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Thread: TNB...Unit Tangent, Normal, Binormal

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    TNB...Unit Tangent, Normal, Binormal

    Can someone explain TNB or unit tangent, unit normal and unit binormal? Why is the unit normal orthogonal to the unit tangent?
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    Re: TNB...Unit Tangent, Normal, Binormal

    Quote Originally Posted by USNAVY View Post
    Can someone explain TNB or unit tangent, unit normal and unit binormal? Why is the unit normal orthogonal to the unit tangent?
    If $R(t)$ is a vector value function. Then recall $\|R\|^2=R\cdot R$ so that $\|R\|=\sqrt{R\cdot R}$.
    From which it follows: $D_t(\|R\|)=\dfrac{(R\cdot R)'}{2\sqrt{R\cdot R}}=\dfrac{R'\cdot R+R\cdot R'}{2\|R\|}=\dfrac{R\cdot R'}{\|R\|}$.
    Therefore, if $\|R\|=c$(a constant) then $\dfrac{R\cdot R'}{\|R\|}=0$ or $R\bot R'$.

     \begin{align*}T' = \frac{{R''\left\| {R'} \right\| - R'\left( {\frac{{R' \cdot R''}}{{\left\| {R'} \right\|}}} \right)}}{{{{\left\| {R'} \right\|}^2}}}\\& = \frac{{R''{{\left\| {R'} \right\|}^2} - R'(R' \cdot R'')}}{{{{\left\| {R'} \right\|}^3}}} \\T' =\frac{R'\times(R''\times R')}{\|R\|^3}\\N &=\frac{T'}{\|T\|}\text{ so recall that }\\\|T\|&= 1\text{ so that }D_t(\|T\|)=0\end{align*}

    Hence $T\cdot N=0\text{ or } T~\&~N\text{ are perpendicular.}$
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    Re: TNB...Unit Tangent, Normal, Binormal

    As long as a curve is differentiable, there exist a "tangent line" at every point- a line that passes through that point and is in "the same direction" as the curve at that point. It can also be thought of as the linear function that "best fits" the curve at that point. If the curve is given by F(x, y, z) = constant, that line is in the same direction as \nabla F at that point. The unit tangent, at that point, is the vector of length 1 pointing in that direction.

    As long as a curve is "differentiable", there is a circle (so of second degree while a line is of first degree) that "best fits" the curve at that point. (Called the "osculating circle"- i.e "kissing circle".) The "unit normal" is a vector of length 1 pointing from the point on the curve to the center of that circle.

    The "unit binormal" is the unique vector of length 1 that is perpendicular to both the tangent and normal vectors.

    The "Frenet-Serret formulas" (https://en.wikipedia.org/wiki/Frenet...erret_formulas) connect them:
    \frac{dT}{ds}= \kappa N
    \frac{dN}{ds}= -\kappa T+ \tau B
    \frac{dB}{ds}= -\tau N

    Where T, N, and B are the unit "tangent", "normal", and "binormal" vectors respectively as functions of the arclength, s, \kappa is the curvature, and \tau is the torsion.
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    Re: TNB...Unit Tangent, Normal, Binormal

    Good.
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