03-13-2021, 01:08 AM

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

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