I am trying to figure out how I would graph the absolute coordinates of the spiral

r = 14 + 0.69θ^1.8, θ = 4π to 0

when the relative system starts with its origin at absolute (0,0) at θ = 4π and drifts towards (-15, -3) as θ approaches zero.


I'm mapping out spiraling Christmas lights on my bedroom ceiling. The strands start in the corners, run along the wall until midpoints (ie x/y axis lines) then begin spiraling towards the center until they merge into a ring near the middle. Here is an example from a previous apartment: enter image description here

My new bedroom has a ceiling fan about 15" x 3" off center, so if I treat the ceiling center as the origin its going to look awkward with the fan off to one corner. Treating the fan as the origin would cause the strands to end unevenly across the 4 walls, and would make it harder since I have to go outside inward.

The solution I came up with is to try an elliptical effect where the coordinates start at the walls with the ceiling center as the origin and gradually shift the origin towards the fan as the strands wrap inward.

I'd really like a way to graph this out to see how it will look before I go through the trouble of putting hundreds of thumbtacks in my ceiling :)

  • 1
    $\begingroup$ And if you're wondering why I'd go through all this effort, I put the lights on dimmer switches and put hundreds of glow in the dark star stickers over the walls and ceiling, which gives the effect of a colored spiral galaxy in space $\endgroup$
    – WebChemist
    Apr 19, 2019 at 0:34
  • 1
    $\begingroup$ +1 for your dedication to holiday decor. $\endgroup$ Apr 19, 2019 at 0:37

1 Answer 1


Interpretation of your question:

You want to plot the graph

$$r = 14+0.69 \theta^{1.8} \qquad \theta \text{ from }4\pi \text{ to }0$$

relative to a moving origin that travels at constant speed in a straight line from $(0,0)$ to $(-15,-3)$.

Let the position of the moving origin be $\mathbf p$, with respect to a fixed origin at $(0,0)$. So $\mathbf p = \mathbf p(\theta)$ moves in a straight line at constant speed from $(0,0)$ to $(-15,-3)$ as $\theta$ varies from $4\pi$ to $0$. It follows that

$$\mathbf p(\theta) = \bigg(\frac {\theta}{4\pi}-1\bigg) (15,3)$$

Next, with respect to this moving origin (which you might think of as a "moving frame relative to an inertial frame"), the Cartesian coordinates of your spiral is given by

$$\mathbf r = (r\cos \theta, r\sin\theta)$$

where $(r,\theta)$ are related by $r = 14+0.69 \theta^{1.8}$. Plug this in to get

$$\mathbf r(\theta) = \big((14+0.69 \theta^{1.8})\cos \theta, (14+0.69 \theta^{1.8})\sin\theta \big)$$

Finally, the $(x,y)$ coordinates of your spiral relative to the moving origin is given by

$$\mathbf p(\theta) + \mathbf r(\theta) = \bigg(15\bigg(\frac {\theta}{4\pi}-1\bigg)+(14+0.69 \theta^{1.8})\cos \theta \, , \, 3\bigg(\frac {\theta}{4\pi}-1\bigg)+(14+0.69 \theta^{1.8})\sin\theta \bigg)$$

for $\theta$ going from $4\pi$ to $0$.

  • $\begingroup$ Yes, this is exactly what I was looking for, thank you! $\endgroup$
    – WebChemist
    Apr 20, 2019 at 0:19
  • $\begingroup$ But why on earth are you putting up Christmas lights at this time of the year? $\endgroup$ Apr 20, 2019 at 16:04
  • $\begingroup$ Because I put the lights on dimmers and with all the glowing stars on the walls and ceiling, its like a spiral galaxy in space above my bed. I'll come back and add a picture link when its finished $\endgroup$
    – WebChemist
    Apr 22, 2019 at 13:24

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