Easy Question: What Is The Size In Mm Of A 1/3" Sensor?

A. 7.6 mm

B. 6.0 mm

C. 3.33 mm

D. none of the above

I'm too proud to ask for another survey, so you must type a letter...

[IPVM UPDATE: Here is a reference chart for various size imagers and actual dimensions:

John Honovich

IPVM

What is the point of this? Do you really want to know the answer? Are you quizzing people?

Rukmini Wilson

1. For me: Knowledge confirmation For member: Knowledge / Demonstration of Mastery.

2. Yes

3. Yes, but in case I'm wrong this gives me an out. (See answer D)

If you try to do PPF calculations yourself where you are using meters for distance, it is natural yet fallacious to rotely convert sensor size from inches to mm. I didn't notice this subtle point in any of your tutorials, perhaps I overlooked it?

John Honovich

IPVM

Calculators usually let you input the sensor size in inches regardless of whether you are measuring distances in meters or feet.

Rukmini Wilson

Don't tell me you believe everything a manufacturer tells you:)

Sensor formats of digital cameras are mostly expressed in the non-standardized "inch" system as approximately 1.5 times the length of the diagonal of the sensor. This goes back to the way image sizes of video cameras used until the late 1980s were expressed, referring to the outside diameter of the glass envelope of the video camera tube. David Pogue of The New York Times states that "the actual sensor size is much smaller than what the camera companies publish – about one-third smaller." For example, a camera advertising a 1/2.7" sensor does not have a sensor with a diagonal of 0.37"; instead, the diagonal is closer to 0.26".[22][23] Instead of "formats", these sensor sizes are often called types, as in "1/2-inch-type CCD - Wikipedia - Image sensor format

Fair to say that I would have taken 10 clams off you in a late night bar bet at the IPVM luau?

John Honovich

IPVM

So you are implying that all FoV calculators are incorrect or?

Rukmini Wilson

No they should be fine because they all use the "real" sizes in their calculations.

I admit I am slightly confused with your position though, since you didn't answer the question.

What you have said to this restated question: Did you know the labeled size of an imager is actually around one third smaller than stated (when stated in inches only)?

If you would have said "No, and I wished you didn't even tell me" then I have a 50" plasma tv that you might be interested in...

Rukmini Wilson

So you are implying that all FoV calculators are incorrect or?

Not incorrect per se but certainly inscrutable.

btw, do you have any insight into this unanswered question? One that seems possibly related to actual sensor size or dimensions...

John, on the graphic for the Tamron CCTV & FA lens calculator, I don't understand why the focal length for the 1/1.8"(.55) sensor size is not somewhere in between the 1/2"(.5) and the 2/3" inch(.6_) size.

Ari Erenthal

Chesapeake & Midlantic

Well, 25.4 divided by 3 is 8.46..., so that.

Rukmini Wilson

In reply to Ari Erenthal

Ari, try using a caliper on a non working sensor of 1/3", measure the diagonal just where the pixels are, see if you see anything close to 8.46...

Alex K

Rukmini Wilson

In reply to Alex K

So Alex what measurement exactly does the 1/4" in your picture refer to, h,w,l,d ?

Brian Rhodes

IPVMU Certified

In other news, a 2"X4" board doesn't really measure 2 inches by 4 inches. But we know it's smaller than a 4"x6", which doesn't measure that way either.

Nominal dimensions have a way of becoming a standard metric over time. Call it right, wrong, or indifferent... the metric is what it is.

I'm not seeing the point here. Maybe we should talk about effective pixels vs. actual pixels. Or why the image sensor is square when the lens is round. OR NOT.

Ari Erenthal

Chesapeake & Midlantic

I just want to know why we drive on a parkway and park on a driveway.

Ari Erenthal

•Jul 21, 2014

Chesapeake & Midlantic

In reply to Ari Erenthal

Now that this thread has been linked to from elsewhere, I just want to say that I've been reading this book, and I now know the answer to this question. A parkway is a road that has been designed for scenic and recreational driving, usually landscaped with trees on both sides and a green strip on the median, and usually exluding trucks and other commercial vehicles. Whereas a driveway was originally a private access road leading from a public road to a stable located on private property.

Reading is fundamental, yo.

Rukmini Wilson

In other news, a 2"X4" board doesn't really measure 2 inches by 4 inches. But we know it's smaller than a 4"x6", which doesn't measure that way either.

Great example Brian! So therefore, using your own analogy, you "knew" the right answer just as surely as a carpenter "knows" the true size of lumber? Good work! I'll mark you down for answer B, because you must of forgotten to type it.

I'm not seeing the point here

Maybe if your carpenter built you a house from plans not knowing the real size of lumber you might get the point. OR NOT.

p.s. May I suggest dropping this easily refuted "we all know" line of argumentation and switch to the "maybe we don't know but who the hell cares" approach as championed most notably by John? Its way more promising.

Brian Rhodes

IPVMU Certified

If I offended you, I apologize for that.

I see no value in raising the question the way it was raised. Fractions are approximations to begin with, it's being treated like a precise number here.

We know (and you do too) that 2x4 is smaller than 4x6 as a general statement of magnitude, just as 1/4" is smaller than 1/3". To that end, cameras should be compared based on effective pixels, not imager size.

This is not a debate, and I'm not arguing anything.

{edit: 1/2" to 1/4"}

Rukmini Wilson

No apology needed, looking back I really sounded wounded, yikes! :)

Maybe I was a bit shaken when to my disbelief I saw that:

1/8" = 2.0 mm

1/6" = 3.0 mm

1/4" = 4.0 mm

1/3" = 6.0 mm

1/2" = 8.0 mm

If only Imperial and Metric could work together so well in the real world!

(One might also have thought that the U.S. is somehow imposing its fractional measurement system on the metric world, when in reality its the other way around...)

Tedor Gligorich

We know (and you do too) that 2x4 is smaller than 4x6 as a general statement of magnitude, just as 1/2" is smaller than 1/3".

this above looks wrong at first glancing

Fractions are approximations to begin with, it's being treated like a precise number here.

U.S.A. only? but over here if I say to good friend Svetozar, "Svet, go to Maxi and get me 1/2.7 kilos of sugar", I better get exactly that! If I say get five kilos, then could be less or more, because its number thats round.

but what i ever get from svet is random because he has half a brain. is that what you mean?

Brian Rhodes

In reply to Tedor Gligorich

IPVMU Certified

Are 1/2.7 kilos bags of sugar common in your area?

Tedor Gligorich

Are 1/2.7 kilos bags of sugar common in your area?

Sometimes i am not sure if you are serious or you are just trying to 'get on my goat'. It is whatever that it is. My area if you must know is Serbia. As for the "bags", you are correct we don't have that sizes of "bags" of sugar.

But when did I say "bags"? Who asks for "bags"? Tourist! Tourist need "bag", Tourist use "bag", Tourist drop "bag" on ground, Tourist want another "bag", Actually tourist want new "bag" or "box" with everything thing he get. That is how he knows it is a good and really "new" one. I really want a "new" tourist.

Locals fill up 'sack' to what exact amount they need, even 1/2.7 kilos, then locals empty 'sack', locals use same 'sack' again.

Try sing this to help remember: Buy-In Bulk-In Balk-an!

Ha! my first IPVM wisecracker! Somebody call Marty Major.

Rukmini Wilson

•Mar 24, 2014

Or why the image sensor is square when the lens is round.

Damn, I thought I found something you hadn't, for a change! ;)

Jim Hall

But I cheated and read the answer because I thought I saw a snake in the grass.

Ten-gallon hat tip to Ari for having enough guts to actually give a real answer!

Carl Lindgren

8.46mm.

Also:

4.121e-5 furlongs.

0.08 hands.

Rukmini Wilson

In reply to Carl Lindgren

Would that be a diagonal or height or width measurement?

Carl Lindgren

That is diagonal, since the 1/3" figure is also diagonal.

Height would be 2.473e-5 furlongs or 0.048 hands. Width would be 3.30e-5 furlongs or .064 hands.

Rukmini Wilson

In reply to Carl Lindgren

IMHO the answer is 6.0mm diagonal, 4.8mm width, 3.6mm height. Alex K was kind enough to Google a chart for us, back a few posts that show different actual sizes. The discussion is now apathetically asking "who cares".

This article is from dppreview (the top digital photography website)

We regularly receive emails from readers confused as to the actual size of sensor used in digital cameras. Sensors (CCD / CMOS) are often refered to with an imperial fraction designation such as 1/1.8" or 2/3", this measurement actually originates back in the 1950's and the time of Vidicon tubes. Those who find the specification sheets for these sensors are then even more confused about the relationship between the fraction and the actual diagonal size of the sensor. Inside you'll find an explanation and a table of common sensor sizes. We'll be adding this information to our glossary for future reference.

Carl Lindgren

Rukmini, what are you using to calculate the conversion from inches to millimeters? My calculator says 1/3" (0.333") equals 8.4582 millimeters. Multiple online calculators also say 0.333" = 8.46mm. For figuring the width and height, obviously it depends on whether the sensor is 4:3 or 16:9 (or 16:10 as a few are). For that, remember that the diagonal measurement equals the square root of the sum of the squares. For 4:3 that would be 3h x 4w = 5d so divide the diagonal measurement by 5 and multiply it by 4 for width or 3 for height.

For 16:9 sensors, 16 squared + 9 squared = 337. The square root of that is 18.36 so width would be diagonal measurement divided by 18.36 and multiplied by 16 and height would be diagonal measurement divided by 18.36 and multiplied by 9.

Rukmini Wilson

In reply to Carl Lindgren

Carl I am not using anything to calculate the conversion. Attempting to do a conversion will not work as the 1/3" designation is only an archaic vestige of a measurement. It only corresponds to the actual imager size in that it is roughly 1.5 times bigger than the actual size, though even this varies quite a bit. It is best to think of it more like a label like ExtraSmall than a measurement. Links above show its vacuum tube origins.

Ever wonder why lens calculators don't let you type in an arbitrary size sensor? Because they have to be 'hard coded' with the real value because there is no straight conversion.

Though for the record I believe, like John and Brian, that it is not important to know what the actual size of a 1/3" imager is, though I would add, as long as one knows its not 1/3"!

Why does that even matter?

Take your own example from a ways back where you stated:

On a similar note, I need to find a new lens calculator. It is simple to calculate lens requirements based on a 1/3" or 1/4" imager when you measure the HFOV and distance to the camera. Not so when you consider the greater variety of modern camera imager sizes (1/1.8", 1/2.5", 1/3.6", 1/5" and 1/6" and the more standard 1/4", 1/3" and 1/2").

And then due to a lack of image size options in your calculator, you just looked up the formula like Tahir did in the same discussion:

But as per their paper http://www.theiatech.com/papers/Resolution_calculation.pdf.
HFOV = Camera Dist * Chip Width / Focal Length

which John then acknowledged as being correct. But if you were to use the listed format size of say 1/3" with the formula you would find you were way off. Just like Chris is in this well meaning but misguided IPVMU verification post:

H Sensor Size: 9.398 mm (1/2.7")
HFOV = Distance / (Focal Length / H Sesnsor Size)
16.75m (55 ft) = 3000m / (1680mm / 9.398mm)
PPF = H Resolution / HFOV
35 PPF = 1920px / 55ft

where the sensor size is really 6.5mm not 9.398.

Carl Lindgren

Hence why I prefer varifocal lenses. I've found lens calculators inaccurate and onerous to use in general.

Horace Lasell