I am a platform diver. I am trying to figure out how deep in the water I go depending on which platform I dive from.

The platforms are 1m, 3m, 5m, 7.5m and 10m above the water surface.

By ignoring air resistance (we assume that I do not do any tricks) I can easily compute the energy that my 90 kg body has when it is hitting the water:

E = mgh = 0.5mv^2

My problem is to find the deceleration in water and thus how deep my feet will go into the water before coming to a halt. We can assume that I stand straight and fall straight down and have a height of 190cm.

How do I compute that?

  • $\begingroup$ I was interested to compute what the effect of the "transition epoch" (when your body enters the water and the buoyancy is increasing as you slide in). I have done it: quite interesting results in my updated answer. $\endgroup$ Nov 12 '14 at 11:20
  • $\begingroup$ I now agree wholly with tpg2114's most remarkable and surprising answer as a valid approximation to your situation. $\endgroup$ Nov 14 '14 at 2:17

The forces slowing you are (1) drag as you note and (2) buoyancy. The former, assuming ram drag is the main one, is given by:

$$F_D = -\frac{1}{2}\,A\,\rho_W\,C_D\,v^2$$

where $\rho_W$ is the density of water, $v$ the velocity of the dragged object, $A$ the cross-sectional area presented to the water as you fall and $C_D$ is a fudge factor called the drag co-efficient. $C_D$ is highly dependent on the object's shape and orientation relative to its velocity through the water. To understand more about ram pressure, see my answer here. So you will need to "calibrate" $C_D$ with an observed depth.

The buoyancy force is the weight of the water you displace. So if your density is $\rho_B$ and your mass $m$, then the buoyancy force is $-\frac{\rho_W}{\rho_B}\,m\,g$ (downwards positive).

At last we have your weight $+m\,g$. Therefore, Newton's second law becomes the following differential equation for velocity $v(t)$ (downwards direction positive)

$$m\,\mathrm{d}_t\,v(t) = m\,g\,\left(1-\frac{\rho_W}{\rho_B}\right) - \frac{1}{2}\,A\,\rho_W\,C_D\,v(t)^2$$

we convert this to an equation for velocity $v$ as a function of depth penetrated $y$ by the identity $\mathrm{d}_t\,v = v\,\mathrm{d}_y\,v = \frac{1}{2}\,\mathrm{d}_y\,v^2$, so we are left with:

$$\mathrm{d}_y\,v^2 = 2\,g\,\left(1-\frac{\rho_W}{\rho_B}\right) - \frac{\rho_W}{m}\,A\,C_D\,v^2=- \frac{\rho_W}{m}\,C_D\,A\,\left(v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right)$$


$$\log\left(v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right) = - \frac{\rho_W}{m}\,C_D\,A\,y + C_I$$

where $C_I$ is an integration constant we must now find. As you know, we have $\frac{1}{2} v(0)^2 = g\,h$, where $v(0)$ is your velocity as you hit the water and $h$ the distance you dive from. If in the above equation we measure $y$ downwards from the water's surface, we have:

$$\log\left(2\,g\,h+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right) = C_I$$

and so we can now work the integration constant out to find:

$$\log\left(\frac{v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}{2\,g\,h+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right) = - \frac{\rho_W}{m}\,C_D\,A\,y$$

and then find the $y$ that makes $v=0$. So at last we have the description of your penetration depth $d$; it is:


Notice how this quantity is negative if your density is greater than that of the water, describing a situation where there were water above the actual surface. This means, of course, that you keep on sinking if you're not buoyant enough.

For very small drags ($C_D\,A\to0$), the above equation becomes:


but I'm almost certain that this will greatly overestimate your penetration depth: it says that your penetration depth will be much deeper than the dive tower is tall.

So you need at least one $d$ observation to work out the value of the unknown $C_D\,A$ - the "fudge factored" effective cross sectional area you present to the water. $C_D$ values for long thin objects are typically about 1. If your cross sectional area (cut through the anatomist's transverse plane) is $0.5\times 0.3=0.15{\rm m^2}$, your mass $90{\rm kg}$, your density with your breath drawn in is $950{\rm kg\,m^{-3}}$ and your drag co-efficient is $1$, then we get, for $d$ and $h$ measured in metres:


yielding $d=2.09{\rm m}$ for $h=1{\rm m}$, $d=2.74{\rm m}$ for $h=3{\rm m}$, $d=3.04{\rm m}$ for $h=5{\rm m}$, $d=3.28{\rm m}$ for $h=7.5{\rm m}$ and $d=3.46{\rm m}$ for $h=10{\rm m}$. These don't seem far off what one observes. These will be underestimates because I didn't correctly describe the "transition epoch" where your body is only partly steeped in the water, and therefore the buoyancy in particular is overestimated.

Moreover, surprisingly, these estimates are not far off tpg2114's answer. Certainly, $d$ is a very weak function of $h$ once $h$ rises above $1{\rm m}$, in keeping with the other answer.

Update: Accounting for the "Transition Epoch"

If we account for the stage where the body is entering the water and model the variable buoyancy as being proportional to the length of body steeped in the water, our basic differential equation becomes:

$$\mathrm{d}_y\,v^2 = 2\,g\,\left(1-\frac{\rho_W\,y}{\rho_B\,L}\right) - \frac{\rho_W}{m}\,A\,C_D\,v^2$$

whose solution (subject to the initial value $v(0)^2 = 2\,g\,h$) is:

$$v(y)^2 = \frac{2\, m\,g\, (A \,C_D\, (L\, \rho_B-\rho_W\, y)+m)}{A^2 \,C_D^2\, L\, \rho_B \,\rho_W}-\frac{2\, g\, \left(-A^2\, C_D^2\, h\, L \,\rho_B\, \rho_W+A\, C_D\, L\, m\, \rho_B+m^2\right)}{A^2 \,C_D^2\, L \,\rho_B\, \rho_W}\,\exp\left(-\frac{A \,C_D \,\rho_W\, y}{m}\right)$$

and when the body is fully steeped ($y=L$) the squared velocity is:

$$v(L)^2 = \frac{2\, g\,\left(m\, (A\, C_D\, L\, (\rho_B-\rho_W)+m)-\left(A\, C_D \,L \,\rho_B \,(m-A\, C_D\, h \,\rho_W)+m^2\right)\exp\left(-\frac{A\, C_D\, L \,\rho_W}{m}\right)\right)}{A^2\, C_D^2\, L \,\rho_B \,\rho_W}$$

so the depth of penetration beyond the body's length is given by the equation:


where now the quantity $h_{eff}$ is given by:

$$h_{eff}=\frac{m\, (A\, C_D\, L\, (\rho_B-\rho_W)+m)-\left(A\, C_D \,L \,\rho_B \,(m-A\, C_D\, h \,\rho_W)+m^2\right)\exp\left(-\frac{A\, C_D\, L \,\rho_W}{m}\right)}{A^2\, C_D^2\, L \,\rho_B \,\rho_W}$$

So now we calculate $h_{eff}$ for the data above ($A=0.15{\rm m^2}$, $m=90{\rm kg}$, $\rho_B=950{\rm kg\,m^{-3}}$, $C_D=1$ and assuming $L=1.9{\rm m}$) with the diving heights of 1, 3, 5, 7.5 and 10 metres:

$$\begin{array}{ll}h=1{\rm m}&h_{eff} = 0.176319{\rm m}\\h=3{\rm m}&h_{eff} = 0.260607{\rm m}\\h=5{\rm m}&h_{eff} = 0.344895{\rm m}\\h=7.5{\rm m}&h_{eff} = 0.450254{\rm m}\\h=10{\rm m}&h_{eff} = 0.555614{\rm m}\end{array}$$

and so, when we put these values into $d=\frac{m}{\rho_W\,C_D\,A}\,\log\left(1+\frac{h_{eff}\,C_D\,A}{m\,\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right)$ we get:

$$\begin{array}{ll}h=1{\rm m}&d = 1.13073{\rm m}\\h=3{\rm m}&d = 1.33494{\rm m}\\h=5{\rm m}&d = 1.48701{\rm m}\\h=7.5{\rm m}&d = 1.63506{\rm m}\\h=10{\rm m}&d = 1.75372{\rm m}\end{array}$$

giving the total depths of penetration of your feet (the above values plus $1.9{\rm m})$:

$$\begin{array}{ll}h=1{\rm m}&d = 3.03{\rm m}\\h=3{\rm m}&d = 3.23{\rm m}\\h=5{\rm m}&d = 3.39{\rm m}\\h=7.5{\rm m}&d = 3.54{\rm m}\\h=10{\rm m}&d = 3.65{\rm m}\end{array}$$

as you can see, a reasonable accounting for the transition epoch adds quite a bit of depth for shallow dives (a whole metre for a 1m dive) but only 20cm for the 10m dive.

On entering the water, the acceleration throughout the transition epoch is:

$$a(y)=e^{-\frac{A\, C_D\, \rho_W\, y}{m}} \left(-\frac{A\, C_D \, g\, h \rho_W}{m}+\frac{g\, m}{A\, C_D \, L\, \rho_B}+g\right)-\frac{g\, m}{A\, C_D\, L\, \rho_B}$$

which is maximum at $y=0$ and given by:

$$g-\frac{A\, C_D\, g\, h\, \rho_W}{m}$$

working out to be about $-15\,g$ for a 10m dive, but only $-0.8\,g$ for the 1m dive.

  • $\begingroup$ The convergence of the penetration lengths to the estimate in my answer as the height increases makes sense -- the approach Newton described becomes more valid as the entry speed becomes faster. At low entry speeds, it is not very valid (imagine dropping a bullet from 3 feet in the air -- it's not going to penetrate very far). I wonder if your expression asymptotes to the $d = L*\rho_1/\rho_2$ value as $ h \rightarrow \infty$. $\endgroup$
    – tpg2114
    Nov 12 '14 at 17:40
  • $\begingroup$ Hey -- if you want to get some more mileage out of your calculations here, a similar question could use your expert analysis: physics.stackexchange.com/questions/167077/freefall-into-snow $\endgroup$
    – tpg2114
    Feb 25 '15 at 23:50

There are bound to be more detailed answers and analysis, but let's look at one of the oldest possible results in the field. Isaac Newton proposed that, for bodies of equal density, the penetration depth during ballistic impact is equal to the length of the penetrating body. Note -- this is independent of the speed (which means independent of the height) of the diver.

Regular water has a density of $1000 kg/m^3$ and the human body is slightly less dense at $985 kg/m^3$. If we assume that this is approximately valid still and we assume the diver is 6 feet tall and extends his or her arms above her head during the dive (and that arms are, roughly, 3 feet long) giving the diver a body length of 9 feet, the penetration depth would be:

$$ d = 9 * 985/1000 $$

which gives an answer of 8.89 feet.

Now, the part that might be confusing -- this is true regardless of the height from which the diver dove. This is counter-intuitive, but ultimately it means that a lower-height dive causes less acceleration upon entry into the water than a higher dive does. So a diver from 1 meter would experience far less g-forces upon entry compared to a 10 meter diver, but they would both go down to approximately the same depth.

  • $\begingroup$ Really interesting historical data: +1. You're exactly right about the accelerations. This is pretty much what one finds with a more detailed analysis - see my answer. I get of the order of $15\,g$ upwards for the 10m dive, but only $0.8\,g$ for the 1m dive. $\endgroup$ Nov 12 '14 at 12:18
  • $\begingroup$ @WetSavannaAnimalakaRodVance It's turns out Newton had some pretty awesome ideas that just took a long time to realize they worked. His ideas about the drag on a body in a flow turned out to be spot on for hypersonic bodies, but didn't do so well on bodies at regular speeds. $\endgroup$
    – tpg2114
    Nov 12 '14 at 17:37
  • $\begingroup$ What I found most interesting about your answer is that I found it hard to believe at first. And, even more interesting, each time I refined my calculations to account for more effects, the closer my results came to the behaviour you describe. I guess that the ram pressure sets up a stronger feedback loop than I would have guessed at first. And the rough momentum transfer argument in the Wiki article makes a great deal of sense. $\endgroup$ Nov 13 '14 at 9:05

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