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Question for my fellow math geeks

Discussion in 'Members' Lounge' started by dohcvtec_accord, Nov 1, 2004.

  1. dohcvtec_accord

    dohcvtec_accord WRX Sellout

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    A co-worker posed this problem to me, and I gave him the following answer, mostly pulled from my ass. Is it the right answer? If not, what is?

    [​IMG]

    OK....you've got a sphere. You want to cut off a portion of the sphere at Line L (I guess it's a plane, reall). How do you figure out the surface area of that sphere?

    My answer: The formula for surface area of a sphere is 4 x pi x radius squared. For the top half of the sphere, the formula is 2 x pi x radius squared. You need a Multiplier for that half of the sphere to find the area of the cut-off portion. Now, in order to find the Multiplier, you draw a line out from the center of the sphere along the "x-axis", and then draw another line from the center of the sphere to the intersection of the plane and the sphere: Chord C. Now, you figure the cosine of Angle A, and multiply it by half of the area of the sphere. There's your answer.

    Correct? My reasoning is, if the "cut-off plane" is right at the center of the circle, your multiplier is 1 (or the cosine of 0). If the plane is tangent to the sphere, right at the top of the sphere, your multiplier is 0 (or the cosine of 90).

    It's tough to explain my thought process on the intraweb. But can anyone prove or disprove my theorem?
     
  2. dohcvtec_accord

    dohcvtec_accord WRX Sellout

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    Sorry for the broken first pic....it's fixed now.
     
  3. pissedoffsol

    pissedoffsol RETIRED

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    i think its the square route of type R * the derriviative of vtec over oil pressure + f20c.
     
  4. dohcvtec_accord

    dohcvtec_accord WRX Sellout

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    :ban:
     
  5. b204dr

    b204dr Senior Member

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    lol ban this seems like a question far beyond here.. u need some one who atleast can blueprint.. i know my basics but you lost me need some more help with ur numbers
     
  6. dohcvtec_accord

    dohcvtec_accord WRX Sellout

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    Nah, I have confidence that either Eric (lsvtec) or Mike (Calesta) will help me out here. :D
     
  7. Battle Pope

    Battle Pope New Member

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    Math makes me poop.

    Seriously though, I'd help you if I could, but I dunnae remember anything from teh math class.
     
  8. civicious

    civicious FüK-VTEC VIP

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    draw it in solidworks, cut off the portion that you want to cut off, and use one of the cosmos engines to find the area
     
  9. dohcvtec_accord

    dohcvtec_accord WRX Sellout

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    But that takes the fun right out of using mathematics to nail down the answer to the question. Besides, this wasn't something to figure out a rea-world situation. It was a "what-if" scenario.
     
  10. MaaseyRacer

    MaaseyRacer Senior Member

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    I would have to think back to sophomore year in high school, and there is this cloud of marijuana smoke preventing me from doing so.
     
  11. TommyTheCat

    TommyTheCat Gonzo Scientist

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    Well, you are wrong. That's the easy part.

    And from now on I'll assume you aren't considering the "circle"-area in the plane L as part of the area. a.k.a we're just looking at a bowl formed from dissecting the traditional "sphere" (sphere = just the surface/skin of the ball; and ball is the technical term for a solid sphere). Anyways...

    This will take me a little bit to write out, so give me a few minutes to post my proof. Lots of integrals.
     
  12. Citizen_Insane

    Citizen_Insane Senior Member

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    well, the equation for the circle that represents that sphere is:
    f(x) = (y+sin(90-a))^2 + x^2 = L^2 this places the part of the sphere that you want to find the area of above the x-axis and the rest of it below.

    now, the equation for the SA of a solid is:
    SA = 2*int(2*pi*f(x)*sqrt(1 + f'(x)^2)dx) from 0 to where ever f(x) intersects the x-axis. Which can be found by finding when f(x)=0

    so, if we solve f(x) for y we get:
    y=±[sqrt(L^2-x^2)-cos(theta)] and
    y'=-x/sqrt(L^2 - x^2)
    so we plug it in:
    SA = 2*int(2*pi*sqrt(L^2-x^2)-cos(theta)*sqrt(1 - x/sqrt(L^2 - x^2)^2)dx) from 0 to wherever 0=±[sqrt(L^2-x^2)-cos(theta)]
    from this we get:
    ....an equation that I cant solve. Damnit. Even my ti89 can't do it. I probably fucked up the algebra somehow. If anyone wants to (read: is crazy enough) to check my math, feel free.

    most simplified I can get it is: int(sqrt(-2x-L^2)/sqrt(L^2-s) dx)
     
  13. TommyTheCat

    TommyTheCat Gonzo Scientist

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    Here goes:

    [​IMG]

    [​IMG]

    [​IMG]

    My thanks to Wolfram Research for their informative articles on spheres and for making Mathematica.
    [sarcasm] Oh how I love Mathtematica 4.1 [/sarcasm]
     
  14. Citizen_Insane

    Citizen_Insane Senior Member

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    wow nice. Its nice to see i was at least on the right track ... kinda
     
  15. TommyTheCat

    TommyTheCat Gonzo Scientist

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    Oh poo...this is the kind of article i wanted to pull up. But it shows that I messed up my equation! Whoops.

    And no, i won't go through it again, you can just read it here: http://mathworld.wolfram.com/Zone.html
     
  16. BigJ

    BigJ I'm just about that action Boss. VIP

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    :concur: Tommy The Cat.

    Derivatives are the essence of the earth.
     
  17. Loco Honkey

    Loco Honkey Banned

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    Math makes me gassy. Uh oh! Hey Jamie, guess what! I just... FARTED! :lol:
     
  18. 90 accord

    90 accord Chicks dig the box Moderator VIP

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    :huh: :blink: that one smells like bean burrito's!!! mmmm
     
  19. Citizen_Insane

    Citizen_Insane Senior Member

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    Hmmm....that required a lot of Calc I don't know yet :lol: I guess half a semester of Calc II isn't cutting it. Oh well, I have 3.5 more years to polish off those math skills....or not.
     
  20. MikeBergy

    MikeBergy Blah blah blah....

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    I hate geometry. I think I'll stick with the Navier-Stokes equations, flow potential, and vorticity problems I have to deal with in aerodynamics. Geometry sucks.
     
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