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Thread Subject:
radius of curvature for wind tunnel

Subject: radius of curvature for wind tunnel

From: will wou

Date: 25 Feb, 2010 00:27:05

Message: 1 of 2

Hi im designing a wind tunnel and i have this curve for a part called the contraction.

Ive got this function F(x)=-1.7068x^3+1.96947x^4-0.60599x^5+0.45

im using the limits of x>=0 and x<=1.3
                            y>=0 and y<=0.5

im trying to find the radius of curvature at two points of the curve to be able to enter the curve into a CAD program and then to be manufactured, but im confused how to do that. The first curve is convex leading to a straight line then to a concave curve. Is there a way of finding the coordinates of the center of curvature for this curve and radius of it?
Thanks for any help!

Subject: radius of curvature for wind tunnel

From: Roger Stafford

Date: 25 Feb, 2010 04:50:20

Message: 2 of 2

"will wou" <ww00039@surrey.ac.uk> wrote in message <hm4g4p$jaa$1@fred.mathworks.com>...
> Hi im designing a wind tunnel and i have this curve for a part called the contraction.
>
> Ive got this function F(x)=-1.7068x^3+1.96947x^4-0.60599x^5+0.45
>
> im using the limits of x>=0 and x<=1.3
> y>=0 and y<=0.5
>
> im trying to find the radius of curvature at two points of the curve to be able to enter the curve into a CAD program and then to be manufactured, but im confused how to do that. The first curve is convex leading to a straight line then to a concave curve. Is there a way of finding the coordinates of the center of curvature for this curve and radius of it?
> Thanks for any help!

  The way you have described the data being entered into CAD worries me. The quintic curve you have defined has a continually varying curvature and radius of curvature along its entire length. Entering the radius of curvature for only two points would not be enough information to allow the curve shape to be manufactured, at least not with any reasonable accuracy. Only in some very coarse sense could it be considered as two circular arcs, each with a constant curvature, and joined tangentially at the center or with an intermediate tangent straight line, if that is what you had in mind.

  Your curve's curvature, (as defined mathematically,) is precisely zero at three points: the beginning of the curve at x = 0, its middle, and the end at x = 1.3. Consequently its reciprocal, the radius of curvature, is infinite at these three points. From calculus the formula for the radius of curvature of a curve y = f(x) is:

  r = (1+f'(x)^2)^(3/2)/abs(f"(x))

and the curvature is upwards or downwards according to the sign of f"(x). When this second derivative f"(x) is zero, as it is at the three points mentioned above, this makes the radius infinite.

  It is easy to derive a formula for the center of curvature for any given point on the curve in terms of this radius formula and the slope f'(x) using well-known analytic geometry formulas. Different points of the curve of course will have different corresponding centers of curvature.

Roger Stafford

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