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148573 2009 年 08 月 23 日 11:22 Reading (loading. ..) Comments (0) Category: Personal Diary
First, this is just to discuss the many erroneous data, rather than comparing the actual results of the two, because we all experience to be real results, if the discussion will lead to more war of words, no meaning. But the principle is necessary to correct the error, or will the proliferation of spread.
Error 1: do not understand each type cmos / mos / ccd of the mark size and the actual size of photosensitive area and the cmos / mos / ccd shape.
I've seen in many forums as the data is referenced.
SONY's XR520V a single pixel area is about
: Panasonic SD300's: 63.6%.
SONY's XR520V a single pixel area is about
: Panasonic SD1's: 11.7%.
and referenced by many forums, such as dvwold.tw,
nike air force one high, has Zhongguancun, the Pacific Forum.
SONY's XR520V a single pixel area is about
: Panasonic SD300's: 63.6%
Is that really true? First, this conclusion has two errors in principle, this calculation is only based on xr520's 1/2.88 inch CMOS and 1/4.1 inch type TM300 (sd300) and to calculate their number of pixels. So mark size and the actual sensor area size and the cmos / mos / ccd shape of the ratio of these two errors. First, is the industry standard size label, we look cmos image, as we look at this passage
inches, then converted into units of approximately 14.1 mm mm. but the 1/1.8 inch CCD with 14.1 mm diagonal do it? true if the actual length of 14.1 mm, then there is no problem, and can be chosen to 1/1.8 inch CCD The actual length of the diagonal is only 8.93 mm. that the rest of the 1 / 3 of 5.17 mm in the end Where are you? In fact, marked in this way, the size and the actual size of the difference is not a 1/1.8 inch the sensor size, but the size of the vacuum tube. 1/1.8 inch form since 1950, to show the size and use of vacuum tubes, so it does not refer to the sensor 1/1.8 inch diagonal length, but it circle around the straight length of the tube. In order to help understand, the following cursory look at it and not feel the vacuum tube and the sensor in the size of any mathematical relationship. but also about the size of CCD practical and effective area of the tube 2 / 3 it still uses a vacuum tube diameter of representation.
the actual sensor area that xr520 not 1/2.88 inch diagonal. TM300 certainly no 1/4.1 inch, which is caused by
SONY's XR520V a single pixel area is about
: Panasonic SD300's: 63.6%
one of the causes of this error data. 1/2.88 1/4.1 inch inch and converted to mm (1 英寸 = 25.4mm) are 8.82mm and 6.195mm, because the process is different for each product, which actually we do not know the size of sensitive areas, and brochures are also not marked, only know to use the pixels, how to count the size of each pixel, how to get their proportion of the two models? ? ? ? In fact, there is an important reference data, and that is the equivalent focal length on the instructions are to 35mm (35mm film used in machine size is 36 x 24mm) conversion, see
www.sony.com.cn specification sheet on the xr520 :
mm with a 35mm focal length and the actual conversion value, you can close like a calculated value of 16:9 diagonal size of the region, because 35mm film is 2:3 type, may say that when you use a 16:9 level of conversion, in fact, the two conversion methods are used by manufacturers, are called
1, diagonal angle equal to the conversion method. 2, the photosensitive surface of the horizontal viewing angle (horizontal line through the midpoint of the photosensitive surface of two endpoints and the connection of the lens into the center of the angle) equal to the conversion method. (To tell the truth,
air force 1 high, and now I have not read this one conversion method,
air force one low nike, this kind of conversion method formula is f = 45 * f1/d1, the 45 is how come? Algorithm formula for the diagonal for f = 43.27 * f1/d1, 43.27 is just the diagonal length of 35mm film, d1 is the digital camera photosensitive surface diagonal length) out of their little difference between conversion. Both conversion methods for a search online for a large, interested to see.
Select here to convert
diagonal, although it was mentioned that the slope of the different aspect ratio problem, but to know the lens is round, a diagonal angle to the local maximum. According to lens imaging formula 1 / f = 1 / u +1 / v, for dv, dc and other such systems, compared to object distance u v, f is much larger under the orders of magnitude, especially u from the set to infinity, the basic imaging is almost in the focal plane,
thing u is precisely because of the number of large multi-level,
nike air force 1 high, approximately the focal length can be seen from the formula V,
air force one shoes, by a similar triangle can be f = 43.27 * f1/d1 the conversion formula, in order to more precise, we take the diagonal length of 35mm film, 43.267.
First calculate the xr520
dynamic time in the length of the diagonal 16:9: Take the telephoto end conversion d1 = 43.267 * 66/516 = 5.534151mm take wide-angle end of conversion: d1 = 43.267 * 5.5/43 = 5.534151mm dynamic model can be seen at this time to use the diagonal 16:9 image area length of 5.534mm.
then calculate TM300, and we know from the parameter data on
Image Sensor Image Sensor
1/4.1 inch MOS x 3 Total pixels
9.15 megapixels (3.05 megapixels x 3) effective pixels Image
6.21 megapixels (2.07 megapixels x 3) [16:9] Still
7.95 megapixels (2.65 megapixels x 3) [3:2]
7.53 megapixels (2.51 megapixels x 3) [16:9]
6.96 megapixels (2.32 megapixels x 3) [4:3 ] brand of the lens part of the lens
Leica Dicomar F value
F1.8 (wide) / 2.8 (telephoto) Optical Zoom
12x digital zoom
30 times - 700 times the focal length
4-48 mm equivalent 35 mm camera
44.9-539 mm (16:9) [Video]
41.3-496 mm (3:2), 40.8-490 mm (16:9),
45.0-540 mm (4:3) [Still ] Filter diameter
43 mm
When the dynamic 16:9
44.9-539 mm (16:9) [Video], telephoto: d1 = 43.267 * 48/539 = 3.85309mm Wide: d1 = 43.267 * 4/44.9 = 3.85452mm. So when you are in a dynamic 16:9 xr520 diagonal 5.534mm, TM300 is about 3.854mm.
we look at the situation with the pixels, xr520 used in a dynamic 16:9 415w pixels, TM300 dynamic 16:9 using 207w * 3 = 621w pixels
now to calculate the size of individual pixels: xr520 16x9x (5.534 ^ 2) / 337 = 13.086mm2 it is 13.086mm ^ 2/415w pixels
TM300 is (16:9 dynamic time) 16x9x (3.854 ^ 2) / 337 = 6.3468mm ^ 2, three in the dynamic use of the total area of 16:9 19.0404mm ^ 2
19.0404mm ^ 2/621W pixels so instead of a single pixel is xr520 larger than 2% TM300 multi-point, basically the same size. For
pixels, and the minimum value of equivalent focal length at this time, see parameter table. So more likely XR520 is a 4:3 type of CMOS, while the TM300 in essence the equivalent focal length as the minimum value of 16:9, 3:2 still images slightly larger point, however, the use of the 795w pixels 3:2 , 16:9 using a 753w pixels. May be used in the last 3 mos TM300 is a 3:2 type. If the total area of 4:3 type and 3:2 models, and combining all the pixels corresponding to the equivalent focal length to count, basic xr520 is slightly larger than the TM300, and 2%.
Error 2: The single output of 200W is not enough 1920 * 1080 pixels (about 200w pixels) image pixel RGB space required, at least X3?
us first look at CCD, Mos, Cmos pixels, what is a pixel, in fact, a primary color R, G, or B, a, is a pixel not to confuse the image pixels, image pixel is a RBG information can be decomposed into the true colors of the pixels,
air force 1 shoes, then the traditional Bayer pattern arrangement GRGB ratio r: g: b is 1:2:1, then the 200 million pixel CCD, cmos is not only a 50w red pixels, 500,000 blue, 100w a green, not just the output that is not enough, you need to do interpolation? Wrong! ! ! Because it involves the pixel multiplexing, and interpolation is essentially different, we look at the arrangement of Bayer pattern:
see red as an example, each pixel can be in and around a red 4 blue and green output four pixel RGB color information is true, that was used 4 times, 50 million pixels combined with 2 other red just a color output of 200 million pixels without interpolation image information. Why do so 200 complete enough
XR520 use to output 4.15 million high-definition images, not 200 enough? Even with electronic image stabilization is also required so many pixels, ah, but the XR520 is not the arrangement of the traditional Bayer pattern, look at sony's official website map:
can see that the basic arrangement of a unit is 1R1B6G, that is all red and blue 1 / 8, Green is 6 / 8, 4.15 million pixels can be seen xr520 about the same, only 50w more red, the same can be in and around a red 4 blue multiplexing, and therefore just need to 200w output high-definition multi-point pixels.
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