View Full Version : A Trig. Question (Tan = Opposite/adjacent etc.).

parabellum

23rd Jan 2009, 04:32

I don't have any four figure tables anymore and am a bit stuck

I have to arrange a concrete slab for a 40' (12M) container and need the diagonal measurement. It is 12 meters long and 2.4 meters wide, would be grateful if someone could come up with the (correct) answer please? Thanks, PB.

birrddog

23rd Jan 2009, 04:45

12.23764683262268 ?

Did you read the one for the mathematicians thread?

I believe the Pythagoras theorem should apply.

KillingTime

23rd Jan 2009, 05:25

http://upload.wikimedia.org/math/2/9/c/29c5a073e6c30188f0a708576a949ccb.png

where i is your diagonal, while c1 and c2 are the length and width of the container.

tinpis

23rd Jan 2009, 06:01

A constipated mathematician worked it out with a pencil

Gelande Strasse

23rd Jan 2009, 06:55

Have you thought about going back to your school and berating them for allowing you to leave without understanding the basics? :cool:

GS

BAMRA wake up

23rd Jan 2009, 07:58

Well for a start 40 feet isn't '12M' it's 12.192m. On the width, more likely to be 8 feet for a standard 'box' which is 2.438m. therefore the diag measurement would be 12.433m rather than 12.238m from your rounded down figures.

chksix

23rd Jan 2009, 08:11

Don't forget to add margins to be able to slide it in and out of the container... :}

Scumbag O'Riley

23rd Jan 2009, 08:32

I always judge an understanding of maths on how somebody uses significant figures. They can try to impress me with all sorts of formulae and other stuff, but if they don't understand the basics they get an E- from me.

Which in todays world would be a pass mark :)

I mean. How the **** do you measure a concrete slab to the precision of the width of a fecking atom as suggested above. Or even to a millimetre.

BombayDuck

23rd Jan 2009, 08:33

(assuming that the 40' figure was for the height....)

17.28 metres would be the longest diagonal in the box. from the bottom left corner on one side to the top right corner on the other side (or bottom right on one to top left on the other). You already have the horizontal diagonal (12.23 m), the vertical diagonals are 17.11 m (vertical and longer horizontal) and 12.43 m (vertical and shorter horizontal).

Mariner9

23rd Jan 2009, 08:47

The width of the concrete slab would also have an impact on the maximum length able to fit in the box diagonally, and indeed depth, if stowed as Bombay Duck proposes (from bottom left to top right though he's wrong in his assumption that 40' is the height of the container)).

Parapunter

23rd Jan 2009, 09:03

A standard ocean going shipping container will have internal dimensions of 39'5" long by 7'6" wide by 7"8' height, give or take an inch or so.

Now what was the question? Ah yes, birddog had it on the first reply. You know side a and side b. What you are trying to find is the hypotenuse, so Pythagoras says a2+b2 = c2, which is (12x12) + (2.4x2.4) = 144+5.76 = 149.76. The square root of that is 12.23m or 40.12ft. You're gonna need a big container.

SoundBarrier

23rd Jan 2009, 09:16

hang on... from which side?

Blacksheep

23rd Jan 2009, 09:41

Getting down to basics, who the hell wants a block of solid concrete 2.4 metres wide by 2.4 metres high by 12 metres long?

...and why does it have to go in a container?

Perhaps we've located a certain missing trade union leader? :suspect:

Captain Stable

23rd Jan 2009, 09:43

Black, in a block that size you could fit at least 24 bodies... :ooh:

Parapunter

23rd Jan 2009, 09:47

Where does anyone say 2.4m high?

UniFoxOs

23rd Jan 2009, 11:24

And why do you want the diagonal , anyway??

UFO

BAMRA wake up

23rd Jan 2009, 11:24

I mean. How the **** do you measure a concrete slab to the precision of the width of a fecking atom as suggested above. Or even to a millimetre.

Crikey! I read parapunter's query as that he was laying a concrete slab on the ground as a base for the container. Of course you wouldn't place the concrete 'to the millimetre' but it's worth calculating everything thus to avoid rounding errors accumulating. The only reason to obtain the diagonal dimension is to achieve squareness. Even so, you don't need to calculate the actual dimension - if the rectangle is set out 'in the field' until both diagonals are equal, it's then square.

You'd want to make the slab bigger than the 'box' base anyway so the edge loads don't damage the concrete. A nice chamfer round the edge is the proper finish.

Parapunter

23rd Jan 2009, 11:29

And why do you want the diagonal , anyway??

Good question. If you're shipping it, it's one question: will it fit in the container or not? The diagonal seems irrelevant.:confused:

parabellum

23rd Jan 2009, 11:56

My thanks to you all, mathematicians and humorists alike.

We are amateurs and members of our local Lions Club.

It was felt that a substantial 'pad' each end would be good enough, supported mid way by mother earth.

My own experience of concrete, (building Twynham huts in Aden, 1963/4), tells me that the diagonals are very important, if they are 'out' and used as a basis to build on you can finish up with something that looks like barley sugar!

Scrubbed

23rd Jan 2009, 12:06

assuming that the 40' figure was for the height....

Who ever heard of a 40 foot high container???

17.28 metres would be the longest diagonal in the box. from the bottom left corner on one side to the top right corner on the other side (or bottom right on one to top left on the other). You already have the horizontal diagonal (12.23 m), the vertical diagonals are 17.11 m (vertical and longer horizontal) and 12.43 m (vertical and shorter horizontal).

WTF??????????? :confused::ugh::bored::ugh:

If you're shipping it, it's one question: will it fit in the container or not? The diagonal seems irrelevant

Man you lot are dumb. Who said he's shipping anything? Maybe he wants to pour some cement in a container for a slab. Maybe he's using the container as a shed to grow his hootch in or house his gimp away from the wife.

Vertical diagonals??? :rolleyes:

A constipated mathematician worked it out with a pencil

She tried to work it out with a pencil but couldn't budget.

Scumbag O'Riley

23rd Jan 2009, 12:35

Why would one want to bother with a huge slab? If it's a shipping container one would imagine it's self supporting at the corners and that would be all you would need to hold up. Thickness of slab and hardcore underneath would be very important to get right.

BAMRA wake up

23rd Jan 2009, 12:50

Spot on S O'R. They are basically 'box beams' so support along the long sides, you could use cuttings of railway sleeper set down into the ground or long strips of hardcore. Best keep the underside off the ground slightly and provide some perimeter drainage so it doesn't flood underneath.

green granite

23rd Jan 2009, 13:21

And why do you want the diagonal

Normally to ensure something is square you measure the diagonals, but they merely need to be equal you don't need to know what they are going to be in advance.

What's wrong with the calculator on the PC, that has tan in the scientific version

UniFoxOs

30th Jan 2009, 10:30

Exactly. You can start off square enough by the 3-4-5 method, this would get you right within an inch or two on this size, then equalise the diagonals at the end to ensure accuracy.

UFO

As noted, the Pythagoras formula will do the job, as long as there are right angles involved. You don't have to use trigonometry if you don't want to, but you can, in case anyone's curious. You start by finding the angle of the diagonal, θ, relative to the adjacent (long) side:

Using the figures from the original post:

Tan θ = opposite / adjacent = 2.4m / 12m = 0.2

taking the inverse Tan of 0.2, you get θ = 11.31 degrees.

Now you have the angle, you can use Cosine to get the diagonal (a.k.a. hypotenuse)

Cos θ = adjacent / hypotenuse

rearranging:

hypotenuse = adjacent / Cos θ = 12 / Cos 11.31°

= 12 / 0.9805 = 12.238 m

This is another case where it might help to draw a picture, instead of trying to do it all in your head... :8

PS: some engineering applications really do require that you're accurate to the millimetre (0.001 m). I'm just starting a course in Surveying (as part of an Engineering degree), and that's what's expected of you - so next time you see a kid standing around holding a pole, take your hat off to him. :ok:

lowerlobe

30th Jan 2009, 21:10

Why would one want to bother with a huge slab

Simply because if you have a pad at each corner you end up with a mess in between which needs mowing every week and weeding as well....

Old railway sleepers will be cheaper but in the end will rot.

If you use Pythagoras' no matter what size you want for the slab you can be sure your slab is a true and not looking like some form of art instead....and there is no need for log tables and working out angles...just a simple piece of paper and a calculator if you want to be sure...

Out of interest parabellum.....what thickness are you intending for the slab and what reo are you using?

As Bamra suggested ensure there is a slight fall in the slab to help with drainage...

Gordy

30th Jan 2009, 21:31

Yet again I am forced to offer my advice:

http://i76.photobucket.com/albums/j35/helokat/8uzn8y1164932566.jpg

parabellum

30th Jan 2009, 22:48

My job is to source the container, now done, and pass on the measurements to others, one of whom is a professional concreter and he will make two slabs, one each end for the container to sit on. As he is a pro I expect he will dig out a fair size hole and lay heavy duty garage forecourt/industrial site type concrete with appropriate reinforcing.

Sleepers were considered but the concrete option seemed easier to level and stay level.

The bottom of a container, both sides and ends, is a solid iron girder so a pad at each end is considered sufficient by those that claim to know! (A 40' container can safely hold up to 57 tons). I originally thought they planned a complete slab the size of the container and plus a bit, which is where the diagonals came in. I suspect the slabs will be flush with the surface to eliminate the weeds and other rubbish problem.

Thanks to all, once again, who have responded.

Oh, cool - is this going to be a piece of the "shipping container architecture (http://en.wikipedia.org/wiki/Shipping_container_architecture)" that's all the rage these days? Examples:

- Container Architecture - shipping container house design and construction in New Zealand (http://www.containerarchitecture.co.nz/)

- Container City | Home (http://www.containercity.com/)

- Corb v2.0 - mutating shipping container condos | Futurismic (http://futurismic.com/2009/01/26/corb-v20-mutating-shipping-container-condos/)

parabellum

31st Jan 2009, 00:49

Nah not a house! Just a store for our stuff to stop the local 'yoofery' from nicking it!

barit1

31st Jan 2009, 02:06

A constipated mathematician worked it out with a pencil

Engineers used to work it out with a slide rule. :}