Bursting Balloons !

Bursting Balloons !
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Sunday, August 29, 2010

Ok well I've been doing more tests and calculations on the subject to try and figure out why we lose so much power as we start going into auto-fp mode and think I just worked it out .
Auto-fp flash logically must stay on for longer for slower shutter speeds .
Let's take 1/1000th sec first , the flash must start 'pulsing' before the first curtain opens - then it must last for 1/280th sec [ if that's how fast the curtain moves ] but once the first curtain reaches the other side it must stay open for the duration of the slit that still has to close !
At 1/1000th sec let's say that slit is 1/4 of the frame = another 1/1120th of a second.

Now we look closer at the slit at speeds closer to normal sync .
1/320th shows a much larger slit [ 3X the size of 1/1000th?]
Now after the first curtain moves across the frame [ 1/280th sec] the flash still has to remain on until the second curtain is closed which will now take 3X as long as at 1/1000th = 3/1120th = 1/370th sec extra  !

This would explain why the first drop into auto-fp mode gives us so much less power than 1/2000th .
When I fire the SB24 at 1/16th on the D90 at 1/250th sec I don't see the curtain starting to close yet though sync speed is back at 1/200th .
This means that when the flash goes into fp mode it is doing so over 1/280th sec plus the 1/250th sec that the rear curtain takes to close cutting the total power in half [approximately ] .
So auto-fp would have to push out around half as much power over twice the time as opposed to around 1/250th sec shutter travel at 1/4000th sec - not counting the power before and after the pulse .

So essentially different shutter speeds would vary how long auto-fp stays on , and how much power it pushes out over that time period resulting in the differences we see between 1/250th and 1/2000th in auto-fp mode !

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