Fun with the Radio Shack EC-4004

Fun with the Radio Shack EC-4004

Complex Number Multiplication

(a + bi) * (c + di) = (a*c – b*d) + (b*c + a*d)i

Store values in the following memory registers before running the program:

K1 = a

K2 = b

K3 = c

K4 = d

Output is stored in registers K5 (real part) and K6 (imaginary part).

Program:

Kout 1

*

Kout 3

-

Kout 2

*

Kout 4

=

Kin 5

HLT

Kout 2

*

Kout 3

+

Kout 1

*

Kout 4

=

Kin 6

Example:

(-8 + 3i)*(6 + 3i)   (K1 = -8, K2 = 3, K3 = 6, K4 = 3)

Result:  -57 – 6i (K5 = -57, K6 = -6)]

Product of Integers

The following program will between and b:  Product = a * (a + 1) * (a + 2) * … * b

P = Π n from n = a to b

Store values in the following memory registers before running the program:

K1 = a

M = b

K2 = 1 (store 1 in K2 each time)

Program:

Kout 1

Kin* 2

1

Kin+ 1

Kout 1

X ≤ M

Kout 2

Example:  Π n from n = 5 to 8:  P = 5 * 6 * 7 * 8

Store 5 in K1, 8 in M (Inv Min), 1 in K2. 

Heron’s Formula

The program calculates the area of the triangle.  Store the length of the sides in registers K1, K2, and K3.  The register K4 is used in the calculation and ultimately have the area.

K4 = (K1 + K2 + K3)/2

Area = √(K4 * (K4 – K1) * (K4 – K2) * (K4 – K3))

Program:

Kout 1

Min

Kout 2

M+

Kout 3

M+

MR

÷

2

=

Min

Kin 4

MR

-

Kout 1

=

Kin* 4

MR

-

Kout 2

=

Kin* 4

MR

-

Kout 3

=

Kin* 4

Kout 4

√

Kin 4

Example:

K1 = 16.4, K2 = 13.8, K3 = 11.4

Area = 77.60164947

Catenaries

The program calculates the length of the wire in a catenary (2*s). 

Input:

K1 = Horizontal Tension

K2 = Weight of the wire (lb-ft, N, etc)

K3 = half of the distance between the poles

Program:

Kout 1

÷

Kout 2

*

(

Kout 2

*

Kout 3

÷

Kout 1

)

sinh

=

*

2

=

Example:

Input:

K1 (H) = 40N

K2 (weight) = 0.227 N

K3 (half distance between poles) = 32.6 m

Result: 65.5726 m

Source:  Chris M. Alley and Brenda M. Cornitius TI-36 Solar Guidebook Texas Instruments. 1985.

Mass Dragged on the Table by a Pulley

The program solves the following systems:

T – M1*a = μ*M1*g

T + M2*a = M2*g

Input:

K1 = mass 1 in kg

K2 = mass 2 in kg

K3 = friction of the table factor (μ)

Output:

K4 = acceleration (a) m/s^2

K5 = tension of the rope (N)

Tension is displayed first, then acceleration

SI units are assumed, where g = 9.80665 m/s^2 (Earth’s gravity constant)

Program:

9

.

8

0

6

6

5

Kin 4

*

(

Kout 2

Kin 5

-

Kout 1

Kin+ 5

*

Kout 3

)

=

÷

Kout 5

=

Kin 5

HLT

Kin- 4

Example:

K1 = 4.3 kg (mass 1)

K2 = 6.4 kg (mass 2)

K3 = 0.1 (friction μ)

Results:

T = 27.7445 N (tension, K5)

a = 5.47156 m/s^2 (acceleration, K4)

Eddie

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