A challenge.....

RutlandMike

RutlandMike

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I'm appealing to anyone who's brain is less addled than mine... I need to support 450kg on two inflatable tubes about 2 meters long - what would the diameter of the tubes need to be??? Whats confusing me is that the tubes need to be only partially submerged - like a rib. Any ideas? /forums/images/graemlins/confused.gif /forums/images/graemlins/confused.gif /forums/images/graemlins/confused.gif
 
30 cms (about 1ft) diameter should float at about half submerged for 2 tubes each 2meters long.

No calculator handy so steam calculation, so disclaim all responsibility if you drown.
 
Radius..

[ QUOTE ]
30 cms (about 1ft) diameter should float at about half submerged for 2 tubes each 2meters long.

No calculator handy so steam calculation, so disclaim all responsibility if you drown.

[/ QUOTE ]Think you mean 30cm radius - ie about 2ft diameter.
 
For your two 2 M long tubes to be half submerged by the 450 Kg mass they would need to be 450 litres each giving a total boyancy of 900 Kg (assuming fresh water). This means that their diameter needs to be about 55 cm.
 
my calc tally with pvb, but I'd allow a bit of a safety margin!

having said that, you have a decent margin given that the figure is based on floating the tubes half way down. what's the application, out of interest?
 
I am humbled by the speed that you all came up with an answer - I've got a headache from trying to remember my pi equations! I'm designing a pontoon type affair to hold a quad bike believe it or not. 60cm diameter tube sounds workable. Thanks awfully!
 
OK, let's try. I'm sure that, if I drop a clanger, someone, somewhere will mention it.... PDQ

1 cubic metre of water has a mass, by definition, of 1000 kg. Your payload of 450kg is, by definition, equivalent to 450 litres of water displaced, or 0.45 of a cubic metre. Therefore your pair-of-tubes assembly must displace 0.45 cu.m.

Therefore one of your tubes must displace 0.225cu.m. or 2250 cu.cm.

As each tube is 2 metres long, the area of the end of each tube will need to be 0.225/2 sq.m. which gives 0.1125sq.m. or 1125 sq.cm. The area of the end-circle of each tube is related to its radius by the expression Area = Pi x r x r ( can't find the right symbols )

So, r = Square Root of 1125/3.142 = 18.9222672 cm.

So, Diameter of each tube is r x 2 = 37.84 cm.

This will just support your payload at neutral buoyancy. You may want to have some reserve buoyancy, so you may care to double the volume of each tube such that each can just support 450 kg.

Reworking as above now gives a tube diameter of 30.1936122 - or 30.2 cm. for cash.......

How does that sound?

Phew!


/forums/images/graemlins/wink.gif
 
Re: Radius..

Did it in my head so your probably right. Although Bilbo got the same answer and he can do hard sums so maybe I was right.

Think the the disclaimer was a good idea all the same.

Hope he didn't drown, although at 450kg he must be a big lad, so may be safer if he did.
 
Radius, again.....

[ QUOTE ]
Reworking as above now gives a tube diameter of 30.1936122 - or 30.2 cm. for cash.......


[/ QUOTE ]Think you mean radius!
 
Re: Radius..

[ QUOTE ]
Did it in my head so your probably right. Although Bilbo got the same answer and he can do hard sums so maybe I was right.

[/ QUOTE ]He got it wrong too - the answer's about 60cm diameter, ie about 30cm radius.
 
[ QUOTE ]
OK, let's try. I'm sure that, if I drop a clanger, someone, somewhere will mention it.... PDQ

1 cubic metre of water has a mass, by definition, of 1000 kg. Your payload of 450kg is, by definition, equivalent to 450 litres of water displaced, or 0.45 of a cubic metre. Therefore your pair-of-tubes assembly must displace 0.45 cu.m.

Therefore one of your tubes must displace 0.225cu.m. or 2250 cu.cm.

As each tube is 2 metres long, the area of the end of each tube will need to be 0.225/2 sq.m. which gives 0.1125sq.m. or 1125 sq.cm. The area of the end-circle of each tube is related to its radius by the expression Area = Pi x r x r ( can't find the right symbols )

So, r = Square Root of 1125/3.142 = 18.9222672 cm.

So, Diameter of each tube is r x 2 = 37.84 cm.

This will just support your payload at neutral buoyancy. You may want to have some reserve buoyancy, so you may care to double the volume of each tube such that each can just support 450 kg.

Reworking as above now gives a tube diameter of 30.1936122 - or 30.2 cm. for cash.......

How does that sound?

Phew!


/forums/images/graemlins/wink.gif

[/ QUOTE ]

I reckon he enjoyed this - dont you ?

And x years ago (where x maybe >40?) Janet and John's son young Bilbo used to jump up and down in class, hand in the air saying "Sir, sir, I've got the answer, sir!". And young Bilbo was correct. Even down to his 7th decimal place! Clever young Bilbo. /forums/images/graemlins/grin.gif John believed that Bilbo took after him, but Janet knew better. Clever Janet.
 
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So, Diameter of each tube is r x 2 = 37.84 cm.

This will just support your payload at neutral buoyancy. You may want to have some reserve buoyancy, so you may care to double the volume of each tube such that each can just support 450 kg.

Reworking as above now gives a tube diameter of 30.1936122 - or 30.2 cm. for cash.......
-----------------------------------------------------------------------

Most impressive but do you mean 30.2 radius?
 
Ah. but you are clearly much cleverererer than I - twas a slight slip and I thank you for your assistance. So, how long would the tubes be if I reduced the diameter to 45cm?
 
[ QUOTE ]
No calculator handy so steam calculation

[/ QUOTE ]

Respec', maaan!

Like, I was using Google calculator.....

I only responded 'cos I'd been up most of the night, writing ruggish, and me old braincell was like sticky toffee. It certainly needed the exercise and - to mix me metaphors - it was like pulling teeth.

May I recommend to RutlandMike and others to kindly have a better sense of decorum, to not split their infinitives, and to refrain in future from such discomforting challenges that strain the synaptic sinews.

Kindly confine yourself to invective, taunts, gibes and conspicuous trollery, that we can all the better and more comfortably engage with without risk of a neural hernia.....

/forums/images/graemlins/grin.gif
 
Just had to get calculating engine fired up for this one

So I offer the following calculation:-

One tube 200cms long with volume of 450 litres to give 50% safety margin
Assuming material has neutral bouyancy.
1 litre = 1000cc
1000cc of water weighs 1 kilo

volume 450 litres = r x r x pi x 200cms [x section area x length of 1 tube]

450,000/200 = r x r x pi [to give x sectional area of tube]

2250/pi = r x r [radius squared]

r = root 2250/pi

= root 715.9

r = 26.756 cms [Radius]

Diameter = 53.5cms [diameter of tube]

Each tube 2 metres long by 53.5 cms diameter

To support 450Kg with 50% safety margin.
 
Oi t'inks thee is quite right, sor. An' I'm right sorree I got out of bed this afternoon, after all.

Oh, well. I got some of it right.... some of the big wurds defeated me! That NVQ in Submarine Design Technology hasn't turned out to be as useful as they said......


/forums/images/graemlins/grin.gif
 
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