Subscriber Discussion
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:
What is the point of this? Do you really want to know the answer? Are you quizzing people?
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?
Calculators usually let you input the sensor size in inches regardless of whether you are measuring distances in meters or feet.
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?
So you are implying that all FoV calculators are incorrect or?
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...
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.
03/11/14 04:21pm
Well, 25.4 divided by 3 is 8.46..., so that.
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.
| 03/11/14 05:41pm
I just want to know why we drive on a parkway and park on a driveway.
| 07/21/14 01:52pm
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.
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.
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"}
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...)
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?
Are 1/2.7 kilos bags of sugar common in your area?
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.
Or why the image sensor is square when the lens is round.
Damn, I thought I found something you hadn't, for a change! ;)
B
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!
8.46mm.
Also:
4.121e-5 furlongs.
0.08 hands.
Would that be a diagonal or height or width measurement?
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.
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.
What a great discussion! Who could have guessed that imager size would be overstated by about 50%? It totally invalidates resolution calculations that ignorantly assume stated imager size bears some relation to a physical dimension. Trying to think of charitable interpretations... Could stated size include blade kerf that arises when separating chips from a wafer? Or perhaps it refers to standard size of the chip mounting package?
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