Volume of a Conical WedgeDate: 03/04/2002 at 17:14:12 From: Roger Staiger Subject: A frustum problem The problem is this: Two porters agree to drink off a quart of strong beer between them, at two pulls, or a draught each, now the first having given it a black eye, as it is called, or drank till the surface of the liquor touched the opposite edge or the bottom, gave the remaining part of it to the other. What was the difference of their shares, supposing the pote was a frustum of a cone, the depth being 5.7 inches, the diameter of the top 3.7 inches. and that of the bottom 4.23 inches? The answer I've been given is 7.05 cubic inches, arrived at by the formula (D^(3/2) - d^(3/2))^2 / (D-d) * (pi h/3) where D = 4.23 d = 3.7 h = 5.7 I approached this problem with the formula for the a frustum of a cone. But apparently, I have made some logic error in switching between volume of the frustum and the area of trapezoids. But I cannot replicate this formula. You assistance is appreciated. Thanks, Roger Staiger, Scotland MD Hi, Roger. I don't recall ever seeing a formula for the volume of a "conical wedge" like this (or a name for it, for that matter); and I can't find one anywhere either. Instead, I had to derive a formula for the volume of the part of a cone left when part is cut away at an angle: pi h R1^(3/2) [R1^(3/2) - R2^(3/2)] V1 = ---- * ------------------------------ 3 R1 - R2 The remaining part of the frustum of height h can be found just by reversing R1 and R2, turning the frustum upside-down: pi h R2^(3/2) [R1^(3/2) - R2^(3/2)] V2 = ---- * ------------------------------ 3 R1 - R2 The difference V1 - V2 is what we want, and subtracting these formulas will indeed give the formula pi h (R1^(3/2) - R2^(3/2))^2 Vdif = ---- * ----------------------- 3 R1 - R2 Let's look at how I derived the formula for the volume of a "conical wedge." First, I approached it in terms of an oblique slice through a complete cone: + ------------------------ /|\\ d ^ / | \ \ | / | \ + | / | \ / | / | + | / | / |\ | / | / | \ | / + | \ | / / | | \ |H / / | | \ | / / | | \ | 2r/----+------+----|------\ | / /| | |h \ | / / | | | \ | / a/ h| | | \ | / / -| | | \ | / / 2| | | \ | / / | | | \ | // | x/2 | | x \ v +------------+------+----+--------------+ ---- | R | R | We know R and H, the radius and height of the full cone, and h, the height of the cut. I'm going to find the volume of the top part of the cone above the slice, by seeing it as an oblique elliptical cone. That is, its base is the ellipse cut from the cone, and its height is d, the perpendicular distance from this base to the apex. I have the find the semiaxes a and b of the ellipse, and the height d. Similar triangles give me the horizontal position of the end of the cut: x/h = R/H x = Rh/H and the radius of the cone at the level of the center of the cut: r/R = (H-h/2)/H r = R(2H-h)/(2H) The Pythagorean theorem gives a^2 = (R - x/2)^2 + (h/2)^2 = [(2R-x)^2 + h^2]/4 where a is half the length of the cut. (We won't actually need to have done this!) To find b, half the width of the cut, consider a horizontal cross-section through the cone at the center of the cut: ********* **+ *** * |\ * * | \r * * b| \ * * |x/2\ * *-----+----+----------* * | / * * b| / * * | /r * * |/ * **+ *** ********* Pythagoras gives b^2 = r^2 - (x/2)^2 Plugging in the values of r and x, b^2 = R^2(2H-h)^2/(2H)^2 - R^2h^2/(2H)^2 = R^2/(4H^2) (4H^2 - 4Hh + h^2 - h^2) = R^2/(4H^2) ((2H-h)^2 - h^2) = R^2(H-h)/H Now we have to find d. I'll redo the drawing with some different lengths labeled for clarity: + /|\\ d / | \ \ / | \ + / q| \ / / | + / | / |\ / | / | \ / + | \ / / | | \ / / | | \ / / | | \ / t / | | \ / / | |h \ / / p| | \ / / | | \ / / | | \ / / | | \ / / | | \ // | | x \ +-------------------+----+--------------+ | R | R | Similar triangles give p/R = h/(2R-x) so p = RH/(2R-x) = RH/(2R - Rh/H) = H^2/(2H-h) and q = H-p = H - H^2/(2H-h) = (2H^2 - Hh - H^2)/(2H-h) = 2H(H-h)/(2H-h) Again, by similar triangles, remembering that 2a is the length of the cut, t/R = 2a/(2R-x) so t = 2Ra/(2R-x) = 2Ra/(2R - Rh/H) = 2aH/(2H-h) Finally, similar triangles give d/q = R/t so d = q*R/t = R(H-h)/a Now we can find the volume of our oblique elliptical cone: Vt = 1/3 Bh = pi/3 abd = pi/3 * b * ad = pi/3 * R sqrt[(H-h)/H] * R(H-h) R^2 * (H-h)^(3/2) = pi/3 * ----------------- H^(1/2) Now we'd like to get this in terms of R1, R2, and h, where R1 is the bottom radius, R2 is the radius of the top of the frustum containing the wedge, and h is the height of that frustum. We have to find H given R2: + ----------------------- /|\ ^ / | \ | / | \ | /R2 | R2\ | +----+----+ | / | \ | / | \ | / | \ | / | \ |H / | \ | / | \ | / h| \ | / | \ | / | \ | / | \ | / | \ | / | \ | / | \ | / | \ v +-------------------+-------------------+ --- | R1 | R1 | Similar triangles give R2/(H-h) = R1/H R1H - R1h = R2H H = R1h/(R1-R2) H-h = R2h/(R1-R2) Plugging these into our volume, R1^2 * (H-h)^(3/2) Vt = pi/3 * ------------------ H^(1/2) R1^2 * R2^(3/2) h^(3/2) (R1-R2)^(1/2) = pi/3 * ----------------------- * ---------------- (R1-R2)^(3/2) R1^(1/2) h^(1/2) R1^(3/2) R2^(3/2) h = pi/3 * ------------------- R1-R2 Now we can find the volume of the bottom part, the actual wedge we are looking for, by subtracting the top part from the whole cone: Vc = pi/3 R1^2 H = pi/3 R1^3 h / (R1-R2) Vb = Vc - Vt R1^3 h R1^(3/2) R2^(3/2) h = pi/3 * ------ - pi/3 * ------------------- R1-R2 R1-R2 R1^(3/2) h [R1^(3/2) - R2^(3/2)] = pi/3 * -------------------------------- R1-R2 I then found the volume of the top part of the frustum by swapping R1 and R2 (you could also subtract Vb from the volume of the frustum), and then found the difference. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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