In this first example we see the plane of motion, the TNB frame, and the circle of curvature associated to the helix
We have
\mathbf r'(t)&=-\sin(t)\mathbf i+\cos(t)\mathbf j+\mathbf k\\
\|\mathbf r'(t)\|&=\sqrt{\sin^2(t)+\cos^2(t)+1}=\sqrt{2}\\
\mathbf T(t)&=2^{-1/2}(-\sin(t),\cos(t),1)\\
\mathbf r''(t)&=-\cos(t)\mathbf i-\sin(t)\mathbf j\\
\mathbf r'(t)\times\mathbf r''(t) &= \det\left[\begin{matrix}\mathbf i&\mathbf j&\mathbf k\\-\sin(t)&\cos(t)&1\\-\cos(t)&-\sin(t)&0\end{matrix}\right]=(\sin(t),-\cos(t),1)\\
\|\mathbf r'(t)\times\mathbf r''(t)\|&=\sqrt{2}\\
\mathbf B(t)&=2^{-1/2}(\sin(t),-\cos(t),1)\\
\mathbf N(t) &= \mathbf B(t)\times\mathbf T(t)=\frac{1}{2}\det\left[\begin{matrix}\mathbf i&\mathbf j&\mathbf k\\\sin(t)&-\cos(t)&1\\-\sin(t)&-\cos(t)&1\end{matrix}\right]=(-\cos(t),-\sin(t),0)\\
\kappa(t)&=\frac{\|\mathbf r'(t)\times\mathbf r''(t)\|}{\|\mathbf r'(t)\|^3}=\frac{\sqrt{2}}{(\sqrt{2})^3}=\frac{1}{2}
\end{align*}
So the circle of curvature has radius \frac{1}{\kappa}=2 and is parametrized by
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Now suppose you are moving around on the curve described above an observing an object moving along the curve (parametrized in terms of the same variable t) given by
So from your perspective the object appears like \mathbf s(t)-\mathbf r(t)=(2\cos(3t)-\cos(t))\mathbf i+(3\sin(2t)-\sin(t))\mathbf j+(\cos(2t)-t)\mathbf k which written in terms of the TNB-frame is
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