Flattening the Frustum of a ConeDate: 10/15/2000 at 23:21:06 From: Jonathan Scholl Subject: Calculating the flat layout from the section of a cone We are building a reception desk that has a curved and canted front, i.e. it forms a section of a cone. We know the radius of this section at its top and at the floor and we know the chord length at each of those locations. Our question is; how do we generate a flat layout of this section? We want to lay out veneer so that when wrapped around the section it will fit the shape exactly. Any help would be much appreciated. Thanks, Jon Scholl Date: 10/16/2000 at 17:10:25 From: Doctor Peterson Subject: Re: Calculating the flat layout from the section of a cone Hi, Jonathan. We receive similar questions from time to time for practical problems, and you can find several answers to this sort of thing in our archives; but not quite exactly what you want. I think the shape you are looking for is the shaded part of the lateral surface of a frustum of a cone, cut radially as shown in the top view below; I'll draw it with the smaller end at the top for convenience, though I think you want it upside-down: Side view: R1 |----->| --- +------+--+---+ ^ /| | |...|\ | / | | |..|.\ H| / | | |..|..\ L | / | | |.|...\ | / | | |.|....\ v / | | ||.....\ --- +------+------+-----++------+ |------------>| R2 Top view: *********** *** /**** *** /.....** * *****/........* * *** /***.......* * ** / **......* * * + a *......* * ** \ **......* * *** \***.......* * *****\........* *** \.....** *** \**** *********** The circumferences of the top and bottom circles are C1 = 2 pi R1 C2 = 2 pi R2 We'll also want the slant height L, which we can find from the actual height H using the Pythagorean theorem as L = sqrt[H^2 + (R2 - R1)^2] Now if we take the surface of the WHOLE frustum (I'll reduce this to the section you want shortly), it will look like this when we flatten it out: *********** ******...........****** C2 *** \.....................*** *** \.......................*** * \.........................* * L\.........................* ** \.........................** * \..........................* * \.*****.C1.................* * *** ***.................* * * \ *.................* * * \ A *................* * * +------*----------------* * * | r1 * L * * * | * * * *** | *** * * **|** * * | * ** |r2 ** * | * ** | ** ** | ** *** | *** ******* | ******* ********* Given two arc lengths C1 and C2 (which are the top and bottom circumferences of our whole frustum), and a segment length L (our slant height), we need to find the two radii r1 and r2 and the angle A in order to draw the right shape. Let's write equations for the numbers we have in terms of the numbers we want: C1 = A * r1 (A is the angle in radians) C2 = A * r2 L = r2 - r1 Now we can solve this for A, r1, and r2: C2 - C1 = A(r2 - r1) = A*L so C2 - C1 A = ------- radians L C1 C1 * L r1 = -- = ------- A C2 - C1 C2 C2 * L r2 = -- = ------- A C2 - C1 Plugging in the formulas for C1, C2, and L in terms of R1, R2, and H, and changing the angle to degrees by multiplying it by 180/pi, we get L = sqrt[H^2 + (R2 - R1)^2] R2 - R1 A = ------- * 360 degrees L R1 * L r1 = ------- R2 - R1 R2 * L r2 = ------- R2 - R1 I can simplify this a bit more by first calculating L H X = ------- = sqrt[(-------)^2 + 1] R2 - R1 R2 - R1 and using that to get A = 360/X degrees r1 = R1 * X r2 = R2 * X Now, that gives you the whole frustum. To get just one part, you have to determine the angle I called "a" on the original top view. You can find this from the chord D2, and the corresponding radius R2 (or the corresponding pair at the other end): ***********+ --- *** /**** ^ *** R2/.|...** | * *****/..|D2/2.* | * *** /***|......* | * ** /a/2 |*......* | * * +-----+*------* | D2 * ** \ **......* | * *** \***.......* | * *****\........* | *** \.....** | *** \**** v *********** --- Then D2/2 sin(a/2) = ---- R2 and a = 2 arcsin(D2/(2 R2)) Now you just have to take a/360 (if you found A in degrees) of the angle in our solution, so that we multiply the angle A by a/360: A = 360/X * a/360 = 2 arcsin(D2/(2 R2)) / X degrees r1 = R1 * X r2 = R2 * X Use these radii and this central angle, and you have a pattern for the surface you need. Note, however, that under my assumptions of the shape, you will find that D1/R1 = D2/R2 If this is not true, then the shape is not what I assumed; in particular, the side edges will not be straight. Let me know if I can help further. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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