The two shapes I have decided to investigate are the cylinder and the Tetrahedron
Cylinders diameter is 3cm , height is 10cm each of the two ends are going to be plastic . so we need to find the length of the of the cylinder when opened out to its net form we know the height is 10cm .
We can do this by working out the circumference which is 3cm x pi (3.14) = 9.42cm
We can discard the ends because these are going to be plastic which gives us a shape of 9.425cm 10cm
9.425cm portrait landscape
+1cm for tabs = 10.425x11
A0 card = 84.1cm x 118.9cm = 9999.49cm² let start with the cylinder shape if we use the shape portrait first, we need to add 1cm to each measurement to give us room for flaps so our measurements change to 11cm x10.425 this gives us an area of 114.67cm² if we divide this into the area of the A0 sheet we get a total number of shapes of
9999.49cm² / 114.67 = 87.20
The only thing is this is not an accurate amount because there has not been taken into account waste and all full cuts you can’t make half of a packaging.
I will test each net shape on the A0 card landscape and portrait
A0 card has a width of 841mm(84.1cm) and the cylinder net has a width of 10.42cm
If I was too divide 10.425 into 84.1 I get 8.067 so I can see I can put 8 across leaving 0.067 remaining (waste)
We then look to see how many we can fit length ways
1189mm(118.9cm) divided by 11cm =10.809 so we can see we can fit 10 whole shapes length ways
And so on. In total we can fit 8x10 potrait =80 whole shapes With 0.067x0.809=0.054 waste .
Lets do the same but this time landscape to see if there is a difference.
84.1/11cm =7.64 so we can use 7 whole shapes across the width
118.9/10.425=11.405 this show we can use 11 whole shapes length ways
Giving us a total of 7x11 = 77 landscape.
Now let’s try the tetrahedron shape we can look at the net of this shape and take it as one big equilateral triangle.using the pythagaros theorem to work out the height A²xB²=C²(height)
13.124 a =
b 15cm 84.1cm(width of A0 )portrait…