PS. The LaTeX file from my previous post after processing

We should mention also the Nash Embedding Theorem that is fundamental to answer the question posed in this thread.

”The Nash embedding theorems (or imbedding theorems), named after John Forbes Nash, state that every Riemannian manifold can be isometrically embedded into some Euclidean space. Isometric means preserving the length of every path. For instance, bending but neither stretching nor tearing a page of paper gives an isometric embedding of the page into Euclidean space because curves drawn on the page retain the same arclength however the page is bent”

(Source: Wikipedia)

Hi, I think you can add a sin term to your original function y=f(x)+eps*sin(w*x+phi). Best

Richard Epenoy is right. f(x)+sinx is the simplest way.

I think the previous answers do not meet your requirements... At each point (x,f(x)) you want to add a point on a sine arc along the normal vector to the curve y=f(x) at that point.

I would agree with José Antonio Vallejo. Let z(x) = (x, f(x)) be a point on the airfoil. The tangent vector to the airfoil at x is z'(x)=(1, f'(x)). The normal (forming a right hind system at x is n(x) = (-f'(x), 1). Let nu(x) denote the unit normal. The "sine" function you are looking for is vector value function w(x)=z(x) + sin(x+phi)*nu(x).

You are right, but I think instead of x we should put the value of the curve length (s) of f(x). The length of the curve (s) is also a function of x. In other words w(x)=z(x) + sin(s+phi)*nu(x). Do you agree?

Truman Prevatt

Dear Mortaza Salehian,

The problem described in this question is an example of a manifold imbedded in Cartesian space. After examining the answer suggested above, I have concluded that it is not correct; the scalar function defined on an imbedded manifold (the airfoil) can not be a vector function in the imbedding Cartesian space.

To Mortaza Salehian, yes one should reparamertize to arc length. This is done by noting that if s(x) is the arc length of the original curve, then s'(x)=norm(z'(x)) in my original notation. Since s'(x)>0, it has an inverse function x(s). By redefining z by z(s)=z(x(s)), z is parametrized by arc-length along the curve and the norm of dz/ds is equal to 1. Now apply my first response. It will only make a difference if the "bow" in the foil is pronounced ( curvature large) I expect. However, you are right since it is easy enough to do, it should be done.

For curves in the plane parametrized by arc-length, the Frenet-Serret formulas result in dT(s)/ds =k(s)N(s), where T is the tangent vector (z'(s)) and N is the unit Normal vector to T and k(s) is the curvature.

Dear Truman Prevatt

The answer to this question can be written in a closed form as follows:

\begin{equation}

f(x,y=y(x))=\sin\bigg(\bigg(\frac{2 \pi}{{\cal L}} \int_0^x\sqrt{1+{y'}^2}ds\bigg)+\phi\bigg)

\end{equation}

where: y=y(x) is a function describing the manifold in Cartesian coordinates (x,y), ${\cal L}$ is a length scale describing the wave perturbation on the airfoil, $y'$ denotes a derivative of $y$ with respect to $x$.

PS. The LaTeX file from my previous post after processing

We should mention also the Nash Embedding Theorem that is fundamental to answer the question posed in this thread.

”The Nash embedding theorems (or imbedding theorems), named after John Forbes Nash, state that every Riemannian manifold can be isometrically embedded into some Euclidean space. Isometric means preserving the length of every path. For instance, bending but neither stretching nor tearing a page of paper gives an isometric embedding of the page into Euclidean space because curves drawn on the page retain the same arclength however the page is bent”

(Source: Wikipedia)

JoaD20210301_1:

@Janusz Pudykiewicz 's formula i very rigorous (congrats!). However it can be applied only to waves, which do NOT change the manifold substantialy. Indeed, since in the case of non-negligible impact of the [sine] wave upon the airfoil, the point (x, y(x)) does NOT lies on the foil. And this restriction does not depend on the physical meaning of the wave (@Peter Beuer 's examples open a few cases to be considered). Janusz's has chosen a scalar NOT influencing the shape of the profile. Thus, the result gives a formula of any quantity like temperature, pressure, brigtheness etc. spread according to sine if treated as a function of the length of the arc along the curve, transformed into dependence on the abscissa of points on the curve (obviously, with the use of the unchangable shape y=f(x) ).

But the picture by @Mortaza Salehian suggests the need a formula for the curve obtained after perturbation of the original one (as it was expressed already by @Richard Epenoy in the first answer, and then discussed by @José Antonio Vallejo, @Truman Prevatt and Peter, who proposed some qualitatively formulated relations or just giving partial sets of formulas which might lead to vectorially formulated perturbations; mostly perpendicularly to the original surface (this would me acceptable also by me for small though non-negligible perturbations - whatevet it means at this stage of analysis).

Below I am trying to give a rigorous description (kinematics, only) of a perturbed curve, by now WITHOUT any further consequences which I thing may lead to a rigorous formulation of the original problem.

The proposal is based on the functions (x(l,t), y(l,t) ) expressing dependence of the coodinates (x,y) of the position of a point occupying the profile-line at distance l from a distinguished at instant t . For unstreachable lines the PARTIAL derivatives satisfy

(dx(l,t)/dl)^2 + (dy(l,t)/dl)^2 = 1 for every t.

It is therefore convenient to present them via trigonometric functions like

y'(l,t) = sin ( V(l,t) ), x'(l,t) = cos ( V(l,t) ) where |V|

Joachim Domsta

Dear Joachim,

Thank you for your comment. The idea of adding the normal extension is not applicable in the context of the current question (how to define a scalar function on the airfoil). However, you can certainly ask another question how to define a perturbation of the existing manifold. The answers to such a query have already been written (drafted) by some participants of this thread.

With my best regards,

Janusz