Find the number of rolls required to wrap a $330$ ft long cylinder with an initial diameter of $1.75$ inches $5$ times. Each Roll is .$408$ inches thick, $24$ inches wide, and $20$ ft long. Each wrap must overlap $2$ inches.
I tried two methods, volume and surface area my surface area method was find the SA of the cylinder at each wrap and divide it by the rectangular area of the roll then add them all together, then add the overlap after (kind of as an afterthought).
Area method is find the area of the new outer wrapped cylinder minus the area of the small internal cylinder
SA method i got $36.83$ with the overlap adding an additional $6.875$ rolls for a total of $~43$ rolls A method i got $40.93$ with the overlap adding an additional $6.875$ rolls for a total of $~48$ rolls
I need to be as accurate as humanly possible with this, and am struggling to understand why the difference between the two is so large. Conceptually, to me, they should both solve the problem.
Any ideas on how to get a very accurate answer here?
I will assume we measure the length of a wrap at the center of the thinckness. This has the appealing feature that the volume of wrapping material stays constant as we wrap. The radius of the first wrap is then $(1.75+.408)/2=1.079$ and the radii increase by $0.408$ each time. Your five wraps then have radii of $1.079, 1.487, 1.895, 2.303,$ and $2.711$. The total length is $2\pi (1.079+ 1.487+ 1.895+ 2.303+2.711)+10=18.95\pi+10$ were the $+10$ comes from the overlaps. The area to cover is then $330\cdot 12 \cdot (18.95 \pi +10)$ and if we can use rolls perfecly we need $\frac {330\cdot 12 \cdot (18.95 \pi +10)}{24\cdot 240}$ rolls. So far, this is perfect if your dimensions are. It comes out about $47.804062$ rolls