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I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

Update

Thanks for all answers. I did some calculations based on the assumption that camera is set straight and there is no distortion effect for simplification.

So let say that object i want to measure is 10 cm width (black continouse line on white background). The distance between object and lens is also 10 cm and focal length is 3.6 mm.

object_width_on_image = 10 [cm] * (3.6 [mm]/ 10[cm]) = 3.6 [mm]

From specification i know that resolution is 3264 x 2448 and image area is 6.18 mm x 5.85 mm. Now, if i want to calculate how many pixels those 10 cm will be i will do simple operations:

object_width_in_pixels = 3264 [px] * (3.6 [mm]/6.18 [mm]) = 1901 [px]

So if 1901 px equals 100 mm, than 1 px is about 0.05 mm which is enough for me.

I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

Update

Thanks for all answers. I did some calculations based on the assumption that camera is set straight and there is no distortion effect for simplification.

So let say that object i want to measure is 10 cm width (black continouse line on white background). The distance between object and lens is also 10 cm and focal length is 3.6 mm.

object_width_on_image = 10 [cm] * (3.6 [mm]/ 10[cm]) = 3.6 [mm]

From specification i know that resolution is 3264 x 2448 and image area is 6.18 mm x 5.85 mm. Now, if i want to calculate how many pixels those 10 cm will be on the image i will do simple operations:

object_width_in_pixels = 3264 [px] * (3.6 [mm]/6.18 [mm]) = 1901 [px]

So if 1901 px equals 100 mm, than 1 px is about 0.05 mm which is enough for me.

I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

Update

Thanks for all answers. I did some calculations based on the assumption that camera is set straight and there is no distortion effect for simplification.

So let say that object i want to measure is 10 cm width (black continouse line on white background). The distance between object and lens is also 10 cm and focal lens is 3.6 mm.

object_width_on_image = 10 [cm] * (3.6 [mm]/ 10[cm]) = 3.6 [mm]

From specification i know that resolution is 3264 x 2448 and image area is 6.18 mm x 5.85 mm. Now, if i want to calculate how many pixels those 10 cm will be i will do simple operations:

object_width_in_pixels = 3264 [px] * (3.6 [mm]/6.18 [mm]) px = 1901 [px]

So if 1901 px equals 100 mm, than 1 px is about 0.05 mm which is enough for me.

I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

Update

Thanks for all answers. I did some calculations based on the assumption that camera is set straight and there is no distortion effect for simplification.

So let say that object i want to measure is 10 cm width (black continouse line on white background). The distance between object and lens is also 10 cm and focal length is 3.6 mm.

object_width_on_image = 10 [cm] * (3.6 [mm]/ 10[cm]) = 3.6 [mm]

From specification i know that resolution is 3264 x 2448 and image area is 6.18 mm x 5.85 mm. Now, if i want to calculate how many pixels those 10 cm will be i will do simple operations:

object_width_in_pixels = 3264 [px] * (3.6 [mm]/6.18 [mm]) = 1901 [px]

So if 1901 px equals 100 mm, than 1 px is about 0.05 mm which is enough for me.

I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

I'm doing a project which involves analyzing an object's width from a photo with given precision of about 0.1 mm. I am to the point where I have to choose which camera to buy in order to obtain the desired photo precision.

Let's say I found a 5MP camera. My question is how can I determine what is the "real" value that corresponds to camera's resolution? I read about DPI which has application in printers, and PPI which has application in displays.

Update

Thanks for all answers. I did some calculations based on the assumption that camera is set straight and there is no distortion effect for simplification.

So let say that object i want to measure is 10 cm width (black continouse line on white background). The distance between object and lens is also 10 cm and focal lens is 3.6 mm.

object_width_on_image = 10 [cm] * (3.6 [mm]/ 10[cm]) = 3.6 [mm]

From specification i know that resolution is 3264 x 2448 and image area is 6.18 mm x 5.85 mm. Now, if i want to calculate how many pixels those 10 cm will be i will do simple operations:

object_width_in_pixels = 3264 [px] * (3.6 [mm]/6.18 [mm]) px = 1901 [px]

So if 1901 px equals 100 mm, than 1 px is about 0.05 mm which is enough for me.