"juky" <juklingos@[EMAIL PROTECTED]
> wrote in message
news:5dd6943a-2f41-4886-b415-4c78906c5b74@[EMAIL PROTECTED]
Apr 6, 12:58 am, "Ken Hart" <kwha...@[EMAIL PROTECTED]
> wrote:
> "juky" <juklin...@[EMAIL PROTECTED]
> wrote in message
>
>
news:9b19cb36-435f-4c4d-a19a-b297b95b1055@[EMAIL PROTECTED]
>
> > Dear all,
>
> > I have a mathematical question regarding relation between zoom
> > factor
> > and magnification. My digital video camera shows me the current
> > zoom
> > factor selected. Focal is 3,5-91 mm. I'm looking for mathematic
> > formulas which give me the zoom factor to be selected so that an
> > object in the shot will appear halved, double, etc. I premise the
> > object distance is unknown.
> > Is there any way?
>
> > Thank you.
> >Juky
>
> If you know what a 'normal' lens focal length is for your camera,
> you simply
> divide the focal in question by the normal focal length, and you get
> the
> magnification. The focal length of a 'normal' lense is basically the
> diameter of a circle that would emcompass the image frame.
>
> For a 35mm camera, 50mm is considered normal. A 28mm lens would be
> 0.56X. A
> 300mm would be 6X. The zoom lens I was shooting with earlier, an
> 85-300mm
> would be 1.7X to 6X. I have no idea what would be a normal focal
> length lens
> for your camera-- a shot in the dark guess would be 20 to 30mm or
> thereabouts, but that's a guess only. The object distance would be
> immaterial.
>
> I'm assuming that you made a typo when you wrote 3 comma 5, and you
> meant 3
> point 5. In which case, a 3.5-91mm zoom is a pretty big range. Are
> you sure
> that's what it says?
Yes it's a big range zoom 3.5-91mm :)
No it's not. Nobody is making a 26:1 zoom lens that I know of.
What is the model number of the camera you are using?
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