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/fun-with-hp-71b.html
Fun with the 71B:
http://edspi31415.blogspot.com/2012/06/fun-with-hp-71b.html
Fun with the 71B III:
71B: Cubic Polynomials:
http://edspi31415.blogspot.com/2012/06/cubic-formula-basic-program-hp-71b.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 non-zero 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^2-B^2) @
Z$ = “A”
64 GOTO 90
70 IF A AND B=0 AND
C THEN 72 ELSE 80
72 S=SQR(C^2-A^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 non-zero. You can test whether a variable is non-zero
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 Bridged-T 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. ISBN-10: 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 Half-Increment
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_I-1/2 + 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/ti-84-plus-and-hp-prime-differential.html
Source: Eiseberg, Robert
M. Applied Mathematical Physics with Programmable Pocket
Calculators McGraw-Hill, Inc: New York. 1976. ISBN
0-07-019109-3
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.
