I'm considering the problem of triangulating an audio source given two receivers a known distance apart that each detect the sound at (possibly) different times. The only piece of information we have is the time difference ($\Delta t$) between when detector 1 ($d_1$) and detector 2 ($d_2$) detected the sound. (I know this cannot be done with two detectors, and I'm ignoring other aspect of sound waves, like dampening with time and distance. I'm interested purely in the geometric solution.)
Speed of sound at sea level is ~340 m/s; let's say $d_1$ and $d_2$ are 340 meters apart. Starting with the easy cases:
If $\Delta t$ is 1 second, and $d_1$ heard it first, then the sound originated from a point on $\overleftrightarrow{d_1d_2}$ leading away from $d_1$.
If $\Delta t$ is 0 seconds, then the sound originated from some point on a line orthogonal to $\overleftrightarrow{d_1d_2}$ situated equidistant from $d_1$ and $d_2$.
My question: what is the class of functions that could contain the source point when 0 $\lt \Delta t \lt$ 1? What we know is that the difference between the distances from the source point to $d_1$ and $d_2$ is constant. I'm struggling to derive the function given this difference.
(Sorry if this is a poorly worded question for this group. I have no math background.)
Let $p$ and $q$ be two given points. We use $d(x,p)$ to represent the distance between $x$ and $p$. We use $v$ to be the speed of sound, so $v \Delta t$ is the difference in distance the sound went. We want to talk about the set of points $X$ where for $x$ in $X$, the $d(x,p) = d(x,q) + (v \Delta t)$. That is, what points are the same fixed distance further from $p$ than from $q$. (Note that $v \Delta t$ can be negative, giving the English language oddity of being "negatively further" = "closer".)
This set is one branch of a hyperbola. At the cited page, the example given is "in radio navigation, when the difference between distances to two points, but not the distances themselves, can be determined". This is equivalent to your problem.
Take your source points to be the foci of the hyperbola. Using the notation of the asymptotes section of the cited page, the distance from $F_1$ to the vertex of the branch closer to $F_1$ is $c -a$. The distance from $F_2$ is $c+a$, so $v \Delta t = (c+a) - (c-a) = 2a$ and the parameter $a$ in your hyperbola is $\frac{v \Delta t}{2}$. The distance between $F_1$ and $F_2$ is $2c$. Also, $$ b = \sqrt{c^2 - a^2} = \sqrt{\left(\frac{d(F_1, F_2)}{2}\right)^2 - \left(\frac{v \Delta t}{2}\right)^2} $$
This reduces the parameters of the hyperbola to the properties of your scenario. The last thing to do is to set your coordinate axes so that the midpoint between the audio sources is at the origin and the $x$-axis runs through the audio sources. Then plot the branch of the hyperbola that is closer to the source that was heard earlier by the receiver.
To do so, we first solve the equation for this hyperbola (with the coordinates chosen as described in the prior paragraph) $\frac{x^2}{a^2} - {y^2}{b^2} = 1$ for $y$: $$ y = \pm b \sqrt{ \frac{x^2}{a^2} - 1 } = \pm b \sqrt{ \frac{x^2 - a^2}{a^2}} \text{.} $$ This radical only produces real values if $x \geq a$ (for $\Delta t$ indicating the receiver is closer to $F_1$) or if $x \leq -a$ (for $\Delta t$ indicating the receiver is closer to $F_2$). Be sure to keep both the $+$ and $-$ in the $\pm$, since both points are possible.