yun seong song - modeling of the supporting legs for designing biomimetic water strider robots

8/20/2019 Yun Seong Song - Modeling of the Supporting Legs for Designing Biomimetic Water Strider Robots

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Modeling of the Supporting Legs forDesigning Biomimetic Water Strider Robots

Yun Seong SongDepartment of

Mechanical EngineeringCarnegie Mellon University

Pittsburgh, Pennsylvania 15213Email: [email protected]

Steve H. SuhrDepartment of

Mechanical EngineeringCarnegie Mellon University

Pittsburgh, Pennsylvania 15213Email: [email protected]

Metin SittiDepartment of

Mechanical Engineering andRobotics Institute

Carnegie Mellon UniversityPittsburgh, Pennsylvania 15213

Email: [email protected]

Abstract — Recent studies on the insect water strider showedthat the insect heavily relies on surface tension force to stay aoat.Inspired by this insect, water strider robots have been developedusing the same locomotive principles as the insect. This paperfocuses on numerically modeling the supporting legs of the insectand the robots. The rigid-leg model as well as the compliant-

leg model is developed using numerical approaches, under anassumption made on the water surface breaking condition. Theeffect of different leg material and geometry are discussed. It isshown through simulations that four 7 cm-long Teon R coatedcompliant supporting legs with optimized shapes can lift up to4.3 grams (0.15 g/cm), while an actual prototype carried 3.7grams. Another prototype using twelve of these legs successfullylifted 9.3 grams. Experiments show that the analyses capturethe important features of the supporting legs. The design rulesproposed in this paper will be useful in understanding the insectstatics and also the robotic water strider supporting leg design.This study will allow a heavier robot to be used for education,entertainment or environment monitoring purposes.

I. INTRODUCTION

Adapting sub-optimized working principles of the biologicalsystems to synthetic technologies is one of the issues inengineering design. Particularly in robotic applications, bio-logically inspired systems can prove highly efcient, as canbe seen from wall-climbing gecko robot [1] to underwater sh-like robots [2]. Many robots adapt designs inspired by insects,such as the 6-legged robots similar to cockroaches [3]-[5], 8-legged robot [6], and the micromechanical ying robot [7].

Recently, the unique characteristics of the locomotion of theinsect water strider have been studied, including the dynamiclocomotion behaviors of the insect [8] and the hydrophobicityof their legs [9]. This insect maintains its body on water almostonly using the surface tension force, and can propel itself atthe peak speed of 1.5 m/s. Hu et al. [8] proposed a robotmimicking this insect that uses elastic energy storage, and Suhret al. [10] built a more robotic water strider robot that usesfour Teon R coated supporting legs.

Dynamic analysis on the robotic water strider robot ispresent [8], [10], but a detailed analysis on the supportinglegs of these robots is yet to be proposed. Since both the insectand the robot rely heavily on the surface tension force of thewater to gain lift force, detailed analysis on the statics on water

Fig. 1. (a) A photo of the water strider, and (b) the robotic water strider. A,B, C and D in (b) are the supporting legs. Other labels in (b) are explainedin [10].

surface can provide useful information on the characteristicsof the insect, which can be applied to the design of therobot. One important feature of the robot (and the insect)

is its loading capacity, and a static model of the legs wouldprovide a way to calculate the maximum weight a robot cancarry. In this paper, a numerical approach to understand thephysics between the supporting legs and the water surface isproposed. Starting from the Young-Laplace equation and someassumptions, characteristics of various kinds of supporting legsare analyzed.

I I . L EG M ODELING

As aforementioned, it is important to know how muchweight the robot can actually carry. And if this information


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Fig. 2. A 3-D illustration of a rigid cylinder partially submerged in water.

Fig. 3. 2-D model of the problem with depth L into y direction. Parametersare dened in Table I.

can be predicted given the design parameters of the supportinglegs, more rigorous planning of the robot fabrication can beperformed. In this section, a numerical model of the watersurface near a partially submerged rigid cylinder is proposed.From this, another numerical model regarding a long exiblesupporting leg is developed.

A. Problem Denition

A long and thin rigid cylinder that represents the supportingleg of the water strider, made of a material with contact angleθc with water, is partially submerged on water as shown inFig. 2. The axis of symmetry of the cylinder is horizontalto the undisturbed surface of the water. Assuming there areno complex water surface behaviors at the two ends of thecylinder, the problem can be understood as a 2-D problemwith a depth of L , which is the length of the cylinder (Fig. 3).Parameters in Fig. 3 are explained in Table I.

The area A in Fig. 3 is proportional to the buoyancy forceacting on the cylinder, and the area B is proportional to thevertical component of the surface tension force [11]. Theweight of the water volume, bound above by the z = 0

surface and below by either the water-air interface or the water-cylinder interface, equals the total lift force the cylinder bodyexperiences. The ratio between the area A and B represents themagnitude of the lift force from the buoyancy effect comparedto that from the surface tension force.

The shape of the air-water interface can be calculated usingthe Young-Laplace equation. Then, by integrating this interfaceprole along the x-axis, the total lift force can be calculated.

1) Governing Equation: From Young-Laplace equation, thedifference of pressure across the water-air interface on a given

TABLE I

D EFINITION OF PARAMETERS IN F IG . 3

Parameters/Labels Notation Typical va lue

Contact angle θc 0◦ − 180 ◦

Submerge angle φ 0◦ − 180 ◦

Length of the cylinder L 10 − 50 mm

Radius of the cylinder R 0 − 1 mmDensity of water ρ 1000 kg/m 3

Water take-off point x0 (= R sin φ) 0 − R mmArea representing thebuoyancy force

A N/A

Area representing thesurface tension force

B N/A

Buoyancy force f B 0 − 10 mNSurface tension force f T 0 − 10 mN

point on the interface, ∆ P , can be written as

∆ P = γ · ( 1R 1 +

1R 2 ), (1)

where γ is the coefcient of surface tension (0.072 N/m), andR 1 and R 2 are the principle radii of curvature of the surface atthe point. In this specic case of a long rigid cylinder ( R 2 =∞ ), (1) can be specied as

ρgh (x ) = γ ·

d2

dx 2h (x )

1 + ddx

h (x )2

32

, (2)

where h(x ) is the surface prole of the water-air interfaceas shown in Fig. 3, ρ is the density of water, and g is thegravitational constant. That is, equation z = h(x ) is thedescription of the water-air interface prole. The boundaryconditions (BC’s) for h(x ) are

dhdx

(x0) = tan( θc + φ − π ) (3)

h (∞ ) = 0 (4)dhdx

(∞ ) = 0 (5)

where x0 = R sin φ is where the water, air and the rigidcylinder all meet together. Since the number of BC’s neededto solve (2) is two, one of the conditions must be removed.In the analysis, the third BC, (5), was ignored. It can be seenlater that the solution to (2) also satises (5).

To solve the problem dened by (2)-(5), the value of φ mustbe known to completely dene the BC. First, assume that thevalue of φ to be given, and then (2) is solved. The increasingvalue of φ implies that the cylinder is pushed deeper toward the− z direction, given a specic value of θc . The exact relationbetween φ and the depth of the cylinder is irrelevant to thisproblem and is not dealt with.


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2) Surface Breaking Condition: The formulation of theproblem is sufcient by (2)-(5), but there is still one issueto be resolved; how far can we descend the cylinder before itbreaks the water surface? In other words, what is the surfacebreaking condition? Only when this question is answered canthe maximum lift force be calculated. Equation (2) does notreveal any other behavior of water to trigger the breaking of thesurface. But the rst BC, (3), can provide some information;if the slope of h (x ) at x 0 is innite, there is no solution to theproblem. Further more, in this case, vertical component of thesurface tension force is in its maximum and cannot withstandany further increase in payload. For the case of θc ≤ 90◦ , thisnever happens, unless the cylinder is above the water surface.But when θc > 90◦ , the slope at x0 becomes innite forφ = 270 ◦ − θc . This specic value of φ is called φcrit in latersections. For cases when θc ≤ 90◦ , the cylinder is completelysurrounded by water when φ = 180 ◦ , and therefore the watersurface must break. Thus, φcrit = 180 ◦ if θc ≤ 90◦ .

The analyses in the following sections are based on thisassumption that φ = φcrit is the condition for water surface

breaking. B. Rigid Leg Modeling

Assuming the supporting legs to be very rigid and stiff, themaximum lift force can be computed using the following twosteps.

1) Coarse Solution: For φ sufciently smaller that φcrit ,we assume that

dhdx

(x ) = s = constant. (6)

Then (2) becomes

d2

dx2 h (x ) =

ρg(1 + s2)32

γ h(x). (7)

The solution to this simpler problem is

h (x,φ,θ c ) = B (φ, θ c ) · e−1c

x , (8)

where

B (φ, θ c ) = − c · e(1c

R sin φ ) · tan( θc + φ − π ) (9)

and

c = γ ρg 1(1 + s2) 32 . (10)Since this solution is only an approximation, it does not

describe the water-air interface correctly. However, this coarsesolution serves as an initial guess to the numerical solutionlater.

2) Numerical Solution: The exact solution to (2) is solvedusing Matlab, given two initial conditions of (3) and (4), andthe initial guess of (8). To implement (4), the innity termmust be replaced by a specic distance that is sufciently farfrom x = 0 . From eye-observation, the maximum width of the water dimple was around 0.02 m, thus 0.01 m wide fromthe axis of symmetry. Here, the innity in (4) is replaced bythe specic distance of 0.02 m, or 2 cm, which is twice the

Fig. 4. Water-air interface prole when θc = 112 ◦ The circles represent thecross-section of the supporting legs (radius of 165 µm).

Fig. 5. Water-air interface prole for various θc

and φ ∼ φcrit

.

maximum size of the dimple. This value can be bigger, makingthe solution more exact, but then the solution fails to capturethe prole of the water-air interface when φ ∼ φcrit .

Fig. 4 shows the numerically solved water-air interfaceprole transition when a rigid cylinder with θc = 112 ◦ islowered on to the water surface. The result is consistent witheye-observations. The water surface is not yet broken if theassumption on the water breaking condition made earlier iscorrect. For the case of φ ∼ φcrit , the total lift force was 3.05mN, and the depth of the center of the cylinder was 3.85 mm.

The radius of the cylinder is 165 µm, which is the radius of the wires used in the actual robot.

Fig. 5 shows the surface prole for various θc when φ ∼φcrit . When θc > 90◦ , the interface prole looks almostidentical regardless of the value of θc . This result implies thatthe maximum lift force will be almost the same if θc > 90◦ ,since most of the lift force come from the surface tensionforce. For the case of θc < 90◦ , the lower the contact angleis (or the more hydrophilic the leg material is), the lower themaximum lift force is. For a 2 cm-long cylinder with θc = 60 ◦ ,


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the maximum lift force was calculated to be 2.57 mN.The numerical solution veried the results of [11] that the

weight of the water that would ll the volume B in Fig. 3would equal the vertical component of the surface tensionforce, f T . The value of ρg ·

∞x 0 h (x ) dx was very close to the

value of γ · sin(tan − 1( dhdx (x0))) . Specically, when θc > 90◦

and φ ∼ φcrit , the value of ρg ·

∞

x 0 h (x ) dx almost exactly

equaled the surface tension coefcient of 0.072 N/m.It was calculated by observing h(x) that when two sup-porting legs that are parallel to each other are pushed downuntil just before they break the water surface, they lose somelift force due to the water dimples created by each legsoverlapping with each other between the legs. For two plainstainless steel legs, they have to be at least 21.0 mm apart inorder to lose no more than 1% of the possible maximum liftforce. To conserve at least 95% of the lift force, the legs haveto be 12.4 mm apart. For the Teon R coated legs, they have tobe at least 21.3 mm apart to conserve 99%, or 12.6 mm apart toconserve 95% of the possible maximum lift force (Table II).All these results are assuming only two legs in parallel. If

there are more than two legs in parallel, the percentage of thelift force conserved drops down due to more than two waterdimples overlapping with one another.

All these results assumes a perfectly rigid cylinder withno considerations of interface prole at the two ends of thecylinder. In actual case, there exists some end-effects that mayor may not be important. Since the legs of the robot are longand thin, these effects are neglected in this simulation. Thesharp edges at the ends of the cylinder may break the watersurface even before the condition φ ∼ φcrit is satised. In theactual experiment, the ends of the cylinder are bent upwardsso that the geometry of the leg at the ends is round enough.

3) Experiments: Fig. 6 shows the numerically solved θcvs maximum lift force relation, for the geometry of specimenshown in the inset, using the experiment setup as shown in

TABLE II

EFFECT OF THE S PACING B ETWEEN T WO S UPPORTING L EG S

Leg Material Spacing Conserved Lift Force

Stainless steel 21.0 mm 99%

Stainless steel 12.4 mm 95%

Teon R 21.3 mm 99%

Teon R 12.6 mm 95%

TABLE III

EXPERIMENTAL DATA

Leg Material θc Maximum Lift Force [mN](Numerical Estimation)

Stainless steel 50◦ − 70 ◦ 2.6 (2.3 − 3.0)Teon R 110 ◦ − 120 ◦ 3.1 (3.6)

Fluorothane TM MP 140 ◦ − 150 ◦ 3.2 (3.7)

Fig. 6. Numerically solved maximum lift forces of the specimen in the inset.

Fig. 7. Photo of the experiment setup using an xyz stage and a load cell,for lift force measurement for the specimen geometry given in Fig. 6 inset.

Fig. 7. The angled ends of the specimen are taken account

into the model in such a way that they are made of cylindersof length dx with different depths of their center into thewater. Actual experiments were performed using three differ-ent leg materials; stainless steel wire, Teon R coated wireand Fluorothane TM MP (Cytronics Corp., model 1.00) coatedwire. The stainless steel wire and the Teon R coated wirehad the same radius, but the Fluorothane TM MP coated wirewas expected to have a larger radius because of the manually-performed coating. Table III shows the experimental data forthe three specimen. They are pushed down onto the watersurface automatically with the speed of 0.3 mm/sec, using adesktop computer. The numerical solution captures the generaltrend of the experimental data fairly well, but has 10-15%

errors for the θc > 90◦

cases. This may be due to the fact thatthe numerical solution assumes a perfectly static case, whereasthe experiments are performed at a certain speed. Also, themodel assumes a perfectly rigid cylinder, whereas the actualspecimen is not so rigid. The angled ends of the specimenmight have forced the water surface to break more easily.

The numerical solution, along with the experimental results,suggests that the maximum possible lift force does not differmuch if the leg materials are hydrophobic. That is, anymaterial hydrophobic enough can make a good supporting leg.


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Fig. 8. Compliant supporting leg divided into N segments. |f i | > | f j | if i < j .

Fig. 9. Parameters used for determining the exibility matrix, F .

C. Compliant Leg Modeling

From the analysis in the previous section, the theoreticalamount of lift force for a rigid cylinder body can now benumerically solved. Applying the result, the robot with four5 cm-long supporting legs (only 4 cm of each legs actuallytouching the water) coated with Teon R will experiencemaximum lift force of 2.44 grams. However, experimentsshowed that the robot can actually hold 1.82 grams includingits body weight [10]. The difference in the values may bedue to the leg being compliant, instead of it being completely

rigid. Fig. 8 shows the compliant supporting leg pushing downthe water surface. Since the amount of lift force per unitlength is a function of the depth of the cylinder, the end-tipof the supporting leg experiences less lift force per unit lengthcompared to the part of the leg that is closer to the base. Tond more realistic value of total lift force, this difference inlift forces along the length must be taken into account.

1) Numerical Solution: To nd the exact deformed shapeof the compliant supporting leg, the leg is divided into N equal-length segments along the length, as shown in Fig. 8.Each segment experience some lift forces, which are assumedto be point forces acting at the center of the segments ( f i ’sin Fig. 8). These point loads are all the external forces thatdeform the leg. Each segment is assumed to be rigid enough sothat it can be assumed as a cylinder whose axis of symmetryis parallel to the undisturbed surface of the water ( z = 0surface). The depth of the position where the load is applieddetermines the magnitude of the point load. Given the radius Rand the contact angle θc of the segments, the exact theoreticalrelation between the lift force per unit length of the cylinderand the depth of the center of the cylinder can be known fromthe results of the previous section. That is, when calculatingthe magnitude of the lift force given φ, the depth of the

center of the rigid cylinder is also calculated, and many datapoints describing the depth versus the lift force relation canbe obtained. These data points were used to t a curve, f (z),that captures this relation. For example, for Teon R coatedwire with the radius of 165 µm and the contact angle of 112◦ ,the lift force per millimeter can be approximated as

f (z) = 1 .1488 · 103 × z3 + 4 .3505 · 10− 2 × z2

− 5.5535 · 10− 2 × z + 3 .1837 · 10− 6 , (11)

where z is the depth of the center of the cylinder in meters,and f (z) is in Newtons. The coordinate system is shown inFig. 8, which is consistent with Fig. 2.

Now, let the xed value z0 be the depth of the base of thesupporting leg, and set an alternative xyz coordinate systemat the base of the supporting leg (Fig. 8). The alternative coor-dinate system is a translation of the original coordinate systemby − z0 in z direction. Dene a vector z = [ z1 z2 · · · zN +1 ]

T ,where z i is the depth of the point where f i is applied, in xyzcoordinate system. Then z describes the depth prole of thesupporting leg. Let the vector g(z ) = [f 1 f 2 · · · f n ]T =[f (z1 − z0) f (z2 − z0) · · · f (zN +1 − z0)]T represent thepoint forces that the segments experience. Then the problemof solving for z can be formulated as

z = l · F · g(z ) (12)

where l is the length of each segment and F is the exibilitymatrix of the leg, whose n -th row describes the depth prole of the leg (in xyz coordinate) when a unit force is applied onlyto the position where f n would be applied. All the elementsin the rst row and the rst column of F are zero, since theposition where f 1 is applied is the base of the supporting leg(Fig. 8). (The position of the base of the leg does not changeregardless of the magnitude of any vertical force applied tothe point.) Having all the elements in the rst column of thismatrix to be zero allows the rows of the exibility matrixto be superimposed and still have the rst element of thesuperimposed vector to be zero. The exibility matrix, F, isdetermined by

F i,j =

x2j6EI

(3a i − x j ) if 0 ≤ x j ≤ a j ,

a 2i6EI

(3x j − a i ) if a j ≤ x j ≤ L ,

(13)

where E is the elastic modulus of the leg material, I is themoment of inertia of the cross section about the x-axis, L is thelength of the supporting leg, a i is the distance between the baseof the supporting leg and the point where f i is applied to, andx j is the distance between the base of the supporting leg andthe point where f j would be applied to (Fig. 9). Note that sincex1 = a 1 = 0 , F i, 1 and F 1,j are all zero, as intended. Equation(13) is derived assuming small deection of the leg. This is afeasible assumption since the length of the leg, which is greaterthan 4 cm, is more than 10 times bigger than the possiblemaximum depth of the base of the leg, which is approximately3.8 mm for Teon R coated wires.


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Fig. 10. Numerically calculated prole of the compliant supporting legmade of Teon R coated stainless wire, when z 0 = 3 .85 mm. The z -axis ismagnied.

Applying the initial guess of z = [0 0 · · · 0]T , (12) issolved. Fig. 10 shows the bended supporting leg prole derivedfrom solving (12). The leg material is stainless steel coatedwith Teon R , and the initial depth of the base is 3.85 mm,which is the maximum depth possible for Teon R coated leg,estimated in the previous section. When the length of the leg is60 mm, the lift force in this case is calculated to be 6.88 mN,or 0.70 grams. Same calculation process was applied for plain

stainless steel legs of same dimensions, and the maximum liftforce was estimated to be 6.43 mN, which is not much of aloss in the loading capacity. Even though plain stainless steelhas smaller contact angle and is considered hydrophilic, themoment of inertia is far greater than that of the Teon R coatedstainless steel wire of the same diameter, making the legstiffer to increase the maximum payload. For four 4 cm-longTeon R coated supporting legs, the maximum payload wasexpected to be 2.32 grams according to this model, whichis about 5% less than the model assuming perfectly rigidcylinder.

Fig. 11 shows the numerically calculated payloads (lift forceminus the weight of the leg) for various lengths of the

supporting legs with the radius of 165 µm unless otherwisespecied. The maximum lift force for the compliant legs are

Fig. 11. Numerically calculated payloads for legs with different geometryand materials.

Fig. 12. Photo of the prototype robot with sixteen 4 cm-long Teon R coatedstraight supporting legs whose bases are 4 mm apart, carrying a 5 gram weight.The body is made of the carbon ber sheet.

capped at around 6.9 mN. When the lengths of the legs areless than 4 cm, no signicant difference is present betweenthe rigid leg and the compliant leg. However, when the lengthof the leg is greater than 5 cm, the difference increases andthe maximum lift force for the compliant leg saturates. Thisresult suggests that the supporting legs should be no longerthan 5 cm in order for them to be most effective. It would bebetter to have large number of short legs instead of having afew long legs. However, if the radius of the legs becomes 0.2mm, the supporting legs become stiffer and they can supportmore weight even when they are longer than 5 cm. The weightadded due to the greater radius was proved to be insignicant.

2) Experiments: As mentioned earlier, the maximum liftforce for the robot with four 4 cm-long supporting legs was1.82 grams [10], which is around 78% of the numerical

estimation of 2.32 grams. Fig. 12 shows the sixteen-leggedprototype robot with 4 cm-long Teon R coated legs. Themaximum lift force in this case was 6.31 grams, or 61.8 mN.Using the results from the previous sections, it is predictedthat this prototype can hold 92.8 mN if the legs are spaced farenough. When the legs are 4 mm apart at the bases and 16 mmapart at the ends as in this case, it is roughly estimated thatthe prototype can hold up to 7.27 grams. The estimation wasdone in the following steps; the legs with 4 mm spacing canonly hold 23.1% of the maximum lift force, whereas if theyare 10 mm apart, they can hold roughly 92% of the maximumlift force. If they are 16 mm apart, as the tips of these legs are,then they can hold almost 100% of the estimated lift force.

Now since the sixteen supporting legs are 4 mm apart only atthe base and 16 mm apart at the ends, the prototype wouldcarry roughly 0.25 · (100+92)+0 .25 · (92+23 .1) = 76 .8% of the possible lift force of 92.8 mN, which is calculated to be71.3 mN, or 7.27 grams. Applying this result, the experimentalvalue is showing 86.8% of the numerical estimation. Moredeliberate estimation of the possible lift force is left for futureresearch.

The difference between the numerical estimation and exper-imental result may be explained in a few ways. Firstly, there


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Fig. 13. Numerically calculated prole of a 7 cm-long supporting legexperiencing maximum lift forces at each segments.

are fabrication errors; all sixteen of the legs must be identicalto one another so that they all touch the water at the same timeand break the water surface at the same time. If the legs werenot consistent, one of the legs would break the surface beforeothers take their possible maximum lift force, reducing thepossible overall maximum lift force. The more you have thelegs, the more this is likely to happen. And some misconductof the experiment could also affect the result. When placing aweight on the test robot, the water surface is suddenly pusheddown and the water surface may become easier to break due

to this dynamic effect.It is clear that all these sources of errors can also exist in

the actual robot. The robot will be able to carry more weightin the future if these issues are resolved.

It was observed that when a 7 cm-long leg was fully loaded,the end tip of the leg was outside of water. At these ends, thewater surface was bended up toward the + z direction, thuspulling the leg downwards. It is a good idea to design the legssuch that this never happens.

D. Optimized Legs

Fig. 11 suggests that compliant legs lose loading capacityat longer lengths. This is because when the leg is pushed

down onto the water surface, it is bended upwards due tocompliance. If the legs could be initially bent downwardsto overcome for this upward bending which decreases thepayload, then compliant legs can be longer and still beeffective.

1) Prole Calculation: The best possible scenario is to havea compliant leg that straightens up at the maximum loading.Fig. 13 shows the bended prole of a compliant leg (7 cm-long) when each of the segments (as described in previoussections) are experiencing − z direction forces of 1.5 mN/cm(or 0.15 g/cm), which is the maximum possible lift forcecalculated for Teon R . A leg with this prole is expected tobe straightened up when pushed down onto the water surface, just before breaking it. Fig. 14 shows numerically estimatedprole change as this leg is being pushed down onto the watersurface.

2) Experiment: Fig. 15 shows the optimized 7 cm-longTeon R coated stainless steel wire, whose prole is as shownin Fig. 13. When pushed down to the limit onto the watersurface, the leg straightened up with the uniform depth asintended, as shown in Fig. 16. Prototype robots were builtusing four and twelve of these legs to test the loading capacity.The one with twelve legs is shown in Fig. 17. The prototype

Fig. 14. Numerical simulation of the optimized supporting leg straightening

up as it is pushed down to the water surface.

with four legs was able to carry 3.7 grams including thebody weight, which is 86% of the theoretical maximum of 4.3 grams. Seeing that a 7 cm-long compliant leg withoutany initial bending can support less than 7 mN (or 0.7grams) from Fig. 11, this optimizing of the leg improvedthe maximum payload by 32%. (Theoretically, it would be53.6% improvement.) The prototype with twelve legs withheld9.3 grams, which is only 71% of the expectation, but stillbetter than the expected performance of the legs without initialbendings (which is 8.4 grams). It is therefore concluded that

longer legs can be used and still be effective if they areoptimized to maximize performance.

When these optimized legs are loaded far less than themaximum possible payload, only the ends of the supportinglegs were touching the water surface, just like the insect waterstrider.

III . D ISCUSSIONS

From the numerical analyses in the previous sections, aspecic design rules for the supporting leg of the water striderrobot can be summarized as follows:

1) Maximum lift force is almost linearly proportional to the

length of the legs, as long as they are shorter than 4 cm

Fig. 15. Photo of the 7 cm-long optimized Teon R coated leg.

Fig. 16. Photo of the 7 cm-long optimized Teon R coated leg, straightenedup just before breaking the water surface.


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Fig. 17. Photo of the prototype robot with twelve optimized 7 cm-longTeon R coated legs, carrying 8.3 gram payload (total of 9.3 gram lift force).

for stainless steel or Teon R coated stainless steel wirewith the radius of 165 µm. To use legs longer than 7cm, optimizing their prole is necessary. In both cases,one can expect 0.15 grams of payload per centimeter forthe Teon R coated legs (Fig. 11).

2) Number of supporting legs are directly proportional tothe maximum lift force, as long as the legs are not soclose to one other. The minimum distance between theneighboring supporting legs coated with Teon R , mustbe at least 10.2 mm apart so that the dimples they createwould not overlap too much and lose possible maximumlift force (Table II).

3) Leg material having θc much greater than 90◦ is desired,but when θc > 120◦ , the advantage diminishes (Fig. 6).

4) The end tip of the supporting legs should be bentupwards so to prevent any singularity of geometry,which may cause early breaking of the water surface.

5) The radius of the supporting leg is insignicant to themagnitude of the surface tension force. However, thesmaller radius of the leg is better for a surface tensiondominated support and to minimize overall weight of the robot.

Abiding by these rules and reducing the fabrication errormay increase the loading capacity of the robot by 20 − 30%.

Throughout the analyses, the numerically calculated maxi-mum lift forces are greater than the experimental results by10− 30%. Considering the errors in the fabrication process andexperiments, these differences are not surprising. Therefore,

it can be concluded that the numerical models capture thephysics behind the supporting legs and the water surface rea-sonably well. Since all the analyses starts from the assumptionthat the water surface breaks when the water-air interface

stands up vertical ( dhdx (x0) = ∞ ), it can be asserted that thisassumption made earlier is correct.

IV. C ONCLUSION

Motivated by the recent studies on the insect water strider,a numerical method to evaluate the synthetic supporting legsis proposed. Under an assumption on the water-breakingcondition, supporting legs of different materials and shapesare analyzed. It is shown that a robot with four 7 cm-longTeon R coated supporting legs with optimized shapes wouldhave loading capacity of 4.3 grams (0.15 g/cm), including itsown body weight. The design rules proposed in the previoussection can be used to design future robots. And the analysesprocedure can be helpful in predicting the maximum loadingcapacity of the robot, or in designing a better geometry orproperties of the legs. In the future, dynamic analysis onthe actuating legs and their behavior on the water surfacewould also be realized. When all the physics of the robot areanalyzed, the robot can potentially be used for the purposesof entertainment, education or environmental monitoring onwater surfaces.

R EFERENCES

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yun seong song - modeling of the supporting legs for designing biomimetic water strider robots

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