Chess And Mathematics Chess is a two-player board game that heavily incorporates strategy and tactics, played on a chessboard. In chess, there are 6-piece types, each of which move differently: king, queen, rook, knight, bishop and pawn. Each player starts the game with 16 pieces consisting of; 1 king, 1 queen, 2 rooks, 2 knights, 2 bishops and 8 pawns. The most important piece in chess is the king, with the most powerful piece being the queen and the least powerful piece being the pawn.
The pawn is the most numerous pieces in chess, with 8 pawns per player at the beginning of the game, with a pawn placed on every square: horizontally across the second row of the player’s respective side. Historically, the pawn symbolizes the army’s infantry, or to put it bluntly, village peasants armed with weapons. A pawn can only move one step at a time, to the square directly in front of itself, although on its starting position, it has the option of moving two squares forward.
Best Ai Chess
A pawn can only capture an opposition piece diagonally forward one square to the right to left. Since the pawn possesses the most limited set of movements and thus, in most circumstances, it is considered the weakest piece. However, in the rare situation that a pawn is able to advance all the way to the opposite side of the board, the pawn is able to be promoted to another piece of the player’s choice: a queen, rook, knight or bishop.
The pawn is the most fundamental piece in chess and it serves as a metaphorical representation of ordinary men and women attempting to travel across the board of life towards eventual initiation into a divine state.
The opposite side of the board represents a higher state, and a pawn reaching the opposing end signifies that the once ordinary man/woman has battled against all odds and triumphantly overcame the trials, adventures and traps of life: allowing them to achieve their dreams and ambitions – attaining greatness in spite of the deprived circumstances they were initially dealt into. Ultimately, the journey of the pawn symbolizes the idea of continuously chasing a dream, despite how unfair, or difficult adversities may be.
Chess sets and pieces come in many different sizes; however, for the sake of consistency, I will use the chess set that I own as a reference for the sizes of the chess pieces explored. The standard size of a pawn in my chess set stands at 4.6 cm tall, with a base-diameter of 2.6 cm. The chess pawn is designed in a very unique way: the top of the piece is spherical, with a short, slender mid-section and a long, wide base. As a result, the pawn’s overall structure resembles that of a crystal decanter used for holding wine or other liquids.
Decanters are beautiful works of craftsmanship, with elegant designs and intricate details carefully crafted out, bringing about a unique artistic flair for every decanter. Despite being used to hold wine and other liquids; decanters are often viewed as playfully intriguing objects with high aesthetic value. Their funky, stunning shapes make them exciting, eye-catching objects that can act as decorations for a living room or dining table (Stimpfig, 2015).
The designs of decanters have always fascinated me as I have always found them to be aesthetically pleasing and always thought of them as truly creative works of art. I have always wanted to own a decanter as means of holding liquids however, I have never
purchased one myself due to their high prices. In addition, I have never seen the point of owning a decanter just for the sake of decoration, since its main purpose is to decant wine (decanting is the process of separating out sediments in a liquid). Nonetheless, this has changed after my 18th birthday; giving me the legal right to consume wine. Now, a decanter appears to be a reasonable purchase as I can now put it to use, in addition to it serving as a beautiful ornament in my living room.
From this interest, I came up with the idea for my Math Exploration. I decided that for my first decanter, I would like its design to resemble something that is relatable and significant to me. To do so, I managed to find a website titled “Make your own wine decanter”, which allows me to design my own decanter which will then be crafted and shipped to me.
Upon attending my weekly chess club, I noticed that the design of chess pawn shares a similar design to some of the decanters I have seen over the years. As playing chess is one of my hobbies and one of my favorite ways to spend my time, I felt that a decanter design inspired by the game of chess would be ideal. Moreover, despite the pawn being the weakest piece in chess majority of the time, I find that it is my favorite piece due to the idea that it embodies – the idea that one should never give up despite hardship or being disadvantaged. As a result, I decided that a glass decanter, designed in the shape of a chess pawn would not only look magnificent, but also serve as a daily reminder to stay resilient no matter what.
Aim:
The aim of this mathematical exploration is to model the shape and size of a Chess Pawn through the use of 2-dimensional graphing techniques, and 3-dimensional volumes of solids of revolution. After modeling the Chess Pawn, I will apply the functions to design a glass decanter in the shape of a pawn. However, part of the challenge will be to ensure that the volume of a decanter is on average, approximately 750 ml. Thus, I will need to ensure that the Pawn decanter has the dimensions needed to store a volume of liquid, close to 750 ml.
Mathematical Analysis:
Modelling the shape of a Pawn
In order to model the pawn as precisely as possible, I took a frontal photograph of a chess pawn that I owned. The purpose of this photograph is to allow me to accurately derive the function for the outline of the pawn. The photograph of the pawn was taken such that the entire shape of the pawn is clearly seen vertically. The outline of the pawn in the photograph was then traced in Adobe Photoshop and split along its axes of symmetry. Figure 1 depicts the photograph I took of a chess pawn, with a height of 4.6 cm and a base-diameter of 2.6 cm.
Figure 1
The outline of the pawn in the photograph was then traced in Adobe Photoshop and split along its axes of symmetry. This is shown in Figure 2.
Using a graphing programme known as Logger Pro, I then inserted the left-half of the pawn’s shape and then rotated it 90° clockwise, so that the height of the pawn was on the x-axis and its width on the y-axis. This is depicted in Figure 2 and was done by inserting the photograph from Figure 1.
Figure 2
Figure 2 shows the model of the pawn, traced by plotting multiple individual points along the outline of the pawn. To attain a high level of accuracy of the overall model, the photograph of the pawn was used to scale the graph, and establish a point of origin. On Figure 2, the bottom-left corner of the shape was set as the origin at (0,0). The origin was then used as the starting point of the graph. In addition, the x-axis of the graph was scaled to the pawn’s height; to 4.6 cm. The y-axis of the graph was scaled to the pawn’s base-radius; to 1.3 cm. This is because only half of the pawn’s shape was traced and thus, only half of the pawn’s base-diameter was plotted.
Due to the fact the graph in Figure 2 does not have a regression line that accurately models the shape of the pawn, the graph has to be split into different segments. After splitting the graph into individual segments, a regression line needs to be produced for each individual segment, in order to model the separate parts of the pawn’s outline. Figure 3 depicts the separate segments in which Figure 2 is split into, based on the appearance of the curves, on first glance.
Figure 3
Manually calculating regression functions
To model the pawn, regression lines have to be fitted along the points that outline the pawn’s shape. To do this, the graph of the pawn was split into 5 different segments where each segment requires a specific regression line to model the curve.
On first glance, the curves in A, D, and E resemble parabolas since they are U-shaped. The curves open downwards and thus, a vertex can be identified on each of the curves. The vertices signify the maximum turning point of the curves, where any point to the left or right of the curves have a lower
y-value than the vertex (the highest point of the graph). Lastly, the curves are symmetrical about a vertical line, where the vertex is the axis of symmetry.
The parabola is a result of graphing a quadratic function and thus, to identify appropriate regression lines for the curves A, D and E, a quadratic equation is to be used. The standard form of a quadratic equation is in the form:
f (x) = ax2 + bx + c
To develop a quadratic function, we have to substitute the x-coordinate from a point on the curve into the x-values in the equation, and the y-coordinate into the f(x)-value. Since the curve resembles a quadratic function, 3 different points on the curve have to be taken: producing 3 separate equations. The values of a, b, and c, can then be determined through simultaneously solving the 3 separate equations. After calculating the values of a, b, and c, a quadratic function can then be produced to form a regression line used to model the curve.
(Note: Figures will all be given to 4 significant figures)
Producing Regression Function for Curve A
Curve A:
| Point | Coordinates | x-coordinate | y-coordinate |
| 1 | (0.04435, 0.5051) | 0.04435 | 0.5051 |
| 2 | (0.6732, 1.260) | 0.6732 | 1.260 |
| 3 | (1.147, 1.033) | 1.147 | 1.033 |
System of Equations:
1.260 = a(0.6732)2 + b(0.6732)+ c
1.033 = a(1.147)2 + b(1.147)+ c
0.5051 = a(0.04435)2 + b(0.04435)+ c
Eliminating c from first 2 pairs:
(0.6732)2a + (0.6732)b + c = 1.260
(1.147)2a + (1.147)b + c = 1.033
(0.6732)2a + (0.6732)b + c = 1.260
−(1.147)2a − (1.147)b − c = −1.033
−0.862411a − 0.4738b = 0.227
Eliminating c from last 2 pairs:
(1.147)2a + (1.147)b + c = 1.033
(0.04435)2a + (0.04435)b + c = 0.5051
(1.147)2a + (1.147)b + c = 1.033
−(0.04435)2a − (0.04435)b − c = −0.5051
1.31364a +1.10265b = 0.5279
Solving for a:
−0.862411a − 0.4738b = 0.227
1.31364a +1.10265b = 0.5279
−0.950937a − 0.522436b = 0.250302
0.622403+ 0.522436b = 0.250119
−0.328534a = 0.500421
a = −1.42319
Solving for b:
−0.862411(−1.52319)− 0.4738b = 0.227
−0.4738b = −1.08662
b = 2.29341
Solving for c:
(0.6732)2(−1.52319)+ (0.6732)(2.29341)+ c = 1.260
c = 1.260 −[(0.6732)2(−1.52319)+ (0.6732)(2.29341)]
c = 0.406383
Therefore,
a = −1.523
b = 2.293
c = 0.4064
After calculating the values for a, b, and c, I was able to produce a regression function for curve A:
f (x) = (−1.523)x2 + (2.293)x + 0.4064
Producing regression lines for Curves D and E
To produce regression lines for curves D and E, I repeated the method used to produce an equation for the regression line for curve A, with 3 separate points from curve D and 3 separate points from curve E.
Curve D:
| Point | Coordinates | x-coordinate | y-coordinate |
| 1 | (2.818, 0.4217) | 2.818 | 0.4217 |
| 2 | (2.988, 0.6974) | 2.988 | 0.6974 |
| 3 | (3.136, 0.5635) | 3.136 | 0.5635 |
Values for a, b, and c :
a = −7.945
b = 47.75
c = −71.04
Regression function for curve D:
f (x) = (−7.945)x2 + (47.75)x − 71.04
Curve E:
| Point | Coordinates | x-coordinate | y-coordinate |
| 1 | (3.280, 0.5861) | 3.280 | 0.5861 |
| 2 | (3.808, 0.7577) | 3.808 | 0.7577 |
| 3 | (4.456, 0.2233) | 4.456 | 0.2233 |
Values for a, b, and c :
a = −0.9758
b = 7.242
c = −12.67
Regression function for curve E:
f (x) = (−0.9758)x2 + (7.242)x −12.67
Producing Regression Function for Curve C
Next, the curve in C, resembles a straight-line graph based on first-glance. This is because a linear gradient can be observed between the first and last points along the x-axis. The equation of a
straight-line graph is in the form:
y = mx + c
To develop a regression function for a straight-line graph, we first use the equation: m = y2 − y1
x2 − x1
where,
m is the gradient of the graph x2 is x-coordinate of the last point x1 is the x-coordinate of the first point y1 is the y-coordinate of the first point y2 is the y-coordinate of the last point
Taking the first point of curve C to be (1.940, 0.4841) and the last point to be (2.818, 0.4217), we can substitute the x and y coordinates of the 2 points into the gradient equation:
m = 0.4217 − 0.4841
2.818 −1.940
m = −0.0624
0.878
m = −0.071071
After finding the value of m, we substitute the value of m into the straight-line equation. Next, we substitute the coordinates of any point (2.375, 0.4421) from curve C into the straight-line equation to determine the value of c :
y = −0.071071x + c
0.4421 = −0.071071(2.375)+ c
c = 0.610894
Therefore, we can produce a regression function for C:
y = −0.7107x + 0.6109
Producing Regression Function for Curve B
Lastly, upon first-glance, the curve in B resembles a cubic function since the curve opens both downwards and upwards. In both directions (upwards and downwards), the curves are slightly U shaped. Moreover, the curve has 2 turning points. Before the first turning point, the curve extends upwards however after the second turning point, the curve extends downwards. Lastly, the curve is not symmetrical and due to having 2 turning points, has 2 vertices.
The equation of a cubic function is in the form:
f (x) = ax3 + bx2 + cx + d
To develop a cubic function, we have to substitute the x-coordinate from a point on the curve into the x-values in the equation, and the y-coordinate into the f(x)-value. Since the curve resembles a cubic function, 4 different points on the curve have to be taken: producing 4 separate equations. The values of a, b, c and d, can then be determined through simultaneously solving the 4 separate equations. After calculating the values of a, b, c and d, a cubic function can then be produced to form a regression line.
Producing Regression Function for Curve B
Curve B:
| Point | Coordinates | x-coordinate | y-coordinate |
| 1 | (1.208, 0.9222) | 1.208 | 0.9222 |
| 2 | (1.393, 0.8254) | 1.393 | 0.8254 |
| 3 | (1.625, 0.7890) | 1.625 | 0.7890 |
| 4 | (1.817, 0.5773) | 1.817 | 0.5773 |
System of Equations:
a(1.208)3 + b(1.208)2 + c(1.208)+ d = 0.9222
a(1.393)3 + b(1.393)2 + c(1.393)+ d = 0.8254
a(1.625)3 + b(1.625)2 + c(1.625)+ d = 0.7890
a(1.817)3 + b(1.817)2 + c(1.817)+ d = 0.5773
Eliminating d:
(1.208)3a + (1.208)2b + (1.208)c + d = 0.9222
(1.393)3a + (1.393)2b + (1.393)c + d = 0.8254
(1.208)3a + (1.208)2b + (1.208)c + d = 0.9222
−(1.393)3a − (1.393)2b − (1.393)c − d = −0.8254
−0.940255a − 0.481185b − 0.185c = 0.0968
(1.393)3a + (1.393)2b + (1.393)c + d = 0.8254
(1.625)3a + (1.625)2b + (1.625)c + d = 0.7890
(1.393)3a + (1.393)2b + (1.393)c + d = 0.8254
−(1.625)3a − (1.625)2b − (1.625)c − d = −0.7890
−1.58797a − 0.700176b − 0.232c = 0.0364
(1.625)3a + (1.625)2b + (1.625)c + d = 0.7890
(1.817)3a + (1.817)2b + (1.817)c + d = 0.7890
(1.625)3a + (1.625)2b + (1.625)c + d = 0.7890
−(1.817)3a − (1.817)2b − (1.817)c − d = −0.5773
−1.70779a − 0.660864b − 0.192c = 0.2117
System of equations without d:
−0.940255a − 0.481185b − 0.185c = 0.0968
−1.58797a − 0.700176b − 0.232c = 0.0364
−1.70779a − 0.660864b −10.92c = 0.2117
Eliminating c :
−0.940255a − 0.481185b − 0.185c = 0.0968
−1.58797a − 0.700176b − 0.232c = 0.0364
−1.70779a − 0.660864b −10.92c = 0.2117
−0.041883a − 0.021434b − 0.008241c = 0.004312
−0.0056405a − 0.002487b − 0.008241c = 0.001293
−0.0073298a − 0.028364b − 0.008241c = 0.009086
−0.041883a − 0.021434b − 0.008241c = 0.004312
+0.056405a + 0.02487b + 0.008241c = −0.001293
0.014522a + 0.003436b = 0.003019
−0.056405a − 0.02487b − 0.008241c = 0.001293
+0.073298a + 0.0028364b + 0.008241c = −0.009086
0.016893a + 0.003494b = −0.007793
Solving for a :
0.014522a + 0.003436b = 0.003019
0.016893a + 0.003494b = −0.007793
0.000051a + 0.000012b = 0.000011
−0.000058a − 0.000012b = +0.000027
−0.000007a = 0.000038
a = −5.42857
Solving for b :
0.014522a + 0.003436b = 0.003019
0.0014522(−5.42857)+ 0.003436b = 0.003019
b = 23.8221
Solving for c :
−0.940255a − 0.481185b − 0.185c = 0.0968
−0.940225(−5.42857)− 0.481185(23.8221)− 0.185c = 0.0968 c = −34.894
Solving for d :
(1.208)3a + (1.208)2b + (1.208)c + d = 0.9222
(1.208)3(−5.42857)+ (1.208)2(23.8221)+ (1.208)(−34.894)+ d = 0.9222
d = 17.8809
Therefore,
a = −5.429
b = 23.82
c = −34.89
d = 17.88
After calculating the values for a, b, c, and d, I was able to produce a regression function for Curve B:
f (x) = (−5.429)x3 + (23.82)x2 − (34.89)x + (17.88)
Manually modelling the shape of the pawn using the regression functions calculated
After separating the graph of the pawn into 5 different segments, I was able to produce a regression function for the curve in each individual segment. I then manually inputted the regression functions, corresponding to each segment of the graph into Logger Pro. I was able to do this by highlighting the set of points that correspond to each segment and then using my regression function to produce a regression line for the particular set of points. Figure 4 depicts the manually produced regression lines for all 5 segments of the pawn, used to model its overall outline.
Figure 4
D
B
C
E
A
Although for the most part the regression functions calculated manually do accurately model majority of the pawn’s outline since the regressions produced do provide a close-fit to most of the points, there are parts of the pawn that are inaccurately modeled; since the regression lines produced do not closely-fit multiple points on the graph. The biggest areas for concern and inaccuracy in Figure 4 are determined to be curves A and E. As can be observed in Figure 4, the curve in segment A fails to accurately provide a fit for majority of the points in that particular area as evidenced by the fact that approximately 15 points do not lie within a proximity to the regression line that is deemed appropriate – a number that is far too large.
In fact, the regression line itself does not appear to even model the curve in segment A accurately since after producing a parabolic regression line, it is apparent that the curve in A has 2 turning points, and the regression line does not produce a close fit for the points in segment A. Furthermore, it can be deduced that the regression line used is incorrect as the curve in A is not a parabola and so, a quadratic function would not produce an accurate, or even correct, line of regression.
As a result, the regression functions calculated manually do not accurately model the entire shape of the pawn due to the fact that the regression lines in segment A do not closely-fit all the points in the segment. This is because in segment A, the nature of the curve was incorrectly identified and thus, an incorrect function was used to produce a regression line.
Using Logger Pro to digitally produce regressions for points in A and E
To solve this discrepancy and to produce a more accurate regression line, I used Logger Pro’s “analyse curve fit” function to digitally produce individual regressions for the points in A. Figure 5 depicts the new regressions for curves A produced digitally, in addition to the previous manual regressions.
B
Figure 5
As shown in Figure 5, using Logger Pro to digitally produce regressions has allowed the points in A to be modeled to a higher level of accuracy. This is shown by the regression line in A providing a closer-fit to the points in each segment, with significantly fewer points in segment A deemed “too far” from the regression (less anomalous points). After digitally producing the regression line for segment A, it became apparent that a quartic function provided the most accurate regression for both segments. This is justified by the fact that the RMSE (Root Mean Square Error) value for the regression A was 0.06703 cm. This is because the RMSE is frequently used to measure the difference between the values predicted by a regression model and the values that are actually obtained.
Thus, a lower RMSE value correlates to a smaller difference in predicted and actual values, representing a higher level of accuracy of a regression curve’s absolute fit. Since the RMSE values for the regression in A was 0.06703, this shows that the function used was able to produce a regression curve that was highly accurate in providing a fit for the data points in A since the RMSE value is close to 0, which is the highest level of accuracy attainable (meaning no difference in predicted and actual values). Thus, the new regression function for the curve in A is:
f (x) = (−7.866)x4 + (21.27)x3 − (20.59)x2 + (8.318)x + 0.008584
3D Graphing the model
Using Autograph, I inserted the equation of each regression line to produce a graph of the regression functions used to outline the pawn. Figure 6 depicts the graphing of the 5 different regression functions.
Figure 6
base-radius (cm)
height (cm)
With reference to Figure 2, it can be observed that positive x-values in Figure 6 represent the height of the pawn in centimeters, and the positive y-values represent the base-radius of the pawn in centimeters. However, since the graph in Figure 6 consists of 5 functions, we need to isolate every function: such that the section isolated, models a specific section of the pawn. This is done by setting limits as follows:
Regression A:
f (x) = (−7.866)x4 + (21.27)x3 − (20.59)x2 + (8.318)x + 0.008584
for
0.000 ≤ x ≤1.161
where 0.000 is the origin, and 1.161 is the final point of intersection between regression A and B.
Regression B:
f (x) = (−5.429)x3 + (23.82)x2 − (34.89)x + (17.88)
for
1.161 ≤ x ≤1.851
where 1.161 is the final point of intersection between regression B and A, and 1.851 is the point of intersection between regression B and C.
Regression C:
f (x) = −0.07107x + 0.6109
for
1.851 ≤ x ≤ 2.813
where 1.851 is the point of intersection between regression C and B, and 2.813 is the point of intersection between regression C and D.
Regression D:
f (x) = (−7.945)x2 + (47.75)x − 71.04
for
2.813 ≤ x ≤ 3.169
where 2.813 is the point of intersection between regression D and C, and 3.169 is the point of intersection between regression D and E.
Regression E:
f (x) = (−0.9758)x2 + (7.242)x −12.67
for
3.169 ≤ x ≤ 4.597
where 3.169 is the point of intersection between regression E and D, and 4.597 is the height of the pawn.
By plotting the regression functions A to E with the limits identified, the outline of the pawn can be accurately produced, as depicted in Figure 7.
Figure 7
base-radius (cm)
(0.399,1.279)(1.851,0.479)
(1.161,0.987)
(0,0) height (cm)
From Figure 7, the coordinate (1.279, 0.399) indicates the highest point of the graph (the maximum y-value achieved). Thus, since the y-axis corresponds to the pawn’s base-radius, the pawn’s base radius is 1.279 cm, giving it a base-diameter of 2.558. The last point on the graph, (4.597, 0.000) indicates that at the intersection of E with x=4.597, the height of the pawn model is 4.597 cm. Subsequently, a solid of revolution can be formed when Figure 7 is rotated through 360°, about the x-axis. This is depicted in Figure 8.
Figure 8
base-diameter(cm)
height(cm)
height(cm)
base-diameter(cm)
From Figure 8, we can see the similarity to the real-life structure of a pawn shown in Figure 9.
Figure 9
Calculating the Volume of Solid of Revolution
To calculate the volume of the solid of revolution formed in Figure 8, by the 5 regression functions and the x-axis, we can use integration. This will also allow us to find out the holding capacity of the 4.5 cm tall pawn (although it is highly improbable that an object of this size will be used to store liquids for regular consumption).
Given that a and b are points on the x-axis, we are able to calculate the volume of solid of revolution between a and b, as limits, using the formula:
b
∫
V = π y2 dx
a
where,
y is the regression function
a is the lower limit of the integral (starting x-coordinate)
b is the upper limit of the integral (final x-coordinate)
As depicted in Figure 8, rotating all 5 regression functions about the x-axis by 360° produces a solid of revolution that models the pawn. To calculate the total volume of solid of revolution of the pawn, the volume of solid of revolution of each segment, from A to E, has to be calculated separately (using their individual regression functions) and then added together.
Volume of solid of revolution of A
1.161
VA = π (−7.866)x4 + (21.27)x3 − (20.59)x2 ⎡ + (8.318)x + 0.008584 ⎣ ⎤⎦2dx
∫
0
VA = 4.226cm3
Volume of solid of revolution of B
1.851
VB = π (−5.429)x3 + (23.82)x2 ⎡ − (34.89)x + (17.88) ⎣ ⎤⎦2dx
∫
1.161
VB = 1.371cm3
Volume of solid of revolution of C
2.813
VC = π [−0.07107x + 0.6109]2dx
∫
1.851
VC = 0.6001cm3
Volume of solid of revolution of D
3.169
VD = π (−7.945)x2 ⎡ + (47.75)x − 71.04 ⎣ ⎤⎦2dx
∫
2.813
VD = 0.4364cm3
Volume of solid of revolution of E
4.597
VE = π (−0.9758)x2 ⎡ + (7.242)x −12.67 ⎣ ⎤⎦2dx
∫
3.169
VE = 1.653cm3
Total volume of solid of revolution of the pawn
V Pawn= 4.226 +1.371+ 0.6001+ 0.4364 +1.653
V Pawn= 8.287cm3
Designing the Decanter
As previously mentioned in the introduction, it is crucial that the decanter possesses the necessary dimensions required to store 750 ml of liquid, since that is the average volume of a decanter. Thus, it will be helpful to use similar dimensions to a decanter that resembles the shape of a pawn. Figure 10 depicts a decanter that is similar in design to the chess pawn modeled, and holds 768 ml of liquid (which I consider to be close enough to 750 ml because the change in liquid quantity is only 18 ml which results in a difference of 2.4%, a difference considered to be a very small margin of error).
28 cm
Figure 10
To design the decanter in the shape of a pawn, it is helpful to use the height and base-diameter of the decanter in Figure 10 for reference, since it will provide the best estimate for the dimensions of a pawn-shaped decanter needed to hold approximately 750 ml of liquid.
Applying the same method as before, a new function for each segment of the pawn can be calculated for the enlarged version of the pawn, using the graph in Figure 2, I altered the scale, such that the new height of the pawn (maximum x-value plotted) is enlarged by a scale factor of 6 of its original height, and the base-radius of the pawn (maximum y-value plotted) enlarged by a scale factor of 7 of its original base-radius. This is because dividing the height of the decanter in Figure 10 by the height of the pawn 28 ⎟ 4.6 = 6.087
( ) will give an approximated value of 6 when
19 cm
rounded off to the nearest one.
In addition, diving the base-radius of the decanter in Figure 10 by the base-radius of the pawn 19 ⎟ 2.6 = 7.308
( ) will given an approximated value of 7 when rounded off to the nearest one.
This is done so that the altered dimensions of the pawn (height and base-radius) are still in a ratio, similar to the original dimensions of the pawn. This allows me to maintain equivalent proportions between data sets, as well as the overall structure/shape of the pawn, but change the coefficients of the regression functions as follows:
Regression function of A after enlargement
f (x) = 7 (−7.866)(x6)4 + (21.27)(x6)3 − (20.59)(x6)2 + (8.318)(x6)+ 0.008584 ⎛⎝⎜ ⎞⎠⎟
Regression function of B after enlargement
f (x) = 7 (−5.429)(x6)3 + (23.82)(x6)2 − (34.89)(x6)+ (17.88) ⎛⎝⎜ ⎞⎠⎟
Regression function of C after enlargement
f (x) = 7 −0.07107(x6)+ 0.6109 ⎛⎝⎜ ⎞⎠⎟
Regression function of D after enlargement
f (x) = 7 (−7.945)(x6)2 + (47.75)(x6)− 71.04 ⎛⎝⎜ ⎞⎠⎟
Regression function of E after enlargement
f (x) = 7 (−0.9758)(x6)2 + (7.242)(x6)−12.67 ⎛⎝⎜ ⎞⎠⎟
Upon calculating the new regression functions for an enlarged pawn for all segments from A to E, a a model of the enlarged pawn can be produced. Figure 11 depicts the new regression functions used to model the shape of the enlarged pawn.
Figure 11
base-radius (cm)
(0,0)
A
B
(2.394, 8.953)
(6.966, 6.909)
(11.12, 3.353)
(16.88, 2.877) C
(19.01, 3.304)
D E
(27.58, 0.00) height (cm)
On Figure 11, the intersection of E with the x-axis at (27.58, 0.000) indicates that the maximum height of the pawn decanter would be 27.58cm, since the x-axis represents the height of the pawn decanter. In addition, the height of the pawn decanter is almost equal to that of the decanter in Figure 10 (negligible difference of 0.42cm), proving that the pawn decanter has been scaled and modeled very accurately in the x-axis against the decanter used for reference. However, the base diameter of the pawn decanter is slightly larger than the base-diameter of the decanter in Figure 10, by a difference of 1.094cm.
The base-radius of the pawn decanter is 8.953cm and so, its base diameter is 17.91cm ( ), indicated by the highest point on curve A (the 8.953⋅ 2 = 17.91cm maximum y-value achieved). Thus, the base-radius of the pawn decanter could have been scaled and modeled to a higher degree of accuracy through vertically enlarging the regression functions by a scale factor rounded off to a greater number of significant figures.
When designing the pawn decanter, it is important to keep in mind that most decanters, including the decanter in Figure 10, have stoppers that can be placed on the mouth of the decanters to prevent liquids from spilling or to prevent substances in the environment from contaminating the liquid being held. Thus, we can assume that the pawn decanter model in Figure 11 includes a stopper attached at its mouth. Referring to the design of the decanter in Figure 10, we can use curve E as the regression function for the pawn decanter’s stopper.
In addition, because curve E is attached to curve D, we can assume that curve D is the regression function for the mouth of the pawn decanter.
Furthermore, since the mouth of a decanter has the purpose of allowing liquid to be poured into or out the decanter, it would be most pragmatic for the mouth of the decanter to have a hole large enough for liquid to comfortably enter the decanter without excessive spillage. In addition, the mouth of the decanter also serves the purpose of holding the decanter stopper in place when it is attached (a part of the stopper is fitted into the mouth of the decanter).
As a result, it is impractical for the mouth of a decanter to hold any of the liquid. It is also highly impractical for the decanter’s stopper to be able to hold any liquid since that would make the stopper heavy to hold and thus, makes it inconvenient for the stopper to be taken on and off the decanter. This would defeat the entire purpose of having a
stopper since the pouring of liquids constantly requires the stopper to be constantly attached and removed from the pawn decanter’s mouth. Therefore, calculating the volume of regression functions D and E is unnecessary since no liquids will be stored in these portions of the pawn decanter. This leaves us with regression functions A, B and C, which are the segments of the pawn decanter that hold liquid. Hence, the volume of liquid that can be held by the pawn decanter is the sum of the volume of solid of revolution of A, B and C.
Volume of A after enlargement
1.161⋅ 6 = 6.966
VA = π 7 (−7.866)(x6)4 + (21.27)(x6)3 − (20.59)(x6)2 + (8.318)(x6)+ 0.008584 ⎛⎝⎜ ⎞⎠⎟ ⎡⎣⎢ ⎤⎦⎥2dx 6.966
Function 1
∫
0
VA = 1242cm3 = 1242ml
Volume of B after enlargement
1.851⋅ 6 = 11.106
VB = π 7 (−5.429)(x6)3 + (23.82)(x6)2 − (34.89)(x6)+ (17.88) ⎛⎝⎜ ⎞⎠⎟ ⎡⎣⎢ ⎤⎦⎥2dx 11.106
Function 2
∫
6.966
VB = 403.0cm3 = 403.0ml
Volume of C after enlargement
2.813⋅ 6 = 16.878
VB = π 7 −0.07107(x6)+ 0.6109 ⎛⎝⎜ ⎞⎠⎟ ⎡⎣⎢ ⎤⎦⎥2dx
16.878
Function 3
∫
11.106
VB = 176.4cm3 = 176.4ml
Therefore, to find the volume of liquid that can be held by the pawn decanter:
VPawn2 = 1242 + 403.0 +176.4
VPawn2 = 1821ml
Conclusion:
I was able to successfully explore functions that accurately model the shape of the Chess Pawn. After modelling the pawn, I was able to produce 5 functions that model an enlarged version to be used as a decanter. Based on functions 1, 2 and 3, the decanter, in theory, is able to successfully hold 1821ml of liquid. In addition, its overall dimensions are similar to the prerequisites given by the exemplar decanter in Figure 10.
However despite having similar dimensions, the model of the pawn decanter can hold a significantly larger volume of liquid than the decanter in Figure 10, at 1821ml in comparison to 768ml. This exploration has shown me that in order for the pawn decanter to have a volume similar to the volume capacity of other decanters, at 750ml on average, the height and base-diameter of the pawn decanter has to be significantly reduced. As a result, I have arrived at the conclusion that the pawn decanter’s practicality may be in question. This is because a pawn decanter standing at a height of 27.58cm, with a base-diameter of 17.91cm, would require a large volume of material to make the pawn decanter.
Thus, this would make the pawn decanter heavy to hold, even when empty. In addition, since the pawn decanter is able to hold a maximum volume of 1821ml, the total mass of the liquid being held would be 1.821kg, the mass of the liquid in the pawn decanter, combined with the mass of the pawn decanter itself, would make the pawn decanter very heavy to carry/hold if the liquid being held by the decanter is close to its maximum volume.
This would make it very difficult to pour liquid from the pawn decanter since it would be very easy for the person holding the pawn decanter to drop the pawn decanter due to its sheer mass. Therefore, accidents would be prone to occur, defeating the main purpose of a decanter. To improve upon this limitation, scaling the height and base-diameter of the pawn decanter by a smaller factor. To reduce the volume of material needed as well as the total volume of liquid that can be held.
Furthermore, the fundamental issue with this design is that I did not account for the thickness of the pawn decanter’s walls. A decanter with these dimensions would require a certain level of thickness in its walls in order for the structure of the decanter to be stable, such that it does not break easily.
The issue is that since I did not take into account the need for having walls that are relatively thick, the pawn decanter would be extremely fragile and thus, would break at the slightest bump. In addition, the pawn decanter would also be unable to hold any liquid since the mass of the liquid would most likely cause the walls of the pawn decanter to break as well, rendering the pawn decanter practically impossible to use.
To actually produce the pawn decanter, I would need to define a function for me to take into account the wall thickness needed for the pawn decanter. I would then have to calculate the volume of material needed to design walls that are thick enough to hold heavy amounts of liquid. Upon calculating the volume of material needed to produce the walls, a new model of the pawn decanter would be created and thus, the pawn decanter would have a new volume capacity to store liquid. In this situation, the volume capacity of the decanter would be reduced.
Overall, the general design of the enlarged Chess Pawn appears to only be suitable as a living room decoration, instead of a decanter.
Written By Max Ong
Sources Cited:
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Lumen Learning. “Recognize Characteristics of Parabolas.” Precalculus 1,
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Royal Scot Crystal. “ROYAL SCOT CRYSTAL – FLOWER OF SCOTLAND (THISTLE) SHIPS DECANTER (GIFT BOXED).” DECANTERS & CARAFES, www.royalscotcrystal.com/drinkware/shop-by-type/decanters-carafs/gift-boxed-flower-of-scotland-ships-decanter.html. Accessed 15 Oct. 2017.
Stimpfig, John. “Modern Decanter Designs.” Home | How To Spend It, Financial Times, 20 Jan. 2015, howtospendit.ft.com/food-drink/74841-modern-decanter-designs. Accessed 15 Oct. 2017.
Talman, Lou. “Volume of a Solid of Revolution.” Application of the Definite Integral, National Curve Bank, web.calstatela.edu/curvebank/volrev/volrev.htm. Accessed 16 Oct. 2017.
