I need an equation, to help me get a world record.

T

TerryTrowbridge

New member
Joined
Aug 27, 2019
Messages
4
Hello,
My name is Terry Trowbridge and I’m the art teacher at McCordsville Elementary School. Last year me and my 5th graders decided to try and get a world record, so we made the worlds largest God’s eye. It’s a wooden cross with yarn woven around it.

It turns out that Guinness takes records very seriously and they want to know quantities of yarn that were used. At the time we were just using all the extra yarn that was in my closet.

I know there is an equation that could be formulated to calculate the quantity, but I’m not a math guy, art teacher here, so I’m curious if anybody here could help me out.
Guinness uses the metric system.
Here are my numbers.
Corner to corner 3.82 meters
Top to bottom 2.63
Left to right 2.71
Yarn is about 2 millimeters thick.

I’m looking for an estimate to give them. Any help would be appreciated. Here is a picture of it. A7ACE219-1C7A-4441-85E8-D0E154BEF96E.jpeg
 
The actual count of layers would be helpful. "about 2 mm" is pretty vague. The problem is that it isn't square. With the same number of laps around the structure, you cover 2.63 m in one direction and 2.71 m in the other direction. Thus:

Vertical: 2.63 m / 4 mm = 657.5 layers
Horizontal: 2.71 m / 4 mm = 677.5 layers

20 layers different! How do we reconcile that?
 
All you can do is to estimate, which they may or may not accept. But for a quick estimate, and supposing you are right that the yarn is on average 2 mm wide, you can find the area covered, about 2.63*2.71 = 7.1273 m^2, and then find the length of a rectangle 2 mm wide with that area: 7.1273 m^2 / 0.002 m = 3564 m. Does an answer of 3.5 km sound good to you?

To get a better estimate of the width of the yarn as used, I would either count all the turns, or just count turns in some specified distance and divide to find the average width. I'd repeat that both horizontally and vertically, to see how different it is; and maybe both near the middle and near the outside. Then just average all those estimates.

Do the students have a math teacher who would be interested in doing this with them?
 
3.5 km sounds like a great number, not that I have a number to compare it too:). I can try to count the turns. That’s a good idea to get a more exact width.
I work at an elementary school and I think this math is above them.
Thanks for the response.
 
I started with 2.63 m x 2.71 m multiplied by 2 to represent the length of the outer border.
Subtracting 8 mm for each subsequent lap around gives roughly 667 laps (depending on how you interpret the non-square shape).
This gives 3,571 m; pretty close to the area estimate given above.
It's just a finite arithmetic series.
 
You must log in or register to reply here.
Top