pixel to object mapping formula

Hi,

I forgot the formula.. something to do with tangent and radians.

Supposed I want to know how big a portion of object a certain distance
away would appear in a pixel in a camera.. with and without lens
magnification, what exact formula do you use?

Supposed the ccd chip resolution is 120x120 with pixel pitch of 25
micron.

And the focal length of the lens is 6.76 millimeters, with field of
view of 25 degrees and F-number of 1.5 with aperture of 4.5
millimeters.

How big in area of the an object say 5 meters away would appear to a
single pixel?

Thanks a lot.

On Sat, 20 Aug 2011 16:53:29 -0700 (PDT), Ron Cuaz
wrote:

Supposed I want to know how big a portion of object a certain distance
away would appear in a pixel in a camera.. with and without lens
magnification, what exact formula do you use?

Supposed the ccd chip resolution is 120x120 with pixel pitch of 25
micron.

And the focal length of the lens is 6.76 millimeters, with field of
view of 25 degrees and F-number of 1.5 with aperture of 4.5
millimeters.

Pixel scale is commonly calculated using

A = 206265 * d / F

where A is the angle per pixel is arcseconds
d is the pixels size (or pitch)
F is the focal length of the lens
(206265 is the number of arcseconds in a radian)

In your example,

A = 206265 * .025mm / 6.76mm = 762 arcsec/pixel = 12.7 arcmin/pixel

(which isn't a very realistic equipment combination for most imagers).

On Aug 21, 8:29*am, Chris L Peterson wrote:
On Sat, 20 Aug 2011 16:53:29 -0700 (PDT), Ron Cuaz

wrote:
Supposed I want to know how big a portion of object a certain distance
away would appear in a pixel in a camera.. with and without lens
magnification, what exact formula do you use?

Supposed the ccd chip resolution is 120x120 with pixel pitch of 25
micron.

And the focal length of the lens is 6.76 millimeters, with field of
view of 25 degrees and F-number of 1.5 with aperture of 4.5
millimeters.

Pixel scale is commonly calculated using

A = 206265 * d / F

where A is the angle per pixel is arcseconds
d is the pixels size (or pitch)
F is the focal length of the lens
(206265 is the number of arcseconds in a radian)

In your example,

A = 206265 * .025mm / 6.76mm = 762 arcsec/pixel = 12.7 arcmin/pixel

(which isn't a very realistic equipment combination for most imagers).

Thanks. So at a distance of 5 meters. How do I know how big an area of
the object it would appear to a single pixel? What's the formula? ow.

On Sat, 20 Aug 2011 18:09:19 -0700 (PDT), Ron Cuaz
wrote:

Thanks. So at a distance of 5 meters. How do I know how big an area of
the object it would appear to a single pixel? What's the formula? ow.

That's just basic trig. For a subtended angle A, a distance D, and an
object of size d,

A = 2 * arctan(0.5 * d / D), or
d = 2 * D * tan(A / 2)

So for a pixel scale of 12.7 arcmin/pixel, and a distance of 5 m,

d = 2 * 5 m * tan(12.7 arcmin / 2) = 0.018 m = 18 mm per pixel.

On Aug 21, 9:36*am, Chris L Peterson wrote:
On Sat, 20 Aug 2011 18:09:19 -0700 (PDT), Ron Cuaz

wrote:
Thanks. So at a distance of 5 meters. How do I know how big an area of
the object it would appear to a single pixel? What's the formula? ow.

That's just basic trig. For a subtended angle A, a distance D, and an
object of size d,

A = 2 * arctan(0.5 * d / D), or
d = 2 * D * tan(A / 2)

So for a pixel scale of 12.7 arcmin/pixel, and a distance of 5 m,

d = 2 * 5 m * tan(12.7 arcmin / 2) = 0.018 m = 18 mm per pixel.

Just for comparison to the human eyes. At the same distance of 5
meters. What area can each retina see? I assume a retina is like a
pixel in camera?

Sorry for being dumb in math.

On Aug 21, 9:36*am, Chris L Peterson wrote:
On Sat, 20 Aug 2011 18:09:19 -0700 (PDT), Ron Cuaz

wrote:
Thanks. So at a distance of 5 meters. How do I know how big an area of
the object it would appear to a single pixel? What's the formula? ow.

That's just basic trig. For a subtended angle A, a distance D, and an
object of size d,

A = 2 * arctan(0.5 * d / D), or
d = 2 * D * tan(A / 2)

So for a pixel scale of 12.7 arcmin/pixel, and a distance of 5 m,

d = 2 * 5 m * tan(12.7 arcmin / 2) = 0.018 m = 18 mm per pixel.

tangent of (12.7arcmin/2) is 0.1113 and multiply it by 2 * 5 is
1.113m. I wonder how you got 0.018 m?

On Sat, 20 Aug 2011 20:35:04 -0700 (PDT), Ron Cuaz
wrote:

tangent of (12.7arcmin/2) is 0.1113 and multiply it by 2 * 5 is
1.113m. I wonder how you got 0.018 m?

tan(12.7 arcmin / 2) = tan(0.1058 deg) = 0.0018. Times 2 times 5
meters is 0.018 m.

On Sat, 20 Aug 2011 19:01:56 -0700 (PDT), Ron Cuaz
wrote:

Just for comparison to the human eyes. At the same distance of 5
meters. What area can each retina see? I assume a retina is like a
pixel in camera?

The retina doesn't behave much like the pixels of an electronic
sensor. Nominally, the resolution of the eye is usually taken to be
about one arcminute- about 13 times finer than the resolution of the
camera system you are using as an example.

On Aug 21, 12:45*pm, Chris L Peterson wrote:
On Sat, 20 Aug 2011 20:35:04 -0700 (PDT), Ron Cuaz

wrote:
tangent of (12.7arcmin/2) is 0.1113 and multiply it by 2 * 5 is
1.113m. I wonder how you got 0.018 m?

tan(12.7 arcmin / 2) = tan(0.1058 deg) = 0.0018. Times 2 times 5
meters is 0.018 m.

but 1 radian is equal to 57.2957795131 degree
so converting 12.7/2 or 6.35/ 57.2957795131 is 0.1108284 degrees

but in your case, you seem to use 1 radian = 60 degree to arrive at
6.35/60 = 0.105833

Why did you use 60 degree per radian? is there a particular
significant like maybe some relativistic correction? Just curious why
you need to use 60 instead of 57 since typing 2 numbers in calculator
requires same effort. Thanks.

On Sat, 20 Aug 2011 16:53:29 -0700 (PDT), Ron Cuaz
wrote:
How big in area of the an object say 5 meters away would appear to a

25 microns divided by 6.76 millimeters and then multiply by 5 meters.
The result will be some 18 millimeters.

And, yes, you must do some unit conversions while calculating this. I
suggest you convert all lengths to millimeters.