Archived posting to the Leica Users Group, 2009/09/05

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Subject: [Leica] math challenged
From: h_arche at yahoo.com (H. Ball Arche)
Date: Sat, 5 Sep 2009 12:22:06 -0700 (PDT)

OK, so I just learned that the M8's have a 0.68 VF. My question is, what 
does that translate into, relative to the crop factor of that sensor?  In 
other words, is it like what an approximately 0.45 VF would show you on a FF 
camera?

Since 0.72 has been the 'standard' VF since the M2, I'd have figured that in 
the M8 they would want to reproduce the same relative field of view, lens 
for lens, narrowing it appropriately. 

If I'd've bothered to wonder about it I would have figured the M8 VF to be 
0.86, or 0.91 like the M3 - something narrower than 0.72. 

The 0.68 field of view VF in the M8 must REALLY favor the wides, taking in a 
much broader field of view than would the 0.58 VF with the same lens on a 
full frame camera.


      


Replies: Reply from mark at rabinergroup.com (Mark Rabiner) ([Leica] math challenged)
Reply from richard.lists at gmail.com (Richard Man) ([Leica] math challenged)
Reply from tedgrant at shaw.ca (Ted Grant) ([Leica] M9!)