# Determining an Appropriate Parabolic Model

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Determining an Appropriate Parabolic Model

In this procedure, I am trying to determine an appropriate parabolic
model that fits the data I collected of whirlybird wing length vs.
time, doing so by using first principles. Also to find out which wing
length would produce longest flight time.

Method:

Firstly, I made a whirlybird model and timed how long it took to reach
the floor from a certain height. This procedure was repeated several
times, each time lessening the wing length and keeping the same
height. For each wing length, the bird was dropped three times for
maximum accuracy. Once this data was collected it was transferred into
a graph, and strange points were excluded. After this an appropriate
parabolic function was introduced, being as close as possible to the
original points. The turning point was the place where the wing length
would produce longest flight time.

Results:

[IMAGE]

Length (m)

Time 1 (s)

Time 2 (s)

Time 3 (s)

Average (s)

16

1.38

1.12

1.25

1.25

14

1.19

1.17

1.5

1.29

12

1.75

1.49

1.85

1.70

10

2.18

2.21

2.41

[IMAGE]2.27

8

2.35

2.19

2.44

[IMAGE]2.33

6

2.12

2.1

2.09

2.10

4

1.62

1.68

1.66

1.65

2

1.03

0.91

1.02

0.99

0

0.79

0.87

0.7

0.79

Above is my original data. In the graph, it can be seen that there are
two significant points that do not fit in a parabolic shape well, and
these are the first point and the last point. I decided to remove
these from my final data. The result is shown below.

MLA Citation:
"Determining an Appropriate Parabolic Model." 123HelpMe.com. 24 Feb 2020
<https://www.123helpme.com/view.asp?id=121396>.

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### Popular Essays

[IMAGE]

Length (cm)

Time (s)

14

1.29

12

1.7

10

2.27

8

2.33

6

2.1

4

1.65

2

0.99

Function:

Y= a(X-b)2+c

I will be referring to the letters in this function to explain
changes.

a<0 then hat shape

a>0 then cup shape

a>1 then thinner

0
(X-b)2 then moves to right

(X+b)2 thenmoves to left

Positive c then moves up

Negative c then moves down

The first function I tried was:

Y= -0.5(X-8)2+2.33

I chose -0.5 because when a<0 it will become a hat shape. I chose -8,
because when b is negative, it moves to the right, and since the
length 8cm and time 2.33s was the highest point on the graph I used
this as the initial turning point, so I used 8 as the x coordinate of
the turning point. I chose +2.33 because when c is positive it moves
upwards, and I used 2.33 as the y coordinate of the turning point. The
result was too thin.

The next function was:

Y= -0.1(X-8)2+2.33

I changed -0.5 to -0.1. As a gets closer to 0 the parabola becomes
wider. The result was still too thin.

The next function was:

Y= -0.05(X-8)2+2.33

I changed -0.1 to -0.05. As a gets closer to 0 the parabola becomes
wider. The result was still too thin.

The next function was:

Y= -0.03(X-8)2+2.33

I changed -0.05 to -0.03. As a gets closer to 0 the parabola becomes
wider. The result was off centre to the left.

The next function was:

Y= -0.03(X-8.5)2+2.33

I changed -8 to -8.5. As b decreases the turning point moves to the
right. The result was a little too wide.

The next function was:

Y= -0.035(X-8.5)2+2.33

I changed -0.03 to -0.035. As a gets farther from 0 the parabola
becomes thinner. The result was a little too thin.

My final function was:

Y= -0.033(X-8.5)2+2.33

I changed -0.035 to 0.033. As a gets closer to 0 the parabola becomes
wider.

Below the final quadratic function, Y= -0.033(X-8.5)2+2.33 is
superimposed onto my original data points.

[IMAGE]

In the tables below, the results when the different lengths are
substituted into the function, can be compared with our initial data
table. You should note the large difference when the length was 12cm.
In the graph above it can be seen that this point does not fit in very
well.

X

Y= -0.033(X-8.5)2+2.33

14

1.33175

12

1.92575

10

2.25575

8

2.32175

6

2.12375

4

1.66175

2

0.93575

Length (cm)

Time (s)

14

1.29

12

1.7

10

2.27

8

2.33

6

2.1

4

1.65

2

0.99

A graphic calculator was used to find out the turning point, and it
was concluded that the wing length that would produce longest flight
time was 8.5000008712cm, which could produce a maximum flight time of
2.33 seconds.

There are a few assumptions that had to be made during this testing.
One was that there was no wind, and that air resistance remained
constant, otherwise this could cause changes in flight time. This
could be improved if the test was done in a vacuum. As well as this,
it must be assumed that distance of the drop remained constant and
that the time of flight was taken accurately. One major problem was
reaction time, which has caused miscalculations, and this can be
easily seen in our first and last data points, in which the time of
flight was shortest. This can be improved by using a type of sensor
device that can accurately measure the flight time.

Conclusion:

In conclusion, I was successfully able to predict through trial and
error and first principals, an appropriate parabolic quadratic
function, which integrated well with my original data points. I was
also successful in finding the wing length that would produce maximum
flight time, by using a graphics calculator to find the max turning
point.