# The Physics of Paper Helicopters

In my physics coursework I intend to study ‘paper helicopters’. A paper helicopter is a piece of plain A4 paper that has been folded, so that when dropped, it spins whilst falling to the ground. Below is a picture of one of the helicopters used.

The top half of the piece of paper is cut down the middle, creating the two wings, and then the rest of the piece of paper is folded up to make the body of the helicopter.

The reason for this choice was that there are many variables that I can change on the helicopter. I intend to have different experiments, with the wing size, mass of helicopter and the height from which it is dropped all being varied. I have done some initial measurements, and found that when dropped down some stairs of height 6m, the helicopter takes about 5 seconds, which is long enough to measure relatively accurately. Although I originally intended to measure the revolutions per second of the helicopter, my initial experiment showed that it would be too difficult to do accurately.

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The basic physics of the paper helicopter

As the helicopter falls through the air, it spins. The basic explanation for this is related to the position of the wings during its flight. As the helicopter falls downward, two forces act on it: its weight and the resistive force. The resistive force is provided in the form of air resistance, almost entirely by the wings.

The wings will be pulled upwards as almost all of the air resistance is acting on them.

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As paper (especially normal A4) is a very flexible substance, the wings bend up. So, in flight, the wings will take a curved shape, with the parts of the wings further from the body bending up more.

When falling, the air will be flowing past the wings, and will be deflected by them. Because the wings are deflecting the air, they are therefore exerting a force on the air. Newton’s third law states that every force has an equal and opposite force acting as a reaction in the opposite direction. From this we can deduce that the air exerts a force on the wings:

There will obviously be a horizontal component to the forces on the wings. Looking at these horizontal forces from the point of view of the above the diagram, they will be acting towards the body of the helicopter. Below is a diagram showing the horizontal components of these forces acting on the wings, from above the helicopter.

Because there are forces acting on different sides of the helicopter, and in opposite directions, the helicopter will spin (anti-clockwise in this diagram).

Method

To take my readings I will be dropping the paper helicopters down a flight of stairs, and measuring the time taken to fall to the bottom. To improve accuracy, I will take each reading three times and take an average. As one of the variables that I will be measuring is the height from which the helicopter is dropped, I will set up a tape measure, and drop the helicopter from intervals of 1 meter. For varying the mass of the helicopter I intend to attach paperclips to the body of the helicopter. To change the size of the wings, I will cut different sized pieces of paper from the wings, and reattach them to the body, so the mass of the helicopter will stay constant.

There are no risks in my experiment.

Experiments

Varying the height

In this experiment I will vary the height from which the helicopter is dropped. The aim of this experiment is to find out how quickly the helicopter achieves its terminal velocity. This happens when the weight of the helicopter balances the air resistance. If the terminal velocity is reached almost immediately, then the graph of time to fall against height will be a straight line and the graph of velocity against time taken will be a horizontal straight line. I will use the same helicopter throughout the experiment.

Height (m)

Time (sec)

1st

2nd

3rd

Ave

Velocity

6.30

4.93

5.13

4.68

4.91

1.28

5.30

3.95

4.15

4.18

4.09

1.29

4.30

3.28

3.22

3.32

3.27

1.31

3.30

2.34

2.46

2.53

2.44

1.35

2.30

1.64

1.82

1.75

1.74

1.32

1.30

0.96

1.01

0.93

0.97

1.34

0.00

0.00

0.00

0.00

0.00

0.00

Evaluation and conclusion

As we can see from Graph 1 in the appendix, there is a straight line. This means that the terminal velocity was reached almost immediately. We can also show this by looking at Graph 2, which also shows that the velocity was consistent for all of the heights, and shows that it was about 1.26 m/s.

Varying the Mass

In this experiment I will vary the amount of extra mass that is on the body of the helicopter. I intend to do this by attaching paperclips to the body of the helicopter. Each paperclip that I will use weighs 1.41 grams, and the mass of the helicopter without any extra mass on it is 4.92 grams. I will keep the height of 6m constant throughout the experiment, and use the same helicopter throughout the experiment.

Total Mass

Time

ln(time)

ln(mass)

(g)

1

2

3

Ave

4.92

5.05

5.26

4.96

5.09

1.63

1.59

6.33

4.53

4.40

4.46

4.46

1.50

1.85

7.74

3.87

3.96

3.84

3.89

1.36

2.05

9.15

3.05

3.16

3.05

3.09

1.13

2.21

10.56

2.87

2.72

2.68

2.76

1.01

2.36

11.97

2.41

2.41

2.43

2.42

0.88

2.48

13.38

2.22

2.13

2.22

2.19

0.78

2.59

14.79

1.91

1.96

2.03

1.97

0.68

2.69

Evaluation and conclusion

If we look at the graph of total mass against time (Graph 3 in appendix), we can see that there is a curve. This means that the relationship between the two is not linear. I expect the relationship to be:

Time = K x Massn

==>ln(time) = ln(K) + nln(mass)

So when we plot ln(time) against ln(mass), the gradient will be n, and the y-axis intersect will equal ln(K).

Gradient = (?ln(time)) / (?ln(mass)) = ((-0.48/0.5) + (-0.88/0.9)) = -0.97

The graph is not big enough to see where it crosses the x-axis, but we can work it out. Taking a point on the line: (1.5, 1.24):

ln(time) = m ln(mass) + c (but ln(time) = 1.24, ln(mass) = 1.5, and m = -0.97)

==> 1.24 = -0.97 x 1.5 + c

==> c = 2.695.

Therefore, n = -0.97 and 2.695 = ln(K)

==> K = e2.695 = 14.806

Therefore, the relationship that we have derived is:

Time = 14.806 x Mass-0.97

Evaluation and conclusion

Another variable that I think might effect the time taken for the helicopter to fall is the placing of any extra mass on the helicopter. As the helicopter falls, it obviously spins. When an object is spinning, it has momentum just like an object moving along a plane. So, when you start to spin a playground ride, for example, it is much harder to spin it to begin with than once spinning, just like pushing a car. This is because it takes energy to give the object its momentum. When something is spinning, we call this rotational momentum. When the helicopter starts to spin, it reduces the air resistance in a vertical direction, as the wings are effectively moving in the direction of that they are pointing. Below is a diagram concentrating on one of the wings, and its movement:

Because the wing is both spinning and falling with the whole helicopter, its net direction is diagonal, as shown by the red arrow. This will mean that the leading edge of the wing, which is what causes most of the air resistance, will be the very thin edge of the paper. This will mean much less air resistance than if it was not spinning, as then the leading edge would be the entire area of the wing.

So we have deduced that when spinning, there will be less air resistance. One way of increasing an object’s rotational momentum is to put most of its mass as far from the center of rotation as possible as this will maximize its speed and therefore give it more momentum. If a spinning object has more momentum when its mass is far from the center of rotation, then it must require more energy to make such an object go the same speed as one with its mass in the center of rotation.

In this experiment I will place two paperclips (one either side of the center) at different distances from the center of rotation. I expect the helicopter with its mass on the outside to take longer as it will take longer for it to spin quickly. The readings will be taken with a constant height of 6m, and mass of 2.84g.

Length from middle

Time (sec)

Velocity

(cm)

1

2

3

Ave

(m/s)

100

2.73

2.43

2.75

2.64

2.28

80

2.75

2.78

2.69

2.74

2.19

60

2.93

2.95

2.51

2.80

2.15

40

2.87

2.61

2.80

2.76

2.17

20

2.78

2.65

2.73

2.72

2.21

Evaluation and conclusion

We can clearly see from the graph of length of mass from middle against time (Graph 5 in appendix) that the position of the mass did not make any visible difference to the time down, as they were fairly constant about 2.7 seconds. While doing the experiment I noticed that the helicopter reached its fastest spinning speed very quickly. This meant that the position of the paperclips made no difference to the times, as the spinning acceleration of the helicopter made up such a small part of the total flight time.

Varying the wing area

The amount of air resistance that there is on the helicopter is obviously highly dependent on the size of the wings. For this experiment I will vary the wing area. I am going to keep the wing width constant, and vary the height of the wing.

There will be a constant mass of 1.41 g on the helicopter and it will be dropped from 6m.

Wing area

Time (sec)

m2

1st

2nd

3rd

Ave

0.0231

4.56

4.59

4.24

4.46

0.0210

4.06

4.12

4.18

4.12

0.0189

3.96

4.03

4.01

4.00

0.0168

3.73

3.65

3.80

3.73

0.0147

3.35

3.32

3.34

3.34

0.0126

3.03

3.12

3.00

3.05

Evaluation and conclusion

As we can see from the graph of wing area against time (graph 6 in appendix) the time is directly proportional to the area of the wing. Also, the line goes through the point (0, 1.1). At this point, the wing area is 0, so this imaginary object has no wings and no air resistance. If we work out the time it takes to fall six meters under the acceleration of gravity using s = ut + 0.5ut2 , it comes to 1.1 seconds.

Varying the ratio of height to width of the wing

Even when the area of the wing is kept constant, I still think that changing the shape of the wing can alter the air resistance. The area of the wing will be kept constant at 6270 cm2, and the height will change at intervals of 10 cm. Again, the height will be constant at 6m.

Wing height

Wing width

h/r ratio

Time (sec)

(mm)

(mm)

1st

2nd

3rd

Ave

110

57

1.93

3.18

2.72

2.72

2.87

100

63

1.59

4.15

4.18

4.09

4.14

90

70

1.29

4.22

4.01

3.93

4.05

80

79

1.01

4.35

4.53

4.91

4.60

70

90

0.78

4.55

4.95

4.91

4.80

60

105

0.57

4.68

4.24

4.32

4.41

50

125

0.40

3.01

3.14

3.23

3.13

Evaluation and conclusion

The graph of wing aspect ratio against time (Graph 7 in appendix) shows that there seems to be an optimum ratio between the height and width of the wing. When the wing is very high and thin, there is very little strength in the hinge on the wing. So, the wing bends up a lot when dropped, and subsequently drops much faster. This is because the wing offers less air resistance when its full area is not facing in the direction of motion. It would seem that the longer the wing is, the more it slows the helicopter down, except when the hinge cannot support the wing. I think this is because when the wing is longer, its speed is effectively greater than if it was short, as it is traveling in a wider arc. As the air resistance depends on the speed, this will mean that there is a larger air resistance, slowing the helicopter down.

Varying the wing angle

One of the reasons for any inaccuracies so far in my experiments is that the wings bend very easily. This means that just by leaving one of the paper helicopters, its wings could droop, and therefore change the results. In this experiment I intend to attach a one end of a paper clip to the body and the other to the wing (one on each wing). The angle of the wing will be my variable. The angle in the table below is taken as the angle between the vertical and the wing, looking from side on:

Angle (?)

Time

Velocity

(degrees)

1

2

3

Ave

(m/s)

90

5.01

5.09

5.10

5.07

1.18

80

5.18

5.15

5.13

5.15

1.16

70

4.20

4.32

4.70

4.41

1.36

60

3.83

3.96

3.83

3.87

1.55

50

3.13

3.42

3.51

3.35

1.79

40

3.07

3.00

3.19

3.09

1.94

30

2.54

2.41

2.50

2.48

2.42

20

1.99

1.98

2.10

2.02

2.97

Evaluation and conclusion

As we can see from the graph, there is a straight line, showing that the wing angle is proportional to the time. When the angle is 90 degrees, the wing is offering its maximum air resistance as it has a much higher cross-sectional area facing the direction of motion. When the angle is 20 degrees, there is a very small cross-sectional area, so there is less air resistance. Another observation that I made is that as the angle got smaller, the helicopter span much faster. If we refer back to the introduction, where I explain why the helicopter spins, we can see that it depends on the horizontal component of the force exerted on the wings by the air. When the angle is 20 degrees, the air is exerting a force on the wings in an almost horizontal plane. This means that the horizontal component is far greater, making it spin quicker.

General conclusion

Although my results have been quite consistent and predictable, I think that there has been one weakness consistent throughout the coursework. This was the flexibility of the paper used. The smallest of touches could deform the paper, and subsequently alter the wings. I would have liked to have been able to measure the revolutions per minute of the helicopter, as I think this would have led to some interesting results. However, the helicopter revolved too fast to get any accurate measurements. 