January 11th, 2018, 05:43 PM  #1 
Newbie Joined: Jan 2018 From: North Salt Lake Posts: 2 Thanks: 0  Unique Circle Area problem
I have a roll (circle) that contains widgets (pins/lugs). I know dimensions about this roll of widgets and I'd like to estimate when I have 5000, 3000, 2000, and 1000 widgets left on the roll. Here is how I set up the problem to solve: If I know how many widgets I have to start (W) and I know the outside radius dimension of the roll (O), and the inside radius dimension where there aren't widgets (I), then I can figure out roughly how many widgets are left based on the area % and solve for (O) the radius. If I know that the area of the circle when full is pi * r^2 or pi *81=254.5 I can calculate the radius if I know the area by SQRT(area/pi) or SQRT(81) or 9. Which I know is a full roll. If I have 2000 widgets, then I know that the roll area would be 66% full. SQRT((area*.66)/pi)= r or 7.31 inches from the center. But.......this is only accurate if the center is filled with widgets. I know that the core is 7.25 in Diameter in this example. I know that I have to subtract the core, but I'm having a hard time wrapping my head around what that does to the radius number at 3000, 2000, and 1000. I've also included 1 widget as a "gut check" because if my formulas are right, 3000 should be at 9 inches and 1 should be just above 3.625 inches. How do I solve for the radius of this roll when 66% is used from the OD and the ID stays the same? I've tried calculating as if the roll was filled to the center (3000 = 3487) and then subtract it out later, but I can't get anything to work. 
January 11th, 2018, 09:28 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
How did you calculate 3487? I get 3580.9 * 0.16223 = 580.9 approximately (because 3000/(1  0.16223) = 3580.9 approximately). 
January 12th, 2018, 04:00 AM  #3 
Newbie Joined: Jan 2018 From: North Salt Lake Posts: 2 Thanks: 0 
Is that right? Math error on my part maybe. I have 3000 widgets in the image for a full circle and I find that the center core takes up about 16%. 3000 x .016 is 487 imaginary widgets living in the core. Now look what we've done, we've got imaginary widgets! But how do I wrap my head around how to calculate a partial radius? When I remove the inner core numbers, things don't add up. Any help is appreciated. 

Tags 
area, circle, problem, unique 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Area of circle  ma78s  Trigonometry  1  March 21st, 2017 12:55 AM 
Finding the area of a circle inside a square only given the area of the square  skimanner  Math  12  July 8th, 2016 07:55 AM 
Area enclosed by a semicircle and a quarter circle  yeoky  Algebra  4  May 3rd, 2014 01:06 AM 
Area of a Circle problem  Jamers328  Calculus  5  August 6th, 2012 01:02 PM 
Area of square with convex parts  area of circle  gus  Algebra  1  April 17th, 2011 04:25 PM 