Expert: Sherry Wallin - 6/7/2010

Question
QUESTION: How many circles of a same diameter can be fitted in a rectangle Please provide a simple formula which we can use regularly

ANSWER: Rahul~
    There must be more to the question than this. Are you talking about exactly fitting as in the edge of the circles are tangent to the sides of the rectangle? (tangent means the circle and the edge of the rectangle touch at a single point. For example 4 circles with a diameter of 2 inches can fit into a rectangle that is 2 inches by 8 inches. Or a rectangle that is m by n and the circle has a diameter of m can fit n/m circles provided n/m is a natural number.  Or a rectangle that is 2 inches by 2 inches can fit just one circle with a diameter of 2 inches. (Yes all squares are also rectangles). Can you please be more specific?

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Thank you.. As suggested i will make my question more specific

We need to know how many drums in the base of the container can be fitted.. for example we have the diameter of the drum say 407 mm, the container floor dimensions are Length - 6096 mm and width - 2370 mm.
The drums can touch the wall of the container as well as they can touch to each other. Height is no concern

Need to have a formula which can help me to find out the no of drums which can be there with different diameter

Thank You

ANSWER: Hi Rahul~
    I actually did provide you a formula...divide the length of the rectangle by the diameter of the drum, this will tell you how many rows of drums you can fit and then divide the width by the diameter of the drum and this will tell you how many columns and then the whole number parts of each answer when multiplied together will tell you how many you can fit. In your last example you have a drum that is 407mm so the length is 6096 divide it by the 407 getting 14.98...this means you can only fit 14 whole drums along the length. Now divide the 2370 by 407 and you get 5.82... so you will only get 5 drums in a width and the total drums you will fit in this rectangular container is 14(5) = 70.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Thank you..
What you explained is correct if we place one next to each in an row and column,
but what if we try to fit between the gaps of 2 drums (like forming a triangle with 3 drums) we need to find the maximum drums can be fitted in the container considering the optimum utilization of the floor area

Answer
Interesting speculations...If you wanted to put in a row of drums and get let's say 4 across each with 1m diameters which means the length is at least 4m but less than 5 mm. Then when you place the next row of drums say in between drum 1 and 2 you are saving a bit of space in the width but you are not going to get 4 in the next row. Call this 1st drum in the second row drum 5, then it would be placed under drum 1 and drum 2 and it's diameter will go from .5m to 1.5m and since all you have is at least 4m but not 5m you would get drum 6 from 1.5m to 2.5m and drum 7 from 2.5m 5o 3.5m and drum 7 from 3.5m to 4.5m assuming you have at least 4.5m. So now instead of getting 2 rows with 4 drums each you have accomplished getting 7 drums in two rows that instead of taking up 2m in width it is taking up slightly less than 2m but not enough significantly to justify staggering the rows for space. I've answered a similar question before about large rolls of paper that vary in diameter but if your drums all have the same diameters you are doing better to just line them up in rows. If you want to calculate the amount of "unused space" between say two rows of two drums that can be done but you will find it is less than the space you lose in staggering your rows. Sorry if this isn't the answer you were expecting.

I can make drawings and show you how you can calculate the space that isn't used in one configuration and the space that isn't used in another if you like. But try drawing this:
Use a quarter and draw two rows of two quarters, draw lines surrounding the quarters that are perpendicular to each other and then call what you have a 2 unit by 2 unit square. Calculate the area of the square: 4 units^2 and calculate the area of the quarters approx
3.142 units^2 and then find the difference being about .858 units^2 (which is your unused space). There are empty spaces that are multiples of each other. Across the top going left to right there is a space that is approx .0536 units^2 and the next is twice that and the next is the same as the first and if you go to the empty spaces between the two rows there is another double and a quadruple and another double and the bottom has a single, a double and a single. You can break each of these up and see that you have 16 equally sized parts of .0536 units^2. Now draw two rows of two quarters and stagger them and you will have a rectangle that is 2.5 units by some number slightly smaller than 2 units. I have calculated that number but I am not sure how much of the math you can do, it requires trigonometry. But now you have empty space of across the top going left to right 4 @ .0536 units^2 plus a big chunk (close to .5 units^2) and then in the next row 2 pieces that are .05236 units^2 and two pieces slightly smaller and then the bottom row has another big chunk and 4 more pieces that are .0536 units^2. Hopefully you can see that what you saved is not anywhere near what you give up!

Upon reading your 2nd version of your question again it may be as simple as taking the measurement of the length and dividing it by the diameter and then take the width and divide it by the diameter and again as I said in an earlier reply multiply the whole number parts of each of the answers to those divisions and that will tell you how many you can place in the space. Let me try once more with a specific example. (hopefully above I have convinced you your best utilization of space is in NOT staggering your drums). Suppose your length is 53 ft and your width is 12 ft and your drums have a diameter of 3.5 ft. Divide 53 by 3.5 = 15.14 so that is 15 drums down the length and now divide 12 by 3.5 = 3.43 and that is 3 across the width but you will be able to carry up to 15(3)= 45 drums of diameter 3.5ft on a trailer that measures 12ft x 53ft. Again create the drawing you will see the most efficient use of space is to make rows that are uniform.


Math Prof