How does rotating a circle create a spherical surface without leaving "gaps"? EDIT: Agreed, this isn't a well formed question. But responses below have at least given me a different way to think about it.
EDIT2: Thanks the answers I discovered the Google S2 library (http://s2geometry.io/devguide/s2cell_hierarchy) which provides methods for tracing a space filling curve on a sphere. I'm working this into my camera position algorithm instead of just gimballing through altitude/azimuth. Thanks!
First post here because I'm not a math guy, but I have a feeling the confusion I have is due to something mathy. Perhaps this isn't even the right forum; if not, I'll delete.
Here's the problem:
I built an apparatus to rotate a sensor through a 360 circle and measure incident light at each point. It's pretty cool, I can automatically orient a sensor on the end of arm to maximize incident light (looks like a radar antenna) using a stepper.
I tried extending this to a sphere, but what I found was: I need to discretize the the second angle to some minimum delta-phi. It seems odd to me that I can sweep through an infinite number of degrees in a circle, continuously as my theta angle, but then I need to choose a delta-phi to rotate that circle through to scan the sphere surface.
How is it that I can rotate through an infinite number of points on a circle in one rotation in finite time, but cannot cover the infinite number of circles that comprise the surface of a sphere in finite time because I can always increment a smaller and smaller phi angle...
Or maybe the question is: how is it that an rotating circle traces out a sphere surface: aren't there always going to be more points "covered" by the circle at the poles, compared to the equator, since the distance covered by an infinitesimal delta-phi is short at the poles than the equator?
Obviously I ultimately have to discretize everything since I'm using a digital circuit, but this sudden appearance if finite/infinite stopped me in my tracks...
I'm not even sure what I'm asking, just puzzled. Anyone know what I'm up against here?
 A: The shorter distance covered at the poles compared to at the equator doesn't imply one of them has to leave gaps, it just means it has a smaller distance to cover in the same amount of time and thus requires less speed. A useful comparison is a disk spinning on a record player. If you draw a line from the center to a point on the circumference the line will be continuously rotated such that it continues to look like a line, and gaps will not appear. For every rotation of the disk, the line will rotate exactly once around the center. Compare this to runners racing on a circular track that each run at the same speed and also start on a line from the center: runners closer to the center will make more revolutions in the same amount of time, so the line one perceived at the start will quickly disappear. Back to the disk example: when watching a line on a disk rotating on a monitor, it is correct that larger gaps will appear near the edge of the disk (equator) while fewer will appear near the center (pole). However, when considering a truly continuous rotation, this will not occur. I hope the analogy provides some insight.
To accomplish your goal similarly to how you traced out a sphere, you would need a tool that scans using a line a rather than a point. While the line on the disk in the analogy covered every point on the disk, to accomplish this starting from a single point, you would need to rotate the point around the disk from every point on the line. You've moved from needing to cover all points in an interval to needing to cover all points on an interval of an interval of points, where the elements of each individual interval are dependent only on one variable.
A: In a nut shell, when swept, a single point will pass through every point on a line even though there are an infinite number of them. But it would take an infinite number of lines to generate a surface, hence anything less than infinity will have gaps.
A: Although a circle is curved and lives inside $2$ dimensional space, the circle itself has no thickness and is actually a $1$ dimension curve.
One way, perhaps too naively, to express this is the formula for the circle is $f(x) = (\cos x, \sin x)$ has only $1$ dependent variable.
Or if we described it in the two dimensions as $x^2 + y^2 = r^2$ we can express one very entirely in terms of the other.  I.e. $y = \pm \sqrt{r^2 -y^2}$.  (Okay, I hand waved that you have two not one choice for $y$ in terms of $x$ but that's minor.  It is a set finit number of choices and there is only one independent variable; the other is completely dependent on the other.)
As such is is very easy to map a $1$ dimensional segment $[0,360]$ to the $1$ dimensional circle via $x \to (\cos x, \sin x)$.
The surface of the sphere is a $2$ dimensional surface.
If we consider it as the equation $x^2 + y^2 + z^2 = 1$ or as the colletions of points where $(x,y)$ are points on a and $h$ is the height on the $z$ access so that $x^2 + y^2 = (1-h^2)$.
It is two independent variables and the third based on the combination of the other two.
We can easily map a 2 dimension patch to the surface of the sphere, say, maybe mapping the rectangle $0\le x \le 360$ and $-1 \le z \le 1$ to $(\cos x\times \sqrt{1-z^2}, \sin x \times \sqrt{1-z^2}, z)$ or maybe mapping $(\theta, \phi)\to (\cos\theta\cdot \cos\phi, \sin\theta\cdot \cos \phi, \sin \phi)$.
But mapping a 1 dimensional patch based on one variable (you assume the angle) to a 2 dimensional surface is not at all as intuitively obvious.  Which isn't to say it's impossible. It's a difficult subject.
You can read it here: Space filling Curves
