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Thread: maths question

  1. #1
    the Afterbirth Tycoon Automotive Encyclopaedia PlacentaJuan's Avatar
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    Default maths question

    i need to find an angle (a)

    i have a standard butterfly type throttle of diameter (x) that i suppose is 0 degrees when closed and 90 degrees when open.

    what angle (a) does the butterfly need to open to equal an open area of (y)?

    this equation is quite beyond me, i know there are smarter poeople here than can probably do it in their heads and hope you can help me.

    sorry i dont have real figures, but the algebraic equation would be good.

    thanks

  2. #2
    Junior Member Backyard Mechanic Hash_ra23's Avatar
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    Default Re: maths question

    y = PI*(x/2)^2 - PI*(x/2)^2*cos(a)
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  3. #3
    Junior Member Too Much Toyota oldcorollas's Avatar
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    Default Re: maths question

    angle of closed throttle is not 0 either... a few degrees off...
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  4. #4
    the Afterbirth Tycoon Automotive Encyclopaedia PlacentaJuan's Avatar
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    Default Re: maths question

    yea thats what i thought, its actually not a throttle and im pretty sure that this particular butterfly is copmletely closed, although it might have a few degrees in it if its not perfectly round to seal.

    either was there is no gap when closed.

  5. #5
    Junior Member Backyard Mechanic Hash_ra23's Avatar
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    Default Re: maths question

    ohhh..well my calculation was for an ideal theoretical throttle....i.e. at 0 degrees fully closed...90degrees fully open. And the throttle valve has no thickness.
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  6. #6
    the Afterbirth Tycoon Automotive Encyclopaedia PlacentaJuan's Avatar
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    Default Re: maths question

    hey what is that equation with relation to a?

    ie a (angle) = .....

  7. #7
    Junior Member Backyard Mechanic Hash_ra23's Avatar
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    Default Re: maths question

    just gotta rearrange equation....

    a = inversecos[ 1 - ( y / PI*(x/2)^2 )]

    i think thats it....
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  8. #8
    Captain Red Grease Monkey SuperDave's Avatar
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    Default Re: maths question

    I get the same.
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  9. #9
    Cunning Linguist Domestic Engineer The Last Streetfighter's Avatar
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    Default Re: maths question

    Atot = Total Area = Pi * r^2
    N = %open = 1 - cos (alpha) {alpha = 0.. 1 - cos (0) = 0; alpha = 90.. 1 - cos (90) = 1: there is your range from 0% open to 100% open. At 45degrees the thing is 29% open}
    Aopen = Open Area = Atot * N = Pi * r * r * (1 - cos(alpha))

    Alpha Radians Cos (alpha) 1 - cos (alpha)
    0 0.00 1.00 0.00 Fully Closed
    10 0.17 0.98 0.02
    20 0.35 0.94 0.06
    30 0.52 0.87 0.13
    40 0.70 0.77 0.23
    45 0.79 0.71 0.29
    50 0.87 0.64 0.36
    60 1.05 0.50 0.50
    70 1.22 0.34 0.66
    80 1.40 0.17 0.83
    90 1.57 0.00 1.00 Fully Opened
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  10. #10
    Toymods member no 341 Domestic Engineer amichie's Avatar
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    Default Re: maths question

    Quote Originally Posted by SuperDave
    I get the same.

    Me too.

    Area of a circle - area of the ellipse.

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