Spinning out of control

Spinning out of Control

Daniel Whittaker

Contents

Page

2 Section 1 - Introduction

2 Section 2 - Flight of a ball in a vacuum

5 Section 3 – Flight of a ball with air resistance

8 Section 4 – Laminar and Turbulent flow

10 Section 5 – Magnus Effect

11 Section 6 – Examples of Magnus Effect in practise

12 Section 7 – Conclusion

13 References

Abstract

In sports, people are always looking how to gain even the slightest advantage over their opponents, whether this is by tactical analysis, or even trying to use the laws of physics to their advantage to manipulate the flight of the ball to benefit them. In this report, we will be examining the trajectories of balls under certain conditions, starting from the very basic model of a ball in a vacuum, then adding air resistance and the Magnus force into our equations. We will see how our equations of motions increase in complexity and how the Magnus effect is produced from a ball’s interaction with the air. We will be using the example of a golf ball to see how the dimples act in a way which may be somewhat surprising and we can then see from what we have learnt and by examining the equations of motions we have gathered, how this can be applied to different sports and how the Magnus effect can be used to benefit, or in some cases, disadvantage the player.

Section 1 - Introduction

In sports, we see and observe many things that we take for granted, such as a free kick in football bending round and over a wall, that golf balls are dimpled and not a simple round shape like a ping pong ball, for example. There is an underlying aerodynamic principle that explains these cases and can be used to a sportsman’s advantage in other sports. This is called the Magnus effect, and in this project, we will be examining its effects on various sports and how it affects the flight path of certain balls.

German physicist Heinrich Gustav Magnus (1802-1870) first described the effect when experimenting on cannonballs [1]. Magnus published a book in 1852 titled “On the deviation of projectiles; and on a remarkable phenomenon of rotating bodies” noting his findings [2]. However, it is believed that the first to notice the effect was Sir Isaac Newton in 1674, while watching a game of tennis at Cambridge College. A British mathematician, Benjamin Robins, also observed the effect before Magnus himself (in 1742) explaining the trajectories of musket balls in terms of the Magnus effect [1].

To understand how spin affects the flight of a ball, we analyse the trajectory of a ball in three different ways. Firstly, we will look into a classic and easy model of the flight of a ball in order to get an understanding of how to derive our equations of motion. Here we will look at the flight of a ball taking into account only the force of gravity. Next, the equations of motions will include the force of air resistance and from here we will already be able to see how this impacts a ball’s trajectory even without any spin modelled yet. Then, we will define what the Magnus force is and how it is applied to the ball and add it to our equations of motion. Then, it will become apparent why certain types of spin benefit one sport and prove to be a disadvantage to another.

Section 2 - Flight of a ball in a vacuum

To begin with, let us model a smooth ball, starting at a point in 2-D space that we will denote as the origin (0,0). The ball will initially be modelled as if it were in a vacuum on earth, therefore we will assume that the only force acting on the ball throughout the flight is gravity.

To derive the equation of motions, we will need to start from Newton’s second law, which is:

F = ma (1)

Where F is the force, m is the mass and a is the acceleration. However, from Figure 1, we see that we need to split the velocity into its x-components and y-components. Taking the initial velocity, which we will call u=Uû (hence U is the initial speed), at the origin at time t = 0, we can model the initial velocity as a straight line; therefore, splitting up its components can be shown in Figure 2 below

Now we consider Newton’s second law in the x-direction. Since there is no acceleration acting on the ball in this direction, we can use (1) to obtain

mx’’(t) = 0

Where x’’ (second derivative with respect to time) is the acceleration in the x-direction. From here, we see that integrating through we obtain

x’’(t) = 0

x’(t) = A (2)

x(t) = At + B (3)

x’ denotes the velocity whilst x shows the distance travelled in the x-direction. A and B are constants. Using our initial condition now

x’(0) = Ucos(θ)

x(0) = 0

We can plug these into equations (2) and (3) and solving for A and B, we obtain

x(t) = Utcos(θ) (4)

If we look at the forces acting in the y-direction from Figure 1, we can see this time we have the force of the weight of the ball acting, so from equation (1) and integration again, Newton’s second law becomes

my’’(t) = -mg

y’’(t) = -g

y’(t) = -gt + C (5)

y(t) =- g+ Ct + D (6)

Similarly, to the other notations, y’’ denotes acceleration, y’ velocity, and y distance travelled in the y-direction, with C and D constants. Our initial conditions only slightly differ to before

y’(0) = Usin(θ)

y(0) = 0

Using equations (5) and (6) to solve C and D we obtain

y(t) = Utsin(θ) - g (7)

Since this is only a 2-D model, there are no forces in the z-direction. Our initial conditions in this case are all equal to 0, hence the resulting equation will simply be

z(t) = 0

Thus, we can put (4) and (7) together to give us a vector equation for the motion of the ball

r(t) = Utcos(θ)i – [Utsin(θ) - g]j

We can also rewrite (4) as

t =

so we can substitute t into (7) so we can obtain a Cartesian form of the equation

y = xtan(θ) - (8)

Figure 3: Motion of a ball in a vacuum [3]

From Figure 3, we can see that the ball’s motion will take a very simple parabolic motion. However, this type of trajectory is unrealistic since of course we are not considering a major force that will in reality be acting on the ball throughout its flight. Air resistance will have an effect and we will see how air resistance impacts the trajectory of the ball and compare it to this model. At this point, the equations are simple enough that we can calculate the range that the ball travels. So in order to do this, we know that when the ball touches the ground again, that will be at the point y=0, so using (8), we can insert this condition in, to get

xtan(θ) =

Since x=0 will get us the starting position, we can cancel the x, hence this equation for the range of a ball with no air resistance is

x = (9)

So for a ball travelling at 30m/s at an angle of 45° (π/4), where the value for gravity is 9.81m/ the ball travels 91.74m. Later on in this project, we will compute and compare the distance our ball travels with and without air resistance so we can see how important it is to the calculations and how our first model for a ball in a vacuum is far too simplistic.

Section 3 - Flight of a ball with air resistance

In our previous example, we obviously ignored any forces acting on the ball apart from gravity. So to model the flight of a ball in a more realistic fashion, we will need to consider air resistance. Air resistance is the drag force acting in the opposite direction to the direction of motion of the ball. Therefore, drag is generated by the ball as it interacts and passes through a fluid, which in our case, is air. As air flows around the ball, the local velocity increases and pressure decreases when the flow of air is halfway round the ball and the velocity is at a maximum whilst the pressure is minimised. When the air flow reaches the rear of the ball, the velocity now decreases whilst the pressure increases, and this varying pressure distribution causes a force to act on the ball [4]. Of course, if the ball is not in motion, there will be no drag, as the drag is generated by a difference in velocity of the ball and the air.

The air resistance can be expressed using the equation

FD = CDρA (10)

Where, FD is the drag force, CD is the non-dimensional drag coefficient, ρ is the density of the fluid (which in our case is air), A is the surface area of the ball and u is the velocity [5].

Now when we model our ball, we can see from Figure 4 that the drag force is applied to our model for the forces acting on the ball, hence our equations of motions will all change. For simplicity, we will first take the drag force to be directly proportional to the velocity, not the velocity squared. This will be done to make our equation of motions easier to solve. Again, we will make a few assumptions throughout these calculations. Since there is a drag force now, we will assume that the air density remains constant. Also, we shall assume there is no spin acting on the ball yet and that there are no external forces acting on the ball such as wind, and any changes in the air conditions due to altitude and humidity can be neglected as well. To tidy up our equations, we shall shorten the drag force. Since the coefficient of drag, air density and surface area all remain constant due to the assumptions made, we can combine all these constants to FD = ku, where k = CDρA. Now we will take the sum of the forces in the x-direction

mx’’ = -kx’

We let x’ = u to obtain

mu’ = -ku

Solving this by variable separable

= -

Where the limits on the integrals are the initial conditions of the ball. We now obtain from this

u(t) = Ucos(θ)[exp(- t)]

Since x’ = u, inserting into the above equation we get

x’(t) = Ucos(θ)[exp[- t)]

x(t) = - Ucos(θ)[exp(- t)] + E; I.C: x(0) = 0

x(t) = Ucos(θ)[1 – exp(- t)] (11)

Now we take the sum of the forces in the y-direction

my’’ = -ky’ – mg

Similarly, to before, we let y’ = v

mv’ = -kv – mg

v’ + v = -g

This equation is a first order, linear, non-homogeneous equation, so an integrating factor exp[dt] will be used

exp( t)(v’ + v) = -g[exp( t)]

exp( t) v = g[exp( t)]dt

exp( t) v = -g [exp( t)] + G

v(t) = - + G[exp(- t)]; I.C: v(0) = Usin(θ)

Inserting y’ = v the above equation becomes

y’(t) = - + [ + Usin(θ)][exp(- t)]; I.C: y(0) = 0

y(t) = - + [ + Usin(θ)][1 – exp(- t)] (12)

Again we will only consider the 2-D motion of the ball, hence the equation of motion in the z-direction will simply be z = 0. So to get a vector equation for the motion of the ball, putting (11) and (12) together yields

r(t) = Ucos(θ)[1 – exp(- t)]i + {- + [ + Usin(θ)][1 – exp(- t)]}j

To calculate the effect that the drag has on the ball, we will eliminate the t’s to get the above equation into Cartesian form so we can compare it with (8)

y = (tan(θ) + )x + log[1 - ]

Previously, we had worked out the range of the ball was 91.74m from (9). Using data from a baseball and keeping our velocity and launch angle the same, the updated distance the ball has travelled is now 89.43m.

However, we can see that there is something wrong with our calculations from (9), since there is only about a 2m difference between the distance that the ball travelled in our example even though they travelled about 90m. The graph above gives a rough visual representation of the difference between the trajectories with and without drag. Although the parameters on this graph vary from what we have been using, we can use figure 5 to deduce that our calculation are not accurate as there is almost a 2m difference even though the ball travelled about 10m.

Why is this the case? Well, throughout our calculations of the ball’s motion with air drag, we took the drag force to be directly proportional to the velocity, which is fine for objects travelling at a low velocity, or low Reynolds number. The Reynolds number is the ratio between inertial forces and viscous forces. It is a dimensionless number, where high values indicate that inertial forces are dominant, leading to a turbulent flow, whilst low numbers are dominated by viscous forces, hence laminar flow generally occurs under these condition [7].

However, since we are looking at the trajectories of balls in sports, golf balls or tennis balls, for example, all have relatively high velocities (high Reynolds Number), therefore in our case, it is inaccurate to take the drag force as FD = ku. The more accurate way of taking the drag force would obviously to go back to the original equation for drag stated in (10), so our drag force is now FD = k. Now are equation of motion in the x direction becomes

mx’’ = -k

and the y-direction

my’’ = -k-mg

Solving for these will yield a more accurate equation regarding the motion of the ball with drag for balls with higher velocities, but solving these equations to get a vector equation for the motion of the ball like we have done previously becomes significantly harder due to the velocity now being squared.

Section 4 - Laminar and turbulent flow

Another assumption made in these equations is that we have taken the drag coefficient (CD) to be constant. This is far from true, as if we rearrange (10) to make CD the subject of the equation, then we can see that it depends on the velocity, which is in constant deceleration due to the drag acting on the ball. The magnitude of the drag coefficient also depends on whether the boundary layer is laminar or turbulent. When the velocity of a moving fluid is low, it is said to have laminar flow conditions.

These types of flow conditions are normally desirable since they reduce drag, however boundary layers in laminar flow are fragile and will separate from the surface of a sphere very easily when encountered by an adverse pressure gradient, where the separation around the ball leaves a large wake of low pressure behind the ball, causing deceleration [9]. Laminar flow is common in cases where a flow channel is relatively small, the fluid is moving slowly and its viscosity is high. This case was used in calculating our equations of motion for the drag force proportional to the velocity, so now we can fully understand why that was a rough estimate for the equation of motion of a ball.

In turbulent flow, the fluid appears to be in chaotic and random motion. A fast moving ball travelling through the air causes a chaotic turbulent zone to develop behind the ball. This turbulence causes energy to be dissipated from the ball, which results in the ball losing energy more quickly when turbulence develops behind the ball [9]. If the ball is smooth in this case, a wide, turbulent, chaotic zone develops which is undesirable. However, if we make this ball dimpled, such as on a golf ball, the rough texture delays the development of these chaotic conditions, therefore a narrower turbulent zone exists.

As seen above, this turbulent flow means that there is a smaller separation of air flow around the ball, creating a smaller wake behind the ball in comparison to the case shown in figure 7. This causes deceleration at a slower rate compared to the smooth ball. Now the Magnus force on a dimpled ball is increased drastically in comparison to that of a smooth ball. The dimples aid the rapid formation of a turbulent layer around the golf ball in flight, giving it more lift. Hence we can start to take into account the Magnus force into our equations of motions.

Section 5 - Magnus effect

But firstly, what is the Magnus force? The Magnus Effect exerts an additional force on a spinning object. This occurs as the speed of the ball relative to the air is different on opposite sides of the ball. Let’s take the case of a ball with backspin applied to it.

It can be seen from the diagram above that the lower side of the ball will have a greater speed relative to the air than the upper side. This therefore creates a larger drag force acting on the lower side of the ball compared to the upper side. So these drag forces are exerting an unequal pressure on the ball and the resulting pressure difference causes a net force on the ball in the direction of the low pressure; where the lower pressure corresponds to lower drag [9]. So in the case of the ball with backspin, the low pressure is on the upper side of the ball, hence we see an upward force acting on the ball, as in the figure 8.

The Magnus force depends on the rotation and velocity of the ball, therefore we express the force as

FM = S(ω x V)

Where S is the spin coefficient, ω is the angular velocity and V is the velocity of the ball [11]. The cross product tells us that the Magnus force is perpendicular to both the balls rotation and its velocity. The cross-product provides the Magnus force in 3 dimensions

S(ω x V) =

We can now rewrite our equations of motion to include the Magnus force along with air resistance. It is important to bear in mind that our initial velocity u, in this case, is in 3-D space, therefore it shall be split up into its x, y and z components. Matching these components with our acceleration components in the same coordinate axis will yield the following 3 equations

mx’’ =- CDρA+ S(uzωy – ωzuy)

my’’ =-mg - CDρA+ S(uxωz – ωxuz)

mz’’ =- CDρA+ S(uyωx – ωyux)

As seen by these equations, the complexity increases yet again as we add more forces to the ball. So solving these equations will prove extremely difficult, since we have velocity in 3 dimensions in each equation, as well as the angular velocity of the ball, which will vary throughout the flight of the ball. Along with air density and the drag coefficient not being constant, the only way to truly map out the flight of a ball is by testing it out in practise or in wind tunnel experiments.

In the graph above, we can see how taking the Magnus force into account dramatically increases the height and distance that a golf ball with backspin can travel compared to our other conditions we took when considering the flight of a ball. This would make sense looking at figure 8, since the Magnus force is providing a force opposite to its weight, so it is generating lift. This is also shown in the graph above, as once the ball has travelled about 40m, the graph then curves slightly upwards, therefore agreeing with the behaviour of the Magnus force acting on a ball with backspin. Now if we look at our equation of motion for the y-direction, we can see that if there is backspin applied, the angular velocity is going to be –ωx therefore giving us a positive Magnus force. This would also agree with the graph since we have a larger force now acting in the y-direction, hence the ball would experience some sort of lift force.

Section 6 - Examples of Magnus Effect in practise

There are other examples of the Magnus effect prevalent in golf that can prove to have a negative impact to the player. “Topping” the ball in golf is when a golfer hits the top side of the ball creating topspin instead of backspin. When topspin is created, the low pressure is now created at the lower side of the ball, so now the Magnus force acts in the same direction as gravity, therefore the ball will drop much sooner and so will not travel anywhere near as far as if it had backspin applied. This proves detrimental to golfers starting off at the tee looking to maximise the distance they can hit the ball. Another downside the Magnus effect has on golfers is when they “slice” a shot. This puts low pressure on the side of the ball now so the Magnus force is now produced along the z-axis now, if we still take the horizontal distance in the x-axis and the vertical distance in the y-axis. This sideways force will most likely bend the ball away from the golfers intended target and possibly leave the ball in a bad position. It may not be that detrimental if the curve of the ball acts against wind but, generally speaking, in golf you do not want to slice the ball as you will lose accuracy.

The Magnus effect spans across all sports and one of the more noticeable examples of this was in a football match played between Brazil and France in 1997. Brazil’s Roberto Carlos took a free-kick from 35m [13] out that curved around “the wall” (players stood blocking a direct path for the shot to the goal) and into the goal, leaving the French goalkeeper bewildered.

Here we can see how Carlos has used the Magnus effect to his advantage by effectively slicing his shot to avoid the wall and score a goal. By striking the right side of the ball, low pressure in this case was created on the left side of the ball, therefore lift was created in this direction and unlike golf, slicing the ball in this case proved very effective.

In table tennis, the Magnus effect is one of the key aspects of the sport and the effect is probably more noticeable in table tennis than any other sport. Due to the ball being very lightweight, the Magnus force has a dominating effect over the balls weight, thus spin exerted on the ball has a greater effect on the trajectory of the flight of the ball. Players are able to hit the ball with topspin but at very high velocities and are still able to get the ball to bounce in the oppositions side of the table due to the overriding effect the Magnus force has on the ball to bring it down in time. Table tennis balls are also more sensitive to changes in altitude and air density, where an increase in altitude causes a ball with top spin to travel further than it would at sea level. From our final equations, this would make sense since at higher altitudes, air density will decrease therefore lowering the force that air resistance has on the ball and since the decelerating force has been lowered, in theory the ball should travel further.

Section 7 - Conclusion

Starting from the most simplified model of a parabolic motion of a ball, we have been able to understand how adding air resistance and spin to the ball leads to a drastic difference in the balls trajectory and how the motion can be manipulated with different spin velocities. Taking the balls motion in a vacuum first, we can see how many assumptions where made and thus the inaccuracy of the model. Then by considering air resistance, we were able to see that there were two different conditions that we could calculate. Whilst we calculated air resistance as being directly proportional to the velocity of the ball, we learned that this was valid for low speeds and should not be used if we want to obtain more accurate results. Finally, we saw how a spinning ball created pressure differences resulting in a lift force on the side of the ball with the low pressure and that this lift force was the Magnus effect. We also saw how smooth balls created laminar flow, whilst adding dimples to the ball created turbulent flow but reduced the wake of turbulence behind the ball. Adding the Magnus force to equations of motion used previously, we saw how complex the equations had become and how much our equations of motion have changed since we initially considered the flight of a ball in a vacuum. If we took into account that factors such as air density and the drag coefficient are not constant during the flight of a ball, then these equations are way beyond the level of what we can calculate, thus making wind tunnel and real world experiments the most accurate way of determining the trajectory of a ball. Finally, we analyse how the Magnus effect was applied in a variety of sports, examining especially how backspin had a big effect on a golf ball and how certain types of spins can be beneficial in some sports, yet prove costly in others.

References

[1] Tietz, T. (2015) Heinrich Gustav Magnus and the Magnus effect. Available at: http://blog.yovisto.com/heinrich-gustav-magnus-and-the-magnus-effect/ (Accessed: November 2015).

[2] Goff, J. E. (2009) Gold medal physics: The science of sports. Baltimore: JHU Press, p.132

[3] Budd, C. and Sangwin, C. (2014) 101 uses of a quadratic equation: Part II. Available at: https://plus.maths.org/content/101-uses-quadratic-equation-part-ii (Accessed: November 2015).

[4] Hall, N. (no date) What is drag?. Available at: https://www.grc.nasa.gov/www/k-12/airplane/drag1.html (Accessed: November 2015).

[5] Hall, N. (no date) Drag on a baseball. Available at: https://www.grc.nasa.gov/www/k-12/airplane/balldrag.html (Accessed: November 2015).

[6] Allain, R. (2012) Projectile motion Primer for FIRST robotics. Available at: http://www.wired.com/2012/01/projectile-motion-primer-for-first-robotics/ (Accessed: November 2015).

[7] Benson, T. (2014) Reynolds number. Available at: https://www.grc.nasa.gov/www/BGH/reynolds.html (Accessed: 12 January 2016).

[8] Ask us - golf ball Dimples & drag (no date) Available at: http://www.aerospaceweb.org/question/aerodynamics/q0215.shtml (Accessed: November 2015).

[9] Kaye, B. H. (1996) Golf balls, boomerangs, and asteroids: The impact of missiles on society. New York: Wiley-VCH Verlag GmbH. p.23

[10] Physics of golf (no date) Available at: http://www.real-world-physics-problems.com/physics-of-golf.html (Accessed: November 2015).

[11] The Magnus force (no date) Available at: http://farside.ph.utexas.edu/teaching/329/lectures/node43.html (Accessed: November 2015).

[12] Tannar, K. (no date) Golf instruction. Available at: http://probablegolfinstruction.com/weather_effects.htm (Accessed: November 2015).

[13] Gill, V. (2010) Roberto Carlos wonder goal ‘no fluke’, say physicists. Available at: http://www.bbc.co.uk/news/science-environment-11153466 (Accessed: November 2015).