Casio fx-50F/Radio Shack EC-4024 Program Collection: October 2026

Casio fx-50F/Radio Shack EC-4024 Program Collection: October 2026

Notes:

* The fx-50F/EC-4024 has a 29 step program space between two program spaces. Program steps can not be edited but can be cleared by [SHIFT] [ C ] (PCL) key sequence.

* Formula calls can not be included as a program step, but scientific constants can be. The values for the constants most likely are from the Japan Standards of 1978 (JISZ-8020-1978), so a little dated to 2026 standards.

* The Casio fx-10F is similar as fx-50F/EC-4024 except that the fx-10F does not have the scientific constants and has a slightly inclined keyboard.

* Mode changes can be included. SHIFT and Kin arithmetic calls are merged.

The programs presented today require inputs to be stored in registers 1-6 using [ SHIFT ] [ Kout ] (Kin) ((K)onstants In).

Angle Between 2 Lines (in Degrees)

Θ = arctan( abs((y – x) ÷ (1 + x * y)) )

abs(x) = √(x^2), key sequence: [ x² ] [ √ ]

x: slope of line 1 (Kin 1)

y: slope of line 2 (Kin 2)

Program Code (20 steps):

MODE 4 (degree)

(

(

Kout 2

-

Kout 1

)

÷

(

1

+

Kout 1

×

Kout 2

)

)

x²

√

TAN^-1

=

Maximum and Minimum of Two Numbers

The two numbers are stored in registers 1 and 2 (Kin 1, Kin 2).

max(x,y) = (x + y + abs(x – y)) ÷ 2 = min(x,y) + abs(x – y)

min(x,y) = (x + y - abs(x – y)) ÷ 2 = max(x,y) - abs(x – y)

Program Code (21 steps):

Kout 2

-

Kout 1

=

x²

√

Kin 3

(

Kout 1

+

Kout 2

+

Kout 3

)

÷

2

=

HLT (show maximum)

-

Kout 3

= (show minimum)

Terminal Velocity of a Spherical Object

vt = √(2 * g * mass ÷ (ρ * cd * exposed area of the object) (units: m/s)

≈ √(34.05656 * mass ÷ (π * radius^2))

The following constants are assumed:

g = 9.80665 m/s² (Earth’s gravitational constant)

ρ ≈ 1.225 kg/m³ (air pressure)

cd = 0.47 (coefficient drag of a sphere)

2 * g ÷ (ρ * cd) ≈ 34.06565 m^4/(s^2 kg)

Program Code (19 steps):

Kout 1

×

3

4

.

0

6

5

6

5

÷

(

π

×

Kout 2

x²

)

=

√

=

Ideal Thermal Voltage and Shockley Diode Equation

Thermal Voltage:

vt = k * T ÷ q * 10^3 (mV)

Shockley Diode Equation:

Id = Is * (exp(vD ÷ (n * vt)) – 1)

Ideal: n = 1

Inputs:

Is: scale current in milliamps (mA) Kin 1

Vd: diode voltage in millivolts (vA) Kin 2

T: temperature in degrees Kelvin (K) Kin 3

Outputs:

vt: Thermal Voltage (mA),

Id: diode current (mV)

Constant used (see comments on the scientific constants above):

k: Boltzmann Constant ([SHIFT] [ 8 ]): 1.380662 * 10^-23 J/K

(q) e: Electron Charge ([SHIFT] [ 4 ]): 1.6021892 * 10^-19 C

Program Code (21 steps):

Kout 3

×

k (SHIFT 8)

÷

e (SHIFT 4)

×

3

10^x

=

HLT (see vt: thermal voltage)

1/x

×

Kout 2

=

e^x

-

1

=

×

Kout 1

=

Source:

“Shockley Diode Equation” Wikipedia. Last Edited November 13, 2025.

https://en.wikipedia.org/wiki/Shockley_diode_equation

Retrieved June 19, 2026.

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.

Python (TI-84 Plus CE) and Swiss Micros DM32: The Integral of y = abs(r * x + s)

Python (TI-84 Plus CE) and Swiss Micros DM32: The Integral of y = abs(r * x + s)

The Integral of y = abs(r * x + s)

This algorithm calculates the integral of ∫ abs(r * x + s) dx, from x = a, x = b), where r and s are constants. For clarity, I am using abs to stand for absolute value instead of the customary pipe characters (|x|).

Let the function y = abs(r * x + s). Then the function can be defined as a piecewise function (without loss of generality):

y =

{ -(r * x + s), x < xc

{ (r * x + s), x ≤ xc

The point x = xc is the critical point because it is the root (zero) of this function:

abs(r * x + s) = 0

Because abs(0) = 0:

r * x + s = 0

r * x = -s

x = -s/r

and:

-(r * x + s) = 0

r * x + s = 0

x = -s/r

Let the critical point xc = -s/r

Taking the indefinite integral of y(x) yields:

∫ y(x) dx =

{ -r * x^2 ÷ 2 – s * x + C, x < xc

{ r * x^2 ÷ 2 + s * x + C, x ≥ xc

and C is an arbitrary integration constant.

Let f(x) = r * x^2 ÷ 2 + s * x and find the definite integral from x = a to x = b.

Case 1: a ≥ xc and b ≥ xc, where both a and b are greater than the critical point. This is the simplest case.

∫ ( r * x + s dx, x = a to x = b)

= (r * b^2 ÷ 2 + s * b) - (r * a^2 ÷ 2 + s * a)

= f(b) – f(a)

Case 2: a < xc and b < xc, both a and b are less than the critical point.

∫ ( r * x + s dx, x = a to x = b)

= -(r * b^2 ÷ 2 + s * b) - -(r * a^2 ÷ 2 + s * a)

= -(r * b^2 ÷ 2 + s * b) + (r * a^2 ÷ 2 + s * a)

= (-r * b^2 ÷ 2 - s * b) + (r * a^2 ÷ 2 + s * a)

= -f(b) + f(a)

= -(f(b) - f(a))

Combining cases 1 and 2, the area can be calculated as:

area = abs(f(b) – f(a))

with (a – xc) * (b – xc) ≥ 0

Case 3: a < xc and b ≥ xc

∫ ( r * x + s dx, x = a to x = b)

= ∫ ( -(r * x + s) dx, x = a to x = xc) + ∫ ( r * x + s dx, x = xc to b)

= -(r * xc^2 ÷ 2 + s *xc) + (r * a^2 ÷ 2 + s * a) + (r * b^2 ÷ 2 + s * b) – (r * xc^2 ÷ 2 + s * xc)

= -f(xc) + f(a) + f(b) – f(xc)

= f(a) – 2 * f(xc) + f(b)

Since area must be positive: abs(f(a) – 2 * f(xc) + f(b)).

Consequently: (a – xc) * (b – xc) < 0.

In summary:

Let xc = -r/s

If (a – xc) * (b – xc) ≥ 0: area = abs(f(b) – f(a))

Else if (a – xc) * (b – xc) < 0: area = abs(f(a) – 2 * f(xc) + f(b))

where f(x) = r * x^2 ÷ 2 + s * x

Please note: ∫ abs(r * x + s) dx ≠ abs(a * x^2 ÷ b * x)

TI-84 Plus CE Python Edition: abslin1.py

Programmed with TI-84 Plus CE Python, but can be used on any calculator with Python since only the math module is used.

# Math Calculations
from math import *

# Python Version
# 2026-01-05 EWS

print("integral of abs(rx+s)")
r=eval(input("r? "))
s=eval(input("s? "))
a=eval(input("lower limit? "))
b=eval(input("upper limit? "))

# critical point
c=-s/r

# integral
f=lambda x:r*x**2/2+s*x
f0=f(c)
f1=f(a)
f2=f(b)

if (a-c)*(b-c)>=0:
  t=abs(f2-f1)
else:
  t=abs(f1-2*f0+f2)

print("area = ",str(t))

Swiss Micros DM32 Program: asblin

Three labels are used: A (172 bytes), Z (20 bytes), Y (17 bytes), total 209 bytes

Text strings can be eliminated.

A01 LBL A

A02 SF 10

A03 “AREA ABS(RX +S)”

A04 INPUT R

A05 INPUT S

A06 x<>y

A07 ÷

A08 +/-

A09 STO C

A10 XEQ Y

A11 STO D

A12 “LOW=A HIGH=B”

A13 INPUT A

A14 XEQ Y

A15 STO E

A16 INPUT B

A17 XEQ Y

A18 STO F

A19 RCL B

A20 RCL- C

A21 RCL A

A22 RCL- C

A23 ×

A24 x≥0?

A25 GTO Z

A26 RCL E

A27 RCL D

A28 2

A29 ×

A30 -

A31 RCL+ F

A32 ABS

A33 STO Z

A34 CF 10

A35 RTN

Z01 LBL Z

Z02 RCL E

Z03 RCL- F

Z04 ABS

Z05 STO Z

Z06 CF 10

Z07 RTN

Y01 LBL Y (Note: f(x) = r*x^2 ÷ 2 + s*x)

Y02 ENTER

Y03 x^2

Y04 RCL× R

Y05 2

Y06 ÷

Y07 x<>y

Y08 RCL× S

Y09 +

Y10 RTN

Examples

Example 1:

y = abs(4 * x + 3)

r = 4, s = 3, xc = -0.75

Lower Limit (a)	Higher Limit (b)	Area
-4	5	87.25
-4	-1	21
0	5	65

Example 2:

y = abs(-3 * x + 6)

r = -3, s = 6, xc = 2

Lower Limit (a)	Higher Limit (b)	Area
-5	5	87
3	5	12
-5	1	72

Hope you find this helpful and have a great day,

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.

RPN: Absolute Value Equations with the HP 11C and DM41X

RPN: Absolute Value Equations with the HP 11C and DM41X

Introduction: Solving |z * w + y| = x

Today’s blog focuses on solving the absolute value equation:

|z * w + y| = x

for the variable w, and the values of z, y, and x are given and are on the Classic RPN stack.

For example, in the problem |5 * w + 7| = 2, the stack would be set up as:

t: (anything, it doesn’t matter)

z: 5

y: 7

x: 2

(This is why I set the variable as w instead of the usual x.)

One approach is to use memory registers and other uses only stack operations. Today’s algorithm focuses on the latter.

Caution: In the above equation, x will always be non-negative. The equation will never be valid if x is negative.

The Algebra

Solve for w:

|z * w + y| = x

This leads us to solve the two equations:

(I)

z * w + y = -x

z * w = -x – y

w = -x/z – y/z

Let w- = -x/z – y/z = (-x/z) + (-y/z)

(II)

z * w + y = x

z * w = x – y

w = x/z - y/z

Let w+ = x/z – y/z = (x/z) + (-y/z)

Then:

w- = -x/z – y/z

w- = (-x/z) + (-y/z)

w- = (-x/z) + (-y/z) + (-x/z) + (x/z)

w- = 2 * (-x/z) + (x/z) + (-y/z)

w- = 2 * (-x/z) + w+

RPN Code: HP 11C (adoptable for other RPN calculators)

LBL A	001	42, 21, 11	Program start
R↓	002	33
R↓	003	33
X<>Y	004	34
R↓	005	33
1/x	006	15
×	007	20
LST x	008	43, 36
X<>Y	009	34
R↓	010	33
×	011	20
LST x	012	43, 36
R↑	013	43, 33
X<>Y	014	34
R↓	015	33
X<>Y	016	34
CHS	017	16
X<>Y	018	34
+	019	40
ENTER	020	36
ENTER	021	36
LST x	022	43, 36
-	023	30
LST x	024	43, 36
-	025	30
RTN	026	43, 32	Program end

RPN Code: DM41X (HP 41C series, no module is required)

LBL “ASBEQ”

RDN

RDN

X<>Y

RDN

1/x

×

LAST X

X<>Y

RDN

×

LAST X

R↑

X<>Y

RDN

X<>Y

CHS

X<>Y

+

ENTER

ENTER

LAST X

-

LAST X

-

RTN

Notes:

* RDN is shown as R↓

* To enter R↑, press XEQ, ALPHA, R, SHIFT, ENTER, ALPHA

Subroutines Used

We used several techniques to manipulate the stack. They are presented below:

Rotate stack from X, Y, Z, T to Z, X, Y, T

R↓

R↓

X<>Y

R↓

Source:

Ball, John A. Algorithms for RPN Calculators John Wiley & Sons: New York. 1978. ISBN 0-471-03070-8. pg. 78

Multiply Y and Z by 1/X. Stack: X, Y, Z, T → X, Y/X, Z/X, T

1/x

×

LAST X

X<>Y

R↓

×

LAST X

R↑

X<>Y

Change X, Y to Y+ X, Y – X

+

ENTER

ENTER

LAST X

-

LAST X

-

The resulting stack is:

z

z

y + x

y – x

Doing LAST X, -, twice is effectively subtracting whatever is in L register twice.

Examples

|4 * w – 6| = 5

Stack:

z: 4

y: -6

x: 5

Results:

y: 2.75

x: 0.25

|2 * w + 5| = 3

Stack:

z: 2

y: 5

x: 3

Results:

y: -1

x: -4

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.