The parametric equations for a simple 3D Archimedes spiral is this:
$z = t$
$r = \frac{d}{2\pi} \theta$
$\theta = 2 \pi t$
But what is the equation of a surface $f(x,y) = z$ spiral that looks like a racetrack that winds upwards (which has a track thickness $d$)?
Thanks for your help :)
I have a good start on your problem and can meet most of your requirements. Let's first of all drop the 3D aspect. It's superfluous and can added later. Now, in the complex plane we can express your spiral as
$$w=re^{i\theta}=r(\cos\theta+i\sin\theta)$$
Noting the $e^{i\theta}$ is a circle, the first thing that comes to mind is to change the circle to an ellipse, to wit,
$$w=r(a\cos\theta+ib\sin\theta)$$
Okay, it's not quite a racetrack, but we have one more trick. We can use a superellipse, which takes the form
$$w=r[a|\cos^{2p}\theta|~\text{sgn}(\cos\theta)+ib|\sin^{2p}\theta|~\text{sgn}(\sin\theta)]$$
where the $\text{sgn}$ function means the sign of the argument. When $p=1/2$ you get true ellipses, when $p>1/2$ you get more of a racetrack design, and when $p<1/2$ you get cusped figures.
The figure below shows a sample calculation with the planiform curve below and the and a 3D rendition above. The OP calls for a uniform spacing, but this model can only give a uniform spacing when $a=b$. But I hope that, at the least, this gives you a way to approach the problem.