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Thread: Cone Surface Area?

  1. #1

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    Frenzied Member Vlatko's Avatar
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    Cone Surface Area?

    How to calculate the height of a cut cone (if it is called that in english), if the surface area is 320pi cm2, R = 10 and r = 5. See picture for explanation.

    First is this correct:
    Code:
    P -surface area
    P=B+B1+M
    
    B= 10^2*Pi
    
    B1=5^2*pi
    
    M=(R+r)*s*Pi
    M=15*s*Pi
    
    so
    P=100*pi + 25*Pi + 15*Pi*s
    320 = 100 + 25 + 15*s
    195 = 15*s
    s = 13
    
    H^2 = s^2 - (R - r)^2
    H^2 = 169 - 25
    H^2 = 144
    H = 12
    It seems correct but i have a solution which says it is not. Can the solution be wrong?
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  2. #2
    jim mcnamara
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    I'm confused.

    I'm going to make some assumptions which may be wrong and go from there. Plus, I really like 'cut cone' better than conic section.

    Math people have no sense of clarity or beauty sometimes. Me, I have no sense at all.

    Code:
    Assume
    The cut is parallel to the bottom of the cone. ie. a circle.
    
    Note:
    You did not give us all the data we need -
    
         We must have either the Length S or an angle-theta
         (the angle made by line S and line R).  Because without one
         or the other H could be infinitely long - if theta is 90 deg
         for instance - an infinitely long rod.
    
    Assume we know s
    
    H = sqr( s^2 - (R-r)^2)
    
    Assume we know theta
    Assume I remember trig which I never use
    
     H = tan(theta) * (R-r)

  3. #3

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    Frenzied Member Vlatko's Avatar
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    Yes, the bottom is parallel to the top.

    S ar an angle are not given. And what do you mean H is infinitely long. It is from the bottom to the top. (it is not infinite for sure).
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  4. #4
    jim mcnamara
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    Let me try again -

    If you do not know S, then you have to know an angle. Period. As far as I can see.

    If the angle were 90 degrees, you would have a soild shaped like a rod. Agreed? You could then cut the rod anywhere, from 1" out to 4 km out from the starting end. The size of the circle's area is the same no matter how long it is. Therefore you have to stipulate an angle less than 90 degrees.

    (I'm making up the results here because I don't have a calculator handy). But. If the angle is 89.99 degrees, H = 50 with r = 1, and
    H=3 with r=1 at 80 degrees. So angle matters.

    The only other way to solve it is with S having a value.

    In other words without any angle and without length S:

    THERE IS NO SOLUTION.

    There are infinitely many solutions.

  5. #5
    transcendental analytic kedaman's Avatar
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    Code:
    |\                  \
    | \
    |  \
    |___\               sector side = cone side
    | 5  \     \
    |     \    cut cone side
    |H     \
    |__10___\  /        /
    - A cone side surface is a Sector
    - A cut cone side surface is a sector with removed inner sector.
    - The sector side is the extrapolated cone side; the crossection of the
    cone can be divided into 2 perpendicular triangles, with
    same angles => equilateral, and ratio is 1:2 (the figure)
    => The sector side is 2 times cut cone side
    => the sector Area/cone side surface Area is 3/4 since,
    <= (the ratio was 2 => the inner sector/outher sector=
    1^2*pi*radius^2/2^2*pi*radius^2=1/4)
    - the area of a sector is theta*radius^2/2 (theta=sector angle)
    - theta = arc/radius (by definition)
    => Sector area= arc*radius/2
    - the arc of the sector is the circumference of the bottom circle which is 
    2*10*pi=20pi
    - the radius is 2 times the cone side, 2*sqr(5^2+H^2) (pythagoras)
    => The Cone side is now:
    3/4 * 20pi * sqr(25+H^2) = (320-5^2-10^2)pi = 195pi (bottom and top surfaces removed)
    => sqr(25+H^2)=13
    => H=12
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  6. #6
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    Answer 20.7

  7. #7
    Frenzied Member HarryW's Avatar
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    I don't really understand Kedaman's post. Anyway I think you have enough information for a solution since you have the surface area. I also would guess that there are 2 solutions of H for any surface area. Perhaps a surface area of 320*pi cm² is the one value which has only one solution.
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  8. #8
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    Since the base radius is 10 and the top radius is 5 we can reason that your looking for the surface area of the lower half of a cone.

    Area of full cone: Pi*R*sqrt((2*H)^2+R^2)
    Area of top cone half: Pi*r*sqrt(H^2+r^2)

    (Full cone)-(short cone) = Pi*(r+R)*sqrt(H^2+(R-r)^2)

    320*Pi = Pi*(15)*sqrt(H^2+25)

    H = 20.7...

  9. #9
    Addicted Member Active's Avatar
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    Kedaman is right ! but I think I will explain it better...


    Code:
                 * . *      *
                 |/|\ \      \
                 | | \ \      \
                /| |  \ \      \
               / | |   \ \      \
              /  | |    \(L-l)   \
             /   | |     \ \      \
            /   H-h|      \ \      \
           /     | |   r   \ \      L
          /------*-|--------\ *      \
         /	    | |         \ \      \
        /        | |          \ l      \
       /         h |           \ \      \
      /          | |            \ \      \
     /___________|_|_____________\ *      *
                         R
    First Let us consider a Right circular cone of Height H
    and Slant height L and base Radius R

    Now the portion formed by the chopping of the top by a
    plane parallel to the base is called the frustum of the cone.
    Let the Frustum have slant height l, top surface radius r
    base radius R and height h.

    Now we see from the figure that

    R/L = r/(L-l) and R/H = r/(H-h)

    Therefore
    R(L-l)=rL ==> L=Rl/(R-r)

    and similarly H=Rh/(R-r)

    Now..

    The total surface area of a cone is

    = Flat surface area + curved surface area
    = pi * (base-radius)^2 + pi * base-radius * slant-height

    Now the Flat surface area of the Frustum is
    = Pi * r^2 + pi * R^2 (since there are two flat suraces)

    The Curved surface area of the frustum

    = Curved surface area of the Entire Cone - Curved surface area of the Tip cone

    = pi.R.L - pi.r.(L-l)

    = pi.R.L - pi.r.L + pi.r.l

    = pi.L(R-r) + pi.r.l

    substituting L = R.l/(R-r)

    = pi.[R.l/(R-r)].(R-r) + pi.r.l

    = pi.R.l + pi.r.l

    Therefore the total surface area of the frustum

    = Flat surface area + curved surface area
    = pi.r^2 + pi.R^2 + pi.R.l + pi.r.l

    T.S.Area = pi(r^2 + R^2 + l(R+r))

    ------

    We also see that

    (H-h)^2 = (L-l)^2 - r^2 (pythagoras theorem)

    Substitute H = Rh/(R-r) and L = Rl(R-r) and simplify

    we get h^2 = l^2 - (R-r)^2

    ------

    Now the given problem...

    T.S.Area = pi * (5^2 + 10^2 + l(10+5)) = pi * 320

    ==> 25 + 100 + 15(l) = 320 or

    l = 13

    therefore h = SQR(13^2 - (10-5)^2)

    h = SQR(144) = 12

    ---
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  10. #10

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    Well it seems active and kedaman have the result 12. (same as me). So i assume the answer is wrong.

    Originally posted by Illuminator
    Area of full cone: Pi*R*sqrt((2*H)^2+R^2)
    The area of a full cone is
    Code:
    P = R*pi *(R + s)
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