Results:
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.
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<a<1 then wider
(X-b)2 then moves to right
(X+b)2 then moves 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.
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.
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.
