From these results I can see that having a larger base and smaller height gives a larger volume. Also the square has a larger volume than all the cuboids suggesting that regular shapes give larger volumes. The results for the cuboids with a 32cm base are shown on the graph on the following page. This graph shows a gradual rise up to when the tube is a regular based cuboid, at which point it begins to fall, helping to show the prediction above.
The next shape I will use as a base for a tube will be pentagon. Working out the area of the base is more difficult than that of the square as more things must be done in order to work out the area. This includes breaking the pentagon into triangles and using tangents. The full method for working out the volume is shown below, firstly for a 24cm base then a 32cm base.
First the pentagons internal angles are calculated using the equation a=180(n-2), where a is equal to the total internal angles and n is the number of sides. This is, therefore
a=180(5-2)
=180x3
=540°
The internal angles are then divided by the number of vertices to get the angle in each corner. The pentagon is then split up into five equal triangles and one of these is then split in half, as shown above. From this only one length is known but in order to work out the area of the base length x is needed. To work this out we need to use the length and angle we know and trigonometry. This stage is shown below
x=tan54x2.4
=3.30(2d.p.)
The final stage is working out the area of the base. This can now be worked out easily by using what was found in the last stage. The height of the triangle is now simply multiplied by the base of the small triangle and the answer is multiplied by the number of sides. So this is
V=(bxh)
=32(5(2.4x3.3))
=32(5x7.92)
=32x39.64
=1268.47cm3
The 32cm pentagonal base would, therefore be worked out as:
x=tan54x3.2
=4.40(2d.p.)
V=(bxh)
=24(5(3.2x4.4))
=24(5x14.09)
=24x70.47
=1691.30cm3
I shall now use this method to work out the volume of hexagonal tubes, then octagonal tubes.
24cm hexagonal based tube.
a=180(6-2)
=180x4
=720°
x=tan60x2
=3.46(2d.p.)
V=n (bxh)
=32(6(2x3.46))
=32(6x6.93)
=32x41.57
=1330.22cm3
32cm hexagonal based tube
a=180(6-2)
=180x4
=720°
x=tan60x2.6
=4.62(2d.p.)
V=n (bxh)
=24(6(2.6x4.62))
=24(6x12.32)
=24x73.90
=1773.62cm3
24cm based octagonal tube
a=180(8-2)
=180x6
=1080°
x=tan67.5x1.5
=3.62(2d.p.)
V=n (bxh)
=32(8(1.5x3.62))
=32(8x5.43)
=32x43.46
=1390.59cm3
32cm based octagonal tube
a=180(8-2)
=180x6
=1080°
x=tan67.5x2
=4.83(2d.p.)
V=n (bxh)
=24(8(2x4.83))
=24(8x9.66)
=24x77.25
=1854.12cm3
The final shape I will try as a base for a tube will be a circle, first a 24cm base then a 32 cm base. To work out the volume for this tube the radius needs to be found first. This is done by the method below
d=c/π d=diameter
=24/π c=circumference
=7.64(2d.p.) r=radius
r=d/2
=7.63/2
=3.82(2d.p.)
Then the area of the base is then worked out as below
b=πr2
=πx3.822
=45.84cm2
Then this is multiplied by the height, in this case 32cm, in order to get the total volume of the cylinder.
V=bxh
=45.84x32
=1466.77cm3
The volume of a 32cm based cylinder is shown below.
r=d/2
=(32/π)/2
=10.19/2
=5.09(2d.p.)
V=h(πr2)
=32(πx5.092)
=32x81.49
=2607.59cm3
The chart below shows the volumes of all of the tubes I have tried.
l=length of base
From this I can clearly see that as the number of sides increases the volume also increases and as a circle has an infinite number of sides it gives the largest volume. Also the tubes which had a 32cm base had a larger volume than the ones with a 24cm base. Therefore a cylindrical tube with a 32cm base is the tube that has the best volume.
The general formula to work out the volume for an ‘n-sided’ shape is shown below and the general formula to work out the volume of a cylinder is also shown below that.
Vn=(l2h)/(4nxtan(180/n))
Vc=(l2h)/4π
The next step is to investigate what shape of paper gives the best volume. To do this I will chose a piece of paper of area Acm2. One length will be xcm and the other A/xcm, as shown below
I will set the value for A and modify the value of x to see what shape paper gives the most effivient tube.
The first value I will try as A will be 100. This is because it is an easy number to divide and multiply. I will start off with x being small and gradually make it larger until I think I have found the optimum size for the paper, and as cylinders are the best tube I will use this as a starting point. The results for this are shown in the table below.
NOTE:x is equal to l and A/x is equal to h
From this table I can see that as the length of the base of the tube increases, so does the volume. More importantly however is the fact that there appears to be no limit to this and the volume will continue to rise until the paper the tube is made from is the shortest and longest it can possibly be.
Therefore in order to make a tube with the largest volume it should have the following properties
- It should have a circular base
- The base should be very long and the height should be small
- If the base cannot be circular then it should be a regular polygon with as many sides as possible
