(41) HP 41  AVIATION  Wind Triangle

03132021, 01:08 AM
(This post was last modified: 03202021 08:39 PM by Gene.)
Post: #1




(41) HP 41  AVIATION  Wind Triangle
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 STT 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 PR 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 STT 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 

03142021, 07:05 PM
Post: #2




RE: HP 41  AVIATION  Wind Triangle
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 

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