Velocity Reviews > Re: Testing magnification. Am I doing something wrong?

# Re: Testing magnification. Am I doing something wrong?

Mark M
Guest
Posts: n/a

 08-18-2003

"Rich Bail" <(E-Mail Removed)> wrote in message
news:5%80b.339\$(E-Mail Removed) .net...
> I bought a "3X" Kenko lens for my F717 and wanted to test if the
> magnification was really 3X. I was surprised at the result, which I double
> checked.
>
> Took a photo using my camera lens at max zoom and got an image of (height
> times width) of 1,900 square inches.
> Did the same with the Kenko and got 307.5 SI.
> Divided 307.5 into 1,900 and got a result of 6.2, thus my 3X lens seems to
> be covering an area that is 1/6.2 of the base lens or a 6.2X lens.
> Seems strange that they call it a 3X if it is really 6.2X. Also strange if
> my simple math is wrong.
> I used yardsticks in the photo so I could not mis-measure.
> If accurate with the Kenko I have a 1,178 mm telephoto in 35 mm

equivalent.
>
> Any thoughts?

"3x" refers to multiplication of the lens' focal length...not the field of
view.
So...If you have a 50mm lens, it will behave as a 150mm lens--exhibiting a
similar field fo view.

Rich Bail
Guest
Posts: n/a

 08-18-2003
OK. Lots of different responses. Let me summarize. A 3X lens is supposed to
triple the focal length of the basic lens whatever that is. On my F717 the
effective focal length at full zoom is 190mm equivalent. I take a photo of
something and then measure the height and width of the scene. Specifically
the scene was 50" by 38" high. This means that the 190mm lens covered an
area of 50 X 38 = 1,900 square inches at some distance.

At the same distance and at full zoom (190mm) my 2X lens came out to 37" X
29" = 1,073 square inches. Dividing 1,900 by 1073 = 1.8X not 2X but close
enough.

At the same distance and still at full zoom using my "3X" lens, the scene
area was 20.5 X 15" = 307 square inches. Dividing 1,900 by 307 = 6.2X.

Someone suggested that the way to do this is linear, by which I think he
wants me to divide the relative lengths. Doing this the lens alone produces
50" length, and the 2X produced 37". 50 divided by 37 is 1.4. Not close to
the 2X claim.

Doing the same for the 3X would result in 50" divided by 20.5" or 2.4X way
under the 3X claim.

Logically, we expect the field of view of a 190mm lens to be twice that of a
380 and 3 times that of a 570. Field of view must mean coverage. Coverage
(of a rectangle) is height times width.

Long winded, but still confused.

"Mark M" <(E-Mail Removed)> wrote in message
>
> "Rich Bail" <(E-Mail Removed)> wrote in message
> news:5%80b.339\$(E-Mail Removed) .net...
> > I bought a "3X" Kenko lens for my F717 and wanted to test if the
> > magnification was really 3X. I was surprised at the result, which I

double
> > checked.
> >
> > Took a photo using my camera lens at max zoom and got an image of

(height
> > times width) of 1,900 square inches.
> > Did the same with the Kenko and got 307.5 SI.
> > Divided 307.5 into 1,900 and got a result of 6.2, thus my 3X lens seems

to
> > be covering an area that is 1/6.2 of the base lens or a 6.2X lens.
> > Seems strange that they call it a 3X if it is really 6.2X. Also strange

if
> > my simple math is wrong.
> > I used yardsticks in the photo so I could not mis-measure.
> > If accurate with the Kenko I have a 1,178 mm telephoto in 35 mm

> equivalent.
> >
> > Any thoughts?

>
> "3x" refers to multiplication of the lens' focal length...not the field of
> view.
> So...If you have a 50mm lens, it will behave as a 150mm lens--exhibiting a
> similar field fo view.
>
>

Mark M
Guest
Posts: n/a

 08-18-2003
> Logically, we expect the field of view of a 190mm lens to be twice that of
a
> 380 and 3 times that of a 570. Field of view must mean coverage. Coverage
> (of a rectangle) is height times width.

No.
You are mistaking area for field of view.
Two different things.
Example: If you double the height or width of a square (same thing)...you
actually QUADRUPLE the **area** of the square.
So... Doubling or tripling the field fo view does not translate to doubling
or tripling the area covered by the rectangle (or square).
Draw a picture. You'll get it.

>
> Long winded, but still confused.

SD
Guest
Posts: n/a

 08-19-2003
Mark M wrote:

>>Logically, we expect the field of view of a 190mm lens to be twice that of

>
> a
>
>>380 and 3 times that of a 570. Field of view must mean coverage. Coverage
>>(of a rectangle) is height times width.

>
>
> No.
> You are mistaking area for field of view.
> Two different things.
> Example: If you double the height or width of a square (same thing)...you
> actually QUADRUPLE the **area** of the square.

Area = height * width

If I double the height or width then

Area = (2*height) * width = (2*width) * height

Where did the quadruple come from?

Charlie D
Guest
Posts: n/a

 08-19-2003
> Mark M wrote:
> > No.
> > You are mistaking area for field of view.
> > Two different things.
> > Example: If you double the height or width of a square (same thing)...you
> > actually QUADRUPLE the **area** of the square.

In article <bhtb9t\$cdc\$(E-Mail Removed)>,
SD <(E-Mail Removed)> wrote:

> Area = height * width
> If I double the height or width then
> Area = (2*height) * width = (2*width) * height
> Where did the quadruple come from?

You come from a .EDU account???

Example; H=2, W=2 Area =4
Double W and H
H=4, W=4 Area = 16

Let's forget about the area measurements.
They have no place in the discussion.
That's where the OP got off base in the beginning.

--
Charlie Dilks
Newark, DE USA

Dragan Cvetkovic
Guest
Posts: n/a

 08-19-2003
Charlie D <(E-Mail Removed)> writes:

> > Mark M wrote:
> > > Example: If you double the height or width of a square (same thing)...you
> > > actually QUADRUPLE the **area** of the square.

>
>
> In article <bhtb9t\$cdc\$(E-Mail Removed)>,
> SD <(E-Mail Removed)> wrote:
>
> > Area = height * width
> > If I double the height or width then
> > Area = (2*height) * width = (2*width) * height
> > Where did the quadruple come from?

>
> You come from a .EDU account???
>
> Example; H=2, W=2 Area =4
> Double W and H
> H=4, W=4 Area = 16
>

Yes, but you doubled both height _and_ the width. In everydays' language,
'OR' is usually exclusive (although in mathematics it is not).

Bye, Dragan

--
Dragan Cvetkovic,

To be or not to be is true. G. Boole No it isn't. L. E. J. Brouwer

Charlie D
Guest
Posts: n/a

 08-19-2003
In article <(E-Mail Removed)>,
Dragan Cvetkovic <(E-Mail Removed)> wrote:

> Charlie D <(E-Mail Removed)> writes:
>
> > > Mark M wrote:
> > > > Example: If you double the height or width of a square (same
> > > > thing)...you
> > > > actually QUADRUPLE the **area** of the square.

> >
> >
> > In article <bhtb9t\$cdc\$(E-Mail Removed)>,
> > SD <(E-Mail Removed)> wrote:
> >
> > > Area = height * width
> > > If I double the height or width then
> > > Area = (2*height) * width = (2*width) * height
> > > Where did the quadruple come from?

> >
> > You come from a .EDU account???
> >
> > Example; H=2, W=2 Area =4
> > Double W and H
> > H=4, W=4 Area = 16
> >

>
> Yes, but you doubled both height _and_ the width. In everydays' language,
> 'OR' is usually exclusive (although in mathematics it is not).

Oops!
At least I don't have a .EDU account.
I didn't read Mark's message well enough to see the "or."
Of course siddharthgdalal was correct.
I believe it has the same meaning in math.
"And" is "and," and "or" is "or."
"And" is both of them, "or" is either (1) of them.

--
Charlie Dilks
Newark, DE USA

Dragan Cvetkovic
Guest
Posts: n/a

 08-19-2003
Charlie D <(E-Mail Removed)> writes:

> In article <(E-Mail Removed)>,
> Dragan Cvetkovic <(E-Mail Removed)> wrote:

> > Yes, but you doubled both height _and_ the width. In everydays' language,
> > 'OR' is usually exclusive (although in mathematics it is not).

>
> Oops!
> At least I don't have a .EDU account.
> I didn't read Mark's message well enough to see the "or."
> Of course siddharthgdalal was correct.
> I believe it has the same meaning in math.
> "And" is "and," and "or" is "or."
> "And" is both of them, "or" is either (1) of them.
>

In mathematics, 'x or y' means: x, y or both. If you want to ensure that
only one is chosen, you use 'exclusive or (XOR)'.

Bye, Dragan

--
Dragan Cvetkovic,

To be or not to be is true. G. Boole No it isn't. L. E. J. Brouwer

Charlie D
Guest
Posts: n/a

 08-19-2003
In article <(E-Mail Removed)>,
Dragan Cvetkovic <(E-Mail Removed)> wrote:

> In mathematics, 'x or y' means: x, y or both. If you want to ensure that
> only one is chosen, you use 'exclusive or (XOR)'.

Is that something new to the "computer age,"
or have I really forgotten that much since my last math class
35 years ago?

--
Charlie Dilks
Newark, DE USA

Dragan Cvetkovic
Guest
Posts: n/a

 08-19-2003
Charlie D <(E-Mail Removed)> writes:

> In article <(E-Mail Removed)>,
> Dragan Cvetkovic <(E-Mail Removed)> wrote:
>
> > In mathematics, 'x or y' means: x, y or both. If you want to ensure that
> > only one is chosen, you use 'exclusive or (XOR)'.

>
> Is that something new to the "computer age,"
> or have I really forgotten that much since my last math class
> 35 years ago?
>

AFAIK, it dates at least back to G. Boole (1844, check eg.
http://www.sjsu.edu/depts/Museum/boole.html). Remember the truth table for
'or':

1 or 1 = 1 or 0 = 0 or 1 = 1
0 or 0 = 0.

whereas the truth table for xor is

1 xor 0 = 0 xor 1 = 1
1 xor 1 = 0 xor 0 = 0

Bye, Dragan

--
Dragan Cvetkovic,

To be or not to be is true. G. Boole No it isn't. L. E. J. Brouwer