Introduction
In my last blog post, I explained how we calculate jump height from takeoff velocity (read here). The takeoff velocity method is what we use for jump assessments where we know the starting velocity of the jumper’s center of mass (COM). These include the countermovement jump, countermovement jump rebound, squat jump, and drop jump. For the first three of these jumps, this is because our software ensures the jumper starts each trial by standing still for at least 1 second before they jump, and this means that we can be confident in knowing what their starting COM velocity will be (0 m/s in this case because they are standing still). By inputting drop height and ensuring that our jumper drops onto (rather than jumps off or steps down from) the drop jump platform (e.g., a box), we can calculate their starting velocity for drop jump trials by understanding the relationship between COM drop height and gravity. However, whenever multiple jumps are performed within a single trial, we estimate jump height for each repetition from the time the jumper spends in the air, the so-called ‘flight time’. This is because during the landing part of the jump, the jumper’s COM position will cross zero, and when the duration of the trial increases, as is the case with multi-rebound jump trials, the integration process (how we convert force to velocity) shifts the baseline. This means that if the jumper stood still at the end of their multi-rebound jump trial, their COM velocity (and displacement) would be way off the zero baseline it should be at. Such a scenario, which we call numerical integration drift, would render the takeoff velocity method of estimating jump height inaccurate. So, I wanted to write this brief article to explain how jump height is calculated from flight time and highlight some considerations associated with this approach.
Calculating jump height from flight time: a (hopefully) simple explanation
When we only know the flight time (time spent in the air, see Figure 2) for any given vertical jump, we can estimate the jump height based on known equations for one-dimensional (uniform) motion. The assumption here, like with the takeoff velocity method, is that air resistance is negligible and so the athlete’s mass is subjected to a constant gravitational acceleration equal to -9.81 m·s-2 during the flight phase of the jump.
As mentioned in my previous article, the displacement between some initial and final instant is equal to the average velocity between the same initial and final instants multiplied by the change in time between the same initial and final instants. This can be expressed in the following equation, as shown earlier, where Δs is displacement, 𝑣̅ is the average velocity and Δt is change in time:
∆𝑠 = 𝑣̅×∆𝑡
When applying the above formula to the flight phase of a vertical jump, the change in time that we need is equal to the difference in time between the athlete leaving the ground at takeoff (termed the initial time [tf] below) and reaching peak vertical displacement (termed the final time [ti] below). The assumption here is that the peak vertical displacement of the athlete’s COM is achieved exactly in the middle of (i.e., halfway through) the flight phase which isn’t entirely correct for reasons to be discussed later, but let’s just go with it for now. Calculating the change in time is, therefore, easy. We simply divide the flight time (shown below as FT) by two to give half of the flight time:
∆𝑡 = 𝑡𝑓 − 𝑡𝑖
So:
∆𝑡 = 𝐹𝑇/2
For example, if the athlete’s flight time is 0.50 s (Figure 1), we would calculate the change in time as follows:
∆𝑡=0.50/2
So:
∆𝑡=0.25s
Figure 1: The center of mass force-time record (top right), force-velocity record (bottom-middle), and force-displacement record (bottom-right) for a countermovement jump. The athlete’s flight time (0.5 s) is shown (top-left) but the corresponding jump height is hidden.
We need to perform a few calculations, however, to calculate the athlete’s average velocity. Firstly, we need to calculate the athlete’s change in velocity (Δ𝑣) between the instant of takeoff (termed the initial velocity [𝑣i]) and the instant of peak vertical displacement (termed the final velocity [𝑣𝑓]). To do this, we simply subtract the final velocity from the initial velocity:
∆𝑣=𝑣𝑓−𝑣𝑖
The final velocity will always be 0 m·s-1 because the peak vertical displacement occurs at the transition point from the upward to downward acceleration (the momentary pause when the maximum jump height is achieved before the athlete descends towards the ground):
∆𝑣=0−𝑣𝑖
So:
∆𝑣=𝑣𝑖
Or, because initial velocity is the takeoff velocity:
∆𝑣=𝑣𝑡𝑜
Therefore, we only need to know what the takeoff velocity is to determine the change in velocity, the problem is, we don’t know what the takeoff velocity is from the flight time alone.
Luckily, we know that change in velocity is equal to acceleration (a) multiplied by change in time. Because, in this case, the change in velocity is identical to the takeoff velocity, this formula will give us the takeoff velocity. Remember, the acceleration due to gravity is a constant of -9.81 m·s-2 and we calculate the change in time just like we did earlier.
∆𝑣=𝑎×∆𝑡
So:
∆𝑣=9.81×(0.50/2)
So:
∆𝑣=2.45 m·s-1
Or, in this case:
𝑣𝑡𝑜=2.45 m·s-1
The next step is to calculate the average velocity by dividing the takeoff velocity by 2 (because the average [mean] velocity is equal to [initial velocity (which is the takeoff velocity) + final velocity (which is 0 m·s-1)] ÷ 2):
𝑣̅=𝑣𝑡𝑜/2
So:
𝑣̅=2.45/2
So:
𝑣̅=1.23 m·s-1
Finally, we go back to the first equation presented in this section and calculate average velocity multiplied by change in time to give peak vertical displacement (i.e., jump height).
∆𝑠= 𝑣̅×∆𝑡
So:
∆𝑠= 1.23×0.25
So:
∆𝑠= 0.31 m
In other words, the athlete’s jump height was 0.31 m (figure 2).
Figure 2: The center of mass force-time record (top right), force-velocity record (bottom-middle), and force-displacement record (bottom-right) for a countermovement jump. The athlete’s flight time (0.5 s) is shown (top-left) and now also is the corresponding jump height of 0.31 m (also top-left).
The above description provided a step-by-step approach to calculating jump height from flight time, but the formula can be simplified to the following (as is often reported in journal articles):
∆𝑠= 𝑣̅×∆𝑡
So:
∆𝑠=((9.81×𝐹𝑇)/4)×(𝐹𝑇/2)
So:
∆𝑠=(9.81×𝐹𝑇2)/8
Please note that the number 4 (step 2 of the above) comes from dividing the sum of acceleration multiplied by flight time by two to give takeoff velocity and then dividing this answer by two to give average velocity in line with the steps outlined earlier.
Going back to the original example of an athlete who achieved a flight time of 0.50 s, we can check that this simplified formula gives us the same answer as earlier:
∆𝑠=((9.81×0.502)/8)
So:
∆𝑠=2.45/8
So:
∆𝑠=0.31 m
So, what’s the problem…the effect of changes in joint geometry
As mentioned above, the flight time method relies on the time taken to travel from the takeoff position to the highest flight position being the same as the time it takes to travel downwards from the highest flight position to the landing position, and for a solid object that’s true. However, we humans aren’t solid objects and so, when applied to us, the accuracy of the flight time method can be compromised by differences in the position of the jumper’s COM between takeoff and landing. This is a consequence of changes in leg joint geometry because the jumper’s COM position is naturally likely to be higher at takeoff than at touchdown because the hip, knee, and ankle joints are (or approach) fully extended at takeoff and this is rarely the case during landing. This means the jumper’s COM displacement is often greater during the downward part of flight and this is reflected by a longer downward travel time. Additionally, jumpers can ‘game’ the flight time method by intentionally flexing their knees and hips during the flight phase in a tuck-jump style to prolong their flight time. All of this impacts the accuracy of the jump height calculated from flight time.
Table 1. A comparison of countermovement jump height data calculated from the takeoff velocity method and the alternative flight time method (recorded from force plate and ergometer data) – created by Professor Jason Lake.
|
Study |
% difference in jump height |
|
Attia et al. (2017)2 |
27.0 |
|
Badby et al. (2023)1 |
-2.9 |
|
Baumgart et al. (2017)1 |
26.8 |
|
Buckthorpe et al. (2012)2 |
23.3 |
|
Chiu & Daehlin (2020)1 |
-3.7 |
|
Hatze (1998)2 |
-3.5 |
|
Reeve & Tyler (2013)1 |
-9.5 |
|
Reeve & Tyler (2013)2 |
-5.1 |
|
Reeve & Tyler (2013)3 |
-15.0 |
|
Yamashita et al. (2020)1 |
-6.3 |
Practical applications for force plate users
Practitioners should consider the effect that using the flight time method can have on the accuracy of jump height. We encourage them to reinforce best practices by coaching their athletes to avoid tucking their legs during flight. We also encourage practitioners to try to avoid comparing jump heights calculated from flight time (or any other methods) to jump heights calculated from takeoff velocity without considering the differences presented in Table 1. However, if athletes are familiar with jumping assessments and are coached well, the flight time method can provide jump heights similar to those obtained from the gold standard takeoff velocity method. Additionally, if the jumper has not stood with the stillness necessary, to obtain accurate velocity data, or they’re unsure of either how data were integrated or when this process began, the flight time method may provide a more robust jump height. Especially where good and consistent coaching sees the jumper keeping their takeoff and landing joint geometry as similar as possible. Luckily, the Hawkin Dynamics system doesn’t let poor vertical jump trials slip through the net, and when this and the calculation methods we use for each of our jump assessment protocols are considered you, the user, can have the utmost confidence in the accuracy of the jump height data provided by our system.
References