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June 13

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Music

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What's the difference between a "blues harp" and a "harmonica"? Carllica4 (talk) 11:30, 13 June 2014 (UTC)[reply]

The Blues Harp would appear to be a technique used in Blues/Jazz music while playing the harmonica. Rojomoke (talk) 12:30, 13 June 2014 (UTC)[reply]
The blues harp is a particular type of harmonica. --Redrose64 (talk) 13:14, 13 June 2014 (UTC)[reply]
"Blues Harp" can refer to both a style of play, or a type of harmonica. E.g. the Hohner Blues Harp [1] can be played in "cross harp" aka "second position" - leading to a blues scale, or it can be played "straight" aka "first position, leading to a major scale. See also Harmonica_techniques. There are only two common "positions" for playing harmonica, but there are at least 5 total, see this nice description here [2]. Finally, I'll add that the Hohner Blues Harps came about far later than playing in the blues harp style of second position. Early bluesmen often used the "Marine Band" model of harmonica. Make sense? SemanticMantis (talk) 15:33, 13 June 2014 (UTC)[reply]

Focal length and HD cameras.

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Back in the bad old days of film cameras, you could calculate the "horizontal field of view" of the camera from knowing the focal length of the lens and that the film was 35mm wide. 2 x atan((35/2)/f)...and knowing that it has a 3:4 aspect ratio, calculate the vertical field of view.

OK - so now a photographer hands me a photo taken on a modern, wide-format digital camera. I need to know the field of view (because I'm mixing some 3D computer graphics into the image) - and he tells me that he used a "60mm lens". How does that translate into field of view?

(I need hard information - guesses and speculation don't help here...I'm more than capable of doing that for myself!  :-)

SteveBaker (talk) 14:59, 13 June 2014 (UTC)[reply]

You need to know the size of the image sensor, then you can use the exact same formula you have above with appropriate substitutions (see also angle of view). You can easily find this out if you know the camera model. 'Wide-format' isn't a normal designation in digital photography, so it's not clear what size sensor it represents. (See also Crop factor#Common crop factors.) TenOfAllTrades(talk) 17:19, 13 June 2014 (UTC)[reply]
I believe the camera is a 1080p HD camera. The problem is that the size of the image sensor isn't frequently specified. In 35mm film, you can grab (say) a 50mm lens and know what kind of a picture you'll get, no matter which camera you're using it with...but with digital cameras, I suppose there might be a wide range of sensor sizes...so is it truly the case that a photographer has to learn a whole new set of lens types for each kind of camera? SteveBaker (talk) 21:50, 13 June 2014 (UTC)[reply]
Well, sort of. In practice, your typical DSLR user with a camera body from one of the Big Two (Canon and Nikon) is going to be dealing with a sensor size that is either the same as a 35mm film frame (so-called 'full frame' sensor; only on the higher-end, more-expensive models) or one that is about two-thirds the size (an APS-C sensor, which is smaller by a factor of 1.5 or 1.6 depending on the manufacturer). The lens mounts are generally designed to be compatible, such that you can put the lens built for a larger sensor on a camera with the smaller sensor; all of your full-frame lenses then effectively are cropped/magnified by the same factor (the so-called 'crop factor'). If I have some older full-frame lenses at 30, 50, 80, and 130mm, say, then moving to a 1.6x crop sensor actually gets me almost the same fields of view using the preceding lens in the series (that is, the field of view of the 50mm on the full-frame sensor is about the same as the 30mm on the crop sensor, and so forth). I then just need to buy a new 20mm (ish) lens to round out the wide end, and I'm good to go—and my old 130mm lens gets to be a pretty long telephoto on the new body.
All that said, if the image(s) you've received haven't been too heavily processed already, there's probably a significant amount of EXIF data embedded in the image file that will tell you the camera make and model, lens information, and sensor size. Open the image in Picasa, then View...Properties. Or open in GIMP, then File...Properties...Advanced. TenOfAllTrades(talk) 22:48, 13 June 2014 (UTC)[reply]
Yeah - the EXIF data is long-gone. This was a video that was split into still-frames. SteveBaker (talk) 16:53, 14 June 2014 (UTC)[reply]
1) Sizes of sensors in DSLR cameras vary indeed. See our article on Digital single-lens reflex camera for the various formats deployed (sometimes by the same manufacturer). The rest is the above trig formula.
2) The 60mm lenses I am familiar with are macro lenses (Canon, Nikkor). This should be obvious from the photo, anyway. In this case, the formulae have to be tweaked (see the article on Angle of view).
3) When combining photos and 3D modes (largely architecture) I generally import the photo as a background and then twiddle the camera´s position / angle of view / etc until “it looks right”. I have to tweak the lighting / insolation anyway so that shadows / reflectivity / turbidity / etc look natural. --Cookatoo.ergo.ZooM (talk) 14:00, 14 June 2014 (UTC)[reply]
The "tweak it until it looks right" approach isn't useful here - this is an automated "augmented reality" system that operates in realtime on video streams - we need to do this by calculation. (Check out http://binocular.io - if you have an iPhone/iPad, you can go to glasses.com and actually run our AR application to let you try on pairs of glasses using augmented reality.) Normally, we get the field of view from the iPhone/iPad - which is easy. But we have some video of some models, shot by a professional photographer that we need to use in the same system - and knowing the FOV is an essential part of the calculations that extract the 3D shape of the user's face.
SteveBaker (talk) 16:53, 14 June 2014 (UTC)[reply]
In which case 60mm may be a suitable focal length to be used for a portrait of the myopic model´s heads.
This site (I am a Pentax user) gives the FOV for a 60mm lens as 27° diagonal, using a crop factor of ~1.5 (I possess two cameras, in one case it is 1.53, in the second case it is 1.54).
Bear in mind that the sensor sizes (see note above) differ, and without knowing the brand / exact model of the camera used the FOV can´t be calculated precisely.
A brief excerpt of our article on APS-C (qv):
Multiplier factors
A crop factor (sometimes referred to as a "focal length multiplier", even though the actual focal length is the same) can be used to calculate the 35 mm equivalent focal length from the actual focal length. The most common multiplier ratios:

• 1.7× — Sigma DP1, Sigma DP2, Sigma SD15, Sigma SD14, Sigma SD10, Sigma SD9, Canon EOS DCS 3†
• 1.62× — Canon EOS 7D, 50D†, 60D, 70D, 600D (T3i/X5), 650D (T4i/X6i)†, 700D (T5i/X7i), 1100D (T3/X50)†, 1200D (T5/X70); Canon EOS M, M2 • 1.57x — Nikon D3100, Nikon D3200
• 1.54× — Pentax K-5 II, K-5†, K-30†, Pentax K-01, K-50, K-500; Samsung NX
• 1.53× — Pentax K-3, Pentax K10D†, Pentax K200D†, Nikon D3300, Nikon D5300
• 1.52× — All Nikon DX format DSLR cameras except D3100, D3200, D3300, and D5300; all Fuji; Sony (except for the full-frame α 850,† α 900,† α 99, α 7 and α 7R); Sigma SD1, Sigma SD1 Merrill, Sigma DP1 Merrill, Sigma DP2 Merrill
• 1.3ׇ — Canon EOS-1D Mark IV†, 1D Mark III†, 1D Mark II† (and Mark II N), EOS-1D†, Kodak DCS 460†, DCS 560†, DCS 660†, DCS 760†, Leica M8, M8.2

Sorry, this is just copy & paste, so the formatting is off. Greetings to Texas. --Cookatoo.ergo.ZooM (talk) 21:11, 14 June 2014 (UTC)[reply]
Just to make sure I get the idea:
1 you have a photo of the model (or a selfie of the prospective customer)
2 you have a poly mesh of the frames available
3 you need to “combine” the two in real time on iOS
4 to avoid spectacular penetrations of your clients prostheses and the customers´s cranium you require a precise fit of the two geometries? --Cookatoo.ergo.ZooM (talk) 21:35, 14 June 2014 (UTC)[reply]