Dear all,
I have a monitor with a resolution of 1920px X 1080 px (121.5 X 68.5 cm). The diagonal size (cm) of the monitor is 139.5 cm. The viewing distance (cm) from the monkey to the monitor is 57 cm. At 57cm of viewing distance, 1 degree should correspond 1 cm. Is it correct? This means that I could present on the monitor stimuli until about 60 degree of eccentricity on the x-axis and about 34 degree of eccentricity on the y-axis. Is it correct? Finally, with my monitor 1 cm should correspond to about 16 pixels. Why, if I add all the information in the ML's panel as above, it calculates 21.706 pixels per degree? Considering this convertion I am able to present stimuli until about 45 degree of eccentricity. Where is wrong my reasoning? What am I not considering? Otherwise, is there some mistake of conversion?
Thanks in advance who will reply
Best
Marco

Your presumption that 1 degree corresponds to 1 cm is not valid. If that is true, a 180-cm display should be able to cover your whole visual field. Obviously 57 * tand(1) * 90 is not the same as 57 * tand(90).

I inherited the calculation from the original ML and I think it is a good approximation.

Dear Jaewon,
just to update the readers.

My assumption comes from the idea that a good approximation to calculate the view distance of the subject from the monitor is by using the trigonometric function as follows: physical distance(s) on the monitor= Viewdistance(d) * tan(alpha in degrees); My advice is to take into account the following paper (https://www.nature.com/articles/s41598-019-57204-1) where the authors estimated the Horizontal blind spot location.

I checked from other several colleagues working in visual neuroscience and they agree on the approximation of that trigonometric function. I will check again by reading more papers. I do not want doubts about this issue.
Thank you for your help.

Bests
Marco

I am not sure if I made myself understood. You are basically saying the same thing that I said. Just your interpretation is not right.

You can say that 1 degree is 1 cm, only when the eccentricity is small. It is not true anymore if the angle becomes larger. Try different numbers to the formula you mentioned, as below.

57 cm * tan(1 deg) = 0.9949 cm
57 cm * tan(10 deg) = 10.0506 cm
57 cm * tan(30 deg) = 32.9090 cm
57 cm * tan(46.5 deg) = 60.0655 cm
57 cm * tan(60 deg) = 98.7269 cm

The difference between the angle and the corresponding distance jumps up after about 20 degrees. So you cannot say that the max eccentricity of your 120-cm monitor should be 60 degrees, just based on the calculation you did for 1 degree. Again, 57 cm * tan(1 deg) * 60 is not equal to 57 cm * tan(60 deg). The correct max eccentricity is about 46.5 deg as the calculation above shows.

Dear Jaewon,
thank you very much for your reply. Now everything is clearer and we are on the same line with the trigonometric approach.

I was wrong in my original argument about the maximum eccentricity of my monitor, firstly because I have always worked with eccentricities within 25-30 degrees. In fact, as you correctly calculated, there is more or less a constant relationship between 1 degree and 1 cm within these eccentricities at a viewing distance of 57 cm. As soon as you move towards more eccentric positions, this relationship is lost.

However, some empirical measurements I made on the 2nd monitor raised my doubts.
In particular, by setting on ML 1) the diagonal in cm (139.5), 2) the viewing distance of the subject (57 cm), I found the following measurements:

1) at 5° degrees ML presents the stimuli on the monitor at 6.7 cm instead of 4.9868 cm
2) at 10° degrees ML presents the stimuli on the monitor at 13.6 cm instead of 10.0506 cm (as you correctly calculated)
3) at 15° degrees ML presents the stimuli on the monitor at 20.4 cm instead of 15.2731 cm
4) at 20° degrees, ML presents the stimuli on the monitor at 27.3 cm instead of 20.7463 cm
5) at 30° degrees, ML presents the stimuli on the monitor at 41 cm instead of 32.9090 cm (as you correctly calculated).

This means that there is some kind of mismatch especially for the smaller eccentricities, which are the most used in my case. In contrast, they should have more or less a constant relationship between about 1 degree and 1 cm.

Hope that this discussion is useful for everybody.
Bests
Marco

It is obvious to use trigonometry, whether you agree or not. It is invented for this kind of calculations.

The discrepancies in your measures are because the pixels per degree (PPD) is a linear approximation and the size of your screen is beyond the range in which the linear approximation is valid. If you want the ratio of 1 cm per degree, just change the diagonal size so that the PPD becomes close to the number you desire. I think a diagonal size of ~310 cm will do.

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See this post for further information.
https://monkeylogic.nimh.nih.gov/board/read.php?2,1560