Sports Biomechanics

If the projectile moves only vertically (for example, a ball thrown straight upwards), its projection speed will determine the height it reaches before gravity accelerates it back towards the Earth. If we don’t take air resistance into account, gravity accelerates all objects at the same rate: 9.81 m·s^-2 barring some regional variations around the planet*. This is about the same acceleration a lion can achieve or twice the acceleration of the fastest humans. To get an idea of how fast it is, drop a small ball from a height of a few metres and watch it accelerate as it falls.

What might position (displacement), velocity and acceleration graphs look like for a ball thrown vertically?

The angle of projection is also an important factor affecting projectile range. If an object is projected vertically, it will land back at its starting point, after gravity has pulled it back to Earth (remember, the object is accelerated positively the whole way if ‘down’ is assigned the positive direction). So, its range is zero. If the object is projected horizontally from ground level, it will not get airborne, so again its range is zero. It can also be projected at angles between 0° and 90°, where it will travel both vertically and horizontally. At a projection angle of 45° the object will have an equal magnitude of vertical and horizontal velocity and its range will be maximised, as you can see in Figure 3.2. However, we need to take into account other factors that influence projectile range in order to answer our question.

You could find v, u or t by re-arranging the equation appropriately (see Appendix B if you are unsure how to do this).

As an example of the use of the second equation, I might ask how far off the ground the ball was at the top of its trajectory, given that it hit the hands at 21.6 m·s^-1 (assuming that the fielder caught the ball only millimetres above the ground):

If we know v, u and a (using the standard Earth value of 9.81 m·s^-2) we can re-arrange the equation thus:

•Step 5: Calculate the flight time. This needs to be done in two parts. First, we calculate the time for the shot to rise to its peak height and back to the starting (release) height; second, we calculate the time to fall the further 2 m to the ground.

•Part 1: There are two things to remember always: (1) the flight time of an object equals time up plus time down, so if it starts and finishes at the same vertical height the total time equals time up multiplied by two, and (2) the vertical velocity or final velocity of an object moving upwards is always zero because it stops briefly at the top of its trajectory before falling back down, so we know that the final velocity, v, also equals zero. Just as we calculated the initial horizontal velocity above, we can calculate the initial vertical velocity using the sin rule. For this calculation, we can use either v = u + at or t = (v – u)/a.

•Part 2: We know the initial vertical velocity of the shot is 8.03 m·s^-1, because if it leaves the hand with this vertical velocity it must attain it again as it falls back past the level of the hand, but we don’t know the final velocity as it is about to hit the ground. We could use the equation s = ut + ½ at² but this requires us to understand how to solve a quadratic equation. If you want to try, have a look at Box 3.2. Fortunately, there is another way: we can use the equation v² = u² + 2as to find the final vertical velocity and then use v = u + at to find the time. (I worked out this method by looking at the equations and thinking about what I already knew. I then realised that if I had v the problem would be easy, so I sought a way to do that. The two-step process isn’t as hard as it might look at first.) Either way:

In the end, it is probably necessary to run biomechanical tests to determine the optimum trajectory for whatever object you need to throw, based on the athlete who is actually going to throw or kick it.

SPECIAL TOPIC: BASIC VIDEO ANALYSIS

We can use relatively simple tools to uncover a lot of information about a person’s performances. Video analysis is one such method.

In this chapter we learned how to use information such as an object’s release angle and speed to optimise performance but we need to find methods of obtaining this information easily. If you don’t have a suite of biomechanical analysis tools, you can use a standard video camera, a television, a sheet of plastic and a marker pen. You’ll also need a protractor (or another instrument to measure angles) and a ruler. You will be recording the athlete from the side, so that you can record the angle and speed of a shot as it is put. Set your camera on a tripod a good distance away from the athlete (at least 6–8 m if possible but the further the better) and side on (that is, perpendicular to the line of the throw) as shown in Figure 3.9.

FIG. 3.9 Set-up for video analysis. The camera is placed to view the thrower side-on (i.e. perpendicular to the line of throw) and at a considerable distance. A rod/line of known length is placed in the direction of the throw near the feet of the thrower.

Objects change their size and shape as they move across or towards/away from the camera, which can cause error s in calculations. The two main errors are: perspective error, which occurs as objects seem to get bigger or smaller as they move towards or away from the camera, and parallax error, which occurs as an object’s size and shape seem to change as it moves across the camera (think of a person at left of camera where you can see their front, then moving to centre stage where you see them side on ... when you see that same person a long way away, you will always see them from side on). You can all but eliminate these errors if you have the camera a good distance from the athlete. You can then zoom the camera so that the athlete fills the screen sufficiently.

Next, place a rod or draw a straight line on the ground in the direction of the forthcoming throw from a point near where the thrower’s feet will be at the time of release. This will allow you to measure the angle of trajectory against a known horizontal line. Measure the rod – if you know its exact length, you can use it to work out how big the objects are or the distances thrown when they are on the television screen. This process is called calibration.

Take video recordings of several throws, capturing the point of release and the first part of the flight of the shot. Only throws where the shot travelled perpendicular to the camera can be used, because if the shot travels towards or away from the camera you will get perspective errors.

Once you have taken the video footage, play the first throw on the television and pause it at the point the shot leaves the hand. Stick the clear plastic sheet on the television and mark the athlete’s toe, hand (to determine the height of release) and the horizontal line or rod that was placed on the ground (as shown in Figure 3.10). Last, mark the point of the shot. Then move the video one frame forwards and remark the shot (you now have four points and one horizontal line).

FIG. 3.10 Determining the angle of trajectory, height of release and speed of release (calculated from the distance travelled by the shot in one frame of video) can be done using a basic video camera and television set-up. First, the important landmarks are located and drawn on a clear plastic sheet (A) and then angles and distances can be measured (B). See text for more detail of the procedures.

Now take your measurements. The angle between the line on the ground and a line joining the marks of the shot is the angle of trajectory. The distance between the shot marks gives the displacement of the shot after release. From the frame rate of the camera, you can work out the time between the two points (for PAL systems this is 0.04 s and for NTSC it is 0.033 s; see Chapter 2) and then find the velocity of the shot using v = s·t^-1 (that is, distance divided by time).

Before you can use the displacement of the shot, you have to know how far it travelled in the real world, not the distance on the television screen. Divide the length of the line or rod as measured on the television screen by its real length, to get a ‘scaling factor’. For example, if its length on the television was 0.3 m (30 cm) and its actual length was 2 m the scaling factor would be 0.3/2 = 0.15. This process is called calibration. If the shot travelled 0.084 m (8.4 cm) across the television, 0.084/0.15 gives you the real distance travelled (0.56 m). Therefore the velocity was 0.56/0.04 = 14 m·s^-1.

So now you have the angle of projection and know the projectile velocity was 14 m·s^-1. If you think you’ll use this process often, you might like to look at freely available software such as Kinovea (http://www.kinovea.org/) or Video4Coach (http://video4coach.com/) to help analyse your data. Software packages such as this make the process of analysing data much easier, and allows for a lot of analysis to be done without needing to do all the hard work. Once you choose your method you can do this for any number of throws, but how do we find the relationship between projectile velocity and angle of projection?

After analysing a number of throws at different release angles, you can put your data into a spreadsheet: the data might look something like the spreadsheet in Figure 3.11.

FIG. 3.11 Release angle and release speed entered into a spreadsheet programme.

You can then create a scatterchart and add a linear regression trend-line (the slope of this line tells you the relationship between the two variables). The slope of the line shown in Figure 3.12 is –0.0853; the equation (at the top of the graph) is in the form y = ax + b, where y is a value on the y-axis (that is, what we’re trying to find), x is a point on the x-axis (that is, what we measured) and 17.412 is the value that the line would cross the y-axis if it continued.

FIG. 3.12 Graph of projection velocity against projection angle. The velocity decreases as the angle increases. The equation to the line (at top) shows that the velocity decreases by 0.0853 m·s^-1 for every degree increase in projection angle. This would be 0.0853 × 57.3 = 4.89 m·s^-1 per radian.

To find y, you simply put in a value of x. For example, the projection velocity at an angle of 35° would be approximately –0.0853 × 35 + 17.412 = 14.9 m·s^-1. The number –0.0853 implies that velocity decreases by this much for every degree increase in angle. In the spreadsheet you created earlier, the units were radians (1.7 m·s^-1 per radian). You can therefore multiply this figure by 57.3 to find the change in velocity for a whole radian: 4.89 m·s^-1, which is significantly larger than the 1.7 m·s^-1 you used earlier. For some reason, our shotputter loses much more velocity as the angle increases. (You can enter 4.89 in cell B3 of your spreadsheet to see how this affects the optimum projection angle for this shotputter.)

For the new thrower, what is the optimum angle of release? How do the original velocity and the relative height of projection affect the results? How might you coach this athlete differently to the shotputter described earlier in the chapter?