Now with eye-popping graphics! (Hey, it's my first rebus.)
Text version:
$$ \begin{matrix} light-speed & & & & here\\ & & ? & & \\ \end{matrix} $$
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asked Jan 16, 2017 at 21:57
5 Answers
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Could it be a
$corpse$
$c$ (speed of light) + $or$ (deciding which ?) + $pse$ (here = pse site)
The title: An untimely rebus may refer to
- late which has a meaning as a deceased body, or theatrical slang for ill-timed stage behaviour
- the same answer 'corpse' to one of the OP's questions posted the same day - very suspicious!
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4$\begingroup$ Great! You might say the second puzzle rose from the ashes of the first ;) $\endgroup$ Jan 18, 2017 at 9:26
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Could it be
CLEFT?
Explanation:
Light speed is usually denoted by $c$, and the fact that "Here" is on the right indicates that that's our frame of reference, i.e. c is on the left, giving "cleft".
answered Jan 16, 2017 at 22:25
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$\begingroup$ Nice, but the question mark is at the heart of the solution, I'm afraid. $\endgroup$ Jan 17, 2017 at 19:26
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$\begingroup$ @HughMeyers Now I get what you meant by "at the heart of the solution". :) $\endgroup$ Jan 18, 2017 at 15:12
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1$\begingroup$ @Ankoganit ... as in "slightly to the left of center". ;) $\endgroup$ Jan 18, 2017 at 15:16
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$\begingroup$ @HughMeyers You might say it filled the CLEFT in the middle of the solution ;-) $\endgroup$ Jan 18, 2017 at 15:17
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Is it -
Future .(as time-travel is there)--thinking about future--
As-
"Here" is front of "speed of light" , so you have crossed speed of light thus you are in traveling to future i.e time travel
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Rayovac, a famous battery brand
Because
What is untimely light speed? Absent time its just distance, an infinite ray. A ray to here over a stick figure with an empty question? The ennui! It's a ray over a vacuous individual or, to stretch it, 'rayovac'.
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Perhaps
Nothing to see here
From
speed of light $c$ = see
nothing between it and here
answered Jan 17, 2017 at 17:02
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$\begingroup$ Another good guess. The answer, however, is a single word. $\endgroup$ Jan 17, 2017 at 17:27
?part of the rebus? $\endgroup$