# Diving into swimming pool

Hi guys,
I have a swimming pool which is 2m deep and was wandering whether I will be safe if I dive from 2nd floor which is around 4m high. [I'm 175cm tall and weigh 65kg]

Working out gravitational energy = mg = (65)*10 = 650
Equating to kinetic energy, (1/2)*(65)*v^2 = 650
v=4.47m/s.

I will be rounding my self like a ball so that my bottom hits the surface of the pool first with most impact creating resistance.

How am I going to get approx figures for drag coefficients and buoyancy and what are the such formulas for working out depth with them?

Help will be appreciated as I'm gonna be jumping off in the weekend MatRobert,

Please be careful. Too many people end up paralyzed or dead by miscalculating dives. In general, I believe that jumping is safer than diving.

you should definitely not attempt to dive into that. It's not even a question of physics, it is a question of common sense. Jumping into it legs first is also kind of pushing it.

you will not find a way to calculate this, there are too many variables that you cannot account for, and besides, uncontrollable things such as your angle at which you strike the water etc. is a major part of the equations.

Danger
Gold Member
Yeah, dude... no math required; it's a dumb idea. It might possibly be safe in 'cannonball' mode, since your water resistance would minimize the penetration depth, but certainly not worth the risk.

Working out gravitational energy = mg = (65)*10 = 650
Equating to kinetic energy, (1/2)*(65)*v^2 = 650
v=4.47m/s.

if you add in air resistance.. that velocity just might drop a little.
What you need to do now is calculate the rate at which you displace the volume in water. This might be a little complicated as when you'll dive, you'll change your pose very quickly just as you enter the surface of the water and this rate of change will be very complex.

Once you are in the water, any position change will not result in a change in the force upon you. It will just differ in the force applied to your different body parts. Since we're talking about safety here, i can say that you should see to it that minimum force is applied on the chest.. to protect your lungs and heart.

Let's say.. you completely enter water without changing your position. If you jump in water something like this:

http://img156.imageshack.us/img156/4521/divingwithlennondb5.jpg [Broken]

Sorry for adding that image.. that's all i could find. Plus.. i did change Yoko Ono to Jessica Alba [ $\textrm{Jessica~Alba} \gg \textrm{Yoko~Ono}$ ].

You are basically a standard ellipsoid where the x-y plane equation is given by:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

You can use the image to find out the approximate eccentricity [I got $\varepsilon$ = 0.896]. Then take a measurement of your height from the head to the hip. This will give you half the circumference of your 'ellipse'. Then using Ramanujan's approximation:

$$C \approx \pi a \left[ 3 (1+\sqrt{1-\varepsilon^2}) - \sqrt{(3+ \sqrt{1-\varepsilon^2})(1+3 \sqrt{1-\varepsilon^2})} \right]$$

you can find out 'a' and 'b' for your 'ellipse'

The forces acting are: i] The force due to gravity, ii] The buoyant force, iii] The drag force [fluid resistance].

The force due to gravity will always be constant. The buoyant force will be given by the amount of water displaced. In the image, i have taken an instant where the body is immersed such that the point just above the surface of the water is 'x' distance from the center of the ellipse.

The volume of that part of the ellipse is given by:

$$V = \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right)$$

I found out this using integration. I'm not doing the proof here now.. but it's easy to find out.

Now, the buoyant force will be given by the weight of fluid displaced:

$$F_b = - \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right) \rho_w$$

where $\rho_w$ is the density of water.

Now, the drag force is given by:

$$F_d = - \sigma v = - \sigma \frac{dx}{dt}$$

In my sign convention, all upward forces are negative hence i've taken negative for both $F_d$ and $F_b$.

Let the acceleration of the body be 'a'. Then,

$$F_d + F_b + F_g = ma$$

giving us:

$$- \sigma \frac{dx}{dt} - \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right) \rho_w + mg = m \frac{d^2x}{dt^2}$$

In this equation '$\sigma$' is a property of the object that travels through a fluid. You can approximate $\sigma = 6 \pi \eta rv$.. but that holds good only for spheres. Try to find the $\sigma$ for ellipsoids.

Solving this differential equation (which i'm not able to) will give you $x(t)$. Once you have $x(t)$, you can use that and it's derivatives to find the impulse imparted [i'm guessing that even that is gonna be time dependent]. This will help you find the pressure that applies on your body as a function of time. You'll need the area of the ellipsoid to calculate the pressure. You can find that at wikipedia. Do remember that the area exposed to the water will be different at different time intervals.

'Safety data' for your muscles and bones is provided in terms of the yield strength i.e. max. pressure they can tolerate. Once you have the data, you can use this data to find out the safety parameters.

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EDIT:

did u say the 2nd floor? Seriously man.. don't even think about it. And if you dive in an ellipsoid like that.. it's certain death. That big an area will result in a huge rate of change of volume which means a huge impulse imparted due to buoyancy. This will act on your skin and will have a 'tearing' effect on your skin.

Also, blood vessels in your body will damage due to sudden change in pressure and can cause clots and internal bleeding. Internal bleeding is a very risky thing to happen.

If you let out a scream or something while jumping off.. your lungs will be in a deflated position... and when the huge pressure hits you.. it'll cause your lungs to collapse under. Also, owing to that.. if you struggle for a lot of air in midwater.. you might as well drown your lungs in fluid.

Seriously man.. don't do it.

P.S: If you jump toes in first or hands in first, you can take your shape approximation as a rod rather than an ellipsoid. That should result in a simple differential equation.

russ_watters
Mentor
Standard olympic sized pools are also 2m deep and they dive off blocks that are under a meter off the water. Do the math yourself...

Integral
Staff Emeritus
Gold Member
Standard olympic sized pools are also 2m deep and they dive off blocks that are under a meter off the water. Do the math yourself...

What do you mean here Russ? Dive used off of the starting blocks is NOT a dive for depth, in fact just the opposite, it is a shallow dive with the purpose to provide as much forward momentum as possible.

A cannonball dive from 4m is not comparable in any way.

My experience says you will be fine as long as you cannon ball. Though you do NOT want to hit on your back (it will hurt) hit on your folded arms then straighten out as soon as you are in the water.

In my foolish youth I jumped from rocks, 4-7m into water of uncertain depth. Rivers in the PNW are green water so you can see if there is anything in the way. 2 m is fine for a 4m cannonball, do not attempt a straight clean dive into the same water from that height.

rcgldr
Homework Helper
If you had prior experience, you could go in feet first and transition into a pike position to create a curved water path, but you'd need to do this with a deeper pool first. I recall as a kid jumping and diving off a 3 meter diveboard into a creek that was only a bit deeper, maybe 7 or 8 feet deep, and you had to curve underwater to avoid hitting bottom.

I'd practic with deeper water first.

4 meters seems a bit high for the floor of a second story.

I do recall being able to step off a 12 foot high platform onto short grass with no damage, but I was young (14 years old) at the time.

russ_watters
Mentor
What do you mean here Russ? Dive used off of the starting blocks is NOT a dive for depth, in fact just the opposite, it is a shallow dive with the purpose to provide as much forward momentum as possible.
Yes, I know - yet they still have the pool 2m deep in case someone screws up. (Competition pools without blocks are as little a 1m deep)
A cannonball dive from 4m is not comparable in any way.
Unless he screws up.
In my foolish youth...
Yes, that's what I mean!

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Astronuc
Staff Emeritus
Hi guys,
I have a swimming pool which is 2m deep and was wandering whether I will be safe if I dive from 2nd floor which is around 4m high. [I'm 175cm tall and weigh 65kg]
Bottom line is that it is NOT safe! One is risking injury.

We cannot and will not enourage or endorse potentially harmful activities or actions.

Moonbear
Staff Emeritus
Gold Member
Professional high divers dive into pools deeper than your standard swimming pool, and take years of training to safely dive from those heights. Doing so into a standard swimming pool without that training is just plain stupid!

There are too many human factors here beyond whatever calculations you may make. You're NOT a perfect sphere or rod, you can slip and lose your footing, push off too hard or not enough and over or undershoot your trajectory, you may or may not get into the tuck position you're aiming for in time, you could over or under rotate while making your jump (even the professional high divers can make this error) and land with an unintended part of your body striking the water first.

The concrete AROUND a pool is a very unforgiving surface to dive into. People end up paralyzed or dead from miscalculating such jumps...all the calculations in the world will not help ensure you actually execute the dive correctly.

RonL
Gold Member
Consider that you do everything right and do no harm to yourself, then some one else thinks there are no risks, and dives in the wrong way and injures themselves, or dies as a results of your foolishness. I think that would be a terrible memory to carry thru a lifetime Hi guys,

Working out gravitational energy = mg = (65)*10 = 650
Equating to kinetic energy, (1/2)*(65)*v^2 = 650
v=4.47m/s.
Actually you forgot to put hight h=4 m into the equation:

E=mgh and not mg

You better don't rely on math when risking your health. In my opinion the only way to jump from hight safely is to increase the hight gradually.

wow rohanprabhu, nice calculations, but that equation will be really difficult to solve.

Actually you forgot to put hight h=4 m into the equation:

E=mgh and not mg

You better don't rely on math when risking your health. In my opinion the only way to jump from hight safely is to increase the hight gradually.

Good point.
Then it gives me 8.9m/s
Giving a bit more pressure on jumping Danger
Gold Member
Giving a bit more pressure on jumping The jumping isn't the problem; the landing, on the other hand, would be messy.

wow rohanprabhu, nice calculations, but that equation will be really difficult to solve.

well.. it is difficult. But as Moonbear said.. no calculations will give you the exact picture. There are too many factors here. Don't risk it.

russ_watters
Mentor
We've discussed this in the mods forum and I was made aware that my first post didn't make sense to some people. The point about competition pools was that the depth of your pool isn't safe for diving from any height above about a meter above the pool (starting block height).

In addition, there is a lot that can go wrong and a lot we don't know about the scenario that makes it impossible to know your odds of not killing yourself. So we can't condone this or allow the thread to continue.