"tricky" math question

IK09

I apologise if this is the wrong area, but here goes. This being the maths area of boards, i figure someone will be able to help.

I am building a floating football pitch....yes i know...what is he thinking.

I need to know how much weight a single keg can keep afloat....

I also need to know weither having 40kegs will hold 40 times the weight that one will.

Im terrible at maths and have no idea how to figure these out.

The kegs will be under crates, and plywood on top of that. How many kegs do you think it would take to keep 40 crates afloat, given that each crate will weigh about 20kg.

:pac::pac::pac:

Thank you in advance if you can help

clived2

Check out this awesome video

https://www.youtube.com/watch?v=jU4oA3kkAWU

IK09

that was the muse for it

professore

IK09 wrote: »

I apologise if this is the wrong area, but here goes. This being the maths area of boards, i figure someone will be able to help.

I am building a floating football pitch....yes i know...what is he thinking.

I need to know how much weight a single keg can keep afloat....

I also need to know weither having 40kegs will hold 40 times the weight that one will.

Im terrible at maths and have no idea how to figure these out.

The kegs will be under crates, and plywood on top of that. How many kegs do you think it would take to keep 40 crates afloat, given that each crate will weigh about 20kg.

:pac::pac::pac:

Thank you in advance if you can help

The 40 kegs will indeed hold 40 times the weight of 1 keg.

To work this out you also need to tell us the dimensions of a keg (assuming it's a beer keg shape, the diameter at the top and the height of it) and how much it weighs. Then I can have a crack at it.

By the way, congrats on your lunatic idea, it's a pity there aren't more of us in the world!

IK09

I will be using the regular (i suppose guinness) kegs.

The dimensions are...

It holds 15.5 gallons

height 23 3/8"
diameter 17"

and empty, it weighs 13.5kg

thanks for the support

professore

Ok I think I have worked it out .... but please note these calculations are for instructional purposes only, I take no responsibility if they are wrong!

Note that I converted your measurements into metres ....

Volume of beer keg V:

V = pi ( D/2) h

V= pi (0.4318 / 2) ^ 2 0.5937 = 0.08694051 m^3

Density of a beer keg:

mass / volume = 13.5 kg / 0.08694051 = 155 kg / m^3

Buoyant mass Mb

Mb = M (1 - (density of fluid / density of object in it))

Mb = 13.5 (1 - 1000/155)) = -73.48 kg.


So I think what this means is that each keg can support a weight of 70 ish kg.

you should therefore assume 40 kg or so as you need to also allow for the plywood and the players and allow a good margin of safety.

professore

And by the way this is actually physics or engineering rather than maths. But we won't hold it against you!

kippy

professore wrote: »

And by the way this is actually physics or engineering rather than maths. But we won't hold it against you!

It's all maths when you break it down.

professore

kippy wrote: »

It's all maths when you break it down.

Not to be pedantic about it but it isn't just that - maths is a tool that can be used to solve physical problems. The problem itself is a physical problem that can be solved with maths.

There are other ways to solve it, like actually attaching the kegs to the crates and see what happens.

However the approach I have used is applied maths. Maths can also be used to solve problems that don't exist in the real world (and often are).

I am sure the OP doesn't care either way!

kippy

professore wrote: »

Not to be pedantic about it but it isn't just that - maths is a tool that can be used to solve physical problems. The problem itself is a physical problem that can be solved with maths.

There are other ways to solve it, like actually attaching the kegs to the crates and see what happens.

However the approach I have used is applied maths. Maths can also be used to solve problems that don't exist in the real world (and often are).

I am sure the OP doesn't care either way!

"Pedantise" away - its maths.

LeixlipRed

Ah leave it out will you!

DanWall

To make it realy simple: 1 litre = 1 kg.
Volume of keg x by number of kegs minus weight of keg = weight it will hold,
Allow a safety margin because it will go lower as the weight of people on it increase

professore

DanWall wrote: »

To make it realy simple: 1 litre = 1 kg.
Volume of keg x by number of kegs minus weight of keg = weight it will hold,
Allow a safety margin because it will go lower as the weight of people on it increase

You're right.

professore

Actually is there a reason to use beer kegs? I would have thought plastic containers of the same size would be cheaper and more effective due to their lower density?

professore

How refreshing to be in a forum where the goal is for everyone to come to the same result and agree with each other and there are definite verifiable answers - beats the hell out of the Politics > Irish Economy forum!

Fremen

You don't even need to approximate the keg as a cylinder to compute the volume. You just need to know how many pints the keg was able to hold when it was full - that's how much air is inside now.

Fremen

professore wrote: »

How refreshing to be in a forum where the goal is for everyone to come to the same result and agree with each other and there are definite verifiable answers - beats the hell out of the Politics > Irish Economy forum!

Try asking about the Monty Hall problem...

IK09

thanks for the replies guys,

your right it probably would be cheaper (maybe easier) to use plastic drums, but firstly, the pitch will be used for a charity event so i dont intend on spending much money.

im gonna put a request into the fix it friday programme on the ray darcy show now so hopefully i ill be able to get the materials without much expence.

i know this probably belongs in the engineering forum, but i figured with all th calculations and formula involved there wouldnt be many takers.

so 1 keg will keep approx. 70kg of weigh afloat?

Fremen

professore wrote: »

Volume of beer keg V:

V = pi ( D/2) h

V= pi (0.4318 / 2) ^ 2 0.5937 = 0.08694051 m^3

This means V = 19 gallons or thereabouts. But the OP said V = 15.5 gallons - this suggests the kegs will float substantially less than 70kg each.

professore

Fremen wrote: »

This means V = 19 gallons or thereabouts. But the OP said V = 15.5 gallons - this suggests the kegs will float substantially less than 70kg each.

Well spotted!

Either my calculations are wrong or there is a significant area of the beer keg that is not empty space ...

The rims with the handles on the top and bottom, that don't actually contain anything, I bet these were included in the height of the keg ...

so looks to be about 57kg in that case. ...

click_here!!!

One way to find out how many kegs you need is to try it.

Be sure to add in a few extra kegs to be on the safe side. And let us know how you get on.

Civilian_Target

The reason for your difference in calculation is that one of you is taking US gallons, the other is taking UK gallons. I would assume that here we need UK gallons... but would be better for the OP to clarify to avoid a sinking football field

But yeah, assuming UK gallons, I guess 48's a good number if you're planning for ~25 players. Make sure you seal up the kegs good and tight, if water leaks in you can lose bouyancy very quickly, particularly if it leads to a sloping pitch!