surface_area_of_cup

cup_surface_area.html

<meta charset="utf-8" emacsmode="-*- markdown -*">
**Surface Area of a Cup with a Straight Outer Edge**

Let's say our cup is defined by the following parameters:

\begin{align*}
    r=\,&\text{radius of the top of the cup}\\
    b=\,&\text{height of the cup}\\
    B=\,&\text{angle of the sloped side of the cup}\\
\end{align*}

*********************
*       B_____C_____*
*    ┌   \    |    /*
*    │   c\   |b  / *
*   h│     \__|__/  *
*    │   ┌  \ |     *
*    │  y│  d\|     *
*    └   └    |     *
*           └─┘A    *
*            x      *
*       └─────┘     *
*          r        *
*                   *
*********************

\begin{align*}
    A=\,&\text{the angle at the tip of the cone that subsumes our cup}\\
    C=\,&\text{the remaining 90 degree angle of our triangle ABC}\\
    b=\,&\text{the height of the cup}\\
    c=\,&\text{the length of the sloped outer edge of the cup}\\
    h=\,&\text{the height of the extended cone}\\
    x=\,&\text{the radius of the bottom of the cup}\\
    y=\,&h - b\\
\end{align*}

From the cross section of half the cup above, there is another triangle we can derive that shares the angle $B$, and has base $r-x$:

***********
*         *
* B'_____C*
* ┌|\    |*
*b│|c\   |*
* └|__\__|*
*         *
*  └───┴─┘*
*   r-x x *
*  └─────┘*
*     r   *
***********
Let's call the outside angle $B'$ the angle that sums to $90\degrees$ with $B$:

\begin{align*}
    B'                       = A = \,& 90\degrees - B\\
    -> \tan{\left(B'\right)} =\,& {{\left(r - x\right)}\over{}b}\\
    -> x                     =\,& -1 \cdot \left(b \cdot \tan{\left( 90\degrees - B \right)} - r\right)\\
    -> x                     =\,& - \left(b \cdot \tan{\left( 90\degrees - B \right)}\right) + r
\end{align*}

So, we've defined $x$ in terms of $b$, $B$, and $r$ which are the parameters we used to define our cup originally.

*************
*           *
* B'_____C  *
* ┌|\    |┐ *
*b│|c\   |│ *
* ├|__\__|│ *
* │    \ |│h*
*y│    d\|│ *
* └      |┘ *
*  └───┴─┘A *
*   r-x x   *
*  └─────┘  *
*     r     *
*************

Next we want to derive $h$ which, along with $r$, defines the cone that subsumes the cup.

We know $c$ now due to Pythagoras' Theorem:

\begin{align*}
    c^{2} =\,& \left(r-x\right)^{2} + b^{2}\\
    c     =\,& \sqrt{\left(r-x\right)^{2} + b^{2}}\\
\end{align*}

Note that in case $x \leq 0$, then you don't really have a cup, you just have a cone.

************
*          *
* B_____C  *
*  \    |┐ *
*   \   |│ *
*   c\  |│ *
*     \ |│h*
*      \|│ *
*       |┘ *
* └─────┘A *
*    r     *
************

In that case, $h$ can be computed in terms of $r$ and $B$ alone:

\begin{align*}
    \tan\left(B\right) =\,& {{h}\over{r}}\\
     h =\,& \tan\left(B\right) \cdot r
\end{align*}

Otherwise, when $x > 0$, we can derive $d$:

\begin{align*}
    \csc\left(A\right) =\,& {{d}\over{x}}\\
    -> {{1}\over{\cos\left(A\right)}}=\,& {{d}\over{x}}\\
    -> d =\,& {{x}\over{\cos\left(A\right)}}\\
\end{align*}

We know $A$ from above, so:

\begin{align*}
    -> d =\,& {{x}\over{\cos\left(90\degrees - B\right)}}\\
    -> d =\,& {{-\left(b \cdot \tan{\left( 90\degrees - B \right)}\right) + r}\over{\cos\left(90\degrees - B\right)}}\\
\end{align*}

And from Pythagoras' Theorem, again, we can derive $h$:

\begin{align*}
    h^{2} =\,& \left({c+d}\right)^{2}+x^{2}\\
    -> h  =\,& \sqrt{\left({c+d}\right)^{2}+x^{2}}\\
    -> h  =\,& \sqrt{\left({{\sqrt{\left(r-\left({-\left(b \cdot \tan{\left( 90\degrees - B \right)}\right) + r}\right)\right)^{2} + b^{2}}}+{{{-\left(b \cdot \tan{\left( 90\degrees - B \right)}\right) + r}\over{\cos\left(90\degrees - B\right)}}}}\right)^{2}+\left({-\left(b \cdot \tan{\left( 90\degrees - B \right)}\right) + r}\right)^{2}}\\
\end{align*}

Now that we have $h$, we can compute the cup's surface area by subtracting the surface area of the cone defining the tip that extends beyond the base (whose triangle cross section has hypotenuse $d$) from the subsuming cone (whose triangle cross section has hypotenuse $d+c$):

\begin{align*}
    \text{subsuming cone surface area}=\,& \pi r \sqrt{r^{2} + h^{2}}\\
    \text{tip cone surface area}=\,& \pi x \sqrt{x^{2} + y^{2}}\\
    \text{cup surface area}=\,& \left(\pi r \sqrt{r^{2} + h^{2}}\right) - \left(\pi x \sqrt{x^{2} + y^{2}}\right)
\end{align*}

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cup_surface_area.py

#!/usr/bin/env python3

import math
import sys

def cone_surface_area(radius, height):
    """Compute the surface area of a cone defined by two parameters:
    
    radius: the radius of the base of the cone
    height: the height of the cone
    
    """
    return math.pi * radius * math.sqrt((radius ** 2) + (height ** 2))

def cylinder_surface_area(radius, height):
    """Compute the surface area of a cylinder defined by two parameters:
    
    radius: the radius of the base of the cylinder
    height: the height of the cylinder
    
    """
    
    return 2 * math.pi * radius * height

def cup_surface_area(radius, height, angle, verbose=False):
    """Compute the surface area of a cup defined by three parameters:
    
    radius: the radius of the top of the cup
    height: the height of the cup
    angle: the angle between the top of the cup and the sloped edge (in radians)
    
    """
    A = (math.pi / 2) - angle
    x = -(height * math.tan(A)) + radius
    b = (radius - x) / math.tan(A)
    c = math.sqrt(((radius - x) ** 2) + (b ** 2))
    d = x / math.cos(A)
    y = math.sqrt((d ** 2) + (x ** 2))
    subsuming_cone_height = math.sqrt(((c + d) ** 2) + (x ** 2))
    subsuming_cone_surface_area = cone_surface_area(radius, subsuming_cone_height)
    tip_cone_surface_area = cone_surface_area(x, y)
    if verbose:
        print(
            f'A={A}',
            f'x={x}',
            f'b={b}',
            f'c={c}',
            f'd={d}',
            f'y={y}',
            f'subsuming_cone_height={subsuming_cone_height}',
            f'subsuming_cone_surface_area={subsuming_cone_surface_area}',
            f'tip_cone_surface_area={tip_cone_surface_area}',
            sep='\n'
        )
    if x <= 0: # in case x is not positive, you have a cone, not a cup
        print(
            'FYI: based on the radius, angle, and height, this cup is actually a cone',
            file=sys.stderr
        )
        return cone_surface_area(radius, math.tan(angle) * radius)
    return subsuming_cone_surface_area - tip_cone_surface_area

if __name__ == '__main__':
    import argparse
    def angle(value):
        value = float(value)
        if (value < 0) or (value >= 90):
            raise argparse.ArgumentError(f'angle must be between 0 and 90 degrees')
        return math.radians(value)
    parser = argparse.ArgumentParser(
        formatter_class=argparse.ArgumentDefaultsHelpFormatter,
        description=__doc__
    )
    parser.add_argument(
        '--radius',
        type=float,
        required=True,
        help='cup radius',
    )
    parser.add_argument(
        '--height',
        type=float,
        help='cup height'
    )
    parser.add_argument(
        '--angle',
        type=angle,
        required=True,
        help='cup angle (in degrees)'
    )
    parser.add_argument(
        '-v', '--verbose',
        action='store_true',
        help='print verbose output',
    )
    args = parser.parse_args()
    cup = args.radius, args.height, args.angle
    print(
        f'''A cup with ...''',
        f'radius={args.radius}',
        f'height={args.height}',
        f'angle={math.degrees(args.angle):0.3f} degrees ({args.angle:0.3f} radians)',
        f'... has surface area of: {cup_surface_area(*cup, verbose=args.verbose)}',
        sep='\n'
    )