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Elliptical (aka ovalised) tubes
I was reading 44’s SO and his use of laterally flattened chainstays got me thinking about the way that shaping tubing alters its characteristics. This, in turn, got me thinking about how we can work out how large the effect actually is. Being me, I decided there’s no substitute for maths, so here we go.
I’m going to model the tube as an ellipse which has been flattened from a tube of standard outside diameter which means the overall circumference remains constant. There is no exact analytical solution which will allow us to compute the lengths of the minor and major axes in this scenario so I will use Ramanujan’s second approximation:
Attachment 91456
where the factor h is given by
h = (a - b)^2/(a + b)^2
The error in this approximation is h^5, so for a/b =2, which gives h = 1/9, the error is 1/9^5 which is about 0.002%, good enough for jazz.
In terms of the original tube diameter, C = pi . D, substituting this into the Ramanujan equation and rearranging gives:
a + b = D / (1 + 3h / (10 + Sqrt(4 – 3h))).
For a given ratio between major and minor axes which I’ll call n:
n = a/b
so a + b = a . (n + 1)/n
substituting and rearranging gives
a = D . n / (n + 1) (1 + 3h / (10 + Sqrt(4 – 3h))) and
b = D / (n + 1) (1 + 3h / (10 + Sqrt(4 – 3h)))
The moment of inertia of an ellipse around the minor axis is given by
(Pi/4) (a^3.b)
and around the major axis is
(Pi/4) (a.b^3).
For a tube of wall thickness t, the dimensions to the inside wall are a1 = a - t and b1 = b – t, so the moments of inertia for the tube cross section are
(Pi/4) (a^3.b – a1^3.b1) and
(Pi/4) (a.b^3 – a1.b1^3)
By the perpendicular axis theorem, the polar moment of inertia is the sum of these two.
Reality check: if a/b is set to 1 (eg the tube is round) the equations give a = b = D/2 = r and the moments of inertia are both equal to Pi/4.(r^4 – r1^4) which is the known formula for an infinitely thin hoop.
To enable practical comparisons of the effect this will have on tube shaping, I wrote a little spreadsheet that calculates the lateral, vertical and torsional stiffness of an elliptical tube compared with those of the same tube before squishing and computes the contributions of said tubes to the overall frame stiffness. The model I used here has already been dicussed in this thread. I used a range of ovalisation ratios from 1 : 1 (eg an unmodified tube) to 1 : 2 .
https://farm2.staticflickr.com/1697/...52952a0e_c.jpg
Ellipses
Note that the elliptical tubes show in increase in lateral stiffness but a decrease in torsional stiffness. In this particular application the net effect is a fairly small increase in overall stiffness which is maximal at the 1 : 1.5 ovalisation ratio, above which the loss of torsional stiffness overcomes the increase in lateral stiffness.
In other applications (such as 44s chainstays) the results will be different. Also please note that ovalisation could be of benefit in other aspects of the design: designers may wish to avail themselves of the reduction in vertical stiffness or it may be used to improve the matching between a medium diameter top tube and a larger diameter seat tube, or simply to improve the “portability” of a cross bike for instance.