Finding the size of a photographed image

[asa!!]

Hi. i have a photographic camerawith a resolution of 640x480.i know that at the focus point the object photographed has a length of 2 cm.i want to find the height h of the object.i was thinking that the ratio h/length would equal the ratio 480/640.am i right?thank you

 

[russ_watters]

Photographs are physically wider than they are high by exactly the ratio of pixels. So a square object in the photo is the same number of pixels high as it is wide.

 

[asa!!]

thanks.i am not sure i understood your answer.do you mean that the ratio of the pixels reflect also the ratio between the width and the height of the object captured?thank you very much and i am sorry

 

[HallsofIvy]

The ratio of height to width is not necessarily the ratio of pixels in each direction: it depends on the pixels size of the medium on which you display it. You would need to know "number of pixels per cm horizontally" and "number of pixels per cm vertically".

Those depend on the resolution your screen (or printer) is set for and can set or determined by some computer program commands.

 

[asa!!]

Thank you for your time. I understood what you say.but i am thinking that the ratio of the pixels per cm horizontally to the pixels per cm vertically should be the same since the same screen is used. So the ratio would not be constant whatever screen i use?

 

[russ_watters]

Well, wait - we may be talking about two different things here: A standard photo is taken at an aspect ratio of 4:3 (1.33333) and 640x480 is also 4:3, so if you're counting pixels, two pixels high and two pixels wide are the same distance for the object you photographed.

If you're actually measuring the photo with a ruler on your monitor, the aspect ratio is not necessarily 4:3 (which is why if you rotate a photo 90 degrees, it may not look right, even if it is.

 

[asa!!]

Hi.no i do not measure the width on the monitor.i have performed an experiment and have calculated the horizontal FOV.From the Horizontal FOV i have calculated the actual width of the captured object not from the monitor.as i do not know the vertical FOV i want to use the ratio 480/640 to calculate the actual height of the photographed object.am i right?thanks

 

[russ_watters]

No, you don't use the ratio 480/640 because the pixels are all the same size - there are just more of them (wider field of view) in one direction than the other. The ratio to use is pixels per inch for the object you captured and already measured. Ie, if it is 6 inches wide and you count out 12 pixels in the photo, that's 12/6=2 pixels per inch. If the object is 5 pixels high, then its 5/2=2.5 inches high.

 

[asa!!]

to tell yo the truth i am completely confused.i have a mobile camera.the resolution of the photos it captures is 640x480.what does this mean?that on the horizontal axis it uses640 pixels and on the vertical 480 pixels.am i right?since i know that the actual width of the object, let's say a wall, that it can photgraph is let's say 5m.the vertical field of view is not the same as the horizontal it is smaller.unless the resolution used does not depict the ratioof width to height than the photo will be distorted.am i right?regardeless of the resolution the ratio of the dispalyed photo should always be 4:3 same as the ratio of width:height of the dimensions of the wall.if not the image of the wall will be distorted.am i right? so if i have photographed 5 meters of the wall in wide i should have photographed also 3.75 meters of height of wall.

the size of the pixel , i think does not change.so even if ai change the analysis of the screen to 1024x768 again the photo will have a ratio of 4:3 even if i have added more pixels per cm.

the width of the wall i do not measure it from the picture taken..i marked the wall in meters took a picture and saw the width of the photographed wall.but i cannot mark the height of the wall.so i said to use the ratio 4:3 to calculate it.why do i need the number of pixels per cm?thanks

 

[DaveC426913]

If you can tell the width of the wall, then you can use that (or fractions of it) as a yardstick to measure anything in the picture - horizontally, vertically, diagonally, whatever - as long as it is in the plane of focus.

If your camera captures a scene exactly 4m wide, then yes, it will capture the scene 3m tall (approximately). Note that all lenses will distort straight lines and distances in images, so your measurements will only be approximate.

Tell you what, why don't you post your pic and we'll take a crack at it. If you'd rather not, then at least give us some numbers.

 

[asa!!]

this foto is when the object is at three cm away from camera.it is blurry because i have added a lens that is blury in front of the camera.i only care for an aproximation of the distances.thanks

 

[DaveC426913]

OK, so, yes. Any object in the same plane as that ruler would use the same scale, regardless of direction.

i.e. a scrap of paper that exactly filled the image would be 2 inches wide and 1.5 inches high. A wall that exactly filled the image and was 4m wide, would be 3m high. (discounting distortion such as foreshortening, etc.)

You can also say that the field-of-view is about 80 degrees. The ratio of distance to field-of-view is 3:5. i.e. if you are 3cm from an object, and it fills the image, the object is 5cm wide. You should check this though, as a 3cm test allows for a lot of error. Try something at 1m.

 

[asa!!]

thanks.one last question.how did you find the field of view that is 80 degrees?i found it around 44 degrees.thanks

 

[DaveC426913]

thanks.one last question.how did you find the field of view that is 80 degrees?i found it around 44 degrees.thanks

Double it. You measured only one of two triangles. FOV is from one edge to the other, not from one edge to the centre.

If in doubt, lay down two pieces of string, each starting at the lens and extending away, just visible within the FOV. You'll see immediately that the angle is nearer 80 than 40.

(44 is awfully narrow for a lens that isn't a telephoto. Most everyday-use lenses are more wide angle than tele.)

 

[asa!!]

actually i have doubled it.may be the reason is that i use a mobile for the capturing of the images?

 

[asa!!]

may be you used the entire width of the picture provided?because a divided with two in order to obtain have of it for my calculations.then the angle i double it

 

[asa!!]

hi again.please help.in order to find the FOV of the camera does the capture photo has to be always in focus?because i have taken a fiew measurements and not all were in focus.then i calculated the FOV as the average value of the field of view of each.please help

 

[DaveC426913]

I don't know how you could possibly get 44 degrees or anywhere near it for the FoV.

Consider: If the angle were 60 degrees (wider than you calced), it would be form an equilateral triangle. Thus, a 2-inch wide obejct in the FoV would have edges that are 2 inches distant, making the distance to the centre \sqrt{5}in (1.75in, 4.3cm) - that's 1.5x the 3cm you reported.

And that's at 60 degrees (wider than your calc)! At 44, your lens would have to be even further away to get all that in.

See diagram.



Oh, and yes, focus would have a small effect on size. Unpredictable how much.

Why don't you just post a pic of it? If you'd like, you can PM me. I promise I won't publicize it, or use it for my own benefit. (Unless it's really cool - like a real-live alien or something.)

 

[asa!!]

at a distance of 12,673m it captures an object with length of 10,350m.so i found that tha fov is 44 degrees.i think i am right
no?thyanks

 

[DaveC426913]

at a distance of 12,673m it captures an object with length of 10,350m.so i found that tha fov is 44 degrees.i think i am right
no?thyanks

OK. Those are measurements you hadn't supplied at first. I was dubious about the first measurements - being so small, there was a wide margin for error.

That begs the question though:

What is the size of the object you are trying to measure for real (as opposed to just tests). Because if it's very small, then your very close tests might be more accurate than your very far tests.