How to calculate airtime in high diving

In high diving, airtime is everything - almost all of the action happens in the air, and a lot of the non-airborne actions serve to maximize airtime.

In many cases it is useful to know how much time you are spending in the air. Most obviously, it is interesting to know how much extra relative time you get from different diving heights, i.e, if I go from 10 to 15 meters, how many percent does my airtime increase by? This would be useful to know when practicing leadups, or when trying to take a known dive to a higher platform.

I talked in a previous blog post about how you can calculate fall time from various heights, and using that you can determine how much additional actual airtime (in seconds) you get from increasing height, but if you don't know how much time you had already, this information is less useful. So we need a way to calculate the actual airtime of a dive, and since this depends on how high you are (or aren't) jumping, we need a bit more math.

A bit more math

First, let's take a look at the displacement function, which I explained in the aforementioned previous blog post. 

As a brief recap, this function describes the displacement of an object with constant acceleration over time. This means that if you plug in values for the variables a, v_0, and s_0, you can then input a value for time (t), which will result in a value for s, which is the distance travelled by the object by that time. Last time however, we simply set v_0 and s_0 to zero, but when we want to consider jump height, this is no longer the case. So let's understand these two terms a bit better.

The two terms represent initial velocity, and initial displacement. What this means is that if you want to consider an object that already has a velocity or displacement before your calculations, you put it in these values. Imagine if you are considering a ball that is travelling at some speed, say 10 m/s, and then suddenly starts to accelerate by constant acceleration. In this case, you would say that v_0 is 10 m/s. This means that the ball's velocity doesnt increase from 0, but it increases from 10 instead. In the same way, if you consider a ball that has already travelled som distance, say 100 meters, you could say that s_0 is 100 meters.

For our purpose of calculating airtime of a high dive, s_0 can still be set to 0, since we are not considering any initial displacement. 

What is very interesting however is initial velocity, v_0, and this is the key to modeling the total airtime of a high dive.

Consider the takeoff of a high dive. At the very split second that the diver is no longer in contact with the ground, they can no longer apply any additional upwards force, and they are only affected by gravity (and air resistance, but we are neglecting this). However, the diver will not immediately start travelling downwards. For a little while first, they are travelling upwards, then for an infinitesimal amount of time they are neither traveling downwards of upwards, and then afterwards, they start falling.

 The reason for this is obviously the fact that the diver is jumping, so they go up. That's what jumping is. What is less obvious is that the effect of the jump is effectively an initial velocity (v_0) to the fall. Recall that gravity is an acceleration, which is a change in velocity over time. If we say that jumping is adding velocity in the upward direction, we can say that gravity is making that velocity smaller, and then negative over time. But before it can become negative, it must equalize the initial velocity v_0. In this way, if we can find the upwards velocity that the diver has, the instant they leave the platform, we can use this to model the displacement of the diver, using the above equation. So let's talk about how we can find this value. Note that since all takeoffs are different, v_0 will differ between not only divers, but also between dives.

Calculating v_0

I will present two methods of determining the specific initial velocity of a dive, depending on what information you have available. The first method uses the time it takes for your center of mass to return to its initial position after a jump, and the second method uses the maximum displacement of your center of mass during a jump. It's not as complicated as it sounds, don't worry. But first, we should talk about this 'center of mass' thing.

Center of mass
So far we have been talking about falling objects and their velocities as a whole, but if we are not precise, it can get confusing when talking about rotating bodies. Imagine a hammer thrown spinning into the air by the handle. If you looked only at a point on the end of the handle, you would see that it is not simply accelerating downward at a constant rate as i have been implying - due to the spinning motion of the hammer, its velocity will change rapidly in a sinusoidal motion, and may even be changing directions at the top of the throw. If you look closely however, you might be able to spot a point somewhere near or on the head of the hammer that seems to be still, or which the hammer seems to rotate about. This is the hammer's center of mass, and you will observe that this point travels in the predicted manner. Here is a great video demonstrating this principle in practice. In all these calculations about entry speed and airtime, what we are actually calculating is the motions of the center of mass of the diver, which can also be thought of as the average motion of any point on the diver.

It is not so complicated to calculate the centers of mass of simple geometric shapes, but it is more complicated to do it for a complicated shape such as a human being. We are going to need some way of estimating the center of mass of a diver, and to get there, let's look at the formula for calculating center of mass in one dimension. 

Here, x_cm is the coordinate for the center of mass, x_i is the coordinate for the particle with the number i, m_i is the mass of this same particle, and M_total is the total mass of all particles. If you've never seen the big sideways M before, it simply means "sum of" (It's a "Sigma" symbol, short for "Sum"). Let's take an example.

Imagine a string with three pearls on it. One pearl is at the very start, the next is 2 centimeters down, and the last is 6 centimeters from the first. The first pearl weighs 2 grams, the second 4 grams, and the third 1 gram.

To find the center of mass of this string of pearls (assuming the string is massless), we can use the above formula. For the coordinates, we can use any coordinate system as input, and the result will be coordinates in that system, so lets say that the coordinate 0 is by the first pearl, then the coordinates for the second and third pearl become 2, and 6 respectively. In this case, M_total is 7 grams. If we plug these values into the formula above we get:

Which means that the center of mass for this particular string of pearls is 2 centimeters down the string, or at the location of the second pearl. If you threw this string of pearls spinning into the air, you would see it rotate about the second pearl (at least if the string is rigid).

You may start to see why the center of mass of a rotating object is interesting to us. While all points on a non rotating object will follow the same predictable path, only the center of mass of a rotating object will be folowing this path. 

I wrote above that this is a one-dimensional example, which means we are only considering one spatial dimension, or axis. In this case, we are only considering the axis parallel to the string - we know all of our pearls will exists on this string. Fortunately, calculating in higher dimensions is just using the same formula for coordinates in other dimensions.

Speaking of dimensions, let me reiterate that when we are calculating airtimes, we are also only calculating in one dimension, the vertical, or up-down dimension. We are not considering any forward-backward, or sideways motion, even though there will be some in any high dive. This also means that we are only interested in vertical movement of the center of mass of the diver. But to get that, we are going to need to analyze it in higher dimensions. 

Center of mass of a diver

In real life, there are three spacial dimension, not just one like the example with the pearlstring. Your center of mass is somewhere in or around your body right now, and it has three spacial coordinates. As briefly mentioned above, calculating center of mass in three dimensions is not much harder than one dimensions, as long as we know the mass and position of all the points in your body.

That's difficult to know, so we will simplify. If we (figuratively) split the body into relatively large parts, that dont move much relative to themselves (such as a head or a foot), we can use those as our points, simplifying the body into a group of bodyparts, which have individual centers of mass, that can be more easily visually estimated. Luckily for our purposes, there have been done studies on the average relative mass of bodyparts, which we can use for our purposes. The linked article has two datasets; I have used the data from Paolo de Lava, since it was allegedly conducted on athletic subjects. This dataset splits the body into 14 parts; 2 hands, 2 forearms, 2 upper arms, 2 feet, 2 shanks, 2 thighs, 1 torso, and lastly a combined neck and head part. Each are given a value as a percentage of the total mass of the person - for example a thigh of a male is given the value 0.1416, which means the thigh makes up 14.16% of the total mass of the person.

Please note that using the relative masses of bodyparts is the same as using the actual masses, when doing center of mass calculations, because the distribution is the same. It's the same reason solid balls of iron and plastic will also have the same center of mass.

Now we have the masses of our points, next is the position of them. If we do another simplification, which is assuming that any dive will not move significantly sideways (perpendicular to platforms edge), we can neglect this dimension, and work only in two dimensions. This is practical, because, while it may not seem true, pictures and videos are two-dimensional.

We can choose any arbitrary coordinate system, and the calculations above will work out, as long as we keep using the same system for any calculation. However, if we want to know something like actual jump height, we have to be a bit more considerate with our choice. Let's do an example.

Here is a beautiful picture of a beatiful diver, doing what i must presume to be a beautiful dive. Note that the picture is not straight from the side, which is not optimal, but all bodyparts are clearly visible and not overlapping, which makes the example more tangible. I have overlayed him with a two-dimensional coordinate system, with entirely arbitrary numbers. The axes are simply named "X", and "Y". What I can now do is plot all of our established "points", using the chosen coordinate system. I will do this by estimating the centers of mass of the individual bodyparts of the diver. For example, his right thigh would get the coordinates (6.25, 5.70), because I estimate that's where that thigh's center of mass is, the left forearm gets the coordinates (2.60, 7.35), and so forth. Doing this for all body parts, we get a list of coordinate sets, and associating them with their relative masses as mentioned above, we have 14 points with coordiates and masses.

Now, to get the center of mass of all the bodyparts in two dimensions, we simply use the equation that we used above in the pearlstring example, but we do it separately for the X- and the Y-axis. This means that we first take all the X-coordinates of all the points, for example 6.25 for the right thigh, and we plug them into the formula along with the masses, like we did with the pearlstring. This gives us the X-coordinate of the center of mass. Afterwards, we plug in all the Y-coordinates, along with the masses, and this gives us the Y-coordinate of the center of mass. Using this method, the X-coordinate of the center of mass becomes 5.39, and the Y-coordinate becomes 6.49.

Here I have plotted all 14 bodyparts as blue circles, with the size of the circle corresponding to the relative mass of the bodypart. The center of mass is denoted by the black X. 

Since we are only considering two dimensions, we can only determine two coordinates in space. Imagine a line going straight through the paper right at the X - we know the center of mass lies on this line. Likely a bit behind the back of the diver.

This illustrates the trouble with only using two dimensions. If the image is not straight from the side, or the diver is twisting, it becomes a bit unwieldy. For the best results, the picture or video used should be as much from the side as possible, and if possible from a distance with a good zoom. 

Calculating v_0 from jump airtime

Now that we have a method for estimating center of mass, let's get back to calculating the initial velocity, v_0, of a dive. As i mentioned, i will present two methods to find this. The first one uses the time it takes for the center of gravity of a diver to return to its initial position. This corresponds to the airtime of a jump on the ground, assuming your center of mass is in the same place when you jump and when you land. Let's use this dive as an example.

I have taken two screenshots of the dive. One at the instant the diver takes off, and one at the point where (by my best estimates) their center of mass has returned to the same vertical position as in the takeoff. Then i put them into the same coordinate system, and calculated the centers of mass of both.

Notice that we are still using arbitrary numbers for the coordinate system (instead of something like meters or feet), because we only need a measure of time, not distance. This is the advantage of this method.

The Y-coordinates of the centers of mass in these two pictures are 5.495 and 5.500 respectively, which is as close as I'm gonna get. I have tried to place the centers of mass of the body parts as precisely as I could, but any change in them would affect the total center of mass. This means I might be off by some frames in the video, but we'll assume it's at least representative.

Using the "Stats for nerds" feature of the YouTube video, we can see that these two frames have the timestamps 2646.33, and 2646.60, which is the amount of seconds after which they appear in the video. This means that, assuming this to be correct timekeeping, they occur 0.37 seconds apart, which is the total airtime of the jump.

Now we can finally do some math.

Let's go back to this equation, which we saw at the start. As mentioned, it describes the displacement over time (t) of an object being affected by a constant acceleration a. The term s_0 denotes initial displacement, which we will not be using, so we set it to 0. The acceleration that the diver will be experiencing while in free fall will be gravity, so we set a to the gravitational acceleration g (9.80665 m/s²). We actually set it to negative g, which sets the down direction to be negative. If we for example get a result of negative 20 meters, that means a displacement (a fall) of 20 meters downwards.

Note that during the takeoff, while the diver is still in contact with the platform, the are experiencing other accelerations than gravity, which is why we only consider the dive from the moment the diver leaves the platform. 

Substituting in the values as explained above we now have a slightly simpler equation:

Now please take a minute to make sure you understand what this equation is saying. What it says is that if you know the initial velocity (v_0) of some object, you can input a time t, and you will get a result that is the displacement, s, of that object after the time t has passed.

If you feel like this makes at least some sense, let's move on.

Let's think about what we know about the dive that we just looked at. We know that the diver must have some initial velocity - she is moving upwards at the start. We also know that after 0.37 seconds, her center of mass has returned to its initial vertical position. This means that her displacement after 0.37 seconds is 0 meters. We can write this fact like this:

This is the same equation as above, but with value substituted in for t, and s. We know from observation that this equation must be true, which means if we can find a v_0 that satisfies this equation, that must be the diver's initial velocity. With a bit of algebra, we can isolate v_0, giving us the result:

Which means that the initial velocity of the diver is 1.81 meters per second (or 6.53 km/h). In general, to find v_0 from the airtime of a jump (the time it takes for the center of mass to return to the position it had at the takeoff) is:

Where t_air is the airtime in seconds.

This method of calculating v_0 can be tedious, if done the way i did here. A much easier way to get a value for your own airtime, and thus initial velocity, is to record yourself doing a flip on the ground, and measuring the time from takeoff to contact. If your flip is good enough, your center of mass should be around the same place when you land as when you take off (read: you land upright), which makes the landing time the time your center of mass returns to its original position. You can then assume that this flip is representative of your takeoffs in your dives, and use that value for v_0. 

A variant of this method is to use the full time from takeoff to entry - this works if you know the total height of the platform. For example, if we assume the platform in the dive we just worked on is 21 meters, you can input that into the displacement equation instead of 0 meters (remember to put negative 21 if g is negative). The entry in the dive is at approximately the timestamp 2648.61, which means the total airtime of the dive is 2.28 seconds. This gives a v_0 of 1.97 m/s. This method however is pretty sensitive to the height of the platform - if we assume the platform to be 21.5 meters instead, the initial velocity becomes 1.75 m/s. To use this method, you can use this formula:

Where t_air is now the total airtime, from takeoff to landing, and h_platform is the height of the platform. You can see if you can derive this formula from the displacement formula above, as an exercise.

Before we talk about what to now do with our newfound initial velocity, let's go over the other method of determining v_0.

Calculating v_0 from jump height

As advertised earlier, the second method of calculating initial velocity is using the maximum displacement of the center of mass of the diver - the maximum height achieved in the jump. We can use the same dive that we just looked at as an example to compare, but now there is a major difference. Since we want to know the exact height the center of mass moves, we can no longer use arbitrary units like we just did, we have to get some useful numbers. Knowing the height of the diver would give us a reference measure, which we can use to determine the dimensions of the footage. If we take the frame where the diver is simply standing, and assume that her height is 1.7 meters while standing on her toes, we can use that to generate a coordinate system of real life units.

Here, the red line is showing the height of the diver, and as you may be able to tell, it corresponds to 1.7 meters on the y axis. I have set the 0 on the y-axis to the base of where the diver is standing, but this would work with any placement of coordinates, as long as the distances are correct - we are only interested in the difference between two centers of mass, and that will not depend on where we place 0.

Please note that I have no idea of the diver's actual height, and 1.7 meters on toes is simply a guess. This does mean that the measurements will be less exact than they could be, but they will serve as an example. If you want to be precise in your own measurements, you can find an object such as a ruler or a plyo box that has a precise dimension, and place that in the shot, at the same depth (z-axis dimension) as you.

Now that we have a coordinate system, we can place centers of mass. After some experimentation, I found a frame that I believe to be the maximum displacement of center of mass for this dive. Comparing this to the initital center of mass, we can get the maximum displacement.

On the left is the same standing frame i used before, and to the right is the aforementioned peak center of mass displacement. As before, the black X'es denote the centers of mass. Since we are still only considering the displacement equation in the vertical direction, we only care about the vertical displacement of the center of mass. In this case, the center of mass have the y-coordinates 0.996 m, and 1.083 m, which means the maximum jump height of this dive is the difference between these, which is 0.087 meters, or nearly 9 centimeters. This seems low to me, which might suggest that the coordinate system isn't well defined, but conversely, the dive is an inward dive, which is arguably the hardest direction to gain height on (barring handstands). It might be pretty standard for an inward dive, I have not tested any others.

Now that we have this, we can again do math. As you may or may not have noticed, the displacement equation is actually a second order polynomial of this form:

Many of you will have worked with this in school, but don't worry if you forgot, I'll remind you of the necessary details. In the case of our displacement function, the variable is t instead of x, y is s, a is a/2 (please note that these two a's represent different things, and are just coincidentally the same letter), and b is v_0. 

This is good, because we can use the math relevant for second order polynomials. Specifically, the term for calculating the toppoint (or vertex) of a parabola, which is the graph drawn by a second order polynomial. Consider this graph of the top part of that same inward dive:

This visualizes some things that we already know; that the displacement over time of the diver can be described by a parabola, and that the vertex of this parabola is 8.7 centimeters (above where it was at t = 0). We also know that the a-value of this second order polynomial is -g/2, and that the c-value is 0 (this corresponds to s_0, which we set to 0). 

We are missing two crucial pieces of information; the b-value, which as mentioned is our v_0, and the time at which the toppoint of the jump is achieved (the x-coordinate of the vertex (x_v)). 

For the latter though, we are in luck. You might remember from math classes that there is a handy formula for finding the x-coordinate of the vertex of a parabola, in terms of the a and b values of the function.  It looks like this:

This might not seem helpful at first glance, since we still don't know the value for b. The reason it is helpful however, is that it allows us to write x_v in terms of b, which means that we can write an equation with only one unknown quantity, b. Remember, we know the y-coordinate of the vertex, 8.7 centimeters. This value is the result of inputting x_v into the second degree polynomial that we are looking for, which means that the following equation must hold:

Notice the presence of the x_v in this equation. Now, if we substitute the formula for x_v into this formula, we can do some algebra with it to make it simpler. 

See if you can follow the math on that one, it's good practice.

Now finally, let's plug in the values for a, b and c, which are -g/2, v_0, and 0, respectively (remember, this is still the displacement equation, the a, b, and c were just placeholders). I will also subtract 8.7 centimeters on both sides, such that one side of the equation becomes 0. The reason for this will become clear soon, maybe you can see it from the equation:

You can try to verify for yourself that this equation is the same as the one above, with the mentioned values plugged in.

Now please, try to follow this carefully; we started with the displacement equation, where the independent variable was t. We then turned to using x as the independent variable instead, for the only reason that it made the formulas seem more familiar. We then also started using the a-b-c values instead of what we were using previously, again for the same purpose. 

At that point we had the general formula for the parabola, with one unknown quantity, b (v_0). Then we input a known solution to this equation, 8.7 cm, and its x-value, in terms of b and a, which left us with a specific equation with one unknown, b.

Then we did some algebra. 

Then we undid what we did in the start, by replacing a, b, and c, with their original values, which leaves us with the exact same specific equation, but this time the unknown is correctly noted as v_0.

Now here is the kicker: This is now a different second order polynomial. Before, we had a formula where the independent variable (x in the general formula) was t. Then we input a known value for t (x_v), in terms of an unknown value, b, or v_0. 

The result of this is that the independent variable of this new second order polynomial is v_0, and the a, b, and c values are 1/2g, 0, and -8.7 cm respectively.

Most importantly, since this is a second order polynomial of the form ax^2+bx+c=0, we can solve it for x, which is exactly what we want. And the way to do it is revisiting an old friend, the quadratic formula:

Here you thought you would never need this formula in real life. Now you know, at least as a diver, you can not live without it.

As you may remember, this formula has two solutions, due to the plus-minus sign in the top of the fraction - it means that you do the calculation twice, once with plus, and once with minus. For a second order polynomial with a b value of 0, which is the case with this new polynomial, the two solutions will be each others negatives (for example 1 and -1). Since we have defined gravity as acting in the negative direction, we know that v_0 will be a positive number, so we can choose the positive solution. Plugging in the a, b, and c values for this new polynomial, and choosing the positive solution, we get the result:

This number is lower than what we got using the other method. I am not confident in saying which of the results is more precise, since I have made a bunch of assumptions in both. We will talk in a bit how much precision in this value changes the end result of calculating airtime of a high dive.

This may have seemed a tedious way to calculate v_0, but this is mostly because have tried going slowly, and explaining how to get to the result. If you know your maximum jump height, you can simply put it into this formula to calculate v_0:

Where h_jump is the maximum jump height. 

This is actually just the quadratic formula from above, but with 1/2g, and -h_jump replacing a and c. See if you can make this derivation yourself- it requires a bit of algebra with exponents, which is good exercise. Note that this is only the positive solution of the quadratic formula.

Calculating the airtime of a high dive

Now that we have a way of finding a value for the initial velocity of a dive, we can finally use this to calculate the total airtime of a dive of any height. What we are going to do is to take that initial velocity v_0, and put it into the displacement equation, and plot it over time. In fact, let's plot all the three values for v_0 that we found. This will show the paths of dives over time, with those initial velocities, and how they differ. I will also plot the path of a simple falling object to compare them to (corresponding to v_0 = 0)

Here we can compare the three different resuts we got for the dive we worked with, along with a simple falling object. As you can see from the zoomed in region, the lines with the higher initial velocity has a higher toppoint, and the one without initial velocity (black), never moves upwards at all. By the time t = 2.28 seconds, which was the time of the dive we recorded from the video, the lines have reached a displacement of -25.49 m, -22.50 m, -21.36 m, and -21.00 m, in order of ascending initial velocity. Note here that the blue line, with initial velocity 1.97, reaches exactly -21 meters, because we defined the initial velocity of this scenario by that fact that it reaches 21 meters after 2.28 seconds.

These results are interesting on their own - depending on which measurement your trust most (jump height, jump airtime, or total airtime+platform height), the resulting displacement differs a bit. For example, it can be argued that the platform is a bit higher than 21 meters, if you trust the jump height measurement and the timekeeping of the YouTube video.

In any case, let's continue with what I think is the most useful calculation to do with this - predicting airtimes for a known takeoff, at a new platform height.

Using v_0 to predict airtimes of dives.

Finally, this is what I think is the primary reason for trying to figure out your own initial jump velocity. Imagine you are learning high diving, and you have learned a double half from 10 meters. Where you train, the only available increment in height is to go to 15 meters, but you don't know how much less power you should give it, how much more time you will gain. If you know the movement equations, you can actually pretty precisely calculate the extra time in seconds, without knowing your v_0. It becomes more precise if you assume some maximum jumping height also. The formula to use is this:

Where t_extra is the extra time you gain in seconds, h_cur is the current height, or the height you are comparing with, and h_des is the destination height. This formula also works if the destination height is lower - it will give a negative answer corresponding to a loss of time. 

As an example, if you go from 10 meters to 15 meters, and you assume you are jumping 15 centimeters up, h_cur becomes 10.15 meters, and h_des becomes 15.15 meters, leading to a t_extra of 0.319 seconds, which means that's the amount of extra time you will have in the air.

Now, the natural question is "How much did I have already?", or more precisely "How much is this extra time do i have, compared to what i already had". The extra time is much more useful if we can compare it to how much time we had from the start. Sure, we can estimate by trying to count while diving, but that will not be so precise. Timing the dive is a good idea. But we can also calculate it, and that's exactly what we are going to do.

With our newfound knowledge of initial velocity, what we can do is isolate the time for a given height in the displacement equation. For this example let's say that the 15 centimeters jump height is precise - using the formulas we found earlier, we can see this leads to an initial velocity of v_0 = 1.72 m/s. Starting with the 10 meter dive, we know that after we jump, at some time after the takeoff, we will have fallen 10 meters from where we started. If we call this time t_10m, we can set this up as an equation, with t_10m as the only unknown:

As we know, this is a second order polynomial. If we add 10m on both sides, we get one side equal to zero, and that means we can solve the equation for t_10m. We can once again use the quadratic equation:

Which means that the total airtime on the 10 meter dive is 1.614 seconds (this is again the positive solution to the quadratic equation). Now, if we do the exact same calculation, but for 15 meters, we find that the total airtime from 15 meters is 1.933 seconds, and sure enough, the difference between these two times is the 0.319 seconds that we found earlier. Now we can find the relative increase in time using this formula:

Where t_des is the airtime from the destination height (in this case 1.933 seconds), t_cur is the airtime from the starting (or current) height (1.614 seconds), and Delta_time is the increase (or decrease) in height, in percent (the triangle is an upper case Delta, which is commonly used to denote change). 

If we plug in the values for t_cur, and t_des, we get the result 0.198, which means that the airtime will increase by 19.8% from 10 meters to 15 meters. 

In my opinion, this information can be useful when learning new heights. To know that the airtime is 20% more is in my opinion easier to understand than some flat seconds value. Its probably best knowing both.

Another case where this can be useful is when doing lead ups - if you know you are going to do 50% more somersaults, perhaps it is good knowing if you have 20% or 50% more time. Using this method, you can get a good estimate. This method is especially useful when considering different takeoffs, since some takeoffs are harder to gain airtime on. In my personal experience, I can jump the highest on back takeoffs, and the lowest on inwards takeoffs. This means that for leadsups of these takeoffs, it would be useful to calculate separate initial velocities to get more precise predictions.

To finish off, here is a final formula which calculates the relative time difference between two platforms, only in terms of v_0 and the heights of the two platforms. It's simply a combination of the airtime formulas, and the above Delta_time formula.

This can perhaps be simplified further, but unfortunately i'm at the limits of my square root algebra. It does however become a bit simpler if you notice the the terms in the square roots are simply the discriminants of the second order polynomials describing a dive for a given height. If you remember, the discriminant for a second order polynomial is:

Which means the above equation can be written much neater as:

Which is for sure neater, but you have to remember what the d is for.

Error margins for v_0

As you may have noticed in the course of this discusison, the values for v_0 I have found are estimates, and they sometimes vary by method. In my opinion, if you want to be really precise in your measurement of your own v_0, you should make several measurements, and take an average. In the end, a somewhat precise result should give you useful info - my primary goal with this post is to argue that you need to include the v_0 in your calculations. In the following graph, I have compared the additional time you gain in percent, based on several values for initial velocity, all compared to 10 meter airtime.

The way to read this graph is by assuming you have just done a dive on 10 meter, and you want to know the extra time you get if you go to 10 or 20 meters. If you know your v_0, you can find your extra relative airtime by going straight up to the line of that v_0. For example, if your v_0 is 1.31 (red line), you gain about 20.5% extra airtime by going to 15 meters, and just shy of 38% by going to 20 meters.

The red, green, and blue lines are the values for v_0 we found for the inward dive earlier, and the orange line is an average of the three. What I have tried to show here is that while your result will vary with your v_0, they will probably be more realistic than not using v_0, which is what the black line is showing.

Note on handstands

For handstand dives, there are some complications in these calculations. First off, for all calculations, we have been calculating the time it takes for the center of mass of a diver to travel a given length. This makes sense for standing dives, since the takeoff and landings occur in roughly the same position (standing position), which means the center of mass will have approximately the same position in the body. However, this cannot be said for handstands, since the initial position is upside down.

Secondly, some handstand dives, like this one, actually has a negative initial velocity at takeoff - when the hands leave the platform, the diver is already moving downwards. Remember, we are only doing calculations for divers in freefall, since the forces involved in the actual takeoff are more complicated.

In general, handstand dives don't have very high initial velocities, since it is much harder to jump from your hands than from your feet. This is an example of an excellent handstand takeoff, and even that barely goes upwards from takeoff. In my opinion you can estimate the initial velocity to be 0 for most handstand dives, at the time of takeoff.

Note on air resistance

Those who read my post about entry velocity will know that considering air resistance in those calculations made a non-insignificant difference in entry velocity of 2.7% at 27 meters (the higher height, the larger difference). In this case, there is a similar discrepancy. But since introducing initial velocity into the equations that consider air resistance makes them considerably more cumbersome, I have not done so in this post (it is already too long anyways). We can estimate the error by calculating the amount of extra time (which is not dependend on initial velocity) using air resistance, and comparing it to the total time of a dive, without air resistance. For example, if we start by calculating how much extra airtime we would gain by going form 10 meters to 27 meters with an initial velocity of 1.5 m/s, we get that the extra airtime is 0.915 seconds. Since the total airtime from 10 was 1.589 seconds, this is an increase in airtime of 57.6%. Comparatively, if we calculate the extra airtime gained from falling from 27, compared to 10 meters, while considering air resistance, we get learn that the extra time is 0.938 seconds (note that in these calculations i added 0.115 meters to the heights, since that is the jump height at v_0 = 15 m/s). If we compare this value to the total airtime from 10, we get that the increase is 59.0%. The difference in result is 2.47%. However, the actual difference should be smaller, since we are comparing both extra times to the total time without air resistance. The actual total time with air resistance should be slightly larger, which means that the relative increase in time would be closer to the result from the non-air resistance calculations.

Conclusion

We can now thoroughly examine the airtime of our dives. While much of the usefulness of these calculations can be accomplished simply by timing your dives on different heights, I think the math and physics behind it all are interesting. I think if you can grasp what's in this article, you should be able to understand a bit more about your diving, and perhaps why some dives feel more different when going up than others.