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thez25 Group Title

Find kappa and tau for the helix r(t)=(acost)i+(asint)j+btk, a,b>0 a^2+b^2 does not = 0. What is the largest value tau can have for a given value of a? give reasons for answer please.

  • 2 years ago
  • 2 years ago

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  1. vf321 Group Title
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    By kappa do you mean curvature?

    • 2 years ago
  2. TuringTest Group Title
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    kappa is curvature and what is tau...

    • 2 years ago
  3. vf321 Group Title
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    Does he maybe mean T, the unit tangent?

    • 2 years ago
  4. thez25 Group Title
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    Yes to both sorry for the confusion

    • 2 years ago
  5. vf321 Group Title
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    Really? I never heard of that use of tau before...

    • 2 years ago
  6. TuringTest Group Title
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    me neither, I've only seen big T

    • 2 years ago
  7. TuringTest Group Title
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    anyway, it's very straightforward with the formulas

    • 2 years ago
  8. vf321 Group Title
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    Well anyway ill handle the tangent u can do kappa, Turing.

    • 2 years ago
  9. TuringTest Group Title
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    you need tangent for kappa, so...

    • 2 years ago
  10. thez25 Group Title
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    its a T in like greek format i believe it is the unit tangent though

    • 2 years ago
  11. vf321 Group Title
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    lol ur right i was thinking about 2D formula for curvature

    • 2 years ago
  12. vf321 Group Title
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    BTW how do you do a vector overscript in this?

    • 2 years ago
  13. TuringTest Group Title
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    ok you don't need T for kappa, but it helps

    • 2 years ago
  14. vf321 Group Title
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    superscript*

    • 2 years ago
  15. TuringTest Group Title
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    \[\vec r(t)=(a\cos t)\hat i+(a\sin t)\hat j+bt\hat k;~~~ a,b>0; ~~~a^2+b^2 \neq0\]most likely

    • 2 years ago
  16. vf321 Group Title
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    \[r(t)=(acost)i+(asint)j+btk\]\[r'(t)=v(t)=D_t(acost)i+D_t(asint)j+D_t(bt)k\]

    • 2 years ago
  17. vf321 Group Title
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    yeah howd u get the arrow and hat superscripts?

    • 2 years ago
  18. TuringTest Group Title
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    \vec r\[\vec r\]

    • 2 years ago
  19. TuringTest Group Title
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    right-click any LaTeX you don't know and you can do show as ->tex commands then see the source code

    • 2 years ago
  20. vf321 Group Title
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    sweet thx was always too lazy to learn latex lol

    • 2 years ago
  21. TuringTest Group Title
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    me too I just pick it up as I go along on here...

    • 2 years ago
  22. vf321 Group Title
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    \[\vec v(t) = <-a \sin(t), a \cos(t), b>\]

    • 2 years ago
  23. vf321 Group Title
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    By what i showed u earlier, @OP

    • 2 years ago
  24. thez25 Group Title
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    thanks im following you just fine thus far!

    • 2 years ago
  25. vf321 Group Title
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    \[\vec T(t) = \frac{\vec v(t)}{\left| \vec v(t) \right|}\]

    • 2 years ago
  26. vf321 Group Title
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    I trust you can handle all that fun algebra yourself...

    • 2 years ago
  27. thez25 Group Title
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    haha yes!

    • 2 years ago
  28. vf321 Group Title
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    And then curvature is pretty easy too. I forgot that it's torsion that's hard. It's just \[\kappa = \left| \frac{d\vec T}{dt}\right|\]

    • 2 years ago
  29. vf321 Group Title
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    And I assume you can take a derivative as well. Do you need explanations for why the formulas for kappa and T are what they are?

    • 2 years ago
  30. thez25 Group Title
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    no just for what the largest value of T is Thanks!

    • 2 years ago
  31. vf321 Group Title
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    ds lol soz

    • 2 years ago
  32. vf321 Group Title
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    \kappa = \left| \frac{d\vec T}{ds}\right|

    • 2 years ago
  33. TuringTest Group Title
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    I usually get curvature with\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

    • 2 years ago
  34. TuringTest Group Title
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    keeps arc length out of it

    • 2 years ago
  35. vf321 Group Title
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    \[\kappa = \left| \frac{d\vec T}{ds}\right|\]

    • 2 years ago
  36. vf321 Group Title
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    I suppose, if you want to avoid chain rule, that TT's method's better.

    • 2 years ago
  37. thez25 Group Title
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    Thats a good idea :p

    • 2 years ago
  38. vf321 Group Title
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    Yeah in fact I'm all for @TuringTest 's method. I was going to do all that messy chain rule stuff.... It's been a while since MVC...

    • 2 years ago
  39. thez25 Group Title
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    Lol I liked calculus till i got into some of this stuff....

    • 2 years ago
  40. TuringTest Group Title
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    this site keeps me fresh on it it's very easy to forget all the formulas there's another way to where you use the cross product and the first and second derivatives also

    • 2 years ago
  41. TuringTest Group Title
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    product of*...

    • 2 years ago
  42. vf321 Group Title
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    Lol wait. I see what happened. We could have gone with mine from the start! I remember doing this in class: \[\kappa = \left| \left| \frac{d\vec T}{ds} \right| \right|=\left| \left| \frac{d\vec T/dt}{ds/dt} \right| \right|=\left| \left| \frac{d\vec T/dt}{v(t)} \right| \right|\]

    • 2 years ago
  43. vf321 Group Title
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    Man its been a while!

    • 2 years ago
  44. thez25 Group Title
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    yea thats right!

    • 2 years ago
  45. TuringTest Group Title
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    ah, thanks for bridging the gap :)

    • 2 years ago
  46. vf321 Group Title
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    but @thez25 the most important thing here is understanding the formulas. Do you get em?

    • 2 years ago
  47. thez25 Group Title
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    yeah im familiar with those its just really the end part thats getting me lol

    • 2 years ago
  48. vf321 Group Title
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    Well the T should be easy... But as for curvature, the best way I can explain that is the following. Curvature, in very lay-terms, is how "curvy" a function is. Thus, the turn of a very elongated ellipse, if you could imagine, should have a high curvature. On the other hand, if you imagine a straight line, we would have zero curvature. The value has to be scalar. If we need to look at something as it travels through space, the best measure of its difference from a straight line path (read: its TANGENT) over how far it travels (read: its ARC LENGTH) is, well, ||dT/ds||

    • 2 years ago
  49. vf321 Group Title
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    Note we can't use time because curvature only cares about shape, and time is simply a parameter we invented to more easily describe a curve. It does not contain info about the curve itself. Also, dT/dt = c*N anyway

    • 2 years ago
  50. thez25 Group Title
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    I got ya thanks :p

    • 2 years ago
  51. vf321 Group Title
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    scratch that last statement tho its only valid in some cases.

    • 2 years ago
  52. thez25 Group Title
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    okay got ya

    • 2 years ago
  53. thez25 Group Title
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    So how do you maximize T for a given value of a?

    • 2 years ago
  54. vf321 Group Title
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    what do you mean 'maximize'? T is a unit vector.

    • 2 years ago
  55. thez25 Group Title
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    The question asks what is the largest value t can have for a given value of a?

    • 2 years ago
  56. vf321 Group Title
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    Strange. |T| = 1 always...

    • 2 years ago
  57. vf321 Group Title
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    if it was velocity, as in \[\vec v(t)\] Then I'd say look at the physics!

    • 2 years ago
  58. TuringTest Group Title
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    well a and b are susceptible to change, so perhaps we need the partial wrt b ??

    • 2 years ago
  59. vf321 Group Title
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    By physics I mean a is the radius, and v is the velocity. And we know from circular motion that mv^2/(a) = acc

    • 2 years ago
  60. vf321 Group Title
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    for 2D circles. the k-component just adds a constant, orthogonal component to motion.

    • 2 years ago
  61. thez25 Group Title
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    I think since were dealing with a helix T could actually be torsion not unit tangent sorry!

    • 2 years ago
  62. vf321 Group Title
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    haha yeah! i forgot. It probably was torsion!

    • 2 years ago
  63. thez25 Group Title
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    Torsion is b/(a^2+b^2) hmm

    • 2 years ago
  64. vf321 Group Title
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    Well it might be i dunno. Let's solve for it, unless your book told you that's it... \[\tau = \left| \left| \frac{d\vec B(t)}{ds} \right| \right| = \left| \left| \frac{d\vec B(t)/dt}{v(t)} \right| \right|\]

    • 2 years ago
  65. thez25 Group Title
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    Lets just do it your way i can follow those formulas better

    • 2 years ago
  66. vf321 Group Title
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    \[\vec v(t)=<−a sin(t),a cos(t),b>\] \[\vec B(t) = \vec T(t) \times \vec N(t)\] \[\vec N(t) = \frac{d\vec T(t)}{ds}\]

    • 2 years ago
  67. vf321 Group Title
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    To explain the torsion formula: Basically the same as my curvature explanation, but torsion is how much your curve "wants" to move out of the 2D plane that it is currently "on" - the 2D plane that is made by the two vectors N and T

    • 2 years ago
  68. thez25 Group Title
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    so we still have to find T lol

    • 2 years ago
  69. vf321 Group Title
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    Hopefully you already know the unit binormal formula and the unit normal formula

    • 2 years ago
  70. thez25 Group Title
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    The T in those formulas is the unit tangent right not torsion?

    • 2 years ago
  71. thez25 Group Title
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    Yeah ive seen all of those before!

    • 2 years ago
  72. vf321 Group Title
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    Yes. and BTW my formula for N should be divided by magnitude, forgot that.

    • 2 years ago
  73. thez25 Group Title
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    right right!

    • 2 years ago
  74. vf321 Group Title
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    So let's start crackin (but b4 i do that, I'll first let |v(t)| be V, since I use it a lot, and it will start cancelling out... \[\vec T(t) = \frac{\vec v(t)}{V} = \frac{<−a sin(t),a cos(t),b>}{V} \]

    • 2 years ago
  75. thez25 Group Title
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    just to make sure its the magnitude of V right?

    • 2 years ago
  76. vf321 Group Title
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    V IS the magnitude - ||v(t)|| = V \[\vec N(t) = \frac{d\vec T(t)/dt}{ds/dt} =\frac{ <-a cos(t), -a sin(t), 0>/V+\vec v(t) *V'/V^2}{\vec v(t)}\]

    • 2 years ago
  77. thez25 Group Title
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    ohh that makes it easier to just cancel rather than figure out T right away good work!

    • 2 years ago
  78. vf321 Group Title
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    Well, wait. Remember, V is the magnitude, which depends on time, so I have to factor in the product rule! HOWEVER, in this case, we are dealing with helical movement, which, because of our physics, we know has CONSTANT velocity magnitude ( you could actually prove this on your own). Thus, V is a CONSTANT and V' = 0. We get: \[\vec N(t)=\frac{<-a cos(t), -a sin(t) ,0>}{V ^2}\] But if you remember I clumsily forgot to make it a unit vector by forgetting to divide by the magnitude. Thus, at the very end, you need something in the direction of <-a cos(t), -a sin(t) ,0> with magnitude 1. We get: <-cos(t), -sin(t) ,0>/

    • 2 years ago
  79. vf321 Group Title
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    Sorry if this is hard to understand. If you don't get it, I can go back...

    • 2 years ago
  80. thez25 Group Title
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    I sort of follow but i can figure out what u mean more when i read over it a few times. U can keep going though dont want to waste your time!

    • 2 years ago
  81. vf321 Group Title
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    Okay. Either way, some way or another, you get: \[\vec N = <-cos(t), -sin(t) ,0>\]\[\vec T = \frac{<−asin(t),acos(t),b>}{\sqrt{a^2+b^2}}\]

    • 2 years ago
  82. thez25 Group Title
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    I got that earlier when i was finding them :p

    • 2 years ago
  83. thez25 Group Title
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    is V^2 in the previous equation equal to 1 somehow?

    • 2 years ago
  84. vf321 Group Title
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    When we cross them, the work for which I won't bother to show, we get \[\vec B = \frac{<a sin(t) cos(t), -a sin(t)cos(t), 0>}{\sqrt{a^2+b^2}}\]

    • 2 years ago
  85. thez25 Group Title
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    sorry didnt see that you explained that at the end of that paragraph

    • 2 years ago
  86. vf321 Group Title
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    wait i messed my cross sry.. it shouldnt have no z value.

    • 2 years ago
  87. thez25 Group Title
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    isnt z the 0 there? so no value right

    • 2 years ago
  88. vf321 Group Title
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    Yes, but that's wrong. This is the right cross product (I checked with mma) \[\vec B = \frac{<b sin(t),- bcos(t), a>}{\sqrt{a^2+b^2}}]\]

    • 2 years ago
  89. thez25 Group Title
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    okay thanks!

    • 2 years ago
  90. vf321 Group Title
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    Now that we have B, we can find the torsion!

    • 2 years ago
  91. vf321 Group Title
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    \[\tau = \left|\left|\frac{d\vec B(t)/dt}{\vec v(t)}\right|\right|\] Also, minor note - i was kind of using improp notation whenever I did that - I should have double bars around both the vector in the num and denom

    • 2 years ago
  92. vf321 Group Title
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    I think that you can handle the torsion from here, right?

    • 2 years ago
  93. thez25 Group Title
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    its alright I got what you meant by it!

    • 2 years ago
  94. vf321 Group Title
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    dB/dt = {b Cos[t], b Sin[t], 0}/V ||v(t)|| = V V = sqrt(a^2+b^2)

    • 2 years ago
  95. vf321 Group Title
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    k i gotta go now im sure u can handle it from there.

    • 2 years ago
  96. thez25 Group Title
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    well if you get a chance later could u explain how to maximize it lol thanks for all your help though!

    • 2 years ago
  97. TuringTest Group Title
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    maximize it with respect to what? if you want to maximize it with respect to t find\[\frac{d\tau}{dt}=0\]

    • 2 years ago
  98. TuringTest Group Title
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    but\[\tau=\frac b{\sqrt{a^2+b^2}}\]and you said you are "given a value of a" so perhaps they want\[\frac{\partial\tau}{\partial b}\]?

    • 2 years ago
  99. thez25 Group Title
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    sorry i left but they want it with respect to any given value of a...

    • 2 years ago
  100. TuringTest Group Title
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    so the only thing to take the derivative with respect to is b. so set\[\frac{\partial\tau}{\partial b}=0\]solve for b

    • 2 years ago
  101. thez25 Group Title
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    alright thanks so that would give the largest value torsion can have for a given value of a then correct?

    • 2 years ago
  102. TuringTest Group Title
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    since tau only depends on a and b, yes

    • 2 years ago
  103. TuringTest Group Title
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    according to the earlier work, which I'm not going to independently verify

    • 2 years ago
  104. thez25 Group Title
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    okay thank you for your help!

    • 2 years ago
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