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Deriving a volume formula, answer seems to be correct but is negative.

  1. Jul 20, 2014 #1
    I'm an engineering student, and I'm making a formula for the volume of liquid in a cylinder that is cut in half diagonally. r is the radius, h is the height from the bottom to the top of the sloped flat bottom, and L is the height of the water within the cylinder. So when L = h you would expect the formula to be equal to h∏r^(2)/2.

    z bar is the plane defining the bottom where the cylinder is "cut." It was derived using the points (-r,0,0), (r,0,h) and (0,r,h/2) so it passes through the xy plane at x = -r and touches h at x = r.

    y bar is the standard equation for half a circle with radius r

    a is the x value of the rightmost edge of the liquid as the tank is filled.

    [itex]v = 2\int_{-r}^{a}\int_{0}^{\bar{y}}\int_{\bar{z}}^{h}dzdydx[/itex]

    [itex]\bar{z} = \frac{hx}{2r}+\frac{h}{2}[/itex]

    [itex]\bar{y} = \sqrt{r^{2}-x^{2}}[/itex]

    [itex]a = \frac{2rL}{h}-r[/itex]

    Solving leads to the following formula:

    [itex]v = \frac{r^{2}h*arcsin(a/r)}{2}+\frac{ha\sqrt{r^{2}-a^{2}}}{2}-\frac{3\pi r^{2}h}{4}+\frac{h(r^{2}-a^{2})^{3/2}}{3r}[/itex]

    [itex]a = \frac{2rL}{h}-r[/itex]

    It's much simpler the leave a as a separate equation, solve for a and then enter it into the larger equation. If this formula were correct, you would expect it to equal zero when L equals zero, which it does. You would also expect it to equal h∏r^(2)/2 when L = h except it equals the negative of that. I've worked hard on this and am trying to get it right. Is something wrong with my setup? Thanks.
     
    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 22, 2014 #2

    verty

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    In what way is the bottom sloped? This I don't understand. Place the cylinder so that the water starts at z=0. If the bottom is sloped, this must be a sharp edge. What is the angle of rotation of the cylinder? What is angle of rotation of the cutting plane?
     
  4. Jul 22, 2014 #3
    The cylinder is vertical. The rise of the slope is h, and the run is 2r, so the angle is arctan (h/2r). It's just cut diagonal from the bottom leftmost edge to the top rightmost edge, then discard the bottom portion.
     
    Last edited: Jul 22, 2014
  5. Jul 23, 2014 #4

    verty

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    Sorry for taking so long to respond. I've gone through the calculations now, I do get the correct sign. Look again at ##sin^{-1}(-1)##, that is the only error I see. In case that doesn't make sense, your third term is incorrect.
     
  6. Jul 23, 2014 #5
    Thanks for the help, I definitely appreciate it. I can make the formula give the correct answer if I choose arcsin(-1) = -pi/2 rather than 3pi/2, but that seems rather arbitrary. How am I supposed to know that the correct answer is the first and not the second?
     
  7. Jul 24, 2014 #6

    verty

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    I made the substitution ##x = r \; sin(u)##. The integration dx runs from -r to a. This gives ##-r = r \; sin(u)##, ##u = arcsin(-1)##. The other boundary gives ##a = r \; sin(u)##, ##u = arcsin(a/r)##. The point is that these need to be within one revolution of each other (or perhaps it is always the nearest solution, not sure) and the lower boundary must be less than the upper boundary. I'm not entirely sure actually but you should be able to work it out.
     
  8. Jul 24, 2014 #7

    verty

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    Let me try and be more clear. X is being translated into an angle, x has a range: -r to a. These need to appear on the same S-shape of the sine curve, the part from -pi/2 to pi/2. They both need to appear in this interval and the bounds must be in the correct order.
     
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