# Calculations when ovalising/ovalizing tube



## 18bikes (Jan 15, 2007)

I'm trying to do some calculations and design work based on ovalising a tube, what I'm needing to know is is there a ratio/formula for working out if I squash a tube by X it will get wider by Y. I would assume that, at least for the first bit the ratio is 1:1, i.e. if a squash a 30mm tube by 2mm it will be 32mm X 28mm but I imagine as the tube gets 'flatter' the ratio would reduce.

I've tried searching the web but can only find info on tubes ovalising when being bent, I want a straight oval tube. Does anyone know how i can work this out?

I'm also assuming wall thickness will only affect how difficult it is to do rather than how it actually shapes the tube.

Does anyone have more experience on this?

Matt


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## smdubovsky (Apr 27, 2007)

wikipedia has all the calculations for an ellipse
http://en.wikipedia.org/wiki/Ellipse

The key is the circumference must stay the same (unless you're stretching the metal). For example: round C1=pi*D1. At the limit (fully flattened) you get C2=2*D2 (D2=long dimension). C1=C2. Solving results in D2=pi*D1/2. Since you're probably not interested in the limit case, you'll need to solve w/ one of the ellipse formulas.


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## dbohemian (Mar 25, 2007)

You should be able to do this in most any 2D CAD software.

Use the ellipse tool and draw first a circle in the diameter you want then just drag the X axis and the corresponding Y axis will change with it. 

Also, in solidworks you could model a tube and use deform/flex tools which I think should work.


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## 18bikes (Jan 15, 2007)

Thanks guys, I'll have a play on ProDesktop tonight, not got it on the shop computer. Failing that I'll try the old fashioned way

Matt


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## smdubovsky (Apr 27, 2007)

ok, back from my run and had time for some breakfast...

An approximation to the circumference of an ellipse thats not too flat is C=2*pi*sqrt((a^2+b^2)/2), where a and b are the major & minor axis radiuses. Setting that equivalent to a circle
2*pi*r=C=2*pi*sqrt((a^2+b^2)/2)
r=sqrt((a^2+b^2)/2)
r^2=(a^2+b^2)/2
2*r^2=a^2+b^2
a^2=2*r^2-b^2
a=sqrt(2*r^2-b^2)

Note: This is the same for diameters. The /2 terms will cancel.
A=sqrt(2*D^2-B^2)

So if you have a 30mm diam tube you want to squish to 20mm wide, A~=37.4mm tall

Any of this assumes that when you squish the tube it forms a real ellipse vs something more like an oval. So an even the general approximation of an ellipse circumference is probably 'good enough' vs a more exact solution.


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## Smokebikes (Feb 2, 2008)

Is this for a bike? I'm just a hack and don't venture to far into the theoretical world but I do wonder what it's like over-there sometimes........and what I may be missing.  I'll go back to my cave now.


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## Live Wire (Aug 27, 2007)

Wouldn't the shape of your form and the method of applying pressure to the tubes introduce enough variables that the equations don't have much real world applicability?

I suppose it all depends on the ultimate use for the tube.


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## Winter Bicycles (Jun 8, 2008)

Squish method will change the accuracy of the final shape (as has been touched on). How accurately do you need to predict? For small scale production it is oft best to just measure the part post "squish".


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## 18bikes (Jan 15, 2007)

I'm wanting to design around the squish so want to work out roughly how wide it will end up to get it as flat as I require.

It will be rolled using a horizontal mill and mandrels I'm making for the job, so again I need the dimensions to know the radius for the mandrels

If the plan does go ahead I will keep people up to date on it

Matt


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## smdubovsky (Apr 27, 2007)

Just because I don't feel like working anymore today, I thought about this again For an oval (round ends, straight sides) : C=pi*B+2*(A-B) Again setting that equal to the circumference of a circle you get: A=pi*(D-B)/2 + B

So w/ the same dimensions in the prior example: D=30mm, B=20mm -> A=35.7mm. I suspect it it would come out somewhere between the two versions as you're likely not to be able to squish into a perfect oval or ellipse.

Curious how you're making eliptical rollers for the mill. Or are you 'simply' approximating w/ a constant radius. You're not the first person I've heard using a horiz as a power roller


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## 18bikes (Jan 15, 2007)

I will be 'simply'appoximating with a constant radius, hence the desire to get everything worked out so I can get something thats pretty close. I've not actually used the mill for rolling yet but it seems like the easiest way of keeping things sturdy

matt


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## Yogii (Jun 5, 2008)

What is wrong with using the tube from a roll of toilet paper?


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## Thylacine (Feb 29, 2004)

18bikes said:


> Thanks guys, I'll have a play on ProDesktop tonight, not got it on the shop computer. Failing that I'll try the old fashioned way
> 
> Matt


I think you'll have better luck the old fashioned way.


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## 18bikes (Jan 15, 2007)

Thanks guys, I've worked it out with the help of some people on Practical Machinist

(the thread is Here

I would thoroughly recommend asking those guys anything when it comes to something non-bike/out of the ordinary. Theres also a lot of stuff archived that is usefull (tube mitring and bending come to mind)

I've ended up with an excel spreadsheet that calculates approximations for me and I think i've worked out what I'll end up using and I've worked out the dimensions of the dies too

Matt


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## smdubovsky (Apr 27, 2007)

:madman: Geez. You do realize thats the EXACT same information I gave you days ago. 

(Channeling PVD: ) So what you're really saying is that you needed help from PM to put it in excel for you?


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## pvd (Jan 4, 2006)

*I work in reverse*

Since the major or the minor diameter is dependent.

B7 =SQRT((2*((C4/2)^2+(B4/2)^2))-((((C4-B4)/2)^2)/2.2))
C7 =B4*2


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## 18bikes (Jan 15, 2007)

smdubovsky said:


> :madman: Geez. You do realize thats the EXACT same information I gave you days ago.
> 
> (Channeling PVD: ) So what you're really saying is that you needed help from PM to put it in excel for you?


yes i realise that and it was a combination of the two that gave me the end result, I had seen your post and was going to sort out a spreadsheet when I thought I would ask elsewhere for information to see if there was a different approach, they came up with basically the same answer so that's what I used. And FYI people are lazy, me included, so if someone has done the work for you and you can copy and paste a formula straight into excel as it's already formatted, I'm going to do that rather than work out how to put it in as I don't use excel that much for more complicated formulas. Maybe my wording wasn't great but I couldn't have done it without the help from here either

I was pointing out that it's nice to get a different view on a problem as those guys pointed out a few different methods that may work better/quicker/easier than rolling in a mill.

Matt


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## Thylacine (Feb 29, 2004)

I'm really dumb when it comes to maths, but isn't a true ellipse not made from 2 (4) radii?


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## smdubovsky (Apr 27, 2007)

Correct. There are two diameters that define it. The minor (the largest circle that will fit inside) and the major (the smallest circle that will fit outside.) Each touches at two points. The curve in between those points does not have a constant radius anywhere.


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