Timeline for How to calculate distance from a subject to get the same field of view with different lenses? [duplicate]
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://photo.stackexchange.com/ with https://photo.stackexchange.com/
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| S Jan 14, 2013 at 22:09 | history | edited | CommunityBot |
insert duplicate link
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| S Jan 14, 2013 at 22:09 | history | closed |
Matt Grum mattdm MikeW Imre Joanne C |
exact duplicate | |
| Jan 14, 2013 at 17:53 | vote | accept | user219882 | ||
| Jan 14, 2013 at 17:11 | review | Close votes | |||
| Jan 14, 2013 at 22:09 | |||||
| Jan 14, 2013 at 16:54 | answer | added | Matt Grum | timeline score: 1 | |
| Jan 14, 2013 at 16:33 | comment | added | user219882 | @BartArondson This equation occured to me too but I haven't found any proof that it's correct so I rather asked here... | |
| Jan 14, 2013 at 16:28 | comment | added | Saaru Lindestøkke |
Isn't it just simply y = x*(35/60)? Not sure about this, hence not posting this as an answer.
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| Jan 14, 2013 at 16:23 | history | asked | user219882 | CC BY-SA 3.0 |