ON THE CURVATURES OF TUBULAR SURFACE WITH BISHOP FRAME

A canal surface is the envelope of a moving sphere with varying radius, defined by the trajectory C(t) (spine curve) of its center and a radius function r(t) and it is parametrized through Frenet frame of the spine curve C(t). If the radius function r(t) = r is a constant, then the canal surface is called a tube or tubular surface. In this work, we investigate tubular surface with Bishop frame in place of Frenet frame and afterwards give some characterizations about special curves lying on this surface


Introduction
Canal surfaces are useful for representing long thin objects, e.g., pipes, poles, ropes, 3D fonts or intestines of body.Canal surfaces are also frequently used in solid and surface modelling for CAD/CAM.Representative examples are natural quadrics, torus, tubular surfaces and Dupin cyclides.
Maekawa et:al: [6] researched necessary and su¢ cient conditions for the regularity of pipe (tubular) surfaces.More recently, Xu et:al: [8] studied these conditions for canal surfaces and examined principle geometric properties of these surfaces like computing the area and Gaussian curvature.
Gross [3] gave the concept of generalized tubes (brie ‡y GT) and classi…ed them in two types as ZGT and CGT.Here, ZGT refers to the spine curve (the axis) that has torsion-free and CGT refers to tube that has circular cross sections.He investigated the properties of GT and showed that parameter curves of a generalized tube are also lines of curvature if and only if the spine curve has torsion free (planar).
Bishop [1] displayed that there exists orthonormal frames which he called relatively paralled adapted frames other than the Frenet frame and compared features of them with the Frenet frame.
This paper is organized as follows.We introduce canal and tubular surfaces in section 2. Section 3 gives us information concerning the curvatures of tubular surface with the Frenet frame.In section 4, we de…ne tubular surface with respect to the Bishop frame.Subsequently, we compute the curvatures of this new tubular surface and give some characterizations regarding special curves lying on it.

Preliminaries
Initially, we parametrize a canal surface via characteristic circles of it.Later, we de…ne a tube as a special case of the canal surface.A canal surface is de…ned as the envelope of a family of one parameter spheres.Alternatively, a canal surface is the envelope of a moving sphere with varying radius, de…ned by the trajectory C(t) of its center and a radius function r(t).This moving sphere S(t) touches the canal surface at a characteristic circle K(t).If the radius function r(t) = r is a constant, then the canal surface is called a tube or pipe surface.
Since the canal surface K(s; ) is the envelope of a family of one parameter spheres with the center C(t) and radius function r(t), a surface point p = K(t; ) 2 E 3 satis…es the following equations.(2.1) Now, we decompose the canal surface into a family of characteristic circles.Let M (t) be center of characteristic circles K(t).For a point p = K(t; ), the vector !C(t)M (t) is the orthogonal projection of !C(t)p onto the tangent C 0 (t) as obtained below.
By Eq (2.1), because (p C(t)) C 0 (t) = r(t)r 0 (t) we get the center M (t) and the radius function R(t) of characteristic circles as where (t) is the angle between !C(t)p and C 0 (t).Thus, the canal surface is parametrized as follows.
2) where N (t) and B(t) are the principal normal and binormal to C(t), respectively.Alternatively, N (t) and B(t) are the basis vectors of the plane containing characteristic circle.If the spine curve C(t) has an arclenght parametrization ( C 0 (t) = 1), then the canal surface is reparametrized as In the event r(t) = r is a constant, the canal surface is called a tube or pipe surface and it turns into the form Let T (s) be tangent to C(s) and let N 1 (s) be arbitrary orthogonal unit vector to T (s).
This means that fT (s); N 1 (s); N 2 (s)g is an orthonormal frame.The frame is called Bishop frame (relatively parallel adapted frame accordance with Frenet frame).If we rotate the Bishop frame by the angle around the tangent vector T , we obtain the Frenet frame as below.The derivative formulas for Frenet frame are given by where and are the curvature and the torsion of the spine curve C(s), respectively.Let k 1 (s) and k 2 (s) be Bishop parameters (normal development).The derivative formulas which correspond to Bishop frame and Bishop parameters are as follows.
In next sections, …rst we will give the curvatures of the tube L(s; ).Afterwards, by taking N 1 (s) and N 2 (s) instead of N (s) and B(s) we will compute the curvatures of this new tubular surface and obtain some characterizations as regards special curves lying on P (s; ).

The curvatures of tubular surfaces with respect to the Frenet frame
For the tubular surface L(s; ), the surface normal vector U and the coe¢ cients of the …rst and second fundamental form are given by ) is a regular tube if and only if 1 r cos 6 = 0.
Proof.For a regular surface, EG F 2 6 = 0.By Eq (3. 2), we have Since EG F 2 6 = 0 and r > 0, L(s; ) is a regular tube if and only if Thus, the Gaussian and mean curvature for a regular tube L(s; ) are computed as ) is generated by a moving sphere with the radius r = 1.
Proof.When K = 0, from Eq (3. 3) cos = 0 and so the normal of L(s; ) becomes Again, when cos = 0 it follows that From the last equation we must have r = 1.
Theorem 3.3.Let L(s; ) be a regular tube.In that case, we have the following.
(1) The s parameter curves of L(s; ) are also asymptotic curves if and only if (2) The parameter curves of L(s; ) cannot also be asymptotic curves.
Proof.(1) A curve lying on a surface is an asymptotic curve if and only if the acceleration vector 00 is tangent to the surface that is U 00 = 0.Then, for the s parameter curves we have From this, we get 2 = 1 r cos (1 r cos ) for s parameter curves.
(2) On account of the fact that U L = r 6 = 0, parameter curves cannot also be asymptotic curves.
Here, the equation 2 = 1 r cos (1 r cos ) is satis…ed for a circular helix C(s).
In the case of general helix, we get the curvature of spine curve C(s) as ; where is the angle between tangent line T and the …xed direction of the general helix.
= tan is a constant for a general helix.Hence, if we substitute this in the equation In this situation, we obtain the curvature as (s) = cos r(tan 2 + cos 2 ) . Because and are constants, it follows that (s) is a constant.Therefore, is also a constant.We see that the general helix becomes a circular helix and …nally the equation is satis…ed for a circular helix.Theorem 3.6 (Line of Curvature).The directions of the parameter curves at a non-umbilical point on a patch are in the direction of the principal directions (line of curvature) if and only if F = f = 0 at the point, where F and f are the respective …rst and second fundamental coe¢ cients [3].
Proof.Weingarten equations are given by Assume that the directions of the parameter curves at a non-umbilical point on a patch are in the direction of the principal directions.In this case, from the Weingarten equations and de…nition of line of curvature we have gF f G EG F 2 = 0 in Eq (3.10).By the last two equations we obtain From this, F = f = 0. ((=) Let F = f = 0 at a non-umbilical point on a patch.By the Weingarten equations it follows that Then, u and v parameter curves are lines of curvature concurrently.We view that the parameter curves of a tube or generalized tube are also lines of curvature if and only if the spine curve is planar.For the tube L(s; ), we have Also, for the generalized tube X(s; ) we have Truthfully, in both two cases, F = f = 0 if and only if the torsion of the spine curve is zero, i.e., the spine curve is planar.

The curvatures of tubular surfaces with respect to the Bishop frame
From this time, we will compute the curvatures of tubular surfaces with Bishop frame and then give some characterizations relative to it.Let P (s; ) be a tubular surface with Bishop frame.By applying Eq (2.6), the …rst and second derivatives of P (s; ) with respect to s and are obtained as Since T N 1 = N 2 and T N 2 = N 1 , the cross product of P s and P is that For this reason, the normal vector …eld U and the coe¢ cients of the …rst and second fundamental form are computed as  In this case, the Gaussian curvature K and mean curvature H for the regular tube P (s; ) are obtained as : (1) Seeing that, U P = 0, parameter curves are also geodesics of P (s; ).
(2) U P ss = 0 if and only if We know that sin and cos cannot be zero at the same time.Now that r > 0, the system of equations above is held if and only if = tan .By the last two equations, cot + tan = 0. From this, it follows that 1 sin cos 6 = 0.This is a contradiction, i.e., the system of equations does not have a solution.Then, the s parameter curves of P (s; ) cannot also be geodesic curves.

Conclusions
In this paper, we de…ned a tube with respect to the Bishop frame.Later, we computed the curvatures of this tube and examined special curves on it.Surprisingly, we viewed that parameter curves of P (s; ) are both lines of curvature and geodesics in other words parameter curves are planar.Furthermore, while a s parameter curve of L(s; ) can also be a geodesic none of the s parameter curves of P (s; ) can concurrently be a geodesic.