Investigation on Freestyle Kicking in Swimming

1. Introduction

I discovered my interest in swimming ever since Grade 4, during which I first learned how to swim. However, swimming has not been my focused sport until Grade 11, when I accidentally broke my meniscus. An injured meniscus means that I could no longer play competitive golf. Being one of the few sports that do not involve knees that much, I shifted my focus to swimming. After swimming regularly for half a year, I figured out that I really enjoy this sport and could participate in competitive tournaments.

Out of the 4 strokes—breaststroke, freestyle, backstroke, and butterfly—freestyle has been my favorite, and the biggest reason is because it is the fastest stroke. I always like to feel myself ‘flying in water’ when swimming freestyle. More importantly, freestyle is a stoke that hurts my knees the least. Breaststroke requires outward, or lateral kicking of the legs, hurting my meniscus. Whereas freestyle only requires the up and down kicks of the legs. The knees move vertically only. I love swimming freestyle especially when I am doing long distances, for it improves my stamina. I also like the feeling after going for a long swim, it is truly energizing.

As a long-distance swimmer who have good stamina, I would like to find out the optimal swimming technique that would allow me to achieve the greater velocity while consuming the least energy. The research on the theory of freestyle involves two physics model—biomechanics and fluid dynamics. Biomechanics includes concerns the study of core muscles, skeletal structure, and body mass. In order to improve these aspects, one would need to create a regimented schedule, controlling diet and training (workout) regularly. The calculation for these areas is intangible as it often times don’t exist. On the other hand, the fluid dynamic approach of analyzing freestyle, by discussing the relationship between the swimmer and fluid (their mutual force), analyze the effect of fluids’ velocity, frictional force, and dragging force, etc. Through these effects, the swimmer’s power, speed, forms, and movement could be drawn, providing scientific evidence for improving swimming techniques.

2.  Fluid dynamics of objects moving in water

The equation for the mutual force between fluid in motion and an object in the fluid is listed below. The force is caused by the fluid pressure difference among the areas ahead and behind the moving object. This could also be called a drag force formula:

“#$%&

= <×>×?×@A

B

(3456789: 1)

“_#$%&  : The fluid’s resistance on an object in motion

‘ : The fluid’s resistance coefficient on the object, this is influenced by shape, stream-line shape objects have a smaller ‘ value. The one for dolphin is 0.05, for human it is 0.58–1.04.

( : The density of the fluid. The density of water, for example, is 1000 +,/./

• : The reference sectional

• ∶ The flow velocity relative to the

Regarding the hydrodynamic drag, this force can be defined as an external force that acts in the swimmer’s body parallel but in the opposite direction of his movement direction. This resistive force is depending on the anthropometric characteristics of the swimmer, on the characteristics of the equipment used by the swimmers, on the physical characteristics of the water field, and on the swimming technique.

3456789: 1 is arguably the most important equation in this investigation as all the forces (derived equations) are created using it.

3.  Drag force of freestyle swimming kicking

Usually, when we talk about freestyle, the stroke requires arms stroking alternatively and legs kicking alternatively. The factors that need to be considered to account for arm strokes involve the angle at which the hand goes in the water, the position of it, the path that the arm takes, how much the arm is bent, the angle of the elbow, etc. However, this essay solely focuses only on kicking. The drag force that is created by kicking simply related to the frequency and the range of kicking.

The important roles played by kicking in freestyle are:

1. Kicking helps the body to maintain a low ‘ value by keeping the body in a stream And makes the area A smaller by lifting the lower body to maintain the whole body in a straighter parallel line.

2. The propulsion gained by kicking can obtain higher

3. Make the velocity more stable in order to promote the efficiency in

Although I understand that as my investigation should cover as many aspects, even to the last detail, as possible, but it is apparently impossible to consider everything in this investigation.

Therefore, for the sake of simplifying, the following assumptions are suggested. Indeed, the assumption will not change the analyzation result much:

Assumption 1: The legs are kicking at the rate of T times per second, forming the range of angle θ

between the leg and the upper body. Although the knees are bent during kicking, we assume that the entire leg stays straight during kicking.

Assumption 2: The moving velocity of legs relative to the body during kicking is always constant.

Although this velocity varies from time to time, a mean value will be suitable for the analysis of a whole kicking periodic (up and then down)

Assumption 3: The notable difference between down kick and up kick is the relative position of the

foot. During the relaxed kicking in freestyle the thigh moves first, causing the shank to move and then the ankle flaps making the foot spread during down kicking and shrink during up kicking. The thigh and the shank are cylindrical and the foots are rectangular. The figures below illustrate this idea.

Let the total length of the leg to be F, the length of the thigh be F_G with diameter H_G, the length of the shank be F_I with diameter H_I. The length of the foot is F_J, with the width of the foot is H_J. On the figure, length F equals to F_G + F_I. In most cases F_G  and  F_I  are almost the same so we

use ^L instead of F_G and F_I. M is the distance between the beginning of the leg to a certain point on

the leg.

According to figure 1, the sectional area in position M along the direction in which the leg is

moving is:

0 = H_G × NM when M is at the thigh or

0 = H_I × NM when M is at the shanks or

0 = H_J × NM when M is at the foot

This is because on Figure 1, M has already reached the shanks, so the diameter is H_I , the width of the shanks. If M < F_G, diameter is H_G, if M > F_G, diameter is H_I. NM is the fraction of M.

Now find the velocity of NM at point M:

As shown in Figure 3, the velocity of position M may be calculated to be:

1_X  = M × Y × Z              (3456789: 2)

Where M is the distance from a position on the leg to the beginning of the leg, Y is the times of kicking per second, and Z is the range of angle at which the leg is kicking from the horizontal line each kicking.

Now that we have got the velocity and the area of contact, we can use the drag force formula in point M to find out the fractional drag force. The knowledge of differentiation, which I learned in Grade 11, may help me do that by differentiating the function:

Substituting 0 = H_G/H_I/H_J ( when x is at the thigh, shanks or foot) × NM and 1_X  = M × Y × Z

into 3456789: 1, we can find out that the fractional drag force at any point M is

‘ × ( × H × NM × (Z × M × Y)B

N”(M) =

Rewrite, taking M outside:

• (where r may be H_G, or H_I, or H_J)

Integrate to get:

N”(M)

‘ × ( × H × (Z × Y)B

=                      2                     MB

NM(where r may be H_G, or H_I, or H_J)

“(M) = ∫

‘ × ( × H × (Y × Z)B

2                     MB

NM(where r may be H_G, or H_I, or H_J)

The reason that this equation is only in terms of M is because the values for all other variables are given constants (could be plugged in).

In calculus, I also learned about integration, which is able to find out the area created between two M values in a function. In this case, by using integration we could determine the mean drag force, which is in a way the area, experienced by the leg in a down kicking.

k  <×>×$l×(m×n)A

“#$%&_#ghi  = ∫ A

MBNM + ∫ L <×>×$o×(m×n)A

MBNM + ∫

LpLq <×>×$q×(m×n)A

MBNM

j                                                                          A                      B                                                 L                                 B

By taking out common factor ^<×>×(m×n)A, the equation can be expressed as:

B

“#$%&_#ghi  =

‘ × ( × (Y × Z)B 2

L

× (H_G ∫ BMB

NM + H ∫ LMB

B

NM + H_J∫

LpLq L

MBNM)

Solving the integration, obtaining:

<×>                      Lr

tLr

/LLA p /LALq pLr

“#$%&_#ghi  =

B ×(H_G × Bs

+ H_I ×

+ H × (     q                              q)) (Y × Z)B

Bs                                             /

For the same method, the drag force by the water in an up kick is:

L ‘ × ( × H_G × (Y × Z)^B  B                               L ‘ × ( × H_I  × (Y × Z)^B B

“#$%&_u>  = ∫ B

2

<×>×(m×n)A

M NM + ∫_L

B                               2

B                           L B

M NM

=           B                 × (H_G ∫ AM

NM + H_I∫k M

A

NM)

= <×> (H

× ^Lr + H × ^tLr) × (Y × Z)B

B          G        Bs           I          Bs

The only difference here is that the part related to the foot disappears according to Figure 2, the foot is as if “shrinked”.

4.     The propulsion and lift force computation

The drag force at a particular position is illustrated in Figure 4. And  “_{#$%&_#ghi}  is the drag force brought about by the water, so “_#ghi  is the force of leg in the down kicking, with the same scaler but opposite direction simultaneously.

“⃗#ghi  = −”⃗#$%&_#ghi

The reason that “_#ghi  is calculated could be explained using the third law of Newton: every force has a counter-acting force with the same value and opposite direction.

Overall, the drag force of the moving legs is:

“#$%&_GgG%x  = “#$%&_u>  + “#$%&_#ghi

Note that the water floatage of the legs is not considered. Because the density of human body is almost the same as water, that means the floatage is cancelled out by the gravity. Therefore, in freestyle swimming, the legs should experience the “_{#$%&_GgG%x}  to do kicking. Now compute for the lift and propulsion forces created by kicking. To do so, the knowledge in vectors can be used. As shown in figure 4, the drag force could be decomposed into two forces—

propulsion force and lift force, “_>$g>uxIygi  and “_xyJG. It could also be said that “_#$%&  is the resultant vector of the two perpendicular vectors.

“⃗#$%&  = “⃗>$g>uxIygi +”⃗xyJG

The resultant force could be calculated by using the knowledge of trigonometry. Given z is the angle between the leg and horizontal, we know that  {8:(z) = |}~ñ}ÄÅoÇñÉ. Therefore,

|Ñ~ÖÜ_ÑñáÉ

“>$g>uxIygi  = {8:(z) × “#$%&_#ghi

However, as mentioned above, the “_>$g>uxIygi  varies with z in kicking period. In order to    find out the average (effective equivalent) propulsion force in the entire down kick, integration shall be used again. Here Z is the range of angle between the leg and the horizontal axis (as body

direction in swimming) in each kick, so ∫ n”

× {8:(z)Nz will give us the “total propulsion

j     #$%&_#ghi

force” generated by each down kick. By using ^è we can get the average (or equivalent) propulsion

n

force, which is like the propulsion force at any angle:

”                                     1     _n

>$g>uxIygi_#ghi  = Z ∫j  “#$%&_#ghi  × {8:(z)Nz

Similarly, the average propulsion force generated by the up kick is:

”                                1     _n

>$g>uxIygi_u>  = Z ∫j  “#$%&_u>  × {8:(z)Nz

The mean propulsion force generated by down kick will be:

“>$g>uxIygi_êë%i  = è ∫n “#$%&_#ghi  × sin(z) N6 − è ∫n “#$%&_u>  × sin(z) N6

n  j                                                                                    n j

= ^è ∫ n(”

− ”               ) × sin(z) N6

n     j           #$%&_#ghi

#$%&_u>

= ^è ×

n

<×>×(m×n)A B

/LLA  p /LALq  pLr             n

× H  × (                             ) ∫

/

sin(z) N6

The reason why the drag force of an up kick is minus from the drag force from a down kick is because the opposite directions as shown in Figure 4 and Figure 5.

In order to make computation easer, I have extracted a common term that is present in the drag force equation, both up and down kick. Let:

í1 =

<×>

/LLA p /LALq pLr

× HJ  × ì                              î  (“H9. “>$g>uxIygi

)   (3456789: 3)

B                                                  /                                                                                   ïñÖÉ

Thus, “_{>$g>uxIygiòêë%i}  can be written as:

“>$g>uxIygi_êë%i

= í1 × YB × Z ∫ n sin(z) Nz

= í1 × YB × Z × (1 − ô9{(Z))

Note that this neat equation benefits from average force used, so the actually position of each leg doesn’t matter for computation. In the same sense, the lift force can be calculated using ô9{(z).

Since ô9{(z) =        |ÅÇql         , the lift force in kicking may be described as:

|Ñ~ÖÜ_ÑñáÉ

“xyJG  = ô9{(z) × “#$%&_#ghi

Now compute for the average lift force generated by kicking down at any angle z :

”                        1     _n

xyJG_#ghi  = Z ∫j  “#$%&_#ghi  × ô9{(z)Nz

Similarly, the average lift force generated by the up kick is:

”                       1     _n

× cos(z) Nz

xyJG_u>  = − Z ∫j  “#$%&_u>

The mean lift force generated by up kick will be:

”                        1     _n

(  )                   1     n

× cos(z) Nzî

xyJG_êë%i  = Z ∫j  “#$%&_#ghi  × ô9{  z  Nz + ì− Z ∫j  “#$%&_u>

As for propulsion:

“xyJG_êë%i

= í1 × YB × Z ∫ n cos(z) N6

= í1 × YB × Z × sin(Z)

Summarizing all the useful equations, which are all derived from 3456789: 1.

“#$%&_#ghi  =

<×> B

×(H × ^L

Bs

+ H_I ×

tLr Bs

/LLA p /LALq pLr

+ H  × (                             )) (Y × Z)

/

”                = ^<×>(H

× ^Lr + H

× ^tLr) × (Y × Z)B

#$%&_u>

B         G        Bs

I          Bs

“#$%&_êë%i  = “#$%&_u>  + “#$%&_#ghi “_{>$g>uxIygi_êë%i}  = í1 × YB × Z × (1- ô9{(Z)) “_{xyJG_êë%i}  = í1 × Y² × Z × sin(Z)

í1 =

‘ × ( 2

3 FFB + 3FBF_J + F/

× H  × õ                                       ú

3

Reflecting on the process of deriving these useful equations, I feel quite proud about coming up with new formulas based on a single formula (3456789: 1) based on my knowledge in calculus (differentiation and integration), trigonometry, and vectors, all of which I have learned in the HL syllabus. I am proud of being able to apply what I learn at school to what I enjoy doing and exploring. This is indeed the charming side of mathematics.

5.     Applying equations to real-life scenario

In order to make the verification method of my formulas to be more convenient, I took myself for example, and obtained measurements of units of myself:

F = 0.9, H_G = 0.13,  H_I  = 0.10, H_J  = 0.11, F_J  = 0.22(âH9. 6:+àû)( 5:87{ 6Hû 6àà 8: .û7ûH{)

The density of water is ( = 1.

As a trained swimmer, I take my constant ‘ be 0.7, slightly less than the average 0.8. This is because I am able to maintain my body better in a stream line than amateurs. As mentioned in Section 2, the ‘ value varies from 0.58—1.04, so a value of 0.7 is not unreasonable.

Using these measurements and substituting them back to í_è, we can get:

í1 =

<×> B

/LLA p /LALq pLr

× H  × ì                           î

/

= 0.7 × 1000 × 0.11 × ì0.1307 + 0.5346 + 0.0106î = 8.67

2                                                       3

Substituting the obtained value into “_{>$g>uxIygi_êë%i}, get:

“_{>$g>uxIygi_êë%i}  = 8.67 × YB  × Z  × (1 − ô9{(Z))

Therefore,

“_{xyJG_êë%i} = 8.67 × YB  × Z × sin(Z)

Now we can calculate the “_{#$%&_êë%i}:

“#$%&_êë%i  = “#$%&_u>  + “#$%&_#ghi

= <×>

B

×(H × ^L

Bs

+ H_I ×

tLr Bs

/LLA p /LALq pLr

+ H  × (                           )) (Y × Z) +

/

<×> B

(H × ^L

Bs

+ H_I ×

tLr Bs

) × (Y × Z)B

= <×>

B

×(H × ^L

èB

+ H_I ×

tLr èB

/LLA p /LALq pLr

+ H  × (                           )) (Y × Z)

/

Substituting my measurements, obtain that:

“_{#$%&_êë%i}  = 26.32 × (Y × Z)B

6.     Data analysis and evaluation

First, I calculated how much the leg force I carry out for kicking (Table 1). Then decomposed the total leg force into propulsion force and lift force created by kicking (Table 2 and Table 3).

I have created 7 tables that shows 21 different kicking frequencies (periods), in terms of times per second, and 9 kicking ranges (groups) in terms of degree, data for leg forces, propulsion forces, and lift forces. The first period(row) and first group(column) are just numbers for location purpose.

Since the tables are too big to be put in the body, I have included them in the Appendix

Section. In my analysis I will also refer to the tables in the Appendix section.

Based on the tables presented above, the following statements can be concluded:

1. The leg force generated by kicking, which is 32 × (Y × Z)B, is directly proportional to

(Y × Z)B. This indicates that Y and Z are of the same importance to the generation of force by kicking. For different values of Y and Z, when the total value of Y × Z is the same, the force generated by the kicking is the same. For example, in Table 1, the value of the sixth period and fifth group is 28.863, Y × Z = 60. It is the same with the value of the ninth period and fourth group, where the value is also 28.863, and Y × Z = 60 as well.

2. The propulsion force that is generated by kicking is of great significant, because the fact that the two legs are kicking in opposite direction cancels out the propulsion force generated. For instance, the value (period 5, group 6) in table is 2.443N, which is only about 8.5% of the force carried out by legs (2.443N out of 28.863N, which is the force generated by the kick).
3. The lift force created by kicking is greater than that of propulsion force. For instance, the value (period 5, group 6) in table 3 is 9.079N, which is 31.5% of the force carried out by legs (9.079N out of 28.863N). More lift fore is created than propulsion force. The function of lift force in freestyle swimming is mainly to allow the body to maintain in a more horizontal position, improving body gesture (better streamline-shape), as well as decreasing the ‘ value, which is the fluid resistance
4. For a fixed value of Y × Z, the greater the Y value, the smaller propulsion is yielded. For instance, the value (period 5, group 6) in table 2 is 2.443N, Y × Z = 60, this value is the same for the value in the period 9, group 4, and the period 13 and group 3. However, the propulsion forces are 1.643N and 1.237N only, respectively. Thus, if Y increases and Z decreases, the propulsion force will decrease. In other words, it is better to kick at a bigger angle than to kick more frequently (a higher Y)

Now we can evaluate kicking’s impact on the velocity of swimming. After measurement, the sectional area 0 is mainly the area of the shoulders, and is found to be around 0.08.B  (Using

obtained length and width of my shoulders to calculate the area as a rectangle). When I am swimming at a velocity of 1 .û7ûH{ (ûH {ûô9:N, according to Equation 1, “_#$%&  can be calculated:

“#$%&  =

0.7 × 1000+, × 0.08.B  × 1B

2./                                          = 0.5 × 0.7 × 1000 × 0.08 × +,/. × 1B

= 281B

(¶)

On the other hand, due to the swinging of the leg, the sectional area 0 should increase theoretically. However, when the body is horizontal to the water surface, there are certain body structures that extrudes, including the area covered by kicking, as illustrated by Figure 6.

Therefore, the actual increase in sectional area in swimming freestyle is very slow, so this essay disregards it in computation.

Figure 6: Effect on sectional area by kicking.

(Although the arm and the legs extrude when swimming, their effects on sectional area are disregarded)

The new velocity 1_è, derived from 3456789: 1, is:

1 2= 2 ×  |Ñ~ÖÜp|}~ñ}ÄÅoÇñÉ_lñlÖÅ  (âH9. 3456789: 1)

è

= 2 ×

<×>×?

28VB + 8.67 × TB × θ × (1 − cos(θ))

0.7 ∗ 1000 × 0.08

561B  +  17.34 × TB  × Z  ×  (1 − ô9{(Z))

=                                  56

561B + 17.34 × TB × Z × (1 − ô9{(Z))

1_è = ™                                   56

Let the original velocities (in meters per second), without leg kicking (swimming parallelly without kicking, using arms only), be 0.75, 1, 1.5, and 2, we can figure out the appropriate 1_è for different additional kicks with frequency and range in the following tables. The tables are again put in the appendix section in order to save space.

According to Table 4 to Table 7, we find out that:

1. Kicking results in a consumption of large amounts of force in comparison to the force generated by movements from other parts of the body. When the initial velocities are 0.75, 1, 1.5, and 2 m/s without moving the legs, using the formula 281B to find out the force generated by are only 15.75N, 28N, 63N, 112N (substituting the initial velocities into 1) respectively, while kicking in 2 times/second at a range of 30 degrees will spent 28.863N (Table 1, period 5 group 6). This shows that the kicking requires much more
2. With the force generated by kicking the same, increasing the degree at which legs are kicking is more effective than increasing the frequency of kicking, for example, when 1 = 0.75, we get 1_è = 0.806 if kicking 2 times/second at 30 degree (Table 4, row 5 column 6). However, we only get

1_è=0.788 if kicking 3 times/second at 20 degree (Table 4, row 9 column 4).

3. As the velocity increases, the relative improvement in velocity produced by kicking

For instance, when 1 = 1, we get 1_è = 1.043 or 4.3% of increase velocity (Table 5, row 5 column

6) by kicking at 30 degree with 2 times/second while if 1=2, we only get 1_è =2.022 or only 1.1% enhancement (Table 7, row 5 column 6.)

Overall, kicking is not so efficient in speeding up free style swimming. In other words, the swimmer will spend too much force just to kick for too little benefits (the velocity increases very little). Therefore, for long-distance, low-speed swimming (200m or longer), it is recommended to just kick a little bit to maintain the body at a horizontal level. Whereas in short-distance races, the velocity can raise to 2.058 from 2 if kicking 20 degree by 6 times/second so that the result could be

48.6 seconds rather than 50 seconds. In short, since no energy needs to be conserved in short distance race, the swimmer should just kick as big (to form a bigger angle) and frequently as possible, because the greatest value of Y × Z will bring about the greatest velocity.

7.    Conclusion

This essay calculates the drag force by fluid dynamics function (3456789: 1), because the kicking leg movement is circular arc we use the integration on the whole leg to figure out the mean force generated from kicking.

The resolution of drag force constitute the propulsion force and lift force. For a whole kicking period, that is one leg up down and the other down up, we use the mean force to reasonably substitute the instant force because only sustainable force is meaningful. To get the equivalent mean force we use the integration method again. The mean propulsion can directly be used in 3456789: 1 to find out the new velocity with kicking.

I applied real life scenario into theoretical calculations. I took myself into account, supposing my coefficient ‘, and got every specific measurement to plug in the equations to obtain data on different forces. These data are technically reliable and certainly important instruct my swimming training in the future.

The implication of the results is significant. By looking at the results generated from the above tables, we found that kicking brings in notable lift force which is the key factor in keeping the good body position that makes the coefficient smaller. Therefore, unlike I predicted before establishing the tables that propulsion force is created from kicking, the lift force is the main effect of kicking. It could thus be inferred that maintaining a streamline is the major impact of kicking instead of increasing the velocity. It will be very costly to expend too much energy in kicking as it does not contribute to much increase in velocity This will be useful information for the swim coaches. What’s more, I believe this essay may be of support and reference for scholars who wish to investigate techniques in swimming.

However, weaknesses exist in this investigation, mainly they are the assumptions made in section 2. The assumptions, though don’t change too much of the result, omit some of the details of the movements in swimming. These details may be added in future investigation and calculated to make the investigation more complete.

I have personally applied the conclusion of the investigation to short and long-distance freestyle swimming in my training session informally (not timed). I believe that kicking mainly using the legs is able to conserve more energy in long-distance swimming (100m or more). However, to verify that statement, I should run timed tests in training. This may be something I could work on in the future.

References:

A .R. Vorontson and V .A. RUMYANTSEV. “Resistance in Swimming”.

Nataphoom Benjanuvatra, Brian A. Morphology and hydrodynamic resistance in young swimmers. Pediatric Exercise Science.

Research Center, Glenn. “The Drag Equation.” NASA, NASA, www.grc.nasa.gov/www/ k- 12/airplane/drageq.html.

Appendix

Table 1: Forces generated by kicking

Table 2: Propulsion force generated by kicking

Table 3: Lift force generated by kicking

Table 4: 1_è with kicking when V=0.75

Table 5, 1_è  with kicking when V=1

Table 6, 1_è with kicking when V=1.5

Table 7, 1_è  with kicking when V=2

Table 7, 1_è  with kicking when V=2