I Pixels Per Inch Calculation Paradox?

[aneikei]

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

 

[Charles Link]

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.

 

[aneikei]

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?

 

[Charles Link]

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.

 

[aneikei]

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.

 

[Charles Link]

$2 \tan(\frac{55.11^{\circ}}{2})=\frac{h_1}{.7 }$ 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).

 

[russ_watters]

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...?

 

[sophiecentaur]

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.

 

[Sanborn Chase]

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.

 

[Sanborn Chase]

I should say a rectangular surface, though other shapes are theoretically possible. Programmers set pixel size in most graphic programs.

 

[cjl]

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.

 

[Sanborn Chase]

Are there other things called "pixels"? The original question said pixels.

 

[Sanborn Chase]

The etymology is derived from an elision of "picture elements". Pixel. They have at least two dimensions.

 

[davenn]

Are there other things called "pixels"? The original question said pixels.

reread post #1 then read post #7 for the correct explanation

 

[drummin]

aneikei, are you trying to measure something with your camera?