I've been reading about the Artin Spin operation. It's defined as taking the classical $n$-knot ($S^n\hookrightarrow S^{n+2}$) to an $(n+1)$-knot. For the $1$-knot case (in $\mathbb{R}^3$), I reproduce the procedure in [knot spinning](https://arxiv.org/pdf/math/0410606.pdf), p. 8, 


1. We manipulate a knot $K$ so that all but a trivial arc $a$ lie in the upper half space $H^3=\{(x,y,z)\>|\>z\geq 0\}$. We remove the interior of $a$.
2. We rotate $H^3$ around $\mathbb{R}^2$ in $\mathbb{R}^4$, inducing a parameterization $(x,y,z)\mapsto (x, y, z\cos\theta, z\sin\theta)$


**-- Question --**
How is this operation similar to the suspension functor (on a topological space $X$) $\Sigma X\equiv X\wedge S=\frac{X\times S}{X\vee S}$?



Is the Artin Spin construction related to the suspension functor?
This question is answered in section 4 of my first paper (with Alex Suciu)


Klein, John R.; Suciu, Alexander I.
Inequivalent fibred knots whose homotopy Seifert pairings are isometric.
Math. Ann. 289 (1991), no. 4, 683–701; [DOI: 10.1007/BF01446596](https://doi.org/10.1007/BF01446596).


The paper is available here:


[http://www.math.wayne.edu/~klein/inequivalent\_knots.pdf](http://www.math.wayne.edu/%7Eklein/inequivalent_knots.pdf) ([Wayback Machine](https://web.archive.org/web/20180720051823/http://www.math.wayne.edu/%7Eklein/inequivalent_knots.pdf))


