MegaSack DRAW - This year's winner is user - rgwb
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My dad has been asked to form some steel hoops for a guy who repairs banjos
He has the rollers etc to form the material but needs to know what length to cut the material before forming.
The material he is using is 1/8th x 1/2" and he needs to form two hoops one with a 4 inch id and one with a 7 inch id
There must be a formula for working it out but cannot find it using google
Any help appreciated
😯
You mean ?d??
(says Rachel, hoping that whatever computer you are using can also display a PI symbol...)
You talking about circumference? 2*pi*radius
Or you talking about the area? radius*(pi^2)
If area it's a case of larger circle - smaller circle I think, your description is a little vague.
Might be missing something but Isn't it just the circumference?
http://en.m.wikipedia.org/wiki/Circumference
For heavens sake - what is the education system coming to !?
I was taught this in primary school for god's sake.
As said before the circumference = pi (3.142) * diameter
Surely you've got to trim the (flat) ends anyway? So cut long and trim to suit.
Presume id = internal diameter?
In which case it's one of the most basic equations of all, taught in primary school. pi (3.141592) x diameter to give the circumference of the circle.
So your first circle of 4", the rod needs to be 12.57" long and it'll bend round to form a perfect circle with internal diameter of 4".
You can now work out the second one all by yourself...
I know how to work out the circumfrence/area but somehow under the illusion that when working with steel you had to allow extra material when bending or rolling
maycontainnuts
That is the method he has used in the end
steel4real
Nothing positive to add ? well they didn't teach you any manners did they
Ah I see. If you need to consider the thickness of the material then just go for the external diameter instead (internal diameter + 2xthickness)
You need to work to the centre line of the material, so 4" plus one material thickness of 1/8" multiplied by pi.
HTH
GrahamS
Tried that and it ends up oversize,this is what got me thinking
steel4reel
Your local to me why not come and show us how its done
kempfab
Thank you
The fab in my name is short for fabricator, but rolling always goes a little wrong, pending how the material stretches and shrinks.
Good luck!
Heh. Today's show is brought to you by engineering, not by GCSE mathematics 🙂
kempfab
I assumed thats what the fab in your name was for.
I do plenty of working out angles and distances on hydraulic cylinders for placement of pipes and port bosses at work but couldn't get my head round this one. 😳
Cheers
engineering, not by GCSE mathematics
Like Americans talking to Englishmen. Separated by a common language.
Heh. Today's show is brought to you by engineering, not by GCSE mathematics
😀 😀
Im just relieved that it isnt an ignorance of pi at the root of the question 😉 . I claim a specification error...
Nah, the question definitely stated the material and the dimensions of the cross-section 🙂

