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Full Version: (41) HP 41 - AVIATION - Wind Triangle
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Bonjour all,

When you are beginning to travel with an aircraft with your instructor between A and B, you have to solve the “Wind triangle”.

In fact, you expect to fly between A and B at speed of 140 knots in a direction where a provisional wind, for example.

To keep the right direction, you have to fly in direction which take into account of Wind Direction vector and Velocitiy.

You have three vectors in head.

First one is one “arrow oriented” with two characteristic dimensions HDG (Heading) and TAS (True Airspeed) (HDG,TAS). Second one is “two arrow” oriented with two characteristic dimensions CRS (Course) and GS (Ground speed) (CRS,GS). Last one is “three arrows” oriented is the Wind direction (WD) and Wind Velocity (WV) (WD,WV).

You may have to solve this “Wind Triangle” in reality or in exam for 4 kind of questions.
Three are vectorial problems and one more is trivial which are nevertheless most common.

Navigation point of view is like this :
(HDG,TAS) + (WD,WV) = (CRS,GS)

Mathematical point of view is little bit different because regular indication of WD is the direction where the wind come from. Other Directions are the direction where you go.

You have,
1> (CRS,GS) – (WD,WV) = (HDG,TAS)
2> (HDG,TAS) + (-WD,WV) = (CRS,GS)
3> (CRS,-GS) – (HDG,-TAS) = (WD,WV)

For the fourth case, there is a trivial computation because the result is not as a vector, but trigonometric computations. You have to fly between A and B on route equal to 315° (CRS=315) with a True AirSpeed of 140 Kts (TAS=140 Kts). These two dimensions are not in the same vector. During this travel there is a wind of (WD,WV). Question is to find the Heading (HDG) and Ground Speed (GS) that is not in the same vector. There is a trigonometric solution for that.

Without HP 41, there is a graphic solution (made with “Aristo 617” for example) which are not easier in flight. Four HP41 PGMs are attached hereafter for HP41: 3 for vectorial computations and one more for trivial computation. An additional for nice display of the result. You can modify as necessary, of course.

PGM 1 lbl “HDG” you know CRS and WND and you look for HDG
PGM 2 lbl “CRS” you know HDG and WND and you look for CRS
PGM 3 lbl “WD” you know HDG and CRS and you look for WND
PGM 4 lbl “HDS” you know CRS and TAS and WND and you look for HDG and GS
PGM 5 lbl DPY for Display SubProgram for the above PGM

DON’T USE THESE PGMs WITHOUT TESTING ALL EXHAUSTIVELY AND YOU ARE CONFIDENT WITH Result.

Feel free to contact me for any question or error; These PGMs are mine and are susceptible of errors not seen by me, best PGMs writiting.

in HP 41, you input first the angle in Y, and the velocity in seconf in X after you push R/S to launch the PGM. Of course if you dont respect correct entries in X and Y always you result will be an error.

For example, you input on hp 41 ;
XEQ "CRS"
first step you input the HDG=250, then push ENTER and input TAS, after correct entries you push R/S. At next prompt, you enter WD in X and push ENTER; Next you input WV=55, then R/S for the result. Immediately after this display you check the answer cohérence.

01 LBL CRS
02 ‘HDG TAS ?’
03 PROMPT
04 P->R
05 ‘WD WV ?’
06 PROMPT
07 CHS
08 P->R
09 ST+ Z
10 X<>Y
11 ST+ T
12 RDN
13 RDN
14 R->P
15 XEQ DPY
16 END

01 LBL HDG
02 ‘CRS GS ?’
03 PROMPT
04 P->R
05 ‘WD WV ?’
06 PROMPT
07 CHS
08 P->R
09 ST- Z
10 X<>Y
11 ST-T
12 RDN
13 RDN
14 R->P
15 XEQ DPY
16 END

01 LBL HDS
02 ‘CRS TAS ?’
03 PROMPT
04 STO 00
05 ‘WD WV ?’
06 PROMPT
07 X<>Y
08 R↑
09 STO 01
10 -
11 X<>Y
12 P-R
13 X<>Y
14 RCL Z
15 /
16 ASIN
17 COS
18 RCL L
19 RCL 01
20 +
21 X<>Y
22 RCL 00
23 x
24 RCL T
25 -
26 XEQ DPY
27 END

01 LBL WND
02 ‘CRS GS ?’
03 PROMPT
04 CHS
05 P->R
06 ‘HDG TAS ?’
07 PROMPT
08 CHS
09 P->R
10 ST- Z
11 X<>Y
12 ST-T
13 RDN
14 RDN
15 R->P
16 XEQ DPY
17 END

01 LBL "DPY"
02 X<>Y
03 360
04 MOD
05 CLA
06 FIX 01
07 RND
08 XEQ 00
09 ARCL X
10 X<> L
11 ├ 2 spaces
12 X<>Y
13 RND
14 XEQ 01
15 ARCL X
16 X<> L
17 AVIEW
18 FIX 04
19 LBL 00
20 100
21 X>Y?
22 ├ 0
23 RDN
24 10
25 X>Y?
26 ├ 0
27 RDN
28 RTN
29 LBL 01
30 10000
31 X>Y?
32 ├ ‘ ‘
33 RDN
34 1000
35 X>Y?
36 ├ ‘ ‘
37 RDN
38 100
39 X>Y?
40 ├ ‘ ‘
41 RDN
42 10
43 X>Y?
44 ├ ‘ ‘
45 RDN
46 END
Bonjour!

In your program I like it that you prompt for input values. There already are wind triangle programs in the libraries section of hpmuseum.org but all/most of them require that you enter all the input values into the stack in specific order before running the program. Who remembers that in programs that one uses only every few weeks?

Regards
Max
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
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