Ice skating became my second hobby 2 years ago. I had never ridden professionally and had no trainer, so I learned everything on my own. At some point I started to wonder how to learn specific jumps and figures and I came up with the idea to try to combine one of my favorite subjects - mathematics with ice skating. The main motivation is that maybe by representing the figures with mathematics I will be able to improve the quality of my skating and above all avoid mistakes and I also believe that my research will help me and maybe others to make a &quot;mathematically&quot; perfect double axel. My goal of the research is to check whether the mathematical calculations are reflected in real skating and if so, what is the accuracy of jumps performed by me in relation to the results that I will calculate.

Axel jump is one of the most difficult jumps in ice skating, however, it&rsquo;s inseparable part of every solo performance. The axel is most commonly performed by: 1. stepping forward on the left foot, outside skate edge; 2. jumping forward off the left foot while the right leg provides momentum by a high knee lift action; 3. initiating rotation with right hip transverse adduction; 4. rotating around the longitudinal axis one and a half revolutions; and 5. landing backwards on the right foot, outside skate edge1 . This means that with a single Axel, the skater must perform 1.5 spins, for comparison, with other skate jumps, the skater usually performs only one spin within a jump. Analogous &ndash; in the double Axel, the skater must perform 2.5 spins in the air.

Although in skating triple and even one quadruple Axel have been recorded, I decided to choose a double Axel, because it has been performed many more times, which means that my possible measurement errors will be smaller. First part of the research will rely on algebra &ndash; finding a proper function to establish a base for a further calculations, and then kinematics and trigonometry will be used to fulfil the aim.

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After calculating the data, I plan to compare it to the data from the performances of mine, despite the prediction being that the biomechanics of jumps performed in natural conditions are not unambiguously similar, there are significant dry jumping parameters that can make a significant difference .The analysis of the double (well-practiced) Axel also allows for an objective examination of the accuracy of the skaters, because a successful jump is not the result of chance.

During the entire jump, various factors affect its correctness. Table 1. Shows the most important aspects in a given part of the jump.

To mathematically define a jump, it has been divided into three parts, with the first two being the most important - take-off and jump itself. This is due to the fact that during landing, the body is affected by the force of gravity, which, as well as air resistance and the weight of the skater, which will not be taken into account.

In ice skating, the speed at take-off has minimal effect on the height of a jump as it is unlikely that much of the horizontal speed is converted into vertical speed. To ensure successful jumps, skaters must time and coordinate their take-off precisely, so that they maintain their speed when launching, and land with the same amount of speed as when they took-off.

The take-off is crucial in determining the height of the jump, as the vertical speed achieved at the moment of take-off will decide how much time the skater spends in the air. Once the skater has left the ice, there is no way to increase the height of the jump, so the take-off must be as powerful as possible for the best results. The more powerful a jump is, the more time the skater has in the air to complete their rotations. The trajectory of the jump should also be balanced (vy=vx), so they create a perfect parabola shape, (gravity force not included). Most of the power for jumping comes from the leg muscles, especially those in the thighs and buttocks, when they are used to flex the knee and hip joints, however, including this data would result in the outcome not being general, but adequate for people with specific leg strength and body build.

The primary goal of the take-off is to get the skater airborne, but it&rsquo;s also used to increase the speed of the skater&rsquo;s rotation to give them enough time to complete the necessary rotations in the air. By pulling in their arms and legs the skater can increase their angular speed, giving them the rotation they need to perform their jumps in the time allotted.

As was mentioned, the perfect jump (vx= vy) creates parabola shape, with two intercepts. There are 2 functions, whose graph is in the shape of a parabola. Table 3 shows the two functions that most accurately show the skater&rsquo;s trajectory. Functions that have been &bdquo;split&rdquo; such as half a circle may also have been included in this study. However, their shape is not similar to the flight trajectory of a skater, so considering them was unnecessary. Coefficients: a, b, c, d,h,k &isin; {R}

In order to find a function that will be used for further research, graphs of two functions were made (Graph 1.). X-axis stands for the time (t) in seconds and Y-axis stands for vertical displacement s(t) in centimeters. To measure the time of the jump, I watched 24 best jumps of women that performer double Axel jump and calculated the mean duration . The data was collected twice, from the video to reduce possible calculation error. The result was rounded to an integer.

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\(\overline{x}=\frac{3 + 2 + 3 + 3 + 3 + 2 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 3 + 2}{24}\)

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\(\overline{x} = 2.75\)

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\(\overline{x} ≈ 3\)

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To make an accurate graph of the functions, it was necessary to find the coordinates of the vertex. Because the x-intercepts are coordinates x1(0,0), x2(3,0) it was possible to find axis of symmetry.

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\(h = \frac{x_1+x_2}{2}\)

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\(h = \frac{0+3}{2}\)

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h = 1.5 sec

Functions - f(x) – quadratic function - g(x) – Sine Function..svg

These two functions have x-intercepts x1(0,0), x2(3,0) in the domain 0 &le; x &le; 3 and vertex V(1.5, 1.5) at the same coordinates. The value k = 1.5&nbsp;is random, but it does not affect the result since both functions have a vertex at this point. The real second vertex coordinate will be calculated later, because I assumed that the height of the function would be very high and stretched, so the graphic differences between the two functions would not be noticeable. Fitting functions to the same principal coordinates graphically shows the difference between these functions. Various variables will be checked in the study &ndash; including the angle and height of the jump. For this purpose, graph with the greatest area of the function should be chosen. The results have been rounded to 3 significant figures. The functions were matched so that the x-intercepts are at the coordinates x1 = (0,0) and x2 = (3,0), have a common axis of symmetry (h = 1.5) and a&nbsp;common height (k = 1.5). This matching allows to explore which function is best suited for the study. Calculation were done with GDC.

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Quadratic function f(x) = &minus;0.68(x &minus; 1.5)2 + 1.5 

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\(\displaystyle\int^3_0(-0.68(x-1.5)^2+1.5)dx\)

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\(=\displaystyle\int^3_0(-0.68x^2-2.04x-3)dx\)

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= &minus;6.12 + 9.18 &minus; 0.09

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= 2.97 cm2

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Sine function g(x) = 1.5 sin(1.047x)

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\(1.5\int (\sin(1.047x))dx\)

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\(=1.5×\frac{1}{1.047}\displaystyle\int^{3.141}_0\sin(u)du \)

 
\(= 1.5 × 0.95510 ...\bigg [−\cos(u)\bigg]^{ 3.141}_ 0\)

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\(= 1.43266 ...\bigg [−\cos(u)\bigg]^{ 3.141}_ 0\)

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= 1.43266 ... &times; 1.9(9)

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&asymp;&nbsp; 2.86532

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&asymp; 2.87 cm2

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2.87 cm2 &lt; 2.97 cm2

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The result and the graph clearly show that the quadratic function is more accurate and efficient, so further calculations will be based on this formula.

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To calculate the real second vertex coordinate of the parabola, I calculated the average height of the jump. The data was collected from videos, from official measurements that are displayed on the screen after the jump and were rounded to an integer, as shown in Figure 4.

Sample Data Displayed On The Screen After Performing A Jump.svg

\(\overline{x}=\frac{38× 3+39× 5+42× 5+40× 6+41× 5}{24}\)

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\(\overline{x}=40.1\)

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\(\overline{x} ≈ 40 cm\)

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Given that the domain of the function : 0 &le; x &le; 3 it can be assumed, that at x = 1.5 skater will reach the highest jump height which is about 40 centimeters. Therefore, x = 1.5 is a turning point, which can be either a local minimum or maximum. Since a is negative and the highest point of the parabola is its vertex, it can be assumed to be a local maximum.

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The next step to calculate the rest of the data is to determine the function that will accurately represent the flight trajectory. Therefore is it&rsquo;s necessary to go back to the formula for quadratic functions.

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s(t) = a(t &minus; h)2 + k

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The calculations made earlier made it possible to determine two variables - h = 1.5 and k = 40.

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s(t) = a(t &minus; 1.5)2 + 40

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To find the coefficient a, I substituted the point (0,0) into the formula above. The result has been rounded to 3 significant figures.

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0 = a(0 &minus; 1.5)2 + 40

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0 = 2.25a + 40

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2.25a = &minus;40

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a &asymp; &minus;17.8

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After performing the calculations, it can be shown that the graph of the function s(t) defining the displacement of the vertical movement with x-intercepts x1(0, 0) and x2(0, 3) is stretched and its equation is given below.

Real Values Of Function S(t). The Graph Was Done Using ‘Desmos’ Calculator.svg

To calculate horizontal velocity (vx) and vertical velocity (vx), the velocity v(t) has to be calculated. Using the data from the previous calculations, the function s(t) can be derivatived and the result will be the function v(t). Using the previously set turning point, one can visualize how the skater&rsquo;s speed behaves using a table.

s&prime;(t) = [&minus;17.8(t &minus; 1.5)2 + 40]&rsquo;

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s&prime;(t) = (&minus;17.8t2 + 53.4t &minus; 0.05)&prime;

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s&prime;(t) = &minus;35.6t + 53.4

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v(t) = &minus;35.6t + 53.4 cm sec&minus;1

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The next step is to determine the velocity of the take-off, which can be easly found by calculating formula given below:

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v(0) = &minus;35.6&times;0 + 53.4

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v(0) = 53.4

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v = 53.4 cm sec&minus;1

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Substituting the value 0 into the function comes from the fact that the coordinate (0,0) is understood as take-off point. The calculation of the velocity (v) gives the basis for the calculation of horizontal velocity (vy) and vertical velocity (vx). Figure 8. Shows the relationship between different velocities, that will be used for the study.

A Graph Showing The Relationship Between Specific Velocities (vx), (vy) And The Angle Of The Take-off (θ).svg

To calculate vertical velocity (vx) and horizontal velocity (vy) formulas given below should be considered.

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vx = vt &times; sin &theta;

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vy = vt &times; cos &theta;

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In the part 2.4. The angle of the take-off (&theta;) different outcomes of jumps, depending on the angle of the take-off were analyzed. Considering that the research aims for establishing perfect double Axel model, only one angle will be used for further calculations, &theta; = 45&deg;.

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vx = 53.4 &times; sin 45&deg;

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vy = 53.4 &times; cos 45&deg;

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\(v_x=53.4 ×\frac{\sqrt2}{2}\)

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\(v_y=53.4 ×\frac{\sqrt2}{2}\)

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vx &asymp; 37.8 cm sec&minus;1

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vy &asymp; 37.8 cm sec&minus;1

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So the assumption, that a perfect jump with an angle of the take-off (&theta;) = 45&deg; has equal vertical and horizontal velocities is true.