Fun with the FX-603P Emulator
Author for the Emulator:
Martin Krischik
Link to Emulator (Android):
https://play.google.com/store/apps/details?id=net.sourceforge.uiq3.fx603p&hl=en
Cost: $5.99 (there is an fx-602P scientific calculator
emulator for $4.99, similar programming language but only 10 programming spaces
instead of 20)
The app is emulates the 1990 Casio fx-603P calculator.
Decibels to Pressure
Program: (29 steps)
“DB?” HLT ÷ 20 = 10^x
* 2E-5 = “Pressure:” HLT
Examples:
DB = 30 dB; Result:
6.32455532 * 10^-4 N/m^2
DB = 120 dB; Result:
20 N/m^2
Turn Performance
Given a plane’s true air speed (TAS in knots), stall speed
(in knots), and required bank turn (in degrees), the following are calculated:
1. G force
2. Normal stall speed
for the plane during the turn (knots)
3. Turn diameter
(nautical miles)
4. Time it takes for
the turn to be complete (in minutes)
Formulas:
G = 1/(cos(bank))
Stall speed = normal stall speed * √G
Diameter = TAS^2 / (34208 * tan(bank))
Time = (0.0055 * TAS) / tan(bank)
Memory Registers:
Input:
M00 = TAS, M01 = Stall speed, M02 = Bank
Output:
M03 = G force, M04 = resulting stall speed, M05 = diameter,
M06 = time
Program: (110 steps)
DEG “TAS?” HLT Min00
“Norm. Stall?” HLT Min01
“Bank?” HLT Min02
MR02 cos 1/x Min03
“G:” HLT
MR03 √ * MR01 =
“Stall Speed:” HLT
MR00 x^2 ÷ ( MR02
tan * 34208 ) = Min05 “Diameter:” HLT
0.0055 * MR00 ÷ MR02
tan “Time:” HLT Min06
Notes:
DEG: [ MODE ] [ 4 ]
Example:
Inputs:
TAS: 123 knots
Norm. Stall: 60 knots
Bank: 44.8°
Results:
G: 1.409302674
Stall Speed: 71.22843498 knots
Diameter: 0.445363387
n.m.
Time: 0.681239424 minutes (about 40.87 seconds)
Source: “Turn
Performance” HP 65 Aviation Pac-1 Hewlett Packard. 1974
.
Sum of a Function
This program uses the subroutine (under P9 with the variable
MinF, or any register M04 or after) to calculate the summation:
Σ f(x) for x = a to b
The sum is stored in M03.
Note: when entering a new f(x), clear P9 (MODE, 3, P9, AC)
first before entering the new function.
It’s a lot cleaner.
Main Program: (34
bytes)
0 Min03
“a?” HLT Min01
“b?” HLT Min02
MR02 – MR01 + 1 =
Min00
Lbl0
MR01 GSBP9 M+03
1 M+01
DSZ Goto0
MR03 “Σ=”
Note:
Lbl0: [ LBL] [ 0 ]
GSBP9: [GSB] [ P9 ]
Goto0: [ GOTO ] [ 0 ]
The character Σ: (in
ALPHA) [SHIFT] [ 7 ]
Memory F: [ Min ], [
MR ], etc. [EXE] for F.
Examples:
Σ n^2 + 3*n – 6 for n = 1 to 8
Subroutine:
Min0F x^2 + 3 * MR0F
– 6 =
Result: 264
Σ (n^3 – 1)/(n^2 + 1) for n = 0 to 11
Subroutine:
( Min0F x^y 3 – 1 )
/div (MR0F x^2 + 1 ) =
Result: 61.6582396282
Combinations: where Repetition is allowed
The program calculates the number of combinations where
repeats are allowed.
nHr = (n + r – 1)! / (r! * (n -1)!)
Program: (39 steps)
“n?” HLT Min01
“r?” HLT Min02
( MR01 + MR02 – 1)
x!
÷ ( MR02 x! * ( MR01
– 1 ) x! )
= “nHr=”
Examples:
Input: n = 5, r = 3.
Result: 35
Input: n = 12, r = 6.
Result: 12376
Aviation: Rate of
Climb
This program calculates the rate-of-climb (ft/min) when
plane increases the elevation (in feet) given the distance to the mountain (in
nautical miles, n.m.) and the true air speed (TAS, in knots).
Formula:
ROC = ( TAS * ΔALT )
/ (60 * √(dist^2 + (ΔALT/6077.1155)^2) )
Program: (88 steps)
6077.1155 Min0F
“TAS (knots)?” HLT
Min01
“CHG ALT (ft)?” HLT
Min02
“DIST (n.m.)?” HLT
Min03
( MR01 * MR02 ) ÷
( 60 * ( MR03 x^2 +
(
MR02 ÷ MR0F ) x^2
) √ = “ROC:”
Example:
Input:
TAS = 87 knots
CHG ALT = 4800 ft
DIST = 13.3 n.m.
Result:
522.3878955 ft/min
Source: “Rate of
Climb and Descent” HP 65 Aviation Pac-1 Hewlett Packard. 1974
Eddie
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Shore. Unauthorized use and/or
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