Determining the Length of a Coil of RibbonDate: 08/31/2001 at 15:28:35 From: George Subject: Using Calculus to determine the length of a coil of ribbon I have a coil of ribbon. The ribbon spirals around 34 times. The interior radius (empty) is 1.12". The exterior radius is 2.22". I solved this problem just by using the average circumference for one loop. I found this by averaging 1.12 and 2.22, 1.67" multiplied by 2pi. By division, I found the width of the ribbon was about .0333". So how will calculus give me the same answer I found with algebra and geometry (29'9")? Date: 08/31/2001 at 16:51:32 From: Doctor Fwg Subject: Re: Using Calculus to determine the length of a coil of ribbon Dear George, Herein and below are a few possibilities. Only the third method involves calculus: TOTAL LENGTH OF SPIRAL GROOVE OR ROLL OF TAPE METHOD I (Area Method): There are several ways to solve this problem. However, a few dimensional measurements are required first. These measurements must provide the outermost radius of the groove (Rmax), the innermost radius of the groove (Rmin), and the distance between successive turns in the spiral groove (D). One must also assume that D is a constant. If not, the following method will not be very accurate. In any case, measuring Rmax and Rmin is pretty straightforward. To estimate D for a spiral groove, mark out the distance between about 11 (or 101) successive groove turns and measure the distance between them with a good ruler (one with fine divisions). If the distance between 11 successive groove turns is X cm, divide by 10 to get the distance between two successive groove turns. If the distance between 101 groove turns is measured, divide by 100 to get the average distance between any two successive groove turns. For better accuracy, use a magnifying lens or other low power optical device to count the number of groove turns. Next, calculate the area (A) of the annulus between Rmax and Rmin, where: A = Pi [(Rmax)2 - (Rmin)2] (Note: Pi is approx. 3.14159...) Next, imagine that it is possible to completely "unwind" the spiral groove so that the thickness of the resulting strip of material is exactly equal to D. The total length (L) of this spiral is the unknown quantity. However, the top edge area (L x D) of this long strip of imaginary material is equal (almost exactly) to A (see above). So: L x D = Pi [(Rmax)2 - (Rmin)2] The only unknown here is L. One can solve for L pretty easily. This is also a pretty neat way to find the total length of material on a long coiled roll of tape. In that case, a micrometer may be used to measure the thickness (D) of a small section of the tape instead of counting a large number of "coils" or turns. METHOD II (Summation Method): 1) Measure distance between the center of a disk and the first turn of its spiral groove. Let this radius be Ro. 2) Measure the distance between 11 (or 101) successive turns in the spiral groove. Let this value be X. Divide this distance X by 10 (or 100). This ratio (i.e., X/10 or X/100) gives the average distance between any two successive turns in the spiral groove. Let this average distance between any two successive turns in the groove be D. 3) Measure the distance between the innermost turn of the spiral groove and the outermost turn. Let this distance be H. 4) Multiply H by X/10 (or X/100 if the distance between 101 successive groove turns was measured). This gives a good estimate for the total number of groove turns (N), so one doesn't have to count them all. Accuracy here will depend on the accuracy of the dimensional measurements. If one can ignore mistakes in counting, there is absolutely no error in counting an integer number of anything. 5) If it can be assumed that each spiral turn is approximately circular, it is possible to calculate the circumference of each "ring," then add them all together to get the total length. Even though this may seem a little tedious, there is a fast way to do this. 6) Let C be the total length of the spiral groove so: C = 2(Pi)Ro + 2(Pi)(Ro + D) + 2(Pi)(Ro + 2D) + 2(Pi)(Ro + 3D) + ...+ 2(Pi)(Ro + [N -1]D). 7) Notice that the total number of 2(Pi)Ro terms is N, so factoring out the 2(Pi)Ro terms and multiplying by N yields: C = 2N(Pi)Ro + 2(Pi)[D + 2D + 3D + ...+ (N -1)D]. 8) Now factoring out the D terms yields: C = 2N(Pi)Ro + 2(Pi)D[1 + 2 + 3 + ...+ [N -1]]. 9) Now if it is noticed that the sum (S) of a series of successive integers (starting from 1 and increasing by 1 unit at a time up to the Nth integer), is: S = (N)(N + 1)/2 or in the case of (N - 1) terms: S = (N -1)(N - 1 + 1)/2 = N(N - 1)/2. 10) So: C = 2N(Pi)Ro + 2(Pi)D[N(N - 1)/2]. 11) Factoring out the 2N(Pi) terms, one gets: C = 2N(Pi)[Ro + D(N - 1)/2]. So, by estimating values for D and N, one can find a good estimate for the Total Length of the Spiral Groove. METHOD III (Calculus Method): 1) Measure the distance between the center of a record and the first turn in the spiral groove. Let this radius be Ro. 2) Measure the distance between 11 (or 101) successive turns in the spiral groove. Let this value be X. Divide this distance X by 10 (or 100). This ratio (i.e., X/10 or X/100) gives the average distance between any two successive turns in the spiral groove. Let this average distance between any two successive groove turns be D. 3) Measure the total time needed to "play" the record. Let this time be T (in minutes). 4) Note that the radius (R) at any point in time is: R = Ro + [D/(2{Pi})]beta, Where beta is the angle (in radians) that the record has turned through at any intermediate time between 0 and T (minutes). Note: when beta equals 2(Pi) radians, the record has turned around once, so the value of R at that point in time is [Ro + D], and so on. Note: it is assumed here that the record is running backward because the play time is independent of rotational direction. In any case, it doesn't make any difference in the answer provided that the distance to the innermost part of the spiral groove can be measured just as accurately as the distance to the outermost turn in the spiral groove. 5) Since we now have R as a function of beta, it is possible to find the total distance (S) that the record needle travels through for any value of beta. The differential form looks like this: dS = R dbeta. 6) The integral limits on S are between 0 and S and on beta between 0 and beta(max). 7) Note that beta(max) = [2(Pi)(33.3)T], and the units here are in radians. It is also assumed here that the angular turning velocity of the record is 33.3... rev/min. 8) Now, rewriting the differential form: dS = [Ro + (Dbeta)/(2{Pi})] dbeta. 9) Now, integrating between the limits one should end up with the following: S = Ro(beta) + [D/(2{Pi})][beta2/2]. 10) Use the value: beta = beta(max) = [2(Pi)(33.3)T] in the equation above to find S. 11) The expression for S (above) can be simplified somewhat, if desired, and it will produce the quantity of interest but one still has to measure or calculate values for certain necessary variables. In this case, one needs to measure or calculate values for Ro, D, and T. However, if the angular velocity is known to be 33.3...rev/min, this value can be used as is shown above. One should also keep in mind that the relative velocity of the needle, with respect to any particular point within the spiral groove, is not constant in this problem even though the angular velocity is constant.The needle can cover a much greater distance per unit time near the outermost part of the spiral groove versus the innermost part of the spiral groove because the angular rotational velocity is constant. I hope this has been helpful. - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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