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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Shore. Unauthorized use and/or
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