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

Discussion in 'Digital Photography' started by Mark M, Aug 18, 2003.

1. ### Mark MGuest

"Rich Bail" <> wrote in message
news:5%80b.339\$...
> 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, Aug 18, 2003

2. ### Rich BailGuest

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" <> wrote in message
>
> "Rich Bail" <> wrote in message
> news:5%80b.339\$...
> > 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, Aug 19, 2003

3. ### Mark MGuest

> 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.

Mark M, Aug 19, 2003
4. ### SDGuest

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?

SD, Aug 19, 2003
5. ### Charlie DGuest

> 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\$>,
SD <> 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

Charlie D, Aug 19, 2003
6. ### Dragan CvetkovicGuest

Charlie D <> 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\$>,
> SD <> 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

Dragan Cvetkovic, Aug 19, 2003
7. ### Charlie DGuest

In article <>,
Dragan Cvetkovic <> wrote:

> Charlie D <> 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\$>,
> > SD <> 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

Charlie D, Aug 19, 2003
8. ### Dragan CvetkovicGuest

Charlie D <> writes:

> In article <>,
> Dragan Cvetkovic <> 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

Dragan Cvetkovic, Aug 19, 2003
9. ### Charlie DGuest

In article <>,
Dragan Cvetkovic <> 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

Charlie D, Aug 19, 2003
10. ### Dragan CvetkovicGuest

Charlie D <> writes:

> In article <>,
> Dragan Cvetkovic <> 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

Dragan Cvetkovic, Aug 19, 2003
11. ### Charlie DGuest

In article <>,
Dragan Cvetkovic <> wrote:

> If you interpret 0 as false and 1 as true, what is your interpretation of
> 2? The usual Boolean functions work on {0,1} domain.

You've dragged me around in circles.
I was correct in the beginning to fault siddharthgdalal
for finding fault with Mark's statement of "length OR height."

Since the camera sensor has a fixed aspect ratio, if
you double, triple, or whatever the length OR height,
the other HAS to double, triple, or whatever also.

--
Charlie Dilks
Newark, DE USA

Charlie D, Aug 19, 2003
12. ### Dragan CvetkovicGuest

Charlie D <> writes:

> In article <>,
> Dragan Cvetkovic <> wrote:
>
> > If you interpret 0 as false and 1 as true, what is your interpretation of
> > 2? The usual Boolean functions work on {0,1} domain.

>
> You've dragged me around in circles.

Don't know what you are talking about.

[snip]

> Since the camera sensor has a fixed aspect ratio, if
> you double, triple, or whatever the length OR height,
> the other HAS to double, triple, or whatever also.

Nobody said anything about a fixed aspect ratio, now you are just adding
constraints.

Anyway, never mind.

Bye, Dragan

--
Dragan Cvetkovic,

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

Dragan Cvetkovic, Aug 19, 2003
13. ### Paul H.Guest

"Charlie D" <> wrote in message
news:-berlin.de...
> In article <>,
> Dragan Cvetkovic <> wrote:
>
>
> > 1 or 1 = 1 or 0 = 0 or 1 = 1
> > 0 or 0 = 0.

>
> That looks like an EXclusive "or."
> If it were INclusive, there should be a "1 or 1 = 2" shouldn't there?

You're joking, right? This is Boolean logic where "1" and "0" are formal
conventions for the "true" and "false" of Aristotelian logic. There is no
"2".

OR:
false or false is false
true OR false is true
false OR true is true
true OR true is true

AND:
false and false is false
false and true is false
true and false is false
true and true is true

An exclusive OR between propositions A and B is equivalent to "(A or B) and
(not (A and B))"

P.S. or should be unless I made stupid typo or other mistake.

Paul H., Aug 19, 2003
14. ### Mark MGuest

"Dragan Cvetkovic" <> wrote in message
news:...
> Charlie D <> writes:
>
> > In article <>,
> > Dragan Cvetkovic <> wrote:
> >
> > > If you interpret 0 as false and 1 as true, what is your interpretation

of
> > > 2? The usual Boolean functions work on {0,1} domain.

> >
> > You've dragged me around in circles.

>
> Don't know what you are talking about.
>
> [snip]
>
> > Since the camera sensor has a fixed aspect ratio, if
> > you double, triple, or whatever the length OR height,
> > the other HAS to double, triple, or whatever also.

>
> Nobody said anything about a fixed aspect ratio, now you are just adding
> constraints.

Oh good gravy!!
Since when would a teleconverter change the aspect ratio!!!???
It does not...so aspect ratio remaining the same is assumed.

You guys crack me up.

Mark M, Aug 20, 2003
15. ### Dragan CvetkovicGuest

"Mark M" <> writes:

> "Dragan Cvetkovic" <> wrote in message
> news:...
> > Charlie D <> 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\$>,
> > > SD <> 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).

>
> Ummm... Show me a tele-converter that ONLY multiplies height and not width,
> and I'll show you a pink elephant.
> Also...Since when can you make a larger square that doesn't add both height
> and width?????
>

It's OK Mark. I haven't seen your post (mentioning tele-converters), I was
answering the following post where that context was lost (article copied
and pasted below). Bye, Dragan

> 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\$>,
SD <> 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