## NotTim Group Title 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. 2 years ago 2 years ago

1. NotTim Group Title

not physics

3. NotTim Group Title

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

5. NotTim Group Title

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

6. NotTim Group Title

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 Group Title

ok...oh well. thx u

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

11. NotTim Group Title

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 Group Title

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 Group Title

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 Group Title

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

18. NotTim Group Title

|dw:1333520833003:dw|

19. NotTim Group Title

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 Group Title

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

21. NotTim Group Title

ok. its alright.

22. bobobobobb Group Title

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

23. NotTim Group Title

hello?

24. NotTim Group Title

computer went wonky.

25. bobobobobb Group Title

Sorry. Were you given the answer?

26. NotTim Group Title

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

27. bobobobobb Group Title

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

28. NotTim Group Title

i drew 2...

29. NotTim Group Title

or do you want what the txtbook has>?

30. bobobobobb Group Title

yeah that would be good

31. NotTim Group Title

|dw:1333522792176:dw|

32. Hero Group Title

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

33. NotTim Group Title

you know, just in case.

34. bobobobobb Group Title

So we are finding the resultant vector?

35. NotTim Group Title

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

36. NotTim Group Title

bump

37. experimentX Group Title

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 Group Title

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

39. experimentX Group Title

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

40. NotTim Group Title

41. experimentX Group Title

let's try it.

42. Dockworker Group Title

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

43. NotTim Group Title

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

44. Dockworker Group Title

theta is angle between resultant and original flight

45. NotTim Group Title

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

46. experimentX Group Title

|dw:1333520171086:dw|

47. Dockworker Group Title

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 Group Title

|dw:1333520250197:dw|

49. Dockworker Group Title

225-135=90

50. NotTim Group Title

its 315...

51. NotTim Group Title

im sure thats not an error in my grammar

52. Dockworker Group Title

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

53. experimentX Group Title

|dw:1333520364868:dw|

54. Dockworker Group Title

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 Group Title

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

56. experimentX Group Title

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

57. Dockworker Group Title

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

58. Dockworker Group Title

the same way its coming from?

59. NotTim Group Title

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

60. Dockworker Group Title

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

61. NotTim Group Title

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

62. Dockworker Group Title

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

63. Dockworker Group Title

draw a line from 315 bearing through the origin

64. NotTim Group Title

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 Group Title

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

66. Dockworker Group Title

just the wording of the problem

67. NotTim Group Title

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

68. NotTim Group Title

i dont know if im making sense here...

69. Dockworker Group Title

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

70. Dockworker Group Title

they are giving it to you from a different reference

71. NotTim Group Title

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 Group Title

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

73. NotTim Group Title

135

74. NotTim Group Title

the wind blowing.

75. Dockworker Group Title

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 Group Title

How did you calculate 135degrees?

77. Dockworker Group Title

315-180

78. NotTim Group Title

oohhhkay.

79. Dockworker Group Title

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

80. Dockworker Group Title

or a straight line

81. NotTim Group Title

ok.

82. NotTim Group Title

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

83. Dockworker Group Title

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 Group Title

we werent taught arctan (to add to my problems)

85. NotTim Group Title

or, that was long before

86. Dockworker Group Title

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

87. NotTim Group Title

oh. trig ratios. tanangle=o/a right

88. Dockworker Group Title

arctan(35/150) is approximately 13.13º

89. NotTim Group Title

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

90. Dockworker Group Title

|dw:1333521664877:dw|

91. NotTim Group Title

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

92. Dockworker Group Title

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

93. NotTim Group Title

yes.

94. NotTim Group Title

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

95. NotTim Group Title

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 Group Title

thx with sticking with me here folks, greatly appreciated.

97. Dockworker Group Title

subtract that angle from the plane's bearing

98. Dockworker Group Title

yw

99. NotTim Group Title

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.