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amistre64

  • 2 years ago

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.

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  1. amistre64
    • 2 years ago
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    |dw:1348235064295:dw|

  2. amistre64
    • 2 years ago
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    i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?

  3. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    i did try it, and since the hypotenuse is smaller than the "leg", it results in a non-real value.

  5. amistre64
    • 2 years ago
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    his assumption is that the wind never blows things downward i spose

  6. amistre64
    • 2 years ago
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    speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s |dw:1348235842574:dw|

  7. amistre64
    • 2 years ago
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    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!
    • 2 years ago
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    exactly.

  9. Algebraic!
    • 2 years ago
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    |dw:1348235955911:dw|

  10. Algebraic!
    • 2 years ago
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    Vy is the same for both

  11. amistre64
    • 2 years ago
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    Im wondering if windspeed is only a horizontal measure? or does directionality not matter?

  12. Algebraic!
    • 2 years ago
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    "assume a horizontal component only"

  13. amistre64
    • 2 years ago
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    there was noting stated prior in the lab to indicate that assumption tho.

  14. amistre64
    • 2 years ago
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    *nothing

  15. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers

  18. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    did you really get 23°?

  20. UnkleRhaukus
    • 2 years ago
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    *measure

  21. amistre64
    • 2 years ago
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    well, it was the best i could do with the little plastic protractor :)

  22. UnkleRhaukus
    • 2 years ago
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    if you were getting <11° you could use the small angle approximation

  23. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    i really dont think i understand this experiment , what results did you get /

  25. amistre64
    • 2 years ago
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    \[\sqrt{.97^2+1.05^2-2(.97)(1.05)~cos(23^o)}\]

  26. amistre64
    • 2 years ago
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    the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.

  27. UnkleRhaukus
    • 2 years ago
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    was it a helium ballon ?

  28. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    reminds me some of walter lewin 's experiments

  30. UnkleRhaukus
    • 2 years ago
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    what were you timing exactly ?

  31. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    so we were calculating the speed along a vector path

  33. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    so your vectors are ; \(\{ (1.05,0°) , (0.97, 23°) \} \) ?

  35. amistre64
    • 2 years ago
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    yes

  36. amistre64
    • 2 years ago
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    those are more like the polar coordinants of the vectors

  37. UnkleRhaukus
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1348240487279:dw|

  39. amistre64
    • 2 years ago
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    the theta is in the wrong spot

  40. amistre64
    • 2 years ago
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    spose to be 23 degrees west of north

  41. UnkleRhaukus
    • 2 years ago
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    ah ,

  42. UnkleRhaukus
    • 2 years ago
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    so theta is 90-23 ?

  43. amistre64
    • 2 years ago
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    in your setup, yes

  44. UnkleRhaukus
    • 2 years ago
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    you can find the resultant right?

  45. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    i thought you were ment to use pythag. ?

  47. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    do you get the same answer both ways /

  49. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    no, you dont get the same answer both ways

  51. amistre64
    • 2 years ago
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    |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
    • 2 years ago
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    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
    • 2 years ago
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    how can i use the pythag thrm with the data collected? i cant.

  54. amistre64
    • 2 years ago
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    \[0.97^2-1.05^2=-0.1616\] \[\sqrt{-0.1616}=nonReal~\#\]

  55. amistre64
    • 2 years ago
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    lol, i just noticed a typo in the original post :) assume a *vertical* component only

  56. amistre64
    • 2 years ago
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    im not sure what to make of demitris reply ...

  57. amistre64
    • 2 years ago
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    the data wasnt "given". I went out and collected the data by performing the experiment

  58. amistre64
    • 2 years ago
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  59. amistre64
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    the 23 im using of course is from a faulty memory.

  60. amistre64
    • 2 years ago
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    but not by use of pythag

  61. amistre64
    • 2 years ago
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    you cant have a hypotenuse shorter than a leg ...

  62. experimentX
    • 2 years ago
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    does wind only have horizontal component?

  63. amistre64
    • 2 years ago
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    experiment: i dont know :/

  64. amistre64
    • 2 years ago
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    the lab just wanted us to determine the resultant vector using the data we collected.

  65. experimentX
    • 2 years ago
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    well .. maybe the wind is blowing from |dw:1348242000106:dw|

  66. amistre64
    • 2 years ago
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    well, magnitude of the resultant that is

  67. amistre64
    • 2 years ago
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    i agree, if the wind has a downward component, than the pythag will not be useful

  68. amistre64
    • 2 years ago
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    nothing stated in the lab said to find a horizontal component ....

  69. experimentX
    • 2 years ago
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    still even we consider wind, this is not just Pythagoras problem ...

  70. experimentX
    • 2 years ago
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    is it @amistre64 ?

  71. amistre64
    • 2 years ago
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    college, yes

  72. experimentX
    • 2 years ago
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    any info given on acceleration?

  73. amistre64
    • 2 years ago
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    no acceleration given. just direction and speed to define the vectors with

  74. amistre64
    • 2 years ago
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    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!
    • 2 years ago
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    of course it does...

  76. experimentX
    • 2 years ago
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    |dw:1348242332765:dw|

  77. amistre64
    • 2 years ago
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    pfft, typoed that

  78. amistre64
    • 2 years ago
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    the speed recorded outside needed to be quicker than the speed inside

  79. amistre64
    • 2 years ago
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    so, shorter time

  80. Algebraic!
    • 2 years ago
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    same diff.

  81. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1348242525949:dw| we can calculate the acceleration on balloon due to wind. \[{d^2x \over dt^2} = -k {dx \over dt} \]

  83. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    the acceleration seems to be a bit outside the scope that the lab was intended for

  85. amistre64
    • 2 years ago
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    yes, becasue I use the cosine rule, and included it in the doc that a wrote up

  86. experimentX
    • 2 years ago
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    not sure ... if this is mathematics |dw:1348243032602:dw| you can simply get that component using cosine rule.

  87. amistre64
    • 2 years ago
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    experiment, I agree :)

  88. experimentX
    • 2 years ago
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    lol ... anyway this was getting interesting. :)

  89. amistre64
    • 2 years ago
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    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.

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  90. amistre64
    • 2 years ago
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    this is just a physics 1 class, very basic stuff so far

  91. experimentX
    • 2 years ago
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    seems plausible. I've been having problems with physics quite a lot lately.

  92. amistre64
    • 2 years ago
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    here in florida, temp outside is usually higher than inside :)

  93. amistre64
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    @demitris, please enable private messaging so i may message you about a post that you'd tagged me in a few minutes ago

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