Advanced: Actual Image Areas, particularly 480 x 360

[Krelmaneck]

At first I was thinking about posting this topic in General conversations, but this topic is kind of advanced newbie to low intermediate.

I think the article "A Quick Guide To Digital Video Resolution and Aspect Ratio Conversions” is a pretty popular article in the internet. IF you do not know about it, Google it.. It has quite a few sampling matrices listing the actual active image area in many popular resolutions for both PAL and NTSC.

Now, here comes the interesting bit. It does not have EVERYTHING. And that is why I have a question or point to bring up.

Question: What is the active image area in 640 x 360 in a NTSC system? DO NOT ANSWER THE QUESTION YET.


I have a theory. But instead of showing you what I came up with, I am going to show you how I came up with the scanlines for WIDTH.

52 59/90 is the line time for one scanline

Here is my work:

First using the 720 x 480 example first:


52 (59/90) x = 720

52 59/90 x =720
(90) (4739/90) x = 720 (90)
4739x = 64, 800
x= 13.67 which is close to 13.5 (NTSC)


52 59/90 (13.5)
4739/90 (13.5)
63,976.5/90
x= 710.85

Now that may work coincidentally for the width
Now, as I am looking at the NTSC chart nothing really changes the height dramatically unless it is less than 480 otherwise it is pretty much 486.

Now I tried to do the same formula for figuring out the height, but it would only equal that as if it were a width. For example, when I do the above formula, I would end up with 710.85 x 473.9 which is wrong according to the chart for a height. The correct actual image area is 710.85 x 486.



using my so called theory does not work for the height.—only the width. That is why I want to bring up 480 x 360

Now if you were using my dumb method, you could find:

52 59/90 x = 480
4739/90 x = 480
4739x = 43200
x = 9.11584723 which is closest to the real sample of 9 MHZ

so:

52 59/90 (9)
4739/90 (9)
x=473.9 which is OK since it is a WIDTH.


So here is the question: What is the ACTUAL HEIGHT of 360? If I do it the incorrect way I get 355.425 and I know the actual active image area of 480 x 360 is not 473.9 x 355.425 because it did not work in the 720 x 480 example.

Where do I get the correct 355.425?

52 59/90x =360
4739/90x=360
4739x=32, 400
x=6.83 which is close to 6.75
so:

52 59/90 (6.75)
4739/90 (6.75)
31,988.25/90= 355.425

And that probably is not the right answer. IF IT IS, I would be amazed. If it were a width it probably would be correct but not as a height. And there is only 1 sampling rate not 2. LOL. But I just thought I would try something.

So my thinking is that if it is ABOVE a height of 480, the actual image area is going to be the constant 486. Therefore, the only thing I could think is that the actual height would be is 255 that is if any height below 480 is 255 which is SLIGHTLY LESS THAN 1/2 of 480.

So my question is does the VERTICAL LINES reduce themselves in ½ if there is less than 480? Looking at a PAL chart (not my area) I notice there is even a 144. But going back to NTSC, why are the lines reduced dramatically in height below 480? Not that I really want to mess around with anything less than VCD (355.425 x 243 actual), but is there a way to figure out the vertical line actual image area?


Please note that actual image areas are not the finalized resolution put into the encoder since most of them are not divisible by 12 or 16.

Lastly, it would have been nice if the original author of “A quick Guide to Digital Video Resolution and Aspect Ratio Conversions” put in 480 x 360 into their article. But since I am asking a few questions, maybe I can get even a better answer.


Just thought of another question: IS the reason why the author of the article did not include 480 x 360 because it does NOT exists on a NTSC system and is only used on computers? Thus, making 480 x 360 both the sampling area and active area? Or in reality, is there still an active part of a sampling matrix regardless of what type of medium it is on?



Summary of questions if you are lost:
1.) if the sampling matix is 480 x 360 what is the actual height in the image area. The actual width is 473.9
a.) by doing my funky incorrect way for finding actual image area I would
get 486 x 355.425
b.) But 355.425 for the height is most likely wrong. It would be OK IF it were a width. But it is not
c.) I am thinking 486 x 243 sounds more likely like the correct image area
Since the height does not follow my pattern

2.) What is the method for figuring out the actual image area for HEIGHT? For NTSC height, do I just write in 486 if it is above 480 and write 255 if it is below? Is there a method? In other words why does the heights reduce below 480?