What is the angle that an Archimedean conical spiral makes with the floor? I have a spiral in the form $$r = r_0(1-{\theta\over2\pi k }) \{r \ge 0\}$$ where $r_0$ is an initial radius, and $k$ is the number of turns. (It is a spiral that decays from $r_0$ to $0$ as $\theta$ increases) and I want to have this mapped out to a cone (apex of $2\phi$ pointing down) so that the gradient of the spiral (and hence the angle between the spiral and the floor) is constant, and $r_0$ matches that of the cone. If this is the case, and when given: $r_0, h, \phi $, and $k$; what is the angle between the right-cones hypotenuse and the spiral track?
 A: You have defined main objective $ (r- \theta) $ relation on cone:
 $$ r = r_0(1-{\theta\over2\pi }) $$  
for Archimedean spiral polar projection. 
$ \theta\over2\pi $ serves the same purpose as $k$ which can be conveniently avoided.
So parametric coordinates of the spiral are: 
$$ (x, y, z) = r_0(1-{\theta\over2\pi }) ( cos\,\theta , sin\,\theta, cot \, \phi) $$
If $\psi$ is angle made between tangent and circumferential direction we derive using differential geometry:
$$ sin\,\phi \,tan\,\psi\, ( 2\pi - \theta ) = 1  $$
The spiral is written on this cone with $ r=0 \,@ \, z=0, r_0 =1,\phi = \pi/6. $ The angle $\psi $ can be computed at any point desired.
EDIT1: Please note it is much easier to go from general case using more comprehensive methods to special ones than the other way round.
EDIT2: 
Here $\psi$ angle is made between tangent and parallel circle or ground.If $\psi$ is constant, it is called a Loxodrome ; the requirement for x- y projection to be an Archimedean spiral conflicts with $\psi$ constancy, i.e., either this or that, mutually exclusive.
It is recommended to learn about geodesics at this stage. Clairaut's Law states $ r \cdot cos\psi $ is constant. If $\beta $ is made with generator, then $ r \cdot sin \beta $ is constant. It is good opportune entry into differential geometry.
EDIT3: If OP meant or implied by gradient (height gained by climbing unit length of arc ) as $ dz/ds = \cos \phi \cdot  \cos \psi $ , then $ \psi$ should be constant as a Loxodrome since $\phi$ is constant for a cone. Its development would be a logarithmic spiral.


A: I think I have an answer, correct me if i'm wrong. Consider a half turn over the spiral:


*

*the vertical displacement will be $h\over 2k$

*the horizontal displacement will be equal to $r_0$

*when a triangle is drawn with these in mind there is a right angled triangle such that $$\tan(c+\phi) = {2kr_0 \over h}$$


then transposing and solving for k and c we get:
$$ k = {h\tan(c+\phi)\over 2 r_0}$$
or
$$ c = \arctan ({2kr_0\over h}) - \phi$$
EDIT this "$c$" is for the angle between the cones edge and the spiral, not as the question originally asked, between the spiral and the floor. The angle between the spiral and the floor will be $\pi - (c+\phi)$
Again, please tell me if you think I am wrong or have any questions.
