## amistre64 Group Title We used vector to find wind speed. A balloon is timed as it rises inside, assume a *vertical component only The balloon is taken outside, and timed. The angle it makes is also recorded. Using the vectors, calculate the wind speed. 2 years ago 2 years ago

1. amistre64

|dw:1348235064295:dw|

2. amistre64

i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?

3. amistre64

Teacher wanted us to use pythag thrm to determine the wind speed. But the data collected gave a sqrt(-#) which would not allow the pythag thrm to be used. I decided to take the more general Law of Cosines.

4. amistre64

i did try it, and since the hypotenuse is smaller than the "leg", it results in a non-real value.

5. amistre64

his assumption is that the wind never blows things downward i spose

6. amistre64

speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s |dw:1348235842574:dw|

7. amistre64

i followed per instructions; he noted on the lab report that the time of the outside balloon cannot be greater than the inside record

8. Algebraic!

exactly.

9. Algebraic!

|dw:1348235955911:dw|

10. Algebraic!

Vy is the same for both

11. amistre64

Im wondering if windspeed is only a horizontal measure? or does directionality not matter?

12. Algebraic!

"assume a horizontal component only"

13. amistre64

there was noting stated prior in the lab to indicate that assumption tho.

14. amistre64

*nothing

15. amistre64

the data recorded showed that it rose slower outside, and the note i got on the lab report afterwards was that the data cannot show that. But he also says that we cannot manipulate the data ... i get the feeling that im getting contradictory advice from him :/

16. amistre64

i wasnt. but if we only assume the wind blows things from a horizontal, or upwards ... then the data would show it to rise faster. Rise is prolly a bad term for the outside part.

17. amistre64

the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers

18. amistre64

the lab was designed to get you working with vectors. The speed of the wind could have been calculated from the outside data alone if we were to assume a horizontal wind speed only. .97*sin(23)

19. UnkleRhaukus

did you really get 23°?

20. UnkleRhaukus

*measure

21. amistre64

well, it was the best i could do with the little plastic protractor :)

22. UnkleRhaukus

if you were getting <11° you could use the small angle approximation

23. amistre64

the teacher suggested the pythag thrm as a means of overcoming the inherent inaccuracies in using the protractors. There was no way to tell if it was "level" and if the person catching the balloon to read the thing hadnt moved it.

24. UnkleRhaukus

i really dont think i understand this experiment , what results did you get /

25. amistre64

$\sqrt{.97^2+1.05^2-2(.97)(1.05)~cos(23^o)}$

26. amistre64

the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.

27. UnkleRhaukus

was it a helium ballon ?

28. amistre64

yes, we used 3 balloons filled with helium. those shiney things you get at a party supply place. I think one of them had Dora the Explorer on it :)

29. UnkleRhaukus

reminds me some of walter lewin 's experiments

30. UnkleRhaukus

what were you timing exactly ?

31. amistre64

i held the balloon on the ground, released it, and timed how long it took it to tighten the string it was attached to.

32. amistre64

so we were calculating the speed along a vector path

33. amistre64

using both vectors (inside and outside) we would have been able to calculate the speed (magnitude) of the resultant vector - which in turn would have been the wind speed

34. UnkleRhaukus

so your vectors are ; $$\{ (1.05,0°) , (0.97, 23°) \}$$ ?

35. amistre64

yes

36. amistre64

those are more like the polar coordinants of the vectors

37. UnkleRhaukus

so converting to cartesian co-ordinates $\{\quad(0,1.05)\quad,\quad\left(0.97 \cos(23°), 0.97\sin (23°)\right)\quad\}$

38. UnkleRhaukus

|dw:1348240487279:dw|

39. amistre64

the theta is in the wrong spot

40. amistre64

spose to be 23 degrees west of north

41. UnkleRhaukus

ah ,

42. UnkleRhaukus

so theta is 90-23 ?

43. amistre64

44. UnkleRhaukus

you can find the resultant right?

45. amistre64

spose we just use this setup, and focus on the vectors |dw:1348240769620:dw| the resultant can be determined by the law of cosines

46. UnkleRhaukus

i thought you were ment to use pythag. ?

47. amistre64

he said we could use whatever method we wanted to find the resultant, and when I used the law of cosines, he said I was wrong for using it.

48. UnkleRhaukus

do you get the same answer both ways /

49. amistre64

which leads me to wonder, is windspeed only a horizontal measurement? if so, then the use of 2 vectors is a bad setup

50. amistre64

no, you dont get the same answer both ways

51. amistre64

|dw:1348241029039:dw| this would have been the resultant using pythag thrm; but my data does not present itself as a right triangle

52. amistre64

there was no "answer" to compare to. So the answer itself is irrelevant. It was the process that I used to find it that he said was wrong. and I dont know why

53. amistre64

how can i use the pythag thrm with the data collected? i cant.

54. amistre64

$0.97^2-1.05^2=-0.1616$ $\sqrt{-0.1616}=nonReal~\#$

55. amistre64

lol, i just noticed a typo in the original post :) assume a *vertical* component only

56. amistre64

im not sure what to make of demitris reply ...

57. amistre64

the data wasnt "given". I went out and collected the data by performing the experiment

58. amistre64

59. amistre64

the 23 im using of course is from a faulty memory.

60. amistre64

but not by use of pythag

61. amistre64

you cant have a hypotenuse shorter than a leg ...

62. experimentX

does wind only have horizontal component?

63. amistre64

experiment: i dont know :/

64. amistre64

the lab just wanted us to determine the resultant vector using the data we collected.

65. experimentX

well .. maybe the wind is blowing from |dw:1348242000106:dw|

66. amistre64

well, magnitude of the resultant that is

67. amistre64

i agree, if the wind has a downward component, than the pythag will not be useful

68. amistre64

nothing stated in the lab said to find a horizontal component ....

69. experimentX

still even we consider wind, this is not just Pythagoras problem ...

70. experimentX

is it @amistre64 ?

71. amistre64

college, yes

72. experimentX

any info given on acceleration?

73. amistre64

no acceleration given. just direction and speed to define the vectors with

74. amistre64

it was noted when i got the lab report handed back to me, that the time outside needed to be quicker than the time inside. But my data did not reflect that.

75. Algebraic!

of course it does...

76. experimentX

|dw:1348242332765:dw|

77. amistre64

pfft, typoed that

78. amistre64

the speed recorded outside needed to be quicker than the speed inside

79. amistre64

so, shorter time

80. Algebraic!

same diff.

81. amistre64

the pythag thrm was suggests as a means to circumvent the inaccuracy of degree measurement. But that only works out if the data collected does not have the wind pushing down on the balloon outside.

82. experimentX

|dw:1348242525949:dw| we can calculate the acceleration on balloon due to wind. ${d^2x \over dt^2} = -k {dx \over dt}$

83. amistre64

thanks for reading it, but my results were calculated already using the cosine rule. I am wondering if it was appropriate to use it in the first place.

84. amistre64

the acceleration seems to be a bit outside the scope that the lab was intended for

85. amistre64

yes, becasue I use the cosine rule, and included it in the doc that a wrote up

86. experimentX

not sure ... if this is mathematics |dw:1348243032602:dw| you can simply get that component using cosine rule.

87. amistre64

experiment, I agree :)

88. experimentX

lol ... anyway this was getting interesting. :)

89. amistre64

we got a new printer in the office, and this ine allows scans to flash drives :) this is the instruction for the lab, just in case you can spot a place that I may have missed.

90. amistre64

this is just a physics 1 class, very basic stuff so far

91. experimentX

seems plausible. I've been having problems with physics quite a lot lately.

92. amistre64

here in florida, temp outside is usually higher than inside :)

93. amistre64

thats a good notice on the outside only part. the balloons were already a part of the lab in class; i had no control on prepping them. the inside part would have been an instruction from the teacher, since its not explicitly stated anywhere in the handout. My only idea as to why we would do the inside part would be to better establish the vertical part of the experiment, but it seems to have added more trouble than it was worth :) Thanks

94. amistre64

thats what i had in mind at the start i prolly could have included that thought in the "results" section of the lab report.

95. hipster

@demitris, please enable private messaging so i may message you about a post that you'd tagged me in a few minutes ago