In  of phone cameras I came across one data point that intrigued me: They appear to write the sensor size as a fraction, in this example 1/5.8". What exactly does that mean?  My hypothesis was that it was (1/5.8)" = 0.1724" = 0.4379cm, is this correct?  And if so, why would you write it as a fraction?
102284
I'm wondering how sensor sizes are calculated. The Sony RX100 reportedly has a sensor size of 1" which is 13.2 x 8.8 mm according to the press release and various sites.  How does that work? That's a pretty small inch.  The Wikipedia page on  has this chart:   which doesn't really help. (The 1" format is labelled Nikon CX. Chart by Moxfyre under the CC-BY-SA license.)   If full frame/35 mm refers to the horizontal size, what does 1" refer to?
92773
How many Decimal Places are Needed For Accuracy to a Given Number of Significant Figures?  For example: $1234/3.141 = 392.8$, but $1234/3.14159$ changes the result to $392.7$, so how do I am using enough digits of $\pi$ for my result to be accurate to a given number of significant figures?
1693
Let's say I have a micro-4/3rd camera and a full frame camera, both set to 1/60 at f/2.8, taking a picture of the same scene in the same lighting. Will the exposure be the same across both cameras despite the different sensor sizes?  The reason why I'm asking is because of the difference in depth of field between micro-4/3 and full frame sensors. I'm finding that, in order to take a picture of certain scenes with the full frame camera at the same depth of field as the micro-4/3rd camera, I have to increase the aperture, which in turn forces me to crank up the ISO.
63628
Here's an example:  Double d = (1/3); System.out.println(d);   This returns 0, not 0.33333... as it should.  Does anyone know?
5370
I was surprised when I heard zero is plural, and even 1.00001 is plural.  Then, what about following numbers ?  (1) 1.00  which means the value measured  between 0.995-1.004, having possibility that it was exactly 1.  (2) 0.99999.....  which is mathematically equal to 1. Also, are others that are mathemattically equal to 1 than this, treated completely same as 1?  (3) -1  Is this singular? Or negative numbers are generally plural?  (4) i = square root of -1  I know it's weird to count things with imaginary numbers, nonetheless, if you were to count, which form, plural or singular, would you be to use?  That's it.
73882
Most mid- and high-end DLSRs offer two or three sizes for RAW capture.  When the camera is generating the medium or small sized RAW files, how does it make them smaller?  Does it capture less information onto the sensor?  Does it capture the full amount of information and then apply some sort of in-camera compression?  Does it do something else that I'm not describing?
8875
I understand that when sampling from a finite population and our sample size is more than 5% of the population, we need to make a correction on the sample's mean and standard error using this formula:  $\hspace{10mm} FPC=\sqrt{\frac{N-n}{N-1}}$  Where $N$ is the population size and $n$ is the sample size.  I have 3 questions about this formula:   Why is the threshold set at 5%? How was the formula derived? Are there other online resources that comprehensively explain this formula besides  paper?
73179
On , $N_A = 6.022\;141\;29 \times 10^{23} \text{ mol}^{-1}$ has 9 significant figures and a standard uncertainty of $0.000\;000\;27 \times 10^{23} \text{ mol}^{-1}$.  First, can it be written as the following?  $$ N_A = 6.022\;141\;29 \times 10^{23} \pm 0.000\;000\;27 \times 10^{23} \text{ mol}^{-1} $$ If yes, consider the following quote from  (emphasis added):     Whenever you make a measurement, the number of meaningful digits that you write down implies the error in the measurement. For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. To record this measurement as either 0.4 or 0.42819667 would imply that you only know it to 0.1 m in the first case or to 0.00000001 m in the second. You should only report as many significant figures as are consistent with the estimated error. The quantity 0.428 m is said to have three significant figures, that is, three digits that make sense in terms of the measurement. Notice that this has nothing to do with the "number of decimal places". The same measurement in centimeters would be 42.8 cm and still be a three significant figure number. The accepted convention is that only one uncertain digit is to be reported for a measurement. In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m.   and the following quote from  (emphasis added):     The absolute error places bounds telling you that the actual value is expected to fall between x - D x and x + D x. The absolute error is an easy way to report the error and makes it easy to check that you are not reporting too many significant figures. For example, suppose you had a measurement that was reported to be 10.02 ± 0.3 m. Because the reported value is uncertain in the tenths place, it is wrong to report it to the hundredths place. It should be reported to the same decimal place as the absolute error: 10.0 ± 0.3 m. Another point is that the absolute error and the reported value must have the same units.   If there is such a convention (cf. Rice's lab guide) and it is wrong to report more significant figures than is warranted by the absolute error (cf. the joint project's guidelines), and NIST should have adhered to such convention and striven to be correct, is it true that NIST adheres and is correct because the standard uncertainty has two least significant digits (i.e., $27$)?  Had the standard uncertainty been $0.000\;000\;2\text{ mol}^{-1}$, would NIST have reported Avogadro's Number as $N_A = 6.022\;141\;3\times 10^{23}$ with 8 significant figures?
109947
I am trying to figure out how big an object of, say, 50cm height at 100m distance and 600mm focal length would appear on a camera's sensor. The ubiquitous formula  object size in image = focal length * object size / object distance   yields  object size in image = 600mm * .5/100 = 3mm   What I am wondering: Are these 3mm entirely independent of sensor size?  E.g., a specific superzoom model has a 1/2.3" sensor (= 4.55mm height according to one source) and a maximum zoom quoted to have a "35mm equivalent focal length of 600mm". Would the object actually cover 3/4.55 = approx. 66% of this sensor's height = 66% of the image, or does the relation between the sensor size and the "35mm equivalent" have to be taken into account somehow?  (My apologies if this has been asked before - I have found many related questions, but not this one.)
103889
When punctuating the dimensions of something, if both dimensions share the same units, does the punctuation occur after each measurement or just after the last one. For example, if I'm looking at a sheet of paper, which is correct: {8 1/2 x 11"} or is {8 1/2" x 11"}?
51170
first post so please be nice. I am doing an image processing project and my problem is now that I would like to know how big my digital image is in the real world.  I have: digital image: 336x256 pixel  camera focal length: 9mm  camara aperture: f/1,25  camera distance: 40m  pixel pitch: 17 micro m  FOV: 35° x 27°1.889 mr  Is there a way to calculate the real width and height of the image?
75667
I never understood what the f really stands for in the f-stop values, like f/1.8. Wikipedia explains it in various articles, but I still find it very confusing. What I understand, though, is that it has something to do with focal length.  Should I assume, therefore, that bigger focal lengths capture more light? For example, should 85 mm f/1.8 lens capture a lot more light than for example 24 mm f/1.8?  References to back up my confusion:  The article about  goes on to claim:     The amount of light captured by a lens is proportional to the area of   the aperture, equal to:            Where f is focal length and N is the f-number.   The "f-number" is mentioned here. But the article about  claims:     In optics, the f-number of an optical system expresses the diameter of the entrance pupil in terms of the focal length of the lens; in simpler terms, the f-number is the focal length divided by the "effective" aperture diameter.   This seems very recursive. Why does the aperture article refers to both the focal length and the f-number, when the latter article claims that f-number already carries the property of focal length?  What's going on here?
80657
I have a pretty high res sensor on my camera, but sometimes I have the feeling my lens doesn't even get to this resolution. How can I tell if my camera sensor has not enough resolution or if the lens is the limit?  For example I have a 5DMarkII with 22 Megapixel Sensor. And a Canon 35mm 1:2 Lens. (For example aperture at 11 shutter at 1/2000)  How can I visually see whether the bottleneck is the lens or the cameras resolution? Is there any place were I can look up the optical resolution of a lens?
55075
I saw a post on  which links to a video where a person . Is he saying a large portion of the numbers wrong? Back in high school my algebra teacher was extremely picky.  If we were going to say the number 135, we would say one hundred thirty-five.  If a person said one hundred and thirty-five my teacher would interpret that to mean 100.35.  Was my algebra teacher wrong, or did Jon just waste over 80 hours counting to 100,000 incorrectly. I tried searching the web, but searching for ‘numbers and’  doesn’t work very well.  Can you point me to a reference or style guide that describes how to properly say large numbers?
113458
Using the equivalency, I got the answer equal to $1$, however the answer is given as $\frac34$.    also gives $\frac34$.  Could someone tell me that the given answer is wrong or I made a mistake!
90213
It is 10 cm on the map. That is 100 km in real terms.   I am proofreading and it seems odd that the writer always uses that in these cases. I would have said this.     Who is right and who is wrong?
73404
I was doing some exercises on error propagation when I came across this problem:  $4\pi^2/(0,034 \pm 0,004 \space s^2/cm)$  I calculated my uncertainty to be $\Delta$ = 136 $cm/s^2$  And so the hole thing gives me: $(1200 \pm 136) \space cm/s^2$.  Now going into my question I rounded $4\pi^2/0,034 = 1161.12993$ to $1200$ which makes sense since my uncertainty is 136 but in the solution to this problem $136$ is rounded to $100$ and that's not making much sense to me.  So should I always round my uncertainty to the nearest hundred (assuming my uncertainty has 3 digits)?