## NotTim 3 years ago A small aircraft, on a heading of 225 degrees, is cruising at 150km/h. IT is encountering a wind blowing from a bearing of 315 degrees at 35km/h. Determine the aircraft's ground velocity.

1. NotTim

not physics

3. NotTim

Draw it out. Find it's components. Add their x componants and y componants separately. Put them back together in a triangle.

5. NotTim

i somehow figured out how to get resultant, but not the angle...

6. NotTim

im not suppose to use component method

Use tan(y/x)

Oh I see. THen I'm afraid I can be of no help to you.

9. NotTim

ok...oh well. thx u

draw it out, and remember when directions are concerned, 0 degrees is North.

11. NotTim

i did....but i didnt draw it here. i got the magnitude, but not the angle. i think though that i might have read the answer wrong...im still double checking again. but thank you

12. NotTim

yeah, i forgot it was bearings

O.K, good luck on this problem it is a good question

Oh yes, note that the wind is FROM a bearing of 315 degrees in other words the force is towards E 45 S.

Note also the angle between the aircraft and the wind vector is 90 degrees. as aircraft is flying on a bearing of W 45 S

16. NotTim

hey, radar, i got 13.1 degrees (not considering bearings), as the angle between the resultant and the aircraft's cruising speed. Um. What should I do with that?

17. NotTim

@Radar Hey....do you have time? I need you to please revisit this...

18. NotTim

|dw:1333520833003:dw|

19. NotTim

thats how i came to my angle. the correct answer is a BEARING of 211.9 degreees. my answer is not a bearing

20. lexi_rocks_lexi

ur question is a hard 1. i tried 8 times and cant figure it out. sorry

21. NotTim

ok. its alright.

22. bobobobobb

Okay ive done this before. Let me dig up my notes.

23. NotTim

hello?

24. NotTim

computer went wonky.

25. bobobobobb

Sorry. Were you given the answer?

26. NotTim

yeah.its 154.0km/h at a bearing of 211.9 degrees.

27. bobobobobb

Draw your version of the diagram. I wanna see if we are using the same one

28. NotTim

i drew 2...

29. NotTim

or do you want what the txtbook has>?

30. bobobobobb

yeah that would be good

31. NotTim

|dw:1333522792176:dw|

32. Hero

I like how "not physics" is the first thing NotTim says.

33. NotTim

you know, just in case.

34. bobobobobb

So we are finding the resultant vector?

35. NotTim

i already found it (somehow...I don't really know how). I'm looking for the angle now.

36. NotTim

bump

37. experimentX

find out the individual horizontal and vertical factors ... add them or substract them according to necessity ... and reconstruct them. this is how these kinds of problems are done

38. NotTim

argghh. that's component method. im not suppose to use component method!

39. experimentX

since they don't have same directions, i don't think you have other choice ...

40. NotTim

41. experimentX

let's try it.

42. Dockworker

$\tan(\theta)=\frac{35}{150}, \theta=13.13\deg$ subtract from 225

43. NotTim

r^2=150^2+35^2-2(150)(35)cos90

44. Dockworker

theta is angle between resultant and original flight

45. NotTim

wait...so trig ratios work, even with non-right angle triangles?

46. experimentX

|dw:1333520171086:dw|

47. Dockworker

it is a right triangle because the plane is flying at 225 degrees, wind coming from 315 degrees or going at a bearing of 135

48. experimentX

|dw:1333520250197:dw|

49. Dockworker

225-135=90

50. NotTim

its 315...

51. NotTim

im sure thats not an error in my grammar

52. Dockworker

blowing from a bearing of 315º is the same as blowing at a bearing of 135º

53. experimentX

|dw:1333520364868:dw|

54. Dockworker

if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135º

55. NotTim

dock, i dont udnerstand. how would i derive that info?

56. experimentX

i think that would be 180 - (315-225) and apply cosine law .. i think that might work

57. Dockworker

if i tell you wind is coming from in front of you, which way is it going to push you?

58. Dockworker

the same way its coming from?

59. NotTim

no; I know the different of to and from (Radar explained that); how would I figure out:|dw:1333531452863:dw|

60. Dockworker

you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector

61. NotTim

yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?

62. Dockworker

|dw:1333520788256:dw| you could do as you did above.

63. Dockworker

draw a line from 315 bearing through the origin

64. NotTim

Yes, I understand that. I want to know how I can derive 135 from 315 when i get the equilibrant vector, like you did.

65. Dockworker

its just intuitive, not really something you should have to prove

66. Dockworker

just the wording of the problem

67. NotTim

no, but how do i figure that out myself? to get 135 from the equilibrant of 315?

68. NotTim

i dont know if im making sense here...

69. Dockworker

they aren't giving you the true bearing of the wind's velocity

70. Dockworker

they are giving it to you from a different reference

71. NotTim

I understand that. Yes. Very much so. The difference between to and from. But the angle, how did you derive your angle like that?

72. Dockworker

its so far up now. what angle did i derive?

73. NotTim

135

74. NotTim

the wind blowing.

75. Dockworker

its just a reference frame. from an observer's point of view, the wind is at a bearing of 315º. from the wind's perspective, the observer is at abearing of 135º

76. NotTim

How did you calculate 135degrees?

77. Dockworker

315-180

78. NotTim

oohhhkay.

79. Dockworker

the observer (plane) and wind, if looking at each other, form a 180º line of sight

80. Dockworker

or a straight line

81. NotTim

ok.

82. NotTim

now that i figured that out, what's our next step?

83. Dockworker

you know the magnitude of the plane's velocity and window's velocity. they form a right angle. the picture i drew above shows this. use arctan

84. NotTim

we werent taught arctan (to add to my problems)

85. NotTim

or, that was long before

86. Dockworker

this will give you the angle between the resultant and the plane's direction

87. NotTim

oh. trig ratios. tanangle=o/a right

88. Dockworker

arctan(35/150) is approximately 13.13º

89. NotTim

how do i know which one is my opposite and adjacent?

90. Dockworker

|dw:1333521664877:dw|

91. NotTim

oh bugger. I thought we were looking for another angle. this clarfies much,.

92. Dockworker

as you can see, the magnitutde can also be calculated with pythagorean theorem

93. NotTim

yes.

94. NotTim

i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.

95. NotTim

ok. we're near the end now. how do i figure out hte angle, as in bearings? we've got 13.13 degrees. wat next?

96. NotTim

thx with sticking with me here folks, greatly appreciated.

97. Dockworker

subtract that angle from the plane's bearing

98. Dockworker

yw

99. NotTim

thank you all. your assistance has been greatly appreciated. I wish you the best of luck in your studies, and that you are able to help others as greatly as you have done for me.