For this project we had to investigate the volumes of open-ended tubes made from a rectangular piece of card with the dimensions, 32cm by 24cm.
Maths Coursework - Tubes
For this project we had to investigate the volumes of open-ended tubes made from a rectangular piece of card with the dimensions, 32cm by 24cm. Either of the two sides could be folded to make the perimeter, with the other side being the height. I made a couple of conjectures before I started. They were:
· the more sides, the bigger the volume
· regular polygons are the best
. For the first question, the volumes have to be investigated from various polygon shapes as the base side.
The first shape I used was a square, with a height of 24cm, and perimeter of 32cm:
To find the area of the base, I used the formula: A = x2, where 'x' is the side of the square. Once the area was found it was multiplied by the height.
A = 82
V = 1536cm3
For a square with height 32cm and perimeter 24cm, it is:
A = 62
V = 1152cm3
From the method I used, a formula can be found for an open ended tubes with square bases. If we let 'h' be the height, and 'p' be the perimeter, it would be:
V = p 2 x h
4
V = p 2 x h
16 essaybank.co.uk
V = p2h
16
The next shape was an equilateral triangle based tube:
2
3
0
h = 24
cm
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2
To find the base area the formula: A = b x h x , with 'b' the base of the triangle, and 'h' the height.
2
A = 10.66 x V(10.662 - 5.332) x wwbb bbw esbbbbs aybb bbba nbb kcbb bbuk;
V = 1182cm3
This is the result for a triangle with height of 32cm, and perimeter of 24cm:
2
A = 8 x V(82 - 42) x
V = 887cm3
From this a formula could be established:
2
V = p x V p 2 - p 2 x x h wwdb dbw esdbdbs aydb dbba ndb kcdb dbuk;
3 3 6
V = p x V p2 xh
6 12
The next shape I used as a base was a hexagon.
each of the sides has a length of 4cm
h = 32
For this, the hexagon was split up into 6 equilateral triangles, with the area of one of those found then multiplied by 6 (the number of triangles). This meant the same formula was used as with the triangle, but multiplied by 6.
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2
A = 4 x V(42 - 22) x x 6
V = 1330cm3
Using the other ...
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The next shape I used as a base was a hexagon.
each of the sides has a length of 4cm
h = 32
For this, the hexagon was split up into 6 equilateral triangles, with the area of one of those found then multiplied by 6 (the number of triangles). This meant the same formula was used as with the triangle, but multiplied by 6.
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2
A = 4 x V(42 - 22) x x 6
V = 1330cm3
Using the other dimensions with 24cm the height, the volume was:
2
A = 5.33 x V(5.332 - 2.662) x x 6
V = 1773cm3
The formula was deduced from the method, which was simplified down:
2
V = p x V p 2 - p 2 x x 6 x h
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V = p x V p2 x h
4 36
The next shape was a cylindrical tube.
h = 24
p = 32
A = pr2
A = 32 2 x p
2p
V = 1956cm3
For the cylindrical tube with a height of 32cm and circumference of 24cm, the volume was:
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A = 24 2 x p
2p
V = 1467cm3
From these two results, a formula could be deduced, using the same method used, but turning it into an algebraic equation, by substituting in 'p' and 'h'.
V = p 2 x p x h
2p
V = p2 h
4p
I collected results from other shaped bases, and they are recorded in the table below:
no. of sides
perimeter
(cm)
height
(cm)
volume
(cm3)
cylinder
24
32 wwab abw esababs ayab abba nab kcab abuk.
467
cylinder
32
24
955
3
24
32
887
3
32
24
182
4
24
32
152
4
32
24
536
6
24
32
330
6
32
24
773
8
24
32
391
8
32
24
854
0
24
32
418
0
32
24
891
2
24
32
433
2
32
24
911
All of these bases were constructed from regular polygons, as they have the optimum area. Showing the volume of a tube with height of 32cm, and a base of 4cm by 8cm can prove this.
A = 4 x 8
V = 1024cm3
This value is more then 100cm3 less when compared to the volume of a tube with height 32cm, and base of a 6cm square.
This coursework from www.essaybank.co.uk
This proves that a regular polygon based tube has the optimum volume.
These results show a pattern, that the more sides there are, the greater the area. However, a cylinder has the greatest volume, as it is infinity sides.
The more sides there are, the greater the volume.
The graph shows the relationship between the no of sides and the volume. It is clear that the 'curve' evens off, but it will never reach a straight line. The lines cannot be joined as there are no polygons with decimal no of sides, eg 5.5 sides. The results are from tubes with a perimeter of 32cm, and height of 24cm.
2. This question involved taking the previous question a step further, and investigating volumes of tubes made from any dimensions of card.
I started off by using the dimensions 10cm by 100cm. I only found the volume of four polygons, as I observed a pattern emerging.
no. of sides
perimeter
(cm)
height wwdc dcw esdcdcs aydc dcba ndc kcdc dcuk.
(cm)
volume
(cm3)
cylinder
0
00
796
cylinder
00
0
7960
3
0
00
481
3
00
0
4811
4
0
00
625
4
00
0
6250
6
0
00
722
6 wwff ffw esffffs ayff ffba nff kcff ffuk;
00
0
7220
To see if my hunch, about the larger perimeter and so larger volume, deduced from the previous results, is constant, I used the lengths 15cm and 40cm:
no. of sides
perimeter
(cm)
height
(cm)
volume
(cm3)
cylinder
5
40
716
cylinder
40
5
910
3
5
40
433
3
40
5
155
4
5
40
563
4
40
5
500
6
5
40
650
6
40
5
732
It is obvious that there is a pattern, where the larger the perimeter, the larger the volume.
This proves that when the folding length is greater, so is the volume.
While collecting results, I have observed a pattern in the process of obtaining the volumes, whereby each polygon is split into triangles by drawing lines for the central point to each of the vertices. These triangles are bisected by dropping a perpendicular to the base, thereby forming two right-angled triangles. Then, the area of the triangle is worked out and then multiplied by twice the number of sides.
i) ii) from www.essaybank.co.uk
iii) iv) The area can then be found by using
angle is 360 divided by twice no. of sides
tangent of the angle.
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Using this principle a general formula can be deduced. However, it will only work for bases with more than one side, as for cylinders you use a different method. For the formulae, 'n' is the no. of sides:
tan 360 = p/2n 'h' is height of triangle
2n h
h = p/2n_
tan 180/n
h = p _
2n x tan 180/n wwga gaw esgagas ayga gaba nga kcga gauk.
Then you need you to multiply the height by half base.
2
A = p x p/2n x
2n x tan 180/n wwdf dfw esdfdfs aydf dfba ndf kcdf dfuk;
Then you turn it into the volume by multiplying it by the no. of sides and the height.
V = p x p x n x h
2n x tan 180/n 2n
V = p2h _
4n x tan 180/n
The general formula for
regular polygons.
V = p2 h
4p
This is the formula
for cylinders.
From the general formula you can see why the bigger the perimeter is, the greater the volume is. This is because 'p' is squared so it increases in a greater proportion to 'h'. Therefore, it is more productive when 'p' is larger than 'h'.
This formula also shows why the volume increases with the more sides there are. It is because when the top of the equation is kept constant, the bottom is: 4n x tan 180/n. For the volume to be large the answer to the bottom has to small. That means that the 'tan 180/n' gets nearer to zero, quicker than '4n', which goes up constantly, increases. The result of this is that the answer to the bottom line keeps getting smaller, even though the speed of reduction slows down. This leads to the increase in volume with the increase in 'n'. This is not actually indirectly proportional, but it is that principle.
3. This question involves investigating the volumes of tubes made from card with an area equal to that of 24cm by 32cm, (768cm2).
These figures are for a square based tube made from a card with area 768cm2.
perimeter
(cm)
height
(cm)
volume
(cm3)
768
48 FhY1e1P from FhY1e1P essay FhY1e1P bank FhY1e1P co FhY1e1P uk
4
92
92
4
48
672
48
4
2304
92
4
9216
768
36864
7680
0.1
368640
76,800
0.01
3686400
These results show an increase in volume when the perimeter is greater. The perimeter length and the volume are directly proportional,
eg: 'p' from 768 to 7680, 'V' from 36864 to 368640. The perimeter is 1000% bigger in one, and so is the volume.
The perimeter length and the volume are directly proportional wwfe few esfefes ayfe feba nfe kcfe feuk;
This graph shows this:
The graph shows the way in which the volume increases as the perimeter increases. Similarly, the volume can get nearer and nearer to zero with the decrease in value of 'p'.
The area of the card is equal to 'ph', perimeter multiplied by height. This can be substituted into the formulae, thereby removing the necessity of having 'h' in the equation. This leaves the final formulae to be:
V = Ap _
4n x tan 180/n
V = Ap_
4p
general formula cylindrical formula
