Pixels Per Inch Calculation Paradox?

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SUMMARY

The forum discussion centers on calculating pixels per inch (ppi) based on a pixel size of 5177 nm and a distance of 0.7 inches. The calculations presented yield conflicting results: 3300 ppi and 4901 ppi. The confusion arises from mixing angular measurements with linear distances, leading to incorrect assumptions about pixel density. The correct conversion indicates that there are 25,400,000 nm per inch, which clarifies the discrepancy in the calculations.

PREREQUISITES
  • Understanding of pixel dimensions and their measurement in nanometers
  • Basic knowledge of trigonometry, specifically tangent functions
  • Familiarity with unit conversions between nanometers and inches
  • Concept of pixels per inch (ppi) in digital imaging
NEXT STEPS
  • Learn about pixel density calculations in digital imaging
  • Study unit conversion techniques between metric and imperial systems
  • Explore trigonometric functions and their applications in geometry
  • Investigate the implications of viewing distance on perceived pixel density
USEFUL FOR

This discussion is beneficial for graphic designers, digital imaging professionals, and anyone involved in pixel density calculations or image quality assessments.

aneikei
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Hello, I could some help. I've reached some sort of paradox. Ultimately I'm trying to calculate the number of pixels per inch, given a pixel size of 5177 nm and extended at a distance of .7inch (.0178 meters)

(A) (60 arcsecs * 0.0178 meters)/206268 = 5177nm pixel per arcmin

(B) 5177 * 60 = 310,620 nm per degree.

(C) tan(55)*.7 = 1 inch (thus 55 degrees = 1 inch)

(D) 310,620 * 55 = 17,084,100 nm per inch

(E) 17,084,100 nm /5177 nm = 3300ppi

However, 5177 nm = .000204 inch
and 1 inch / .000204 = 4901

Thus, 4901 x 5177nm = 25,377,450 nm per inch
25,377,450 nm /5177 nm = 4901 ppi
which is way different from 17,084,100 nm per inch
17,084,100 nm / 5177 nm = 3300 ppi

What did I do wrong? Which is correct? Thank you in advance
 
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I can not follow what you are trying to do. If there is a lens involved with focal lengths and other parameters, please supply some more detail.
 
Charles Link said:
I can not follow what you are trying to do. If there is a lens involved with focal lengths and other parameters, please supply some more detail.

I'm trying to calculate If given 55 degrees and a run of .7 inches what's the number of (5177nm) pixels I can fit in that height?
 
aneikei said:
I'm trying to calculate If given 55 degrees and a run of .7 inches what's the number of (5177nm) pixels I can fit in that height?
It would help to have a diagram. Most likely this is a fairly straightforward and simple geometry problem. A good diagram would be very helpful to see what is needed.
 
Charles Link said:
It would help to have a diagram. Most likely this is a fairly straightforward and simple geometry problem. A good diagram would be very helpful to see what is needed.

Thank you for your help btw.

In the image, the angle is a static 55 degrees. At .7 inches there is less height than at 1 inch. At both, I want to know how many pixels that are 5177 nm in size can fit vertically at each position.
 

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## 2 \tan(\frac{55.11^{\circ}}{2})=\frac{h_1}{.7 }## and ## 2 \tan(\frac{55.11^{\circ}}{2})=\frac{h_2}{1} ##. ## \\ ## Number of pixels=## \frac{h}{d} ## where ## d= ## pixel size. ## \\ ## The height ## h ## is in inches, so to convert to nanometers : ## h_{nm}=h_{in} \cdot 2.54 \cdot 10^7 ##. ## \\ ## Number of pixels=## \frac{h_{nm}}{d_{nm}} ##. (You need to have the units be the same for both ## h ## and ## d ## ). ## \\ ## ## h_1 ## is the case of a distance of .7", and ## h_2 ## is the case of 1". ## \\ ## I don't really see the application of exactly what the purpose of this calculation is, but perhaps what I gave you is helpful. ## \\ ## In the future, I would recommend that this type of calculation belongs in the homework section (even if it isn't exactly homework). ## \\ ## I am going to ask the Moderators to move this post to the Homework section. In the Homework section, it is a requirement to fill out the homework template. In addition, the student is required to work to the solution, rather than the Homework Helper providing the solution. (See Physics Forums Rules).
 
aneikei said:
In the image, the angle is a static 55 degrees. At .7 inches there is less height than at 1 inch. At both, I want to know how many pixels that are 5177 nm in size can fit vertically at each position.
What is confusing here is that you are mixing angles and distances when there doesn't seem to be any actual connection. 5177nm per pixel is 4906 pixels per inch. Period. It doesn't matter if it's .7 inches away or .7 miles away.

...unless there is something else you aren't telling us, like that this is a projected image...?
 
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Several people have mentioned the advantage of including a diagram. Everyone has some way of producing something adequate with the standard applications that you find on computers. Failing that, a photo of a felt tip on paper diagram would be better than nothing. PF can only work on the information you have provided. A diagram is essential in many situations.
 
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A pixel is a description of a surface. They have height and width, which don't have to be the same. If you know the dimensions you'll need two arc lengths: the base and the height setting the limits.
 
  • #10
I should say a rectangular surface, though other shapes are theoretically possible. Programmers set pixel size in most graphic programs.
 
  • #11
aneikei said:
Hello, I could some help. I've reached some sort of paradox. Ultimately I'm trying to calculate the number of pixels per inch, given a pixel size of 5177 nm and extended at a distance of .7inch (.0178 meters)

(A) (60 arcsecs * 0.0178 meters)/206268 = 5177nm pixel per arcmin

(B) 5177 * 60 = 310,620 nm per degree.

(C) tan(55)*.7 = 1 inch (thus 55 degrees = 1 inch)

(D) 310,620 * 55 = 17,084,100 nm per inch

(E) 17,084,100 nm /5177 nm = 3300ppi

However, 5177 nm = .000204 inch
and 1 inch / .000204 = 4901

Thus, 4901 x 5177nm = 25,377,450 nm per inch
25,377,450 nm /5177 nm = 4901 ppi
which is way different from 17,084,100 nm per inch
17,084,100 nm / 5177 nm = 3300 ppi

What did I do wrong? Which is correct? Thank you in advance

What you're doing wrong is that there aren't 17,084,100 nm per inch. There are 25,400,000. Somehow, both of your calculations are getting this wrong, though one is much closer than the other. You don't need to use angles or viewing distances at all.
 
  • #12
Are there other things called "pixels"? The original question said pixels.
 
  • #13
The etymology is derived from an elision of "picture elements". Pixel. They have at least two dimensions.
 
  • #14
Sanborn Chase said:
Are there other things called "pixels"? The original question said pixels.
reread post #1 then read post #7 for the correct explanation :smile:
 
  • #15
aneikei, are you trying to measure something with your camera?