To what extent can volumes of revolution be used to calculate the volume of a hot air balloon?

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

• The height of the hot air balloon not including the basket is 18.30 metres ("Ballooning Information")
• The circumference of the balloon at its widest point is 18.59 metres (R, Alexia)
• The circumference of the balloon at its most narrow point is 13.41 metres (R, Alexia)

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

 

r_n = 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

 

r_w = 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 -

• (2,959,11.18) which is the widest point of the balloon
• (0,18.30) which is the maximum height of the balloon, which can also be said to be the vertex of the curve. These points represent k and h in the quadratic formula in its vertex form.

 

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 -

• The amplitude (a) is 18.3 since this is the maximum height of the balloon, and therefore the maximum point the cosine graph will reach.
• The period (b) can be calculated with the equation \(b=\frac{2\pi}{period}\) We know that one half of the cosine penod graph will be the distance of the curved section of the balloon, which is 18.30 - 11.18 = 7.120 therefore, we multiply 7.120 by 2 to find the full period: 7.120 x 2 = 14.24 Now we know that \(b = \frac{2\pi}{14.24}\) and this is equal to 0.4419
• The phase shift (c) is O because the balloon is centred around the y-axis
• The vertical shift (d) is O because there is no need for the curve to be vertically shifted since the maximum already intersects the straight line.

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.

• The horizontal radius (a) of the ellipse is the radius of the widest section of the balloon. We calculated previously that the radius at the widest point is 2.959. Therefore, a= 2.959
• The vertical radius (b) of the ellipse is the difference between where the balloon changes from straight to curved, and all of that divided by 2. We divide by 2 to find the radius instead of the diameter. b = 18.30 - 11.18. Therefore, b = 7.120
• The centre points of the ellipse (h, k) are (0,11.18) because when the eccentricity is equal to one, a parabola is achieved. ("Eccentricity")

 

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.

For the piecewise function to be created, the equation of the straight line needs to be rearranged so that they are in terms of 'y'. The equations need to be in terms of y because the hot air balloon will be rotated around the y-axis to find its volume.

 

\(y=-\frac{11.18}{0.825}x-28.92\)

 

\(y \,+ \,28.92=-\frac{11.18}{0.825}x\)

 

0.825y + (0.825 × 28.92) = -11.18x

 

0.825y + 23.86 = -11.18x

 

\(\frac{0.825y \,+ \,23.86}{-11.18} = x\)

 

We also need to rearrange the formula of the ellipse so that it is equal to x and in terms of y. We will rearrange the formula so that xis the subject.

 

\(\frac{x^2}{2.959^2}+ \frac{(y-11.18)^2}{7.120^2}=1\)

 

\(\frac{x^2}{2.959^2}= 1-\frac{(y - 11.18)^2}{7.120^2}\)

 

\(x^2=2.959^2-\frac{2.959^2(y-11.18)^2}{7.120^2}\)

 

\(x = \sqrt{2.959^2-\frac{2.959^2(y-11.18)^2}{7.120^2}} \)

 

Now that both sections of the model are complete, a piecewise function can be formulated.

 

\( x = \begin{cases} \frac{0.830y + 23.86}{-11.18}, & 0 \leq y \leq 11.18 \\ \sqrt{\frac{2.9592(y - 11.18)^2}{7.120^2}}, & 11.18 \leq y \leq 18.30 \end{cases} \)

The lower bound of the first section of the piecewise function is O and the higher bound is 11.18, therefore those will be substituted into the a and the bin the volume of revolution equation.

 

\(V=\pi \displaystyle\int^{11.19}_0 \bigg(\frac{0.830y \,+ \,23.86}{-11.18}\bigg)^2dy\)

 

So fi.rst I w.ill i.ntegrate the

 

\(V =\bigg(\frac{0.830 \,+ \,23.86}{-11.18}\bigg)^2\)

 

To begin, I will take the integral and factor out the constants -

 

\(V=\frac{1}{-11.18}^2\displaystyle\int(0.830 + 23.86)^2\)

 

\(V=\frac{1}{-11.18}^2 × \frac{1}{0.830}× \frac{(0.830y + 23.86)^3}{3}\)

 

\(V=0.009639 × \frac{(0.830y + 23.86)^3}{3}\)

 

V = 0.003213 × (0.830y + 23.86)³

 

Now I can find the volume of revolution by substituting in the bounds and subtracting them.

 

V = (π × (0.003213 × (0.830(11.18) + 23.86)³) - (π × (0.003213 × (0.830(0) + 23.86)³)

 

V = (π × (0.003213 × 36390)) - (π × (0.003213 × 13580)

 

V = (π × 116.9) - (π × 43.63)

 

V = 367.3 - 137.1

 

V = 230.2

 

This can be confirmed on the Tl-84 plus graphic calculator -

I will do the same method as I did for section 1 to find the volume of revolution for section 2.

 

Using the same method as previously, the integral and volume of revolution will be calculated.

As a consequence of having the volume of both sections of the piecewise function, I can now find the total volume of the hot air balloon. I will add both the individual volumes.

 

Rounding error from preview page rectified.

 

V = 230.3 + 130.6

 

V = 360.9m³

In conclusion, I have achieved my aim of modelling a hot air balloon, not including the basket, and finding the volume of gas it holds, which was 360.9m³. I achieved this through creating a piecewise function and finding the volume of revolution to find the volume of gas that is holds. I successfully achieved this.

In order to ameliorate my results, I could have used one continuous function instead of a piecewise function to improve the accuracy of the results. In addition, my model can be used to find the volume of gas required to make it airborne instead of just the total volume of the hot air balloon on the ground. One factor to consider is if my function would be applicable to different sizes of hot air balloons.

"Ballooning Information." A Beautifiil Balloon, www.balloon-rides.com/atics-facts.htm. Accessed 16 Nov. 2022.

 

"Eccentricity." Mathisfun, www.mathsisfun.com/geometry/eccentricity.html. Accessed 12 Dec. 2022.

 

Hot Air Ballooning Is a Relaxing Peaceful Pastime That Many People Enjoy." Clzegg, www.chegg.com/homework-help/questions-and-answers/project-2-hot-air-ba Iloons-hot-air- ballooning-relaxing-peaceful-pastime-many-people-enjoy--q4279 l 529.  Accessed 16 Nov. 2022.

 

Kaushik, Nimisha. "Difference Between Circle and Ellipse." Difference Between Similar Terms and Objects, 13 Mar.  2018,  www.differencebetween.net/science/mathematics-statistics/difference-between-circle­ and-ellipse/. Accessed 23 Nov. 2022.

 

R, Alexia. "How Wide Is a Hot Air Balloon?" OutdoorFizz, 27 June 2021, outdoorfizz.com/how-wide-is-a­ hot-air-balloon/. Accessed 16 Nov. 2022.

 

TI-SmartView Emulator Software for the Tl-84 Plus Family. 4.0.0.113 ed. Texas Instruments, educatioo.ti.com/en/software/details/en/ffea90ee7f9b4c24a6ec427622c77d09/sda-ti-smartview-ti-84- plus. Accessed 10 Mar. 2023.