Fun with HP 71B IV
For the
previous entries:
Links to other
HP 71B Programs:
Fun with the 71B:
http://edspi31415.blogspot.com/2012/06/funwithhp71b.html
Fun with the 71B:
http://edspi31415.blogspot.com/2012/06/funwithhp71b.html
Fun with the 71B III:
71B: Cubic Polynomials:
http://edspi31415.blogspot.com/2012/06/cubicformulabasicprogramhp71b.html
Rake Wall
The program RAKEWALL
calculates:
* The positions
and lengths of studs on a rake wall
* Angle of the
incline
Program RAKEWALL
192 Bytes,
12/22/2016
10 DESTROY B,R,L,N,A
15 DESTROY I,T,O
50 INPUT “BASE:”; B
52 INPUT “RISE:”; R
54 INPUT “RUN:”; L
56 INPUT “O.C. :”; S
74 N = INT(L/S)
76 DEGREES
78 T = ATAN(R/L)
80 DISP “ANGLE =”;
T; “°” @ PAUSE
90 FOR I=L TO 0 STEP
–S
92 A = I*TAN(T)+B
94 DISP I; “, “;
A @ PAUSE
96 NEXT I
98 DISP 0; “, “; B
Notes:
Degree symbol (°): [ g ], [RUN] (CTRL), [ A ]
Example:
(amounts are in feet)
BASE: 4
RISE: 3
RUN: 6
O.C.: 16/12
(1 foot, 4 inches)
Output:
ANGLE =
26.5650511771 °
(Position from
where the incline meets the base, length of studs)
6, 7
4.66666666667,
6.33333333334
3.33333333334,
5.66666666667
2.00000000001,
5
0.66666666668,
4.33333333334
0, 4
Automotive Cylinders: Calculating Displacement and Piston Speed
The program
supplies default values which can be accepted or changed.
Program AUTOCYN
217 Bytes,
12/22/2016
10 DESTROY
N,B,S,R,E,P
15 INPUT “#
CYLINDERS:”, “6”; N
20 INPUT “BORE (IN):”,
“4”; B
25 INPUT “STROKE
(IN):”, “4”; S
30 E = PI/4 * B^2 *
S * N
35 INPUT “RPM:”; R
40 P = S * R / 6
45 DISP “ENGINE
DISPLACEMENT =” @ WAIT 1
50 DISP E; “ IN^3” @
PAUSE
55 DISP “PISTON
SPEED =” @ WAIT 1
60 DIPS P; “FPM”
Example 1:
N = 6
cylinders, B = 4 in, S = 4 in, RPM = 3500
Output: E ≈ 301.59290 in^3, P ≈ 2333.33333 FPM
Example 2:
N = 6
cylinders, B = 4 in, S = 3 in, RPM = 4500
Output: E ≈ 226.19467 in^3, P = 2250 FPM
Right Triangle Solver
The program
RIGHTTRI is a solver for the sides A, B, and C.
To solve for the third side, the desired side needs to have a 0 value,
while the other two sides have nonzero values.
Each time a solution is found, the values of A, B, and C are reset (set
to 0).
Program RIGHTTRI
502 Bytes,
12/22/2016
10 DESTROY A,B,C,Z$
11 ! LOOP
12 DISP “A, B, C,
Solve, Exit”
14 DELAY 0, 0
16 Z$ = KEY$
18 IF Z$ = “A” THEN
30
20 IF Z$ = “B” THEN
40
22 IF Z$ = “C” THEN
50
24 IF Z$ = “S” THEN
60
26 IF Z$ = “E” THEN
96
28 GOTO 12
30 INPUT “A = “; A
32 GOTO 12
40 INPUT “B = “; B
42 GOTO 12
50 INPUT “C = “; C
52 GOTO 12
60 IF A=0 AND B AND
C THEN 62 ELSE 70
62 S=SQR(C^2B^2) @
Z$ = “A”
64 GOTO 90
70 IF A AND B=0 AND
C THEN 72 ELSE 80
72 S=SQR(C^2A^2) @
Z$ = “B”
74 GOTO 90
80 IF A AND B AND
C=0 THEN 82 ELSE 86
82 S=SQR(A^2+B^2) @
Z$ = “C”
84 GOTO 90
86 DISP “MUST HAVE
ONE 0” @ BEEP @ WAIT .5
88 GOTO 10
90 DISP Z$; “ = “; S
@ PAUSE
92 GOTO 10
94 DISP “DONE”
Notes:
The line of IF
A=0 AND B AND C… tests whether A is zero, B and C are nonzero. You can test whether a variable is nonzero
by just typing the variable in an IF condition.
The function
SQR is the square root function of the HP 71B.
Anything
following an exclamation point (!) is a comment.
Basic BridgedT Notch
Filter
The program
NOTCH calculates the required capacitor and resistor to null out an undesired frequency.
Inputs: inductance (H), frequency to nullified (Hz), resistance
of the required coil (Ω)
Source: Rosenstein, Morton. Computing With the Scientific Calculator
Casio: Tokyo, Japan. 1986. ISBN10: 1124161430
Program NOTCH
244 Bytes,
12/22/2016
10 DESTROY L,F,I,C,R
20 DISP “INDUCTANCE”
@ WAIT 1
22 INPUT “in H: “; L
24 DISP “FREQUENCY” @
WAIT 1
26 INPUT “in Hz: “;
F
28 DISP “COIL’S
RESISTENCE” @ WAIT 1
30 INPUT “in Ω: “; I
40 C = 1/(2 * PI^2 *
F^2 * L)
42 R = (PI * F *
L)^2 / I
44 DISP “NOTCH
CAPACITOR” @ WAIT 1
46 DISP C; “ F” @
PAUSE
48 DISP “NOTCH
RESISTENCE” @ WAIT 1
50 DISP R; “ Ω”
Notes:
Capital Omega
Character (Ω): [ g ], [RUN] (CTRL), [ Q
]
Example: L = 0.12 H, F = 1170 Hz, Coil Resistance =
30 Ω
Output:
NOTCH CAPACITOR
≈ 3.08402 * 10^7 F
NOTCH
RESISTENCE ≈ 6485.04070 Ω
Differential Equations and HalfIncrement
Solution, Numerical Methods
The program HALFDIFF solves the numerical differential equation
d^2y/dt^2 = f(dy/dt, y, t) given the initial conditions
y(t0) = y0 and dy/dt (t0) = dy0
In this notation, y is the independent variable and t is the
dependent variable.
The Method
Let C = f(dy/dt, y, t). Give the change of t as Δt.
First Step:
With t = t0:
h_1/2 = dy0 + C * Δt/2
y1 =
y0 + dy0 * Δt
Loop:
t = t0 + Δt
h_I+1/2 = h_I1/2 + C * Δt
y_I+1 = y_I +h_I+1/2 * Δt
Repeat as many steps as desired.
For more details, click here:
http://edspi31415.blogspot.com/2016/11/ti84plusandhpprimedifferential.html
Source: Eiseberg, Robert
M. Applied Mathematical Physics with Programmable Pocket
Calculators McGrawHill, Inc: New York. 1976. ISBN
0070191093
Program HALFDIFF
250+ bytes, 12/22/2016
Edit f(A,Y,T) at line 5 where
A = y’(t),
Y = y(t), T = t
5 DEF FNF(A,Y,T)
= insert function y’(t), y(t), and t here
10 DESTROY
A,Y,D,N,H,T
12 RADIANS
20 INPUT “dY/dX
(0) = “, “0”; A
22 INPUT “Y(0) =
“, “0”; Y
24 INPUT “DELTA
T = “,”1”; D
26 INPUT “TMAX =
“,”5”; N
28 T = D
44 H = A +
FNF(A,Y,T) * D/2
48 Y = Y + H * D
50 DISP D; “, “; Y @ PAUSE
70 FOR I = 2 * D
TO N STEP D
72 T = I @ A = H
76 H = H +
FNF(A,Y,T) * D
78 Y = Y + H * D
80 DISP I; “, “;
Y @ PAUSE
82 NEXT I
Example: y’’(t) = y’(t) + 2 * t with y’(0) = 0, y(0) = 1, Delta t = 1, Max
t = 6
FNF(A,Y,T)
= A + 2 * T
Results:
T

Y

1

2

2

8

3

26

4

70

5

168

6

376

This blog
is property of Edward Shore, 2016.
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