The image sensor of a particular digital camera contains
2016 × 3024 pixels. The geometry of this sensor is identical to
that of a traditional 35mm camera (with an image size of 24 × 36
mm) except that it is 1.6 times smaller. Compute the resolution of
this digital sensor in dpi
this looks like pure math question how it is related to programing?
DPI is dot per inch, inch is 2.54mm so you divide resolution in dots (pixels) by size of chip in inches.
chip size is:
(24 × 36)/1.6 mm
(24 × 36)/(1.6*2.54) inch
(24 × 36)/(4.064) inch
The DPI is then
(2016 × 3024) / [(24 × 36)/(4.064)]
(2016/24 × 3024/36) * (4.064)
(84 × 84) * (4.064)
(341.376 x 341.376) DPI
So the camera pixel density is 341.376 DPI
Related
I have a MRI data array with the shape 121 × 145 × 121 and voxel size 1.5 mm × 1.5 mm × 1.5 mm. I want to find the regions of AAL atlas in my data. How can I do that in python?
My camera is HIKIROBOT MV-CE060-10UC 6MP USB Camera and its lens is HIKIROBOT MVL-HF2528M-6MPE 25MM lens
Camera Resolution : 3072 x 2048 == 6 MP
more details about camera
more details about lens
and the operating distance 370 mm.
My need is find the Dimensions of object using python. I try some method its not give accurate value.
I search in online to calculate the ppi link of website.I follow the steps to calculate ppi but i don't know the diagonal in inches
diagonal image
That website give some examples to calculate the ppi like
They calculate the ppi of computer screen here thay give diagonal inches of screen, but in my case don't know the daigonal inches. What to do ?
I'd like to calculate camera bandwith. Main question: "How can transmit GigE more than 1 Gbit/s data ?"
----- Camera specs --------
Resolution (HxV) :2590 px x 1942 px
Frame Rate : 14 fps
Mono/Color : Color
Interface : GigE
Pixel Bit Depth : 12 bits
---- Bandwith calculation ---
bit/s = Resolution x ChannelSize(Color) x fps x BitDepth
bit/s = 2590 x 1942 x 3 x 14 x 12
bit/s = 2.535.009.120
Gbit/s = 2.3609
Where am I wrong?
Thanks a lot
Either the data from the camera is compressed already or the pixels are laid out in a Bayer Pattern so there's only 12 bits per pixel, not 36.
I am using OpenCV to generate images with depth of 1 bit to cut in a laser cutter (Github repo here). I save them with:
cv2.imwrite(filepath, img, [cv2.IMWRITE_PNG_BILEVEL, 1])
Each pixel corresponds to 0.05mm (called "scan gap" in the laser cutter). A sample image has 300 x 306 pixels and appears in the laser cutter software (LaserCut Pro 5) with size 30 mm x 30 mm. This corresponds to a resolution of 254 pixels per inch and the uncommon value could be from the software. I want a size of 15 mm x 15.3 mm and want to set a higher resolution to achieve that. I could resize by hand, but if I make a mistake, the pixels are no longer exactly aligned with the scan gap of the laser, resulting in inaccuracies in the engraving.
Does OpenCV have a way to set the resolution or final size of the image?
I'm working on something where an admin puts in a threshold for PPI of an image, for example 35. If the uploaded image has PPI of greater than 35 then return true or else return false.
So I'm finding out the PPI of an image using imageMagick:
identify -format "%x x %y" myimg.png
This gives me numbers, for example, 5.51 PixelsPerCentimeter and I convert them to PixelsPerInch by 5.51 * 2.35
This all works fine. However, I am curious as to how the PPI relates to the zoom factor of an image.
Questions
Does a low resolution (say, 10 PPI) image mean it can't be zoomed in as much as a high resolution image can (say, 72 PPI)?
Well I'm sure a low resolution can be zoomed in at a high percentage but the image quality won't be as good i.e. it will be pixelated?
Is there a better metric that I should be looking at rather than PPI to determine whether an image is high resolution or low resolution.