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User avatar #3 - DragonTurk (04/03/2013) [-]
Awkward moment that there's a time delay that light takes to move from the watch to your eye, so the time actually reads "Now minus CxDistance between your eye and the watch"... ******* Relativity...
#16 to #3 - theJENK (04/04/2013) [-]
Wouldn't it be distance/c ?   
d/c= m / (m/s) = s   
   
c*d= m/s * m = m^2/s   
Since you're not looking for rate, you just want seconds    
   
   
Also wouldn't it be "Now + d/c", not minus? because, say it said exactly 3:00, by the time the light reached you it would actually be 3:00 + d/c seconds   
Using your formula, say the distance between your face and the watch was 0.3 meters and the watch said exactly 3:00pm = 15 hours = 5400 seconds    
c*d= (3*10^8 m/s)(0.3m) = 9*10^7 m(m/s) which is a unit of acceleration   
5400 sec - 9*10^7 m(m/s) doesn't really make sense.    
   
Or maybe I'm just retarded and you're right, I dunno, it's late. I'm sorry.
Wouldn't it be distance/c ?
d/c= m / (m/s) = s

c*d= m/s * m = m^2/s
Since you're not looking for rate, you just want seconds


Also wouldn't it be "Now + d/c", not minus? because, say it said exactly 3:00, by the time the light reached you it would actually be 3:00 + d/c seconds
Using your formula, say the distance between your face and the watch was 0.3 meters and the watch said exactly 3:00pm = 15 hours = 5400 seconds
c*d= (3*10^8 m/s)(0.3m) = 9*10^7 m(m/s) which is a unit of acceleration
5400 sec - 9*10^7 m(m/s) doesn't really make sense.

Or maybe I'm just retarded and you're right, I dunno, it's late. I'm sorry.
User avatar #7 to #3 - hukedonfonics (04/04/2013) [-]
don't forget the small delay for your mind to process what you are seeing. It's about 80milliseconds
#4 to #3 - notafunnyguy (04/04/2013) [-]
wouldn't that make watches and clocks around the globe technically innaccurate?
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