Disney World’s Magic Kingdom with Dan Cockerell, Former Director of the Park
Sandcastle Builder Guide: How to build the strongest foundation for the sandcastle? As for problem one, assuming that the speed and direction of the wave are certain, the height of each wave is independent and obeys a certain distribution. We only consider the erosion effect of the water on the mound.
The erosion effect is a long process, the number of waves impacting the mound is very large. Using the large number theorem, we can approximate the number of the wave impacts at different heights X . This number estimates the degree of erosion by water at different heights. Based on this point of view, if the distribution function of wave height obeys F (x), the curve of the side view of the mound is a constant multiple of 1 − F (x).
By using the related conclusions of hydrodynamics, we can get the velocity distribution and pressure distribution of water flowing through the soil pile. By using this result, we can get that the pressure on the front section of the soil pile is larger and the pressure on the side is smaller. Therefore, the section of the soil pile at each height needs flattening, and the best section is an ellipse with larger eccentricity.
According to these two conclusions, we obtain the optimal three-dimensional shape of the mound.
As for problem two, we firstly define the normal saturation states of sand-water mixture and its corresponding internal tension based on physical structure and soil-water function and soon establish the water-defense-humidification model, in which each single wave may humidify a certain area of the sand foundation, thus influencing its ability to stand against out forces.
In order to find the optimal distribution of water content in the foundation, we vertically cut the foundation into 100 slices, then simulate the whole process by the cellular automata algorithm, and has finally found the optimal water-ratio distribution in the whole foundation is about 26% to 32% in the core while about 30% to 36% surrounding by simulated annealing search ,though the result is related to our assumption, but we find a quite interesting fact that the less water-ratio of the core than the surrounding makes the foundation more resistant to the impact and erosion of waves.
As for problem three, we continue to use the model established in problem two, while we horizontally cut each vertical piece of the sand-castle foundation and take the rainwater infiltration into account, thus establishing the rainwater-infiltration model. Since we have proved that the optimal foundation shape is an ellipse for each horizontal cross-section in the first problem, we then creatively convert the influence of rainwater on the foundation structure into the degree that ellipse cross-section in the first problem’s result be away from the balanced status under the condition of rainfall. With the help of cellular automata algorithm, we finally come to the conclusion that the foundation will become a flat ellipsoid, and will be more likely to be broken by the waves compared with the circumstance without the rainwater.
Moreover, for problem four, we can consider two aspects to increase the firmness of the mound. The first is to reduce the erosion of water on the mound. We consider adding a hydrophobic layer on the surface of the mound, which can reduce the infiltration of water, so it can reduce the erosion of water. On the other hand, we consider reducing the speed of water and adding a stone column in front of the soil pile to resist the water flow. Furthermore, through the related formula of hydrodynamics, we can deduce that the larger the radius of the stone column is, the smaller the distance between the pile and the stone column is, the smaller the velocity of water flow to the pile is.
Finally,we write an informative, one- to two-page article describing our model and its results for publication in the vacation magazine: Fun in the Sun, whose readers are mainly non-technical. Hope that they will like it!
Keywords: Hydrodynamics, Cellular automata, Optimization model, Structural mechanics
1.1 Problem Background
People often build castles on the beach. People want their castles to be strong enough to last long enough without collapse. The main content of this paper is to explore how to make a castle stronger and present the results in the form of answering several questions.
1.2 Variables
Symbol Meaning
ai Foundation slice i
W(ai) Water-ratio of slice ai
D(ai) Water-defense ability of slice ai
θS Saturated water-ratio
θM Crashing water-ratio
A impact of each single wave
L Wave-humidification range
F Evaporation coefficient
v∞ Velocity of water flow
p∞ Pressure of water flow
h Height of foundation
2.1 Assumptions
1. When waves hit the sand pile, the speed and direction of the water flow are constant.
2. Each time the wave immerses the sand pile, it will erode the surface of the sand pile, taking away part of the sand on the surface of the sand pile as well as making the sand pile decrease the same thickness.
3. The effect of the water flow on each erosion of the sand pile is small enough that sand pile erosion is a long-term process.
4. The height of the sand pile is constant as hand the maximum height of the wave is also h
5. The height of each time the wave immerses the sand pile is Xand each height is independent and identically distributed. The distribution function is F (x)
2.2 Model building
Considering side view of the mound
Suppose that the mound is impacted n times by waves, and the heights of the waves are X1, X2, , Xn, then n random variables are distributed with the same independence. We assume that their distribution function are F (x),so that
Fn (x) = 1n∑n i=1
I{Xi≤x}
. Then Fn (x) is the empirical distribution function of samples. We noticed that if Fn (x) = kn,then the sand at height x impacted n − k times. According to the law of large numbers, we infer that
Fn (x)P−→ F (x) as n −→ ∞
Because the wave’s erosion effect on the sand pile is small enough, the erosion effect is a long-term process. Therefore, when n is large enough, We can use F (x) to approximate Fn (x)
At this point, the number of times the sand pile at the height of x received impacts by the wave approximately expressed as n (1 − F (x)) time.
Discuss the side curve of the sand pile based on the result above, which is the side view of the sand pile. As shown by figure
Assume that the side curve of the sand pile is f (x), and that the sand pile height is h,As shown in the figure below
while the length from the front of the sand pile to the castle is M.
{f (0) = M
f (h) = 0
Assume that the thickness of the sand pile which is taken away by one shock is α, then the side view curve of the sand pile after n shocks is like:
f (x) − αn (1 − F (x))
Under the assumption that sand pile is only exposed to wave erosion, keeping the sand pile last long is equal to withstanding a sufficient number of shocks. The condition for
the castle to collapse is that the sand pile below the castle is eroded which means the foundation becomes hollow, and after n shocks ,sand pile side curve, the function, has points less than 0. Therefore, to prevent the castle from collapsing, the condition should be met that
f (x) − αn (1 − F (x)) ≥ 0 for any x in [0, h]
Which means
n = inf
{f (x)
α (1 − F (x)) : xϵ[0, h]
}
Modify f (x) in order to make n the maximum In fact,if x = 0,then α (1 − F (0)) =Mα
f (0)
Therefore n ≤Mαand whenf (x) = M (1 − F (x)) ,
α (1 − F (x)) =M (1 − F (x))
f (x)
α (1 − F (x)) =Mα
In conclusion, when f (x) = M (1 − F (x)), the maximum number of attacks the sandpile can withstand.
Considering the top view of the mound:
It is assumed that the currents brought by the waves are direct currents and are non-rotating. According to the knowledge of fluid mechanics, when the top view of the sand pile is an ellipse, the total resistance it suffers is the smallest. Based on this conclusion, for any sections with a sand pile height of x, it is always an oval. Because the height of the sand pile is small, the influence from the vertical direction of the water flow can be ignored, and it is approximately assumed that the force situation at each height of x is exactly the same.
We supposed that the direction as well as the size of water flow is constant. Under the assumption, we can calculate the velocity distribution and pressure distribution on the surface while considering the direct current flow around the ellipse on the coordinate axis.Let the direct current be
ψ1 = v∞y
And the elliptical dipole streamlinesis like
ψ2 = −M2πy
y2 +x2
a2
The streamlines after superimposed is like
ψ = v∞y −M2πy
y2 +x2
a2
The surface of the oval has zero streamlines, so bringing in ψ = 0 gives y2 +x2
a2=M
2πν∞
If R =
M
2πv∞, the equation of the ellipse is x2
a2+ y2 = R2
Finding the partial derivative of the convection function can get the speed in the x direction and the y direction:{υy = −∂ψ
∂x = −2xy
∂y =2y2
a2R2 υ∞
vx =∂ψ
R2 υ∞
Thinking about the points on the ellipse and transforming them into polar coordinates, according to the Euler integral equation without a spin field, we can get the equation.
p − p∞
2ρυ2∞= 1 − 4 1
(sin2θ cos2θ
a2+ sin4θ
)
According to the pressure distribution on the ellipse, it can be known that as θ becomes larger, the pressure on the ellipse becomes smaller, which means the pressure on the front part of the ellipse is the largest, and it is easier to be eroded, while the pressure on both sides is smaller, which refers that the impact of erosion is also small. According to this conclusion, the shape of the top view of the sand pile should be flat, which means the elliptic eccentricity should be large. At this time, the sand pile is stronger and less susceptible to erosion by flowing water.
2.3 Model result
Accroding to the model build above, If we put vinf = 2 , a = 2 , R = 1 , h = 1.And suppose that the wave height X obeys the uniform distribution from 0 to h.The shape of the sand pile is shown in the figure below
Figure 1: the 3-dimensional geometric shape of sand pile
3 Problem 2
3.1 Analysis of problem 2
As in our model of problem 1, we default that the sand-water ratio keeps the same and under the condition of constant wave speed. While in this problem,we need to
consider an optimal sand water ratio, which is not necessarily constant. In order to achieve the best protection effect, the water content of different positions of the foundation may be different, thus gaining different ability of preventing seawater erosion and possibility of being washed out by seawater. Therefore, we establish a water-defense humidification model so as to describe the erosion of seawater on sand foundation. In order to simplify the solution, we cut the base into many slices, each slice shares the same water content. Under different sea water destructive forces, we use cellular automata algorithms to simulate the solution, and then get the foundation’s optimal water content distribution structure.
3.2 Preparations for the model
Water-ratio
We first assume that the sand is of the same shape and size without water swelling, then we can define the water-ratio of an area a as the volume of water in the smallest cube containing all the sand particles in this area divides the whoe volume of the cube. We recorded a’s water-ratio as W(a).
Saturated state
Figure 2: unsaturated-saturated-oversaturated state
For the sand in a certain area, we define the following three states: unsaturated, saturated and supersaturated. If we keep adding water into a piece of sand. The sand in this area is initially unsaturated, as shown in figure 1.1 above. Continue adding water until the gap between the sand particles is just filled. We call this state saturated state, we assume the water-ratio now as θS and it is saturated as shown in Figure 1.2 above. Then continue adding water, as shown in Figure 3, we call it an oversaturated state.
Foundation into slices
Figure 3: cutting into slices
Since the direction of sea water is fixed, it can be assumed that the sand foundation on the same vertical plane has the same water-ratio W(a). In order to simplify the
model, we cut the foundation evenly into 100 parts along the direction of sea water action, which is recorded as ai(i = 1, 2, …, 100) , as shown in the figure, assuming that each piece has the same initial water content, a 1 ∗ 100 water-ratio matrix.
W = (W(a1), …, W(a100))
3.3 Water-defense-humidification model
3.3.1 Water-defense model
Figure 4: Water-defense process
The ability of the structure to resist seawater erosion mainly depends on the viscosity inside the structure. After consulting the data, it is found that the viscosity of sand is mainly determined by its water-ratio, and obeys the following formula:
θS − θr= F(ψ) = 1
θ − θS
[1 + ψab](1 −1b)
Generally, C is about 1. Which means,it can be regarded as a linear relationship before being saturated.
Therefore, the concept of water-defense D is derived from the abstract soil viscosity. The relationship between the structural defense of area a and its water-ratio W(a) obeys the following function,in which θS is the saturated water-ratio.
W(ai) =
θaiθai < θS
θS θS < θai < θM 0 θai ≥ θM
Assuming that the destructive force of each wave impact is constant when the sea water speed is fixed, it is recorded as A. Each time the wave passes a certain slice ai, the total A will lose the same amount of the corresponding slice’s water-defense ability D(ai) . When A is reduced to 0, the impact of a sea wave is regarded as completely offset. After recording the number of slices taking part in the defense as L, (A can be quantitatively described by the analysis in the first question). We also record the water-defense ability of each slice as a matrix.
D = (D(a1), …, D(a100))
Humidification and erosion model
Figure 5: Humidification
Each wave will undoubtedly humidify L slices while participating in the wave defense process, and because the humidification effect of the sea water on the foundation is obviously gradually weakened. It is advisable to set the humidification effect of the wave on the first slice participating in the water-defense process as M, which means we add M to its water-ratio. Then the humidification effect linearly reduce it to 0, and then we get a new water-ratio matrix W.
W(1, start + i) = W(1, start + i) + M ∗ (L − i)
L(i = 0, 1, …, L).
In which ’start’ is the serial number of the first slice whose water-defense ability is not zero.
For each slice,if its own water-ratio W(ai) exceeds an upper limit θM (greater than the saturated water content θS), then this slice is considered to be washed out by the waves and the structural defense force is cleared.
3.3.3 Evaporation process
Since there is ainterval between two waves, we can assume that the evaporation coefficient in this period is F, and before the next wave reaches, the elements in the water-ratio matrix W will all evaporate at the rate F, as the following formula shows.
W(1, i) = W(1, i) ∗ (1 − F)(i = start, start + 1, …, 100).
3.3.4 Determination of the optimal result
Since we need to find out a most powerful foundation, which means it will be able to defend as many waves as possible. Although the number of waves k in this model is not directly equal to the real times of the wave the sand foundation can bear ,we can just use it as a reference and find out an optimal initial water-ratio matrix W which maximizes the wave number k in this model.
max
Wk(W)
3.4 Simulation based on Cellular Automata
Firstly, Step 1: Set index ′start′ = 0, set the wave number k = 0.
2: Input the initial water-ratio matrix W, then calculate the corresponding water defense ability matrix D.
3: Put in a wave, calculate the range L that the wave can influence(from the start layer) according to the recent water-defense matrix D.
In addition, Step 4: Update the wave count k = k + 1.
5: Calculate the humidification of the slices which take part in this water-defending process.
6: Renew the water-ratio matrix W, find the slices whose water-ratio is higher thanθM.
7: Get a new starting point ’start’ for the next wave.
Almost lastly, 8: Calculate the evaporation process and then renew the water-ratio matrix W again.
Lastly, 9: Calculate the new water-defense matrix D according to the recent water-ratio matrix W.
Step 10: Turn to step2 while W(1, 100) < θM.
3.5 Model result
By using the simulated annealing algorithm,we get the optimal initial water-ratio matrix W under different evaporation rates, link the node drawing, and get the optimal initial sand water structure, as shown in the figure below.
Figure 6: Optimal initial water-ratio distribution
Problem 3
4.1 Analysis of problem 3
From the former results of the first problem, we have known that the optimal foundation shape is an ellipse, and every transverse piece of the foundation is an ellipse, which generally combines to a flat ellipsoid. We also establish the water-defense humidification model to describe the change of water-ratio distribution in problem 2. And find that the optimal water-ratio distribution of the foundation. Moreover, to be relatively dry in the core while being a little bit wetter (near the saturated bound) outside. For this problem, we include the rain factor into our consideration. We need to consider the impact of rainwater on the humidity of the foundation, since the rainwater may penetrate downward. So we then established the rainwater-infiltration model based on the water-defense-humidification model by adding certain factors caused by rainwater and connect the size of the ellipse piece with the piece’s average water-ratio.
4.2 Preparations
Segmentation operation
Just like the way we operate in problem 2,we need to cut the whole foundation into pieces in order to improve the later simulation efficiency . However, as the former result of the previous problem shows, the optimal distribution of water-ratio does not changes gentle enough, so we still decide to vertically cut it into 100 slices a1, a2, …, a100, meanwhile, we horizontally cut each slice ai evenly into 10 layers ai1, ai2, …, ai10 . Till now, we cut the sand foundation into 1000 pieces, and then record it as a 10∗100 matrix.
Ellipse layer’s optimal size
a11 a21· · · a1001
a12 a22· · · a1002 ............ a110 a210 · · · a10010
As known from the result of problem 1, each horizontal layer is an ellipse and obeys the following formula, in which b represents the length-width ratio.
P V − P∞
2ρv2∞= 1 − 4sin2θ(sin2θ + b2cos2θ)
1
After we take the solid viscosity V in account (use the water-defense ability to replace viscosity). And fix the value b, we then find out that the optimal size of each layer are proportional to its weighted-average water-defense ability.
Rainwater-infiltration model
4.3.1 Infiltration rule
We assume the sand foundation only has vertical infiltration ,but no water penetration in the horizontal direction. According to the permeability speed formula: (vrisaf ixedspeedinnaturalworld, p is a constant)
vr= ( θθS)p
v
Then the rule of infiltration could be easily derived, we will then discuss the physical meaning of it.
Wtrans = ( θθS)p
s
Since we have assumed that the interval between two waves is constant,we define Wtrans(aij) as the amount of water-ratio that transfers from aijto aij+1 during one interval (consider the ground as the last horizontal layer).
4.3.2 Crash and vertical filling
Continue to use the same judging method in the water-defense-humidification model of problem 2, when the water-ratio of aijis greater than the crashing boundary θM, we view aij be washed away, and the cubes ait, t = (1, 2, …, j − 1) (if exist) will go down ward and automatically fill the place.
4.3.3 Water-defense-humidification process
This part is almost the same as the previous model. The only difference we need to consider is that this time the matrix has a width of 10, so we need to divide the sea water power into parts as the following process shows.
Water-defense process: We recorded the seawater impact as A, and vertically divide it into ten parts A = (A1, A2, …, A10. Each time the wave Ai passes a certain slice aij, the total Ai will lose the same amount of the corresponding slice’s water-defense ability D(aij) . When A is reduced to 0, the impact of a sea wave is regarded as completely offset. After recording the number of slices taking part in the defense as L, (A can be quantitatively described by the analysis in the first question). We also record the water-defense ability of each slice as a matrix D.
D11 D21· · · D1001
D12 D22· · · D1002 ............ D110 D210 · · · D10010
Humidification:
Each wave layer will also humidify Lilayer i slices which participate in the wave-defense process. And the humidification effect of the sea water on the foundation is obviously gradually weakened. It is advisable to set the humidification effect of the wave on the first slice participating in the water-defense process. As M, which means we add M to its water-ratio. Then the humidification effect linearly reduces it to 0, and then we get a new water-ratio matrix W.
W(j, start + q) = W(j, starti + q) + M ∗ (Li − q)
Li(q = 0, 1, …, Li, j = 1, 2, …, 10).
In which startiis the serial number of the first slice in layer i whose water-defense ability is not zero.
4.4 Simulation based on Cellular Automata
Firstly, Step 1: Set index starti = 0,(i = 1, 2, …, 10), set the wave numberk = 0.
Secondly, Step 2: Input the initial water-ratio matrix W, then calculate the corresponding water defense ability matrix D.
Thirdly, Step 3: Put in a wave, calculate the range Li,(i = 1, 2, …, 10) that the wave can influence(from the start layer) according to the recent water-defense matrix D.
Fourthly, Step 4: Update the wave count k = k + 1.
Fifthly, Step 5: Calculate the humidification of the slices which take part in this water-defending process.
Sixthly, Step 6: Renew the water-ratio matrix W, find the slices whose water-ratio is higher thanθM.
Seventhly, Step 7: Calculate the infiltration of layers then renew the water-ratio matrix W.
Step 8: Get a starting points starti,(i = 1, 2, …, 10) for the next wave.
Lastly, Step 9: Calculate the new water-defense matrix D according to the recent water-ratio matrix W.
Step 10: Turn to step2 while W(i, 100) < θM,(i = 1, 2, …, 10).
4.5 Model result
4.5.1 Distribution
By using the simulated annealing algorithm, we get a beat-suited initial water-ratio matrix W and put different layers as a distinct curve as the picture shows.
Figure 7: Water-ratio distribution of different layers
4.5.2 Ellipse layer’s optimal size
From the result above, we find that the distribution structure of each layer is similar to the problem two’s result, while the weighted-average optimal water-ratio of the deeper layer is lower than the less deeper layers, which means, the deeper horizontal ellipse foundation layer’s optimal size is expected to be relatively larger than they are in the problem’s result while the less deeper ellipse layer is expected to be smaller than they are in the problem 1’s result after we consider the rainwater influence.
5 Problem 4
In order to make the sand pile stronger, we should consider the following two aspects
5.1 Reduce water erosion
The erosion of water is mainly to change the proportion of sand and water on the surface of the sand pile, so that the proportion of sand and water on the surface of the sand pile increases, so that the tension between the sand decreases. Therefore, as long as the permeability is reduced, the erosion caused by flowing water can be reduced.
Consider covering the surface of the sand pile with a hydrophobic layer to reduce infiltration. Due to the effect of the hydrophobic layer, the water penetration will be greatly reduced, so the sand water ratio of the sand pile can be kept as constant as possible. Under the condition of the best sand water ratio in the second question, the sand pile can last longer without considering the wave deformation.
5.2 Reduce water flow speed
The above assumption is only discussed in the case of small waves. At this time, the impact of waves on the deformation of the mound can be ignored, only the erosion of water.
The actual situation is quite different from the ideal situation. In different places of the beach, the size of the tide is very different, which is closely related to the location, precipitation and season. Therefore, according to the specific situation of the local beach, build a sand pile high enough to make the height of the wave smaller than the height of the sand pile, so as to avoid the wave directly damaging the castle.
When the sand pile is high enough, if the water speed is fast, the deformation of the sand pile caused by the wave cannot be ignored. The most direct idea is to reduce the flow rate of water. For this reason, it is considered to place a solid stone cylinder in the direction of water flow.
In an ideal case, that is, the stone cylinder is strong enough to be incompressible, the cylinder does not rotate under the action of water flow, and the water flow around the cylinder does not rotate. Based on the knowledge of hydrodynamics, the change of flow velocity after the flow bypasses the cylinder is deduced.
Similarly, it is assumed that the flow velocity is v∞, and the radius of the cylinder is R. The stream function can be written as follows:
(
ψ = v∞
Consider polar transformation:
(
ψ = v∞
The flow function changes to:
y −R2y x2 + y2
y −R2y x2 + y2
) )
(
ψ = v∞
r −R2 r
)
sin θ
Find out the radial velocity vrand the tangential velocity vΘ
vr =∂ψ
(
1 −R2
)
r∂θ = v∞ cos θ
(
vθ = −∂ψ
∂r = −v∞ sin θ
r2
1 + R2 r2
)
Assuming that the distance between the front section of the sand pile and the end of the stone column is s
the velocity of the front section of the sand pile is
(
v = vr = −v∞
1 −R2 (R + s)2
)
According to this formula, when R is large enough or s is small enough, the flow veloc ity in the front section of the sand pile can be greatly reduced by the stone column, so as to reduce the flow velocity and prevent the deformation of the sand pile. Further more, the thermodynamic diagram of velocity distribution can be drawn, assuming that R = 1 , v∞ = 2
6 Problem 5
When we mention summer, what first comes into your mind? Maybe lemonade, ice or something else. There must be someone who would think about the beach. As summer is approaching, the beach is about to become the most popular spot. Lie on the beach, sunbathe, drink cool soda and enjoy the tranquility of your holiday. Various sandcastles on the beach became a beautiful landscape. Have you ever thought about how to build a sandcastle which can stay the longest on the beach? Today, we will show you how to construct the perfect sandcastle.
First of all, the sandcastle cannot resist the impact from the ocean, like waves or tides, by itself. In order to prevent the sandcastle from directly exposure to the water shock, build a strong foundation.
Firstly, it is common knowledge that the waves always rush towards the sandcastle from directly in front, therefore the pressure from the water on the front side is large, and the pressure on the side is small. Secondly, the erosion on the front side is more serious than the front side. As a result, which means the front needs more thickness.
Thirdly, to withstand such shocks, the foundation will be a flat oval-like shape, just like a pudding.
As we all know, proper mixing of water with sand can increase the adhesion between particles which can make the sandcastle stronger. Therefore, the water ratio should be considered in order to make the sands harder. In addition, in order to find the optimal distribution of water content in the foundation, let’s assume that the the foundation consists of slices.
We vertically cut the foundation into 100 slices. After the cellular automata simulation. We finally found the optimal water-ratio distribution in the whole foundation is about 26% to 32% in the core while about 30 % to 36 % surrounding. In a brief, less water-ratio of the core than the surrounding makes the foundation more resistant to the impact and erosion of waves.
In addition to the impact of the tide on the sandcastle, rainfall will also cause the sandcastle to collapse. Waves and tides only impact sandcastle forwards, but rainfall will make sandcastle suffer rain erosion from top to bottom as well as from outside to inside. According to the assumption mentioned above. Rain fall degrades the foundation, layer by layer. Therefore, the optimal moisture content distribution decreases gradually from top to bottom. While we horizontally cutting each vertical piece of the sand-castle foundation and the rainwater infiltration into account, we finally come to the conclusion that the foundation should become a flatter ellipsoid.
There are also many other measures to construct a durable sandcastle, like change the material of sands or alter the interior structure of the sandcastle. For instance, we can add a hydrophobic layer whose function is to speed up the outflow of water. Due to the effect of the hydrophobic layer. The water’s ability of penetration will be greatly reduced, so the water ratio of the sand pile is able to be kept as constant as possible.
Moreover, what’s more, how about build a gigantic sandpile to protect the castle from erosion?
According to the specific condition of the beach, maybe we can construct a foundation which is higher than the waves. In order to protect the castle further, a hard stone pillar can be deposited in front of the castle, which can reduce the speed of water flow greatly.
Lastly, true knowledge comes from practice. While you are enjoying your leisure time on the beach and watching this magazine, have a try, making your sandcastle is the most durable. In conclusion, good luck!
7 Strengths and weaknesses
7.1 Strengths
1. Using hydrodynamics to describe the velocity distribution and pressure distribution makes the model more accurate
2. It is relatively reliable to simulate the sea water impact on the sand foundation by cellular automata.
3. The qualitative conclusion of the model has practical significance and needs a solid explanation.
4. A creative model established for practical problems, and it has certain value.
7.2 Weaknesses
1. Only when the velocity and direction of water flow are constant is considered, and the influence on the model caused by the change of velocity and direction is not considered.
2. The simulation accuracy with cellular automata is relatively low because of the limited number of layers and slices.
3. No consideration of the lateral permeability of water.
4. The model is sensitive to some initial conditions such as evaporation rate and sea water velocity.
Sandcastle Builder Guide: How to build the strongest foundation for the sandcastle? References
[1] A model for the dynamic of sandpile surfaces J.-P. Bouchaud (*), M. E. Cates, J. Ravi Prakwh (**) and S. F. Edwards Cavendish Laboratory, Madingley Road, Cambridge CB3 OHE, U-K- (Received 31 March 1994, accépted in final form 30 June 1994)
[2] How to construct the perfect sandcastle Maryam Pakpour1,2, Mehdi Habibi1 , Peder Møller3 & Daniel Bonn2,3 1 Institute for Advanced Studies in Basic Sci ences, P.O. Box 45195-1159, Zanjan, Iran, 2 Van der Waals-Zeeman Institute, IoP, University of Amsterdam, Science Park 904, 1098XH, Amsterdam, the Nether lands and, 3 Laboratoire de Physique Statistique de lENS, 24 Rue Lhomond, 75231 Paris Cedex 05, France.
[3] THEORETICAL AND EXPERIMENTAL STUDIES OF STRESS DISTRIBUTION IN WEDGE-SHAPED GRANULAR HEAPS Thirapong Pipatpongsa1 Tadaki Mat sushita2 Maho Tanaka3 Shinichi Kanazawa4 Katsuyuki Kawai5 ( 1Global Scientific Information and Computing Center, Tokyo Institute of Technology, 2-12-1 O-okayama, Meguro-ku, Tokyo 152-8550, Japan) ( 2Department of International Development Engineering, Tokyo Institute of Technology, 2-12-1 O-okayama, Meguro-ku, Tokyo 152-8550, Japan) ( 3Maeda Corporation, 2-8-8 Sarugaku-cho Chiyoda-ku, Tokyo 101-0064, Japan) ( 4Nishimatsu Construction Co., Ltd., 20-10, Toranomon 1-chome, Minato-ku, Tokyo 105-8401, Japan) ( 5Research Center for Urban Safety and Security, Kobe University, 1-1 Rokkodai-cho, Nada-ku, Kobe, Hyogo 657-8501, Japan) Received 29 February 2012, revision received 10 April 2013
