I could use some help with the following situation.
Two physically displaced speakers need to arrive on time, in order to achieve summation, for a given position (green dot) within a listening plane (orange dashed line).
To compensate for the difference in path length ergo time (speed of sound) electronic delay is added to the closest speaker.
$a$, $b$, $d$ & $e$ (at starting temperature) are known.
But when the temperature changes, so does the speed of sound and the same amount of electronic delay won't be suitable anymore.
What I'm trying to calculate is the new position WITHIN the constraint of the listening plane (constant height of both speakers, listening plane and position) AFTER temperature change with the INITIAL AMOUNT of electronic delay.
By observation I know that the point moves to the left when temperature drops and to the right when temperature rises.
At the original position there will be misalignment and as a result comb filtering and the new synchronized position has moved.
AFAICT there's only one solution within the constraint of the listening plane where the time-of-flight to the farthest speaker equals the time-of-flight to the nearest speaker PLUS electronic delay AFTER temperature change.
$a$ & $d$ must now have decreased or increased accordingly
The formula for speed of sound is $c [\mbox{m}/\mbox{s}]=331,3+(0,606\cdot T[°C])$
Your help is sincerely appreciated!
Known quantities: $a, b, d, e, v_1, v_2$
We have the following equalities for the times: $$ t_{\tiny\mbox{blue}} = t_{\tiny\mbox{red}} + \Delta t \iff \frac{c}{v_1} = \frac{f}{v_1} + \Delta t \\ \Delta t = \frac{c-f}{v_1} $$ which gives the time delay. The initial position is $a$. We have $$ c = \sqrt{a^2 + b^2} \quad f = \sqrt{d^2 + e^2} $$ so we can calculate the delay from the known quantities. The new time relation after temperature change is $$ \frac{c'}{v_2} = \frac{f'}{v_2} + \Delta t \\ $$ and these geometric relations: $$ c' = \sqrt{a'^2 + b^2} \quad f' = \sqrt{d'^2 + e^2} $$ with $$ d' = a' - (a-d) $$ Now we can try to solve for $a'$, the only unknown which is left: $$ v_2 \Delta t = \sqrt{a'^2 + b^2} - \sqrt{(a' - (a-d))^2 + e^2} \quad (*) $$ This gives $$ (v_2 \Delta t)^2 = a'^2 + b^2 + (a' - (a-d))^2 + e^2 - 2 \sqrt{(a'^2 + b^2)((a' - (a-d))^2 + e^2)} $$ and $$ (a'^2 + b^2)((a' - (a-d))^2 + e^2)= \frac{(a'^2 + b^2 + (a' - (a-d))^2 + e^2 -(v_2 \Delta t)^2)^2}{4} $$ which leads at most to searching zeros for a fourth degree polynomial in $a'$. Looking at the highest potences we see that the $a'^4$ terms equal out, so it is a third degree or less problem and analytically solvable by roots, if one puts in the effort.
Putting it in a computer algebra system, like WolramAlpha here gives indeed two solutions.
I would implement these results into a small program to see values for different inputs, using a numerical solver (for zeros or fixed points) on the easy $(*)$ or implementing the longer algebraic solutions.