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1/2 inch vs 35mm lens field of view?


Guest Matti Poutanen

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Guest Matti Poutanen

Hi,

 

first of all, I´m sorry if this is discussed to death in various topics, but I didn´t get a definitive answer with earlier topics.

 

I´m exploring different zoom lenses with wider and longer focal lengths than stock lens for Sony EX3, a 1/2 inch chip HD camera.

 

The 35mm format I´m familiar with, and I have a pretty good notion what I would get with, let´s say 35mm motion picture lens.

 

I´m puzzled with these 1/2 inch ENG video lenses: I´m having a hard time figuring out what kind of range I would have, for example 4.6 to 74 mm zoom in 1/2 inch world. I understood the forum rule of "50mm lens is a 50mm lens is a 50mm lens": the focal length stays the same with different formats, but the field of view is a other matter?

 

So my question is: how I can calculate the field of view of a 1/2 inch video lens so I can do a comparison with 35mm lens field of view?

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Just set up a simple proportion equation. For example, if you want to find the lens that is equivalent in horizontal angle of view to a 50mm lens on S35 you would set the equation up in the form:

 

lens on S35/width of S35 frame=lens on 1/2"/width of 1/2" frame

 

so we get:

 

50mm/24.89mm=X/6.4mm

 

X=12.86mm lens

 

You can also just look at a field of view table, though I'm having trouble in my brief search finding one for 1/2" video. Super 8 is pretty close and a table for that would ballpark it, but not be perfect.

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From what I can gather searching online, the best guess is that the sensor is 6.97 x 3.92mm. Of course, the active pixel area may lie inside that.

 

That makes it a roughly 3.5X magnification difference with Super-35 (24mm wide). So you'd use a 14mm lens on a 1/2" camera to get the same field of view as a 50mm lens on a Super-35 camera.

 

In comparison, a 2/3" sensor is 9.58 x 5.39mm and has a 2.5X magnification factor compared to Super-35.

 

So a 4.6 to 74mm zoom on the Sony EX would be like a 16.1 to 259mm zoom on a Super-35 camera in terms of field of view (a 16:1 zoom seems rather extreme...)

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I found a size of 6.4mm x 4.8mm for 1/2" from this site. This is what my math above is based on. I did a quick search and don't really know about its accuracy.

 

Yep, AFAIK that is correct but only for 4:3 1/2" sensors. But the EX series have native 16:9 sensors... no idea there...

 

Cheers, Dave

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A "normal" lens (for any given format) is the lens providing the same FOV as the naked eye.

A "normal" lens (for any given format) is the focal lens that equals the diagonal of the format.

 

I don't know if one can or how to "calculate" the FOV, but:

 

7.9967 (hence 8mm) is the "normal" for 1/2", as well as the 35 is "normal" for motion picture and "50" is "normal" in still photography film.

Now, the FOV of a "50" is "normal" when used on a 24/36mm frame and the same lens gives you the FOV of an 85 when used on a motion picture camera (due to the crop factor)

35 lens is "normal" on an ARRI/Panavision and "wide" on a film-still Nikon, Canon, Minolta, etc.

I hope this helps.

Cheers,

Dan Diaconu.

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A "normal" lens (for any given format) is the lens providing the same FOV as the naked eye.

A "normal" lens (for any given format) is the focal lens that equals the diagonal of the format.

 

I don't know if one can or how to "calculate" the FOV, but:

 

7.9967 (hence 8mm) is the "normal" for 1/2", as well as the 35 is "normal" for motion picture and "50" is "normal" in still photography film.

Now, the FOV of a "50" is "normal" when used on a 24/36mm frame and the same lens gives you the FOV of an 85 when used on a motion picture camera (due to the crop factor)

35 lens is "normal" on an ARRI/Panavision and "wide" on a film-still Nikon, Canon, Minolta, etc.

I hope this helps.

Cheers,

Dan Diaconu.

 

I wouldn't go so far as to say a normal lens provides the same angle of view as the naked eye. We could argue all day as to what our angle of view is.

 

It's better to say it provides a normal-looking sense of depth as our vision. The depth dimensions looks neither expanded nor flattened.

 

 

 

You can calculate a field of view according to the formula:

 

AOV=(2arctan)D/2F

 

Where D is the dimension in millimeters of the sensor dimension you want to calculate the angle of view of (horizontal angle of view is pretty usual). F is the effective focal length. This will be the actual engraved focal length except in macro situations, in millimeters.

 

 

 

Your normals are also a bit long, BTW. 30mm is exact for half frame 35 and 43mm for full frame. A lot of people use the 32mm prime as their normal for 35mm shooting. It doesn't seem like a lot but each is about 15% shorter than your numbers, which is noticeable.

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Thanks for correcting the definition.

Some set primes include 32, others...35. Some sets have a 40 (aside from 50) and yes, the lens that doesn't change the "normal" perspective is the "normal".

32 is "more normal" :lol: (mwuaaah) than 35 is.

Human vision seems to be the same for all (if you want to be included or excluded)

http://webvision.med.utah.edu/Facts.html

and I must confess: I do not know how many of my cones are still working :blink: (hell... don't even know how many I had to begin with :lol: ).

Cheers,

Dan

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  • 2 months later...

I have a Angeniux 12-120mm lens that came with my Beaulieu R16 . I have not been using this camera for a long while. I used the lens on a ECLAIR ACL II for my last film project.

 

Moving away from film I am increasingly filming on Video these days. Will be great if I could find a way to adapt the C mount lens to fit the Sony EX-3. Any suggestions?

Thanks best

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Is the EX3 just a standard 1/2" B4 mount or something specific that that camera? If it's B4 you should fairly easily be able to find an adapter from somewhere like visual products.

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Apologies if this conversion has already been done in this thread; I didn't see it though. . .

 

Assuming you are using the post-1959 redefinition of the international inch (which is close enough for our purposes here to be used even with pre-1959 lenses), a 1/2-in. lens has the same focal length as a 12.7mm lens (indeed, I have a "5-in." lens that is actually still sold as a 127mm. The name changes, but the focal length remains the same.

 

Something else to keep in mind: a lot of metric lens lengths are rounded customary units. The 28mm is actually 1.1". 180mms are actually 7-1/8". I'm pretty sure 50mm's are a true 2" as well.

 

Hope this helps!

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Math not being my strongest skill, I hit google and found this: http://www.digified.net/focallength/

 

It appears to be talking about still cameras, with "digital compact with 1/2" (8.00) sensor", but half inch sensor is half inch sensor no matter what the camera is, right?

 

Unfortunately due to standards differential, the definition of an inch has varied throughout history.

 

The standard used to be the Imperial Yard bar in London. Unfortunately it was shrinking at a rate of a couple millionths (1/1000000") of an inch every year though, so the British inch that was used in the United States, Canada, South Africa, and Austral-Asia basically started to differ from the comparable standards in all of those countries.

 

Due to treaty, in 1959, because of problems encountered between schematics that were supposed to be identical in international manufacturing, the inch was redefined as exactly 25.4mm. Before this time, the British inch was slightly smaller, the U.S inch (still used technically by surveyors as the "survey inch" was fractionally larger, and the South African, Indian, and Austrailian versions differed ever-so-slightly as well.

 

Hovever, due to Anglo-American cooperation as far back as WWII, the 25.4mm approximation was used prior to '59.

 

Even in the printing industry, which basically is almost entirely non-metric stilll, the reason why a pica is 1/72.27 of an inch is due to this differential between British and American inches when they came up with this unit of measure i the printing industry.

 

So the use of millimeters is basically stable and there is no confusion as to which version is being used. Even meters have changed ever-so-slightly in their definitions. Originally a meter was a millionth (and a mm was a billionth) the distance between Paris, France and the North Pole. Now it is ever-so-slightly smaller, but only because the original distance was improperly measured because fo the assumption that the Earth was a perfect sphere; it is not.

 

But, in any case, perhaps this is too much of a nitpick. Lenses are regularly close to 10% off from their stated focal-lengths anyway. So there is flub room built in there anyway. . .

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Unfortunately due to standards differential, the definition of an inch has varied throughout history.

 

The standard used to be the Imperial Yard bar in London. Unfortunately it was shrinking at a rate of a couple millionths (1/1000000") of an inch every year though, so the British inch that was used in the United States, Canada, South Africa, and Austral-Asia basically started to differ from the comparable standards in all of those countries.

 

Due to treaty, in 1959, because of problems encountered between schematics that were supposed to be identical in international manufacturing, the inch was redefined as exactly 25.4mm. Before this time, the British inch was slightly smaller, the U.S inch (still used technically by surveyors as the "survey inch" was fractionally larger, and the South African, Indian, and Austrailian versions differed ever-so-slightly as well.

 

Hovever, due to Anglo-American cooperation as far back as WWII, the 25.4mm approximation was used prior to '59.

 

Even in the printing industry, which basically is almost entirely non-metric stilll, the reason why a pica is 1/72.27 of an inch is due to this differential between British and American inches when they came up with this unit of measure i the printing industry.

 

So the use of millimeters is basically stable and there is no confusion as to which version is being used. Even meters have changed ever-so-slightly in their definitions. Originally a meter was a millionth (and a mm was a billionth) the distance between Paris, France and the North Pole. Now it is ever-so-slightly smaller, but only because the original distance was improperly measured because fo the assumption that the Earth was a perfect sphere; it is not.

 

But, in any case, perhaps this is too much of a nitpick. Lenses are regularly close to 10% off from their stated focal-lengths anyway. So there is flub room built in there anyway. . .

 

I'm not sure how that is practically helpful to us, but it was kind of interesting anyway. :lol:

 

On the subject of focal length flub room: 10%? That seems like way too much. That's noticeable as in, "Is that really the 32mm? Are you sure that's not the 28?" Have you any way to cite that big a margin of error? Perhaps someone has done a chart test of a bunch of 50mm lenses with a locked off camera or something?

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I'm not sure how that is practically helpful to us, but it was kind of interesting anyway. :lol:

 

On the subject of focal length flub room: 10%? That seems like way too much. That's noticeable as in, "Is that really the 32mm? Are you sure that's not the 28?" Have you any way to cite that big a margin of error? Perhaps someone has done a chart test of a bunch of 50mm lenses with a locked off camera or something?

 

Well, in theory, the focal length is the actual distance from the front tothe rear element. A simple ruler will show what the true focal length is.

 

Yeah, 5-10% margins are pretty normal with all the lenses I've seen. A lot of 200mm lenses are actually 190mm (7-1/2").

 

YOu can also count it by simply measuring the diameter of the F/stop and then plugging that back in to the formula where focal length equals diameter of F/stop as a ratio to the focal length of the lens.

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Well, in theory, the focal length is the actual distance from the front tothe rear element. A simple ruler will show what the true focal length is.

 

Hi-

 

I believe it's more like the distance from the optical center of the lens to the focal point.

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Well, in theory, the focal length is the actual distance from the front tothe rear element. A simple ruler will show what the true focal length is.

 

 

I believe it's more like the distance from the optical center of the lens to the focal point.

 

You're usually spot on but that's not even close, Karl. What you're saying would mean that the rear element of a 200mm prime would be 200mm from the film, which would be impossible with most lens mounts.

 

Patrick's pretty close. That works for simple lenses. The lenses we use (complex lenses, that is) don't have an optical center but really an optical space that is between two nodal points.

 

The focal length of a lens is the distance from the rear nodal point of the lens to its focal point, which is the point at which parallel rays are focused to convergence.

 

There are several ways to calculate or measure the focal length and none of them are just measuring with a ruler, though that gets you 95% of the way on view camera lenses if you just estimate the rear nodal point to be the back surface of the lensboard. Most lenses, especially the lenses we use that are optimized in a dozen different ways, are much more complicated than that.

 

In fact, telephoto lenses (in the optical lens design term, not in the way that we term a long focal length lens a telephoto lens) are designed in such a way that the rear nodal point is actually in front of the lens. That allowed lensmakers to design long lenses and macro lenses that took less bellows extension than a normal design lens of the same focal length. Now it allows designers to make whole series of lenses that share physical dimensions and fit into the same lens mount. It's quite amazing, actually.

 

 

The other way you mention will work fine, though. I've done it a few times when trying to figure out the make and model of old mystery view camera lenses.

Edited by Chris Keth
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You're usually spot on but that's not even close, Karl. What you're saying would mean that the rear element of a 200mm prime would be 200mm from the film, which would be impossible with most lens mounts.

 

Geez, Chris, do I have to pull out my physics textbook on you? :P

 

No, I was saying it is the distance between front and rear element of the lens, not the rear element and the film.

 

I think being 1/2 off, according to Patrick is pretty good, considering I haven't opened a book on the subject in almost 6 mos.

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No, I was saying it is the distance between front and rear element of the lens, not the rear element and the film.

 

It couldn't be that either, though. That would mean that every simple lens has a focal length of zero, wich we know is not true.

 

In any case, I like tossing the info out there. I figure after a couple of years of that I can just give people search keywords rather than answering any questions.:P

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  • 2 weeks later...
From what I can gather searching online, the best guess is that the sensor is 6.97 x 3.92mm. Of course, the active pixel area may lie inside that.

 

That makes it a roughly 3.5X magnification difference with Super-35 (24mm wide). So you'd use a 14mm lens on a 1/2" camera to get the same field of view as a 50mm lens on a Super-35 camera.

 

In comparison, a 2/3" sensor is 9.58 x 5.39mm and has a 2.5X magnification factor compared to Super-35.

 

So a 4.6 to 74mm zoom on the Sony EX would be like a 16.1 to 259mm zoom on a Super-35 camera in terms of field of view (a 16:1 zoom seems rather extreme...)

 

Hello: I just did a little calculation for the 2/3" sensor using your dimensions of 9.58X5.39mm, and came up with a diagonal of 10.99mm.

If this is accurate, does this mean that a 16mm focal-length 2/3"video lens will not cover the standard 16mm film-frame? My approximation for the standard-frame diagonal is 13.25mm.

Therefore, only a 1"format video lens could cover the standard film frame; but would it cover Super 16?

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Yep, AFAIK that is correct but only for 4:3 1/2" sensors. But the EX series have native 16:9 sensors... no idea there...

 

What they do is keep the diagonal constant. That way the same lenses cover both aspect ratios. For instance:

 

2/3" 4:3 is 8.8 mm x 6.6 mm. 2/3" 16:9 is 9.6 mm x 5.4 mm. The diagonal is 11.0 mm.

 

1/2" 4:3 is 6.4 mm x 4.8 mm, 1/2" 16:9 would be 6.97 mm x 3.92 mm. The diagonal is 8.0 mm.

 

 

A rough do it yourself way to measure focal length: Set the focus scale to infinity and the aperture wide open. Pick a nice easy to measure place on the outside of the lens. In bright sunlight, focus the sun on something, and measure from that plane to the chosen place on the lens. Then turn the lens around 180 degrees, focus the sun again and measure again to the same place on the lens. Add the two numbers, divide by two, and that's your approximate DIY answer.

 

 

 

 

 

-- J.S.

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Hello: I just did a little calculation for the 2/3" sensor using your dimensions of 9.58X5.39mm, and came up with a diagonal of 10.99mm.

If this is accurate, does this mean that a 16mm focal-length 2/3"video lens will not cover the standard 16mm film-frame? My approximation for the standard-frame diagonal is 13.25mm.

Therefore, only a 1"format video lens could cover the standard film frame; but would it cover Super 16?

 

You're correct about the 2/3" video lens not covering 16R. Though the safe action area of 8.4x6.3mm

has a diagonal of 10.5mm. So it could work for a TV Xfer. The 11mm diagonal will cover the 1.78/1

area on 16R. Your dimensions for 2/3" sensor are also the same as 1.78/1 16R. But using a 2/3" CCTV lens for a 35mm blow-up? Maybe on a crash camera.

 

The diagonal for a 16R frame is more like 12.7mm. AKA 1/2".

 

The 1" video lens has a diagonal of 16mm, more than enough for S16.

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You're correct about the 2/3" video lens not covering 16R. Though the safe action area of 8.4x6.3mm

has a diagonal of 10.5mm. So it could work for a TV Xfer. The 11mm diagonal will cover the 1.78/1

area on 16R. Your dimensions for 2/3" sensor are also the same as 1.78/1 16R. But using a 2/3" CCTV lens for a 35mm blow-up? Maybe on a crash camera.

 

The diagonal for a 16R frame is more like 12.7mm. AKA 1/2".

 

The 1" video lens has a diagonal of 16mm, more than enough for S16.

 

Gentlemen:

Thank you for your replies to my inquiry, and thank you for the correction of my faulty measurements and arithmetic; this is the first time I've seen in print anywhere the diagonal-length of the 16R frame, and that includes my 1973 edition of "American Cinematographer Manual".

 

I've been out of the active world of film for a long time, but thought I'd ask my question, just in case...

(Incidentally, does anyone anywhere do projection of 16mm film, anymore, and has anyone ever projected Super16 film prints for exhibition?)

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