Pixels Per Inch Calculation Paradox?

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Discussion Overview

The discussion revolves around calculating the number of pixels per inch (ppi) based on a given pixel size of 5177 nm and a distance of 0.7 inches. Participants explore various calculations and methods to reconcile differing results, while also seeking clarification on the geometry involved in the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a series of calculations leading to two different ppi results: 3300 ppi and 4901 ppi, questioning the source of the discrepancy.
  • Another participant requests additional details about any lenses or focal lengths involved, indicating a need for more context to understand the calculations.
  • Several participants suggest that a diagram would clarify the geometry of the problem, emphasizing the importance of visual aids in such calculations.
  • One participant asserts that the pixel size of 5177 nm translates to 4906 pixels per inch, independent of distance, unless additional context is provided.
  • Another participant points out a potential error in the initial calculations regarding the number of nanometers per inch, suggesting that the correct figure is 25,400,000 nm per inch.
  • There is a discussion about the definition of pixels, with some participants noting that pixels have dimensions and can vary in shape.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and the relevance of angles and distances in determining ppi. There is no consensus on the correct approach or the validity of the calculations presented.

Contextual Notes

Some calculations appear to depend on assumptions about geometry and the relationship between angles and distances, which remain unresolved. The discussion highlights the complexity of the problem and the need for clear definitions and diagrams.

Who May Find This Useful

This discussion may be of interest to individuals involved in optics, imaging, or graphic design, particularly those seeking to understand pixel density calculations and the factors that influence them.

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?