Determining Cone's Original Dimensions from a SliceDate: 05/18/99 at 11:33:33 From: Mike Branson Subject: Truncated cones revisited Dr. Math, You have helped us tremendously with the truncated cone problem I presented you several weeks ago. Thank you. I have another question for you along the same lines. We have a customer who has provided us with some dimensions, but does not know how to measure the angle. We have a large arc length of 64", a small arc length of 41", and a length of 26" between these two arc lengths. Do you know how we can make this work? Thanks again, Mike Branson of Sefar America Date: 05/18/99 at 16:00:36 From: Doctor Rob Subject: Re: Truncated cones revisited Thanks for writing to Ask Dr. Math! I think the situation you are describing is as follows: 41 __,,---''''---..__ _,--' `--._ _,-' `-._ o--------------------------------------o `. 26 ,' `._ _,' `._ _,' `-._ _,-' ``--..__,,--'' 64 You can use the formulas on the following Ask Dr. Math FAQ web page to find out what you need to know about these two segments with a common chord: Circle Formulas: Segment of a Circle http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment If this is not what you meant, write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/18/99 at 16:53:33 From: Mike Branson Subject: Re: Truncated cones revisited Dr. Math, I think I might need to explain my scenario a little better. The dimensions I gave you could be most easily visualized as a truncated cone which is cut down the seam and laid flat. We can draw it on autocad by first drawing a circle and offsetting it with another circle by 26". Then we draw a 90 degree angle from the center and remove the portion of the two circles outside of the ninety degree angle. Next we needed to play with the angle to allow us to have one arc of 64" and one arc of 41". It looks sort of like this: + + 64 + + + + + + + + 41 + + + + + + +------+ +------+ <- 26 -> <- 26 -> My question is: Given the arc length, can you determine the circumference of a circle? And, if given the above figure, can you calculate the angle (or an angle) that is needed? Date: 05/18/99 at 17:04:54 From: Doctor Peterson Subject: Re: Truncated cones revisited Hi, Mike. I see Doctor Rob got to you before me, without the knowledge of your earlier question. Here's my understanding of your problem: *********** ******...........****** S2=64" *** \.....................*** *** \.......................*** * \.........................* * L=26"\.........................* ** \.........................** * \..........................* * \.*****.S1=41".............* * *** ***.................* * * \ *.................* * * \A *................* * * +------*----------------* * * | R1 * L=26" * * * | * * * *** | *** * * **|** * * | * ** |R2 ** * | * ** | ** ** | ** *** | *** ******* | ******* ********* Given two arc lengths S1 and S2, and a segment length L, you need to find the two radii R1 and R2 and the angle A. Three numbers to find three numbers - sounds feasible! Let's write equations for the numbers you have in terms of the numbers we want: S1 = A * R1 (A is the angle in radians) S2 = A * R2 L = R2 - R1 Now we can solve this for A, R1, and R2: S2 - S1 = A(R2 - R1) = A*L so S2 - S1 A = ------- radians L S1 S1 * L R1 = -- = ------- A S2 - S1 S2 S2 * L R2 = -- = ------- A S2 - S1 With your numbers, 180 deg A = (64 - 41)/26 = 0.8846 radians * ---------- = 50.68 degrees pi radians 41*26 R1 = ----- = 46.35 inches 64-41 64*26 R2 = ----- = 72.35 inches 64-41 As before, of course, this doesn't include any allowances for seams. I think the only real difference from the last time is that you have given the circumference rather than the diameter of the end circles of your cone. The formulas this time should be a little easier to follow. Incidentally, if you are interested in the vertex angle of the cone itself (after it is rolled up), that will be the arcsin of the ratio of the base radius to the slant height, or S2 A S2-S1 A_vertex = arcsin(-------) = arcsin(----) = arcsin(------) 2 pi R2 2 pi 2 pi L = arcsin(0.8846/6.28) = 8.09 degrees - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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