimate the area of the normal curve

∫( e^(-x^2 / 2) dx, a, b) / √(2 * π)

Steps:

1. Enter the programming editor: [ <| ] [ R/S ] {PRGM}

2. Load the integration program: [ <| ] [ ← ] {LOAD} [ e^x ] { B }. The screen shows “int”.

3. Go one up one step to get to Step 58: [ <| ] [ 8 ] { ↑ }

4. Enter the program after 58: 61, 41, F (LBL F)

51, 11	x^2
45	÷
2	2
74	=
32	±
12	e^x
45	÷
33	(
2	2
55	×
61, 22	π
34	)
11	√
74	=

5. Store the lower limit in register 5: a [ STO ] [ 5 ]

6. Store the upper limit in register 6: b [ STO ] [ 6 ]

7. Enter the number of intervals, it must be an even integer, and execute label A: n [ XEQ ] [ √ ] { A }

Example

At n = 20 intervals, estimate areas (ALL setting)

a = 0, b = 3; area ≈ 0.498649878

a = -3, b = 3; area ≈ 0.997293118

a = -1, b = 2; area ≈ 0.818595675

HP 20S: Direction Cosines

The direction cosines of 3D vector v = [x, y, z] are:

a = arccos(x / norm(v))

b = arccos(y / norm(v))

c = arccos(z / norm(v))

where norm(v) = √( x^2 + y^2 + z^2 )

The following program sets the angle mode to degrees, however, a change in the second step will allow the user to use radians or grads instead. The program uses the rectangular to polar conversion to obtain the norm.

Math note:

Find the magnitude of (√(x^2 + y^2), z).

magnitude

= √( [√(x^2 + y^2)]^2 + z^2 )

= √( x^2 + y^2 + z^2)

= norm(v)

Executing the →P command gives the angle first. Obtaining the magnitude requires a swap. ( [ <| ] [ INPUT ] {SWAP} ).

61, 41, b	LBL B
61, 23	DEG (61, 24 for RAD, 61, 25 for GRD)
22, 1	RCL 1
31	INPUT
22, 2	RCL 2
51, 21	→ P
51, 31	SWAP
31	INPUT
22, 3	RCL 3
51, 21	→ P
51, 31	SWAP
21, 4	STO 4
22, 1	RCL 1
41, C	XEQ C
26	R/S
22, 2	RCL 2
41, C	XEQ C
26	R/S
22, 3	RCL 3
41, C	XEQ C
61, 26	RTN
61, 41, C	LBL C (subroutine)
45	÷
22, 4	RCL 4
74	=
51, 24	ACOS
61, 26	RTN

Store x in register 1, y in register 2, and z in register 3. The angles are shown in order

Examples (FIX 4):

x = 4, y = 8, z = 5

Direction Cosines: a ≈ 67.0231°, b ≈ 38.6734°, c ≈ 60.7941°

x = -3, y = 8, z = 6

Direction Cosines: a ≈ 106.6992°, b ≈ 39.9807°, c ≈ 54.9217°

Source:

“Direction Cosine” Wikipedia. Accessed November 5, 2024. https://en.wikipedia.org/wiki/Direction_cosine

HP 20S: Determining the Coefficient for Friction Loss

When fighting fires, the friction loss of a hose lay can be determined by the formula:

FL = C * (flow rate/100)^2 * (hose length/100)

where:

C = coefficient

flow rate = the rate of water in GPM (gallons per minute)

hose length = length of the hose in ft (feet)

FL = friction loss in PSI (pounds per square inch)

This formula assumes a single line is used.

Solving for C:

C = FL / ((flow rate/100)^2 * (hose length/100))

The friction loss was determined by using various flow rates and hose lengths by using the FireCalc Pocket Calculator. You can see my spotlight on the FireCalc Pocket Calculator here: https://edspi31415.blogspot.com/2024/11/spotlight-akron-brass-firecalc-pocket.html

Friction Loss Table:

1” Hose Size

GPM ↓ / Length →	100 ft	150 ft	200 ft
100	150	225	300
150	338	506	675
200	600	900	1200

1.5” Hose Size

GPM ↓ / Length →	100 ft	150 ft	200 ft
100	24	36	48
150	54	81	108
200	96	144	192

2” Hose Size

GPM ↓ / Length →	100 ft	150 ft	200 ft
100	8	12	16
150	18	27	36
200	32	48	64

The coefficient is built in to the FireCalc. I used the HP 20S to extract the coefficient by the following program:

61, 41, A	LBL A
33	(
22, 2	RCL 2
45	÷
1	1
0	0
0	0
34	)
51, 11	x^2
55	×
33	(
22, 3	RCL 3
45	÷
1	1
0	0
0	0
34	)
74	‘=
15	1/x
55	×
22, 1	RCL 1
74	=
61, 26	RTN

Values are stored in the following registers:

Register 1 = friction loss (PSI)

Register 2 = flow rate (GPM)

Register 3 = hose length (ft)

Fortunately, running the program with various data points above, I obtain the coefficient as:

1” Hose Size: coefficient = 150

1.5” Hose Size: coefficient = 24

2” Hose Size: coefficient = 8

Source:

Task Force Tips. “Hydraulic Calculations Every Firefighting Needs to Know” Firefighter Trending Report. 2024. Retrieved November 10, 2024. https://tft.com/hydraulic-calculations-every-firefighter-needs-to-know/

Enjoy!

Eddie

All original content copyright, © 2011-2025. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.