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Zoom and magnify relation

 
 
juky
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      04-05-2008
Dear all,

I have a mathematical question regarding relation between zoom factor
and magnification. My 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
 
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Jürgen Exner
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      04-05-2008
juky <(E-Mail Removed)> wrote:
>I have a mathematical question regarding relation between zoom factor
>and magnification.


Those two have nothing to do with each other.
The zoom factor is simply the longest focal length divided by the
shortest focal lenght of a zoom lens. For your example below the zoom
factor would be 26x which seems to be rather large.

The (apparent) magnification on the other hand depends on the focal
length and the size of the sensor. Commonly 50mm focal length is
considered a 1x magnification or 'normal' lens on a 36mm sensor
(correspondingly ~33mm focal length on DX-size sensors).

> My video camera shows me the current zoom factor
>selected. Focal is 3,5-91 mm.


That doesn't make much sense because the zoom factor is a constant for a
given lens and does not change. Do you mean it shows you the current
focal length?

>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.


Assuming you are talking about the focal length rather than the zoom
factor, then the magnification and the focal length are 1:1. A doubling
in focal length also doubles the apparent magnification.
If you know the size of the sensor or the 'normal' focal length (like in
50mm for 36mm sensor) for your camera then you can also directly convert
a given focal length to an apparent magnification factor, e.g. a 400mm
lens corresponds to a 8x magnification on a 36mm sensor.

jue
 
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juky
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      04-06-2008
On Apr 6, 12:40*am, Jürgen Exner <(E-Mail Removed)> wrote:
> juky<(E-Mail Removed)> wrote:
> >I have a mathematical question regarding relation between zoom factor
> >and magnification.

>
> Those two have nothing to do with each other.
> The zoom factor is simply the longest focal length divided by the
> shortest focal lenght of a zoom lens. For your example below the zoom
> factor would be 26x which seems to be rather large.
>
> The (apparent) magnification on the other hand depends on the focal
> length and the size of the sensor. Commonly 50mm focal length is
> considered a 1x magnification or 'normal' lens on a 36mm sensor
> (correspondingly ~33mm focal length on DX-size sensors).
>
> > My video camera shows me the current zoom factor
> >selected. Focal is 3,5-91 mm.

>
> That doesn't make much sense because the zoom factor is a constant for a
> given lens and does not change. Do you mean it shows you the current
> focal length?
>
> >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.

>
> Assuming you are talking about the focal length rather than the zoom
> factor, then the magnification and the focal length are 1:1. A doubling
> in focal length also doubles the apparent magnification.
> If you know the size of the sensor or the 'normal' focal length (like in
> 50mm for 36mm sensor) for your camera then you can also directly convert
> a given focal length to an apparent magnification factor, e.g. a 400mm
> lens corresponds to a 8x magnification on a 36mm sensor.
>
> jue


The additional information I can get right now about video camera is
the image sensor=1/4" CCD
 
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Paul Furman
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      04-06-2008
David Ruether wrote:
> "juky" <(E-Mail Removed)> wrote in message news:(E-Mail Removed)...
>
>> I have a mathematical question regarding relation between zoom factor
>> and magnification. My 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

>
> ???
> Well, there is the basic (and most important, for practical purposes...)
> visual effect of zooming as seen in the viewfinder. I can't see much
> use for anything else, especially since any "zoom factor" indicator
> needed to set the "ZF" on the camcorder would be only an
> approximation, if present at all. Further, applying a zoom factor (if
> you could do it accurately) to a FL gives a good idea of lens coverage
> and changed magnification only at long focal lengths. At short FLs, it
> doesn't (I.E., going from a 35mm equivalent FL of 400mm to 800mm
> would almost exactly double the size of the image, and reduce to very
> nearly one forth the area covered at a given distance - but going from
> a 35mm equivalent FL of 20mm to one of 40mm would not do the
> same thing, due to simple geometry).


I drew a little diagram in CAD for an AP-S sensor doubling the focal
length & got these numbers assuming the angle of view is a simple
x/hourglass geometry:
10mm 135deg 1.4x
20mm 100deg 1.6x
40mm 62deg 1.6x
80mm 33deg 1.8x
160mm 17deg 1.9x
320mm 9deg 2.3x <-maybe rounding errors?
640mm 4deg 2.0x
1280mm 2deg

> Additionally, the focus distance
> must be kept constant since some lenses change their FLs with focus.
> So, fergit it...


 
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Don Stauffer in Minnesota
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      04-06-2008
On Apr 5, 3:21 pm, juky <(E-Mail Removed)> wrote:
> Dear all,
>
> I have a mathematical question regarding relation between zoom factor
> and magnification. My 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


There are several meanings for the term "magnification" in cameras.
In pure optics it means the ratio of the image size to the object
size, and unless you are into macro and microphotography, the number
is less than (usually MUCH less) than one.

I think the meaning you are questioning is the ratio of the current
focal length to the shortest (lens zoomed all the way out). Just
divide the current focal length by the shortest possible focal length.
 
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Chris Malcolm
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      04-10-2008
David Ruether <(E-Mail Removed)> wrote:

> So, if we assume a 24mm x 36mm FF film
> frame, the diagonal (for finding the angle of coverage on the
> format diagonal for a particular focal length lens) is 24mm
> squared (576) plus 36mm squared (1296) added together (to
> get 1872) - and the square root of this is 43.27mm. Let's call
> it 44mm... To find the angle of coverage of a lens with this
> format, a right triangle can be made by drawing a vertical line
> perpendicular to the frame diagonal line, and intersecting it in
> its center (as for a non-shifted lens of a given focal length, with
> the focal length being the length of that new line). That gives us,
> for an 18mm lens, a vertical line side of the triangle of 18mm,
> and a horizontal line side of the triangle of 1/2 of 44mm, or
> 22mm. We now have everything needed except a calculator....
> In a right triangle, the opposite side divided by the adjacent
> side (not the diagonal hypotenuse) gives the "tangent" for the
> angle in degrees between the vertical line and the hypotenuse,
> or 1.22. The tangent angle for that value is 50.66 degrees,
> which is one half of the full angle (remember that the frame
> diagonal was split in two to solve for the angle in a right
> triangle?), so the full angle is 101.32 degrees...
> Continuing --


> 9mm = 135.50 degrees, or 1.34x more than an 18mm lens.
> 18mm = 101.32 degrees, or 1.62x more than a 36mm lens.
> 36mm = 62.86 degrees, or 1.85x more than a 72mm lens.
> 72mm = 33.98 degrees, or 1.96x more than a 144mm lens.
> 144mm = 17.37 degrees, or 1.99x more than a 288mm lens.
> 288mm = 8.74 degrees, or 2x more than a 576mm lens.
> 576mm = 4.37 degrees, or 2x more than a 1152mm lens.
> 1152mm = 2.19 degrees, or 2x more than a 2304mm lens.


> So you can see that as focal lengths are halved or doubled, the
> angle of view and image magnification does not follow those ratios
> toward the wide angle end of the possible lens focal length range,
> though it does for long focal lengths for the format...


Tangents and sines approximate to linearity and each other as angles
become small (you could have done your calculation using sines instead
of tangents).

--
Chris Malcolm http://www.velocityreviews.com/forums/(E-Mail Removed) DoD #205
IPAB, Informatics, JCMB, King's Buildings, Edinburgh, EH9 3JZ, UK
[http://www.dai.ed.ac.uk/homes/cam/]

 
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Paul Furman
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      04-11-2008
David Ruether wrote:

>
> 9mm = 135.50 degrees, or 1.34x more than an 18mm lens.
> 18mm = 101.32 degrees, or 1.62x more than a 36mm lens.
> 36mm = 62.86 degrees, or 1.85x more than a 72mm lens.
> 72mm = 33.98 degrees, or 1.96x more than a 144mm lens.
> 144mm = 17.37 degrees, or 1.99x more than a 288mm lens.
> 288mm = 8.74 degrees, or 2x more than a 576mm lens.
> 576mm = 4.37 degrees, or 2x more than a 1152mm lens.
> 1152mm = 2.19 degrees, or 2x more than a 2304mm lens.
>
> So you can see that as focal lengths are halved or doubled, the
> angle of view and image magnification does not follow those ratios
> toward the wide angle end of the possible lens focal length range,
> though it does for long focal lengths for the format...


OK so field of view doesn't change as quick for the wide angles.
Starting with a super long lens, halving the focal length pretty much
doubles the field of view down to 150mm, then it doesn't get as much
wider as one would think.


 
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Dave Martindale
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      04-11-2008
Paul Furman <(E-Mail Removed)> writes:

>> So you can see that as focal lengths are halved or doubled, the
>> angle of view and image magnification does not follow those ratios
>> toward the wide angle end of the possible lens focal length range,
>> though it does for long focal lengths for the format...


>OK so field of view doesn't change as quick for the wide angles.
>Starting with a super long lens, halving the focal length pretty much
>doubles the field of view down to 150mm, then it doesn't get as much
>wider as one would think.


However, the sizes of objects in the image *is* accurately predicted by
the focal length ratio, all the way to the widest wide angle, as long as
you're still talking about rectilinear lenses (not fisheyes).

Dave
 
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Chris Malcolm
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      04-11-2008
Dave Martindale <(E-Mail Removed)> wrote:
> Paul Furman <(E-Mail Removed)> writes:


>>> So you can see that as focal lengths are halved or doubled, the
>>> angle of view and image magnification does not follow those ratios
>>> toward the wide angle end of the possible lens focal length range,
>>> though it does for long focal lengths for the format...


>>OK so field of view doesn't change as quick for the wide angles.
>>Starting with a super long lens, halving the focal length pretty much
>>doubles the field of view down to 150mm, then it doesn't get as much
>>wider as one would think.


> However, the sizes of objects in the image *is* accurately predicted by
> the focal length ratio, all the way to the widest wide angle, as long as
> you're still talking about rectilinear lenses (not fisheyes).


Except for the widening towards the edges in wide angle rectilinear
perspectives, which has nothing to do with the lens, being a simple
consequence of pinhole projection geometry when the image is viewed
from a narrower perspective than it was taken from (as it usually is
with wida angles).

--
Chris Malcolm (E-Mail Removed) DoD #205
IPAB, Informatics, JCMB, King's Buildings, Edinburgh, EH9 3JZ, UK
[http://www.dai.ed.ac.uk/homes/cam/]

 
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Dave Martindale
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      04-15-2008
A long time ago I wrote:
>>> However, the sizes of objects in the image *is* accurately predicted by
>>> the focal length ratio, all the way to the widest wide angle, as long as
>>> you're still talking about rectilinear lenses (not fisheyes).


"David Ruether" <(E-Mail Removed)> writes:
>> How can this be? If you shoot a very large grid of small squares from a
>> constant distance (assuming constant sensor size and distance - and with
>> the subject grid parallel with the sensor), and keep halving the lens focal
>> lengths, the angles of view will progressively not follow the FL reduction
>> ratio (as I showed earlier), resulting, I would think, in less than the
>> number of squares visible in any direction in the image than a straight
>> following of the ratio would predict. In other words, the size of subjects
>> as rendered in the image is also not proportional to the FL ratio when it
>> comes to short FL lenses (they are larger...). Am I missing something?


>Ah, I just finished a MUCH more detailed coverage of this, which
>is going up now on my web page, listed in the "Articles" page as,
>"On Lens Angles Of View, Magnification, And Perspective. The
>direct URL is --
>http://www.donferrario.com/ruether/l...erspective.htm


Sorry I didn't comment sooner - I never saw this reply. But you are
wrong when you say "resulting ... in less than the number of squares
visible ... than a straight following of ratio would predict".

Say you go from 36 to 18 mm lens focal length. The FOV goes from 62.86
to 101.32 degrees, less than a factor of 2 increase. But, as long as
you are shooting a flat wall with a camera aimed perpendicular to that
wall, you will still get twice as many squares visible. The true field
width, on the plane containing the grid, *is* doubled.

If you don't believe me, draw a scale diagram and measure. Or do the
math using similar triangles. You've decreased the lens-image distance
by a factor of two, while keeping the image width the same, so the
tangent of half the FOV has *exactly doubled*, even though the angle
itself has not doubled. On the subject side, the triangle is similar
to the image-side triangle, so the subject-side tangent of half the FOV
is *also* doubled. But the lens-subject distance has not changed, so
the amount of the subject visible has exactly doubled.

Or the handwaving argument: you haven't doubled the FOV angle, but the
additional squares you can see are being increasingly foreshortened by
the very wide angle they are off-axis, so you get more of them in each
degree of extra visual angle. The two effects cancel, and you get
exactly twice as many squares in not twice as many degrees.

By the way, your argument would be correct if the squares were drawn on
a sphere centered on the lens, since the squares would always appear a
certain number of degrees wide everywhere in the field. But we're
assuming a flat subject, not a spherical one.

As it happens, I write camera software for video games, so I can create
any wideangle lens I want, no matter how extreme, as long as it is a
distortion-free rectilinear lens. When I want, say, "50% more field of
view", I do *not* multiply the camera FOV by 1.5. I divide the
effective lens focal length by 1.5 instead. (This is actually
implemented by multiplying the tangent of half of the FOV angle by
1.5). This gives the desired effect: making all objects 2/3 the size
they were, and showing 1.5 times more width of any plane perpendicular
to the line of sight - even though it does not double the FOV angle.
I just recognize that FOV is nonlinearly related to what I really want
to control, which is magnification.

Dave
 
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