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Partially filled frustum of cone

  1. Dec 18, 2006 #1
    1. I have a cone frustum with known height (h) and known base diamter (d) and cap diameter (D). From this, I can calculate the volume. Question: if the frustum is half full of liquid, how can I calculate the height of the liquid?

    2. V = pi/12 x h x (D2 + d2 + D x d)

    3. I guess I need to calculate the cone angle to solve this but after that I'm lost
  2. jcsd
  3. Dec 18, 2006 #2


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    The answer depends on how you interpert "half full" .

    a water level of h/2 could be seen as "half full" or

    A volume of water = V/2 could be seen as "half full".

    if you are looking for the latter. Replace V with V/2 in your relationship then solve for h.
  4. Dec 18, 2006 #3

    Hi integral,

    It's half volume I'm trying to find. I understand what you are saying:

    V/2 = [pi/12 x h x (D2 + d2 + D x d)] / 2

    d is a known but if I have V/2, there are two unknowns (h and D). I think the key is that they are linked by a ratio.
  5. Dec 18, 2006 #4


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    Draw a picture, you will find that the difference of 2 radius involved (I do not understand your notation) and the height form a right triangle. you can eliminate one of the radius using the pythagorean theorem.
  6. Dec 19, 2006 #5
    Sorry you have lost me. I understand what you are saying - the angle is a constant and it must fit in there but I am totally lost. Here is the formula and an image.


    In case anybody else is reading - I have a value for D, d and h. The above formula gives me Volume, V. Now, if I halve the volume of liquid in the cone frustum, how can I obtain the new height?
  7. Dec 19, 2006 #6


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    use the fact that

    [tex] Tan \theta = \frac h {D - d} [/tex]

    [itex] \theta [/itex] is the wall angle.

    Since this is constant you must also have

    [tex] \frac h {D - d} = \frac {h_2} {D - d_2} [/tex]

    where the supscripted quanities are the half volume values.
  8. Dec 20, 2006 #7
    I have to say that I don't understand that. I'm going to go away and have another look but surely:

  9. Dec 20, 2006 #8
    I've been looking and I can now see that your expression is for the total inclusive angle of the side walls with the top of the cone (i.e. 90 degrees minus the angle with the base of the cone, as in my expression above).

    I had sat and worked through the idea of using the 90 degree triangle after your second reply. I can see there are two triangles with a common angle. My problem is that I just don't see how this fits in with the Volume expression.

    At the end of the day, I still have two unknowns (D and h). If I rearrange (and boy have I tried), I don't seem to be able to isolate them on one side (and even if I did, I don't know what I'd do next). I even tried solving the two Volume expressions (i.e. full and half-full) as simultaneous equations but it was exactly the same story.

    At the end of the second reply, you said "eliminate one of the radius using the pythagorean theorem". I know this is the key to it but I just cannot see a way through :frown:
  10. Dec 20, 2006 #9


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    Yes, you are correct, I drew my pic the other way round.

    Use the expression for the Tangent to eliminate one of your variables. So rearange the relationship to express [itex] d_2 [/itex] in terms of D, d, h, and [itex] h_2 [/itex]. Now you can write an expression for the volume in terms of your knowns and [itex] h_2 [/itex]
    Last edited: Dec 20, 2006
  11. Dec 20, 2006 #10
    Thanks again.

    It's not pretty but I've found that if I rearrange so that d2 is the subject, it still leaves h2 on the other side.


  12. Dec 21, 2006 #11


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    Now take that expression for [itex] d_2 [/itex] and plug it in for [itex] d_2 [/itex] in the expression for the volume when the cone is half full.
  13. Dec 22, 2006 #12
    More progress

    Thanks integral. I feel like I am getting somewhere although there is a way to go yet.

    I have arranged everything so that the equations better represent the specific problem I am trying to solve:

    Equation for partially-filled volume

    Equation for partially-filled diamater

    So its just a case of inserting the expression for D2 in the Volume (partially-filled) equation and re-arranging for h2 (which is a tough job)...I'm working on it :approve:
    Last edited: Dec 22, 2006
  14. Dec 22, 2006 #13
    More progress

    Hi again,

    Been doing a bit more work. I decided that I prefered working with radius values and the side wall angle (rather than diameter and inclusive angle):


    This means that:

    Volume = pi/3 x h x (R^2 + r^2 + Rr)


    Tan of angle = R - r / h

    I also realised that you can substitute the above into the Volume expression.

    So this is as far as I got:


    I'm still going to carry on working on it later today/tomorrow but any help on isolating getting h2 on the right hand side would be appreciated
  15. Dec 23, 2006 #14
    I'm thoroughly beaten

    The equations I derived in the last post are slightly wrong. They should be:


    I am now fundamentally stuck at this point. For clarity, I have represented the the above expression thus:

    a = h2
    b = Tanθ
    c = r
    x = V2 / 0.333pi



    I need to express the above in terms of a and I am stumped. One half of me says it can't be done. I've put a lot of work into this, can anyone help me get over this hurdle?
    Last edited: Dec 23, 2006
  16. Dec 25, 2006 #15


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    You know that

    Tan [tex] \theta = \frac h {R-r} [/tex]
    Perhaps that replacement will cancel some factors.

    You have a cubic in [itex] h_2 [/tex] there is no easy closed form method of factoring this. One way would be to put in numbers and plot it.
    Last edited: Dec 25, 2006
  17. Dec 26, 2006 #16

    I have decided to use approximation. I converted the frustum to a whole cone to make the approximation leg-work easier. It isn't what you'd call satisfactory but it works.

    1: You know the volume of the frustum (V1) so the volume of half-filled frustum is half that value (V2).

    2: Using all your knowns, you calculate the height of the remaining portion of the cut-off cone (h3). You can then work out the volume of that portion (V3).

    3: Now you know that V2 + V3 = pi/3 x R^2 x (h2 + h3). Only R and h2 are not known.

    4: You re-arrange the eqution so that all the knowns (V2, V3 and pi/3) are on one side and work out that value.

    5: You estimate the height of the liqiud in the half-full-frustum (I reckoned about 75% of the full frustum). From this you get a value for radius at that height. Plug these values in and compare with the value calculated at step 4. You keep going using succesive approximation until both sides of the eqution agree.

    To check all this, I plugged all my new values back into the Frustum Volume equation and it all works out fine. There is probably a cleaner mathematical way of doing the above so any suggestions would be graciously received!!!
    Last edited: Dec 26, 2006
  18. Feb 25, 2007 #17
    If anyone is interested, I have a proper mathematical solution to this problem.
  19. Sep 22, 2008 #18
    Hi Polyperson,

    I would be interested in your mathematical solution to the cone volume with 2 unknowns.


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