**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.

[up to five/six programs]

**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

All original
content copyright, © 2011-2018. Edward
Shore. Unauthorized use and/or
unauthorized distribution for commercial purposes without express and written
permission from the author is strictly prohibited. This blog entry may be distributed for
noncommercial purposes, provided that full credit is given to the author. Please contact the author if you have
questions.

Thanks for the informative article. This is one of the best resources I have found in quite some time.Nicely written and great info. I really cannot thank you enough for sharing.

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