The functionality of hot air balloons has intrigued me ever since being in a hot air balloon myself and looking down and realizing how high up from the ground I really was. I visited the hot air balloon festival in Chateau D'oex in Switzerland, and I went in a hot air balloon in Myanmar, and it was these events that commenced my interest for hot air balloons.
A+ aim is Stated
This IA aims to model the volume inside of a hot air balloon not including the basket and how much gas it holds. For the sake of this IA, all calculations will be rounded to 4 significant figures unless it is not possible.
B+ significance level chosen
B- rationale for level nct stated
The volume inside of the hot air balloon will be modelled by creating a piecewise function and then using that function to find the volume inside the hot air balloon with volumes of revolution. The piecewise function will be created through creating an equation of a straight line to model the straight section of the hot air balloon and then I will use multiple different methods to model the curved section of the balloon, once these are all modelled, I will evaluate which function suits the hot air balloon most appropriately. The methods I will use to create a function for the curved section are a quadratic equation, a cosine graph, and the equation of an ellipse.
In order to create the piecewise functions, data of the height and width of a standard hot air balloon is needed.
+ research for
standardised measurements
Using this information, an approximate model can be created to estimate the shape of the balloon.
A- calculation fully shown, but
This is assumed knowledge
Using the circumference equation, we will work out the radius of the narrowest and widest point of the balloon. First, we need to rearrange the equation so that the radius (r) is the subject. Then we need to input the values of the circumference (c) and calculate the radius.
\(c = 2\pi r\)
\(r= \frac{c}{2\pi r}\)
For the narrowest point, the circumference is 13.41m therefore the radius at the narrowest point (rn) is -
\(r_n = \frac{13.41}{2\pi}\)
Therefore
rn = 2.134m
For the widest point, the circumference is 18.59m therefore the radius at the widest point (rw) is -
\(T_w = \frac{18.59}{2\pi}\)
So therefore
rw = 2.959m
To find the height of the straight section of the hot air balloon, I have done some research to work out the ratio of the straight section compared to the curved section of the balloon. A previous study of hot air balloons was conducted and suggested that the height of the straight section was 44 feet, and the curved section has a height of 28 feet. This will be used to find the ratio of the heights.
It can be seen that the curved section of the balloon is not a hemisphere because the dimensions of the height and width do not match that of a hemisphere or a circle. Therefore, I will attempt to create a model that best fits this previously modelled balloon in order to find its volume.
A+ Create an accurate model
To find the ratio, we must divide the smaller section by itself, and we must divide the larger section by the smaller section.
\(\frac{28}{28}=1\)
and
\(\frac{44}{28}=1.571\)
The total height of the balloon that I am modelling is 18.30 metres. Therefore, we need to divide the total height by the sum of the ratios.
\(height \, \, \,of \, \,curved \, \,section \,=\frac{total \, \,heght \, \, of \, \,ballon}{sun \, \,of \, \,ratios}\)
\(\frac{18.3}{1+\frac{44}{28}}=7.116m \,(height \, \,of \, \,curved \, \,section)\)
This is the height of the curved section of the hot air balloon and the height of the straight section is calculated by multiplying the 7.116 by the ratio.
\(7.116 × \frac{44}{28}=11.18m \, \,(height \, \, of \, \,straight \, \,section)\)
Now that all the components of the balloon are collected, the hot air balloon can be partially modelled because the equations of the lines have not yet been determined.
For section 1 of this model, the equation of the line will be found using the straight-line equation. First, we can find "m" in the equation which refers to the gradient of the line. A gradient of a line is found using the coordinates along the line (y1, y2) and (x1, x2). The equation to find the gradient is \(m=\frac{Y_2-Y_1}{ X_2-X_1}\) Therefore, the coordinates of the straight line are (2.13,0) and (2.96,11.19). Therefore, we can find the gradient.
\(m=\frac{11.18-0}{2.959-2.134}\)
\(m=\frac{11.18}{0.825}\)
Therefore, our gradient is \(m=\frac{11.18}{0.825}\).We can input this into the straight-line equation, which is y = mx + c
\(y=\frac{11.18}{0.825}x+c\)
To find "c" we substitute the x and y coordinates from one of the points along the line, so for this case we will substitute the coordinates (2.134,0).
\(0=\frac{11.18}{0.825}(2.134)+c\)
0 = 28.92 + c
-28.29 = c
This can be proved to be correct since the c in the equation is where the line intersects the y-axis, and in this case, it is known that the line intersects the y-axis at 2.134 because that is the radius of the narrowest point on the balloon, hence the opening at the bottom.
Consequently, we know that the first section of the piecewise function will be
\(y=\frac{11.18}{0.825}x-28.92\)
And this is valid for the regions 0 ≤ y ≤ 11.18
For the other side, the same equation applies, however, the gradient and the intercept are negative.
Therefore, the equation is represented as -
\(y =-\frac{11.18}{0.825}x-28.92\)
Now I will begin on the curved section of the hot air balloon. This section will need to be curved therefore I will be modelling a quadratic equation to see if it fits the model of the hot air balloon.
I will model the curve using the vertex form because it is not possible to model it using the general form since I only have two points. The known coordinates are -
From these two points, a quadratic equation in the vertex point can be found since we have the constants k and h. The coordinates from the vertex point will be input into the equation as k and h along with substituting the coordinates from the point of the maximum height. Once these are inputted, the equation can be rearranged to find the constant "a".
y = a(x - k)2 + h
Input the coordinates (0,18.3) of the vertex point into "h" and "k":
y = a(x - 0)2 + 18.30
Substitute the coordinates from the widest point of the balloon -
11.18 = a(2.959 - 0)2 + 18.30
Rearrange to find "a"
11.18 = a (2.959)2 + 18.30
11.18 = 8.756a + 18.30
-7.12 = 8.756a
a = -0.8132
Accordingly, the equation of the quadratic parabola can be generated because all the constants are known -
y = -0.8132(x - 0)2 + 18.30
This equation can be put into combination with the equations of the straight line to analyse whether it adheres the model.
he highlighted regions show the sections of the curves that are relevant for the model of the hot air balloon. This equation does not fit the model of the hot air balloon because the slope of the quadratic
D+ reflection of proposed model ond reattempt
equation is too steep thus deeming it as not acceptable because it would suggest that the balloon is not proportioned correctly. In addition, the curve is not circular enough for it to model the shape of the top of the balloon. Consequently, it is necessary to try another equation that will fit the curve.
In order to construct a cosine graph, a few variables need to be known. The general equation of a cosine graph is -
y = acos (bx + c) + d
Where a is the amplitude, bis the period, c is the phase shift and dis the vertical shift.
Components needed -
Once all the components are collected, the equation can be created by substituting in the points in their relevant positions.
y = 18.3 cos(0.4419x - 0) + 0
Evaluation
As seen from 'Fig 3' the cosine curve does not model the shape of a hot air balloon accordingly because the gradient of the curve is still too steep.
D+. A further attempt to mottel
Since the other equations did not create a curve that was circular enough, I will formulate an equation using the equation of an ellipse to find a different parabola that will match the model better. This equation will adhere to the model more accordingly because the curve of the parabola is constant as well as the gradient. An ellipse is different from a circle because in a circle, all the diameters are the same size, but in an ellipse, there are major and minor axes which are of different heights and lengths. (Kaushik) This suggests that it will fit the model better because it will be more proportional. The general form of an ellipse is -
\(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)
Where a is the horizontal radius, bis the vertical radius and (h, k) are the coordinates of the centre of the ellipse or otherwise known as the vertical and horizontal shifts.
Hence, the equation of the ellipse can be formed since all the necessary components are collected. The coordinates (h, k) will be substituted into the original equation as well as the horizontal and vertical radius. This will all be equated to 1.
\(\frac{(x-h)^2}{a^2}+\frac{(y -k)^2}{b^2}=1\)
\(\frac{(x-0)^2}{2.959^2}+\frac{(y - 11.18)^2}{7.120^2}\)
This can be plotted on a graph to model the shape of the balloon.
As predicted, this curve fits the model the best because it has an elliptical shape and intersects the straight lines at the correct points (-2.959, 11.18) and (2.959, 11.18) and it intercepts the y-axis at the point (0, 18.30) which adheres to the model of the hot air balloon because this is the height of the hot air balloon. However, it can be seen that the ellipse intersects the straight lines at 2 different points, this could be because of the rounding of the significant figures, but the equation of the ellipse fits the model the best.
D+ reflection.