1.78:1 (1 Viewer)

[AlanBrom]

Question. I read a lot of different DVD reviews on the net. I notice that when titles are released at 1.66:1 or 1.85:1 on their packaging, some reviewers list the title as being 1.78:1. Can anyone explain this, as well as how wide 1.78 is as compared to 1.66 and 1.85?

 

[Patrick Sun]

Your typical 4:3 TV set is 1.33, so on a 4:3 TV, 1.66 would show up with thin black bars on the top and bottom. Then 1.85 would show up with thicker black bars (the picture looking more and more like a rectangular, less squarish), and 2.35 would have even thich black bars (and the picture looks like a shoe box).

Aspect ratio is just telling you the width relative to the height of the video image of the film. The smaller the aspect ratio, it looks like the typical 4x3 TV set (not the newer 16x9 TV). So with 2.35, the width is 2.35 longer than the height of the video image. To create such an aspect ratio, you'll get thicker black bars on the TV displays to preserve the video composition of the film.

Here's a link to the Beginner's FAQ that addresses this issues as well:

http://www.hometheaterforum.com/htfo...908#post511908

 

[John Berggren]

1.78 or 1.77:1 is roughly the AR of a 16x9 television. You'll find many modern "TV on DVD" products in this ratio.

 

[AlanBrom]

Thanks guys. I kind of already knew what you stated. What I want to know is why some DVD reviewers classify 1.66 and 1.85 as 1.78 in their reviews, and how it compares to both of them in terms of ratio height and width?

 

[Josh Steinberg]

1.78:1 (16x9) is only slighty less wide than 1.85:1. For the purpose of home DVD viewing, it's about the same thing.

When a 1.66:1 film is presented in 1.78:1 (16x9), to take advantage of the extra resolution offered by 16x9 encoding, they pillarbox the film slightly (put black bars to the left and the right of the image) so it fits within the 16x9 frame without any cropping of the image.

 

[Scott Kimball]

There are a few reasons...

1. It's sometimes so darned hard to get the facts on what the OAR is, and how it relates to what is actually on the disc.
2. Often, a 1.85:1 film is zoomed / cropped to 1.78:1, with no indication on the box.
3. The difference in ratios from 1.85:1 and 1.78:1 can be offset by overscan on some display devices.
4. Some studios don't include the aspect ratio on the package.

Items (2) and (3) above make it so 1.78:1 and 1.85:1 are virtually interchangeable... whether that's right or wrong is fodder for another debate.

-Scott

 

[Patrick McCart]

Actually, 1.78:1 transfers are usually less cropped than they would be if matted to 1.85:1.

1.78:1 simply adds a few extra lines of resolution and sometimes keeps an image from being overmatted.

 

[Bill Burns]

Alan --

Lemme see if I can muddle my way through an explanation of my own here: occasionally, a film's OAR (Original Aspect Ratio) is listed on a home video release, but the transfer itself offers only an MAR (Modified Aspect Ratio). This is rare for films that have been P&Sd or from which mattes have been removed (most companies correctly label a film as widescreen or full screen, in other words), but when 1.66:1 films are modified to 1.78:1 (presumably for ease of transfer) the correct 1.66:1 ratio may be listed, but a monitor with little or no overscan will reveal that the film is actually presented at 1.78:1.

A recent example of just what you describe is The Christopher Lee Collection. I haven't seen the release, so I don't know if the films are presented at their correct ARs or not, but the box lists three of the four included films at 1.66:1 (oddly, the back cover scan at DVDEmpire doesn't offer an info field for ratios, but when I looked at the box on a Best Buy shelf, this info was presented on the back ). These are 16x9 encoded transfers, and this is perfectly feasible with unaltered 1.66:1 -- the film must simply be windowboxed within a 1.78:1 frame, leaving small black bars to the left and the right, instead of the top and the bottom. When converted to 4x3 by your player (if your monitor is 4x3 and the player properly set), the 1.66:1 image will be displayed in a 1.33:1 box, thus leaving small black bars at the top and the bottom. Displayed at full 16x9 resolution, those bars (narrower, in fact, because the differential in ratios is less) appear instead to the left and the right. And because of the specific pixel resolution of DVD (which I haven't memorized ), anything wider than about 1.5:1 (which in the film world generally means anything 1.66:1 or wider) will benefit from 16x9 encoding. Anything more narrow should be encoded as 4x3. It's worth noting that the "anamorphic" spec for DVD just means the image is encoded (by means, if I've understood correctly from David Boulet, of alternately shaped pixels, but in the same resolution configuration as 4x3*) as a 16x9 window (1.78:1), and anything wider than this leaves black bars at the top and the bottom. When downconverted for a 4x3 set, those black bars are "thicker," because more dead space is left at the top and bottom of a 4x3 window when you place, say, a 2.40:1 image inside of it than is left in a 1.78:1 window when that same 2.40:1 image is configured within it.

But back to your question: in my example, The Christopher Lee Collection, three films are presented, according to the box, as 1.66:1 and encoded as 16x9 (the fourth is probably a 2.39:1 'Scope film, listed on the box as the generic "2.35:1"; if it complies with the later CinemaScope and current wide Panavision formats, it's 2.39:1). Now, according to the recent DVDTalk reviews of each of these three 1.66:1 films, the presentation is actually 1.78:1. Did DVDTalk make a mistake and fail to check the box, relying on what their eyes saw on a 16x9 television with overscan, or did Blue Underground (the DVD company) make the mistake, listing these films as 1.66:1 when they've actually been matted to 1.78:1 (a decision I presume either offers the manufacturer some ease in creating the discs, or, alternately, stems from a misinformed notion that black bars to the left and right of a 1.78:1 screen will bother owners of such screens; overscan means few rear projection television owners will see anything noticeable, and front projection owners tend to be very serious about movies, generally a good indication that OAR is a priority ... so I'd say it never really makes sense to alter these films, but some DVD manufacturers seem to disagree)? I haven't seen the discs, so cannot say. But it's precisely the predicament you describe, Alan.

In determining whether a 1.66:1 film has actually been presented as 1.66:1, on-line reviews are at their most unreliable. Far too many automatically list 1.78:1 if the image fills their widescreen monitor, ignoring issues of overscan. On my own television, overscan could be minimized or eliminated, but my DVD player (a Sony carousel) clips the Avia test chart for overscan to right around 3%; minimizing the image any further reveals dead space, both on Avia and of course on any film disc I play, so I'm forced to leave my television's overscan to 3%. Perhaps I can reduce it further when I next upgrade my player. But at 3% (it's actually a bit more than this horizontally and a bit less vertically, as I recall), I can just see a black bar to the right and/or left on 1.66:1 material (such as Universal's Rear Window). 1.78:1 material fills the screen completely (it's a 4x3 set with a 16x9 chip that automatically compresses the scan lines of the tube into a 16x9 window when it detects 16x9 encoded material). So ... I just have to bite the bullet on films that should be 1.66:1, whether the box says 1.66:1 or 1.78:1, and hope for the best. I have yet to find a 100% reliable review source for this particular quirk.

I'm not sure how the math quite works out -- there's a 12% difference between 66 and 78, but you haven't lost 12% of the image in converting 1.66:1 to 1.78:1, because reducing it to 2.66:1, by 100%, would still leave image on the screen, and so you of course haven't lost 100% of the original image, even though there's a 100% differential between the original value of 1.66:1 and a new value of 2.66:1. The answer is probably simple, which is why I can't seem to find it at the moment. Perhaps someone who has a ready formula for converting ratios such as this into percentage values could share it and post just how much image area is lost when a film intended for 1.66:1 projection is transferred at 1.78:1. At any rate, image is lost, and so the practice is one I discourage.

I hope all of that was of some help, Alan. Earlier explanations offered in the thread are often the cause as well (and because these films use mattes to compose a wide image from an original piece of film measuring around 1.37:1, Patrick's quite right; if a DVD opens its matte from 1.85:1 to the slightly narrower 1.78:1, it can do so, as suggested, by "opening" or reducing the matte, rather than clipping the sides; this is a fairly negligible alteration to original composition that actually increases the amount of exposed image area, though fully "open matte" 4x3 transfers of course drastically recompose any film intended for 1.85:1; while not as objectionable as panning-and-scanning across a wide image -- say a hard matted effects shot, or of course a 'Scope film, or indeed any 1.37:1 aperture film in which production equipment, such as boom mikes, intrude on the image and must be covered -- any significant recomposition of a film's intended ratio should be discouraged, and for this reason "overmatting" 1.66:1 to 1.78:1 is something I never like to see). Each release is its own question mark in this sense.

* As a result, while 4x3 films and 16x9 films have the same full resolution on the format, any film wider than 4x3 (actually anything wider than 1.5:1, because of the pixel structure of the format) should be presented in a 16x9 box, so as to minimize the number of pixels (or, if you like, on a CRT, scan lines) "wasted" to create those black bars at the top and the bottom of the screen. They're actually dead space left over by placing a wide image inside a narrower window, so by creating a 16x9 window, you minimize the amount of dead space, using more of the available pixels to define the image. It isn't, in other words, that a 4x3 image is less defined than a 16x9 image on the format, which I think is a common misperception. It's that any image wider than 1.5:1 wastes pixels in dead space, and you minimize that waste by encoding your wide film in a 16x9 window, rather than a 4x3 window (films that fill the 16x9 window have no dead space, just as films that fill the 4x3 window have no dead space).

Convoluted enough for ya'? Heh. I tend to overcomplicate simple matters in working through them, but I believe the above is accurate. I'm sure Dave or others will correct anything I've improperly conveyed.

 

[Matthew_Millheiser]

*scratches head*

OK, so the Architect was saying that there's been more than one Neo in the Matrix, or that there's been more than one Matrix but with the same Neo?

 

[Adam Tyner]

I'm not sure how the math quite works out -- there's a 12% difference between 66 and 78

(1 - (1.66 / 1.78)) * 100 = 6.7%, not 12%

That should mean that a 1.66:1 transfer is a little over 93% the size of a 1.78:1 image, although I could be doing the math completely incorrectly.

 

[Dan Rudolph]

Adam, your figures are correct.

 

[Bill Burns]

This is what I was getting at: a 1.66:1 rectangle is 66% wider than a square (1:1). A 1.78:1 rectangle is 78% wider than a square. A 2:1 rectangle is twice as wide as a square (100%). Thus there is a twelve percent difference in the width of a 1.66:1 rectangle and a 1.78:1 rectangle (.78-.66 = .12).

Where am I going wrong with your equation, Adam?

(1 - (1.66/1.78)) * 100

1 - 1.66 = -.66. When that is multiplied by 100, one is left with -66. 1 - 1.78 = -.78, which when multiplied by 100 produces -78. So far as I can see, the strict solution to that equation is -66/-78 (see asterisk below), which are actually -6600% and -7800% (-66%/-78% are only accomplished if the entire equation is expressed as the numerator and "100" the denominator, i.e. 66 is 66% of 100, but 66 is 6600% of 1, or 66 times greater)! Even 66% and 78% are of course nothing like the figures you've generated, much less their negative counterparts as per above.

Perhaps the Matrix has corrupted my algebraic reasoning. It's been a few years since I last poured over a formula ....

* Unless the slash is treated as a division, i.e. .66/.78, which yields ~.846. 1 - .846 yields .154, which multiplied by 100 is 15.4. Nothing I can crunch gets me anywhere near that 6.7% figure, based on this formula. What am I overlooking?

 

[Dan Rudolph]

* Unless the slash is treated as a division, i.e. .66/.78, which yields ~.846. 1 - .846 yields .154, which multiplied by 100 is 15.4. Nothing I can crunch gets me anywhere near that 6.7% figure, based on this formula. What am I overlooking?

The slash does mean division. What you're overlooking is that it's 1.66/1.78, not .66/.78.

 

[Adam Tyner]

Thus there is a twelve percent difference in the width of a 1.66:1 rectangle and a 1.78:1 rectangle (.78-.66 = .12).

I think the decimals are throwing you off. You can't get percentages by subtracting decimals like that.

Think of it like this:
1.78 multiplied by ??? is 1.66
(or 1.78 * X = 1.66)

Divide both sides by 1.78, and that'll give you:
??? is 1.66 / 1.78
(X = 1.66 / 1.78)

That works out to that unknown number being around 0.932584, or 93.2584%

So, plugging that number back in the equation:

1.78 times 0.932584 is 1.66
(1.78 * 0.932584 = 1.66)

That means that 1.66 is 93.2584% the size of 1.78, or conversely, that 1.78 is 6.7416% larger than 1.66:1.

 

[David Von Pein]

 

[Gordon McMurphy]

Uh.... I failed math. I SO failed math!

Personally, I blame that girl with the big -- uh, oh man what was her name again?


Gordy

 

[Joseph DeMartino]

uh, oh man what was her name again?

Rhonda. I failed math because of her, too. (Or, "because of her two" depending on how you look at them - it. Depending on how you look at IT!)



Joe

 

[Bill Burns]

Dominique, Danielle, a few other visions to dazzle the eye and heart ... but I never failed math, only to get anywhere with them. All for the best -- I may not be here today otherwise; one of them was engaged!

Eh, but I'm finally up to speed, Adam. I knew the ratios and proportions I described earlier before couldn't be right, because they didn't account for screen area, only the ratio itself. As I mentioned, a rectangle of 1.66:1 is indeed 66% wider than a square of 1:1, and a rectangle 2:1 is twice as wide (100% wider). So there's a difference of 12% in 1.66:1 and 1.78:1, but not 12% less screen real estate (just as there's a difference of 12 cents between $1.66 and a $1.78, but there isn't 12% less money; the reason is because you're starting from a base of $1.00, just as you're starting from a base square of 1:1; you have 12% fewer "cents" in the first example, but not 12% less money, because getting down to 0 cents, 0% of "cents," still leaves you with a dollar. Getting down to 0% in the ratio differential still leaves you with that square of space, 1:1, so even though there is no longer any difference between the values -- the height of the square is exactly the same as its width; there is no difference, i.e. 0% -- you still have a square; you're value isn't 0, only your differential between values. So I knew that wouldn't get me to a correct figure, but I couldn't see the formula).

Your explanation put it right in view, though -- many thanks. In short, the question that must be asked is "what percent of 1.78:1 is 1.66:1?" This will give one a true representation of screen real estate relationships. As a mathematical word problem:

1.66:1 is what percent of 1.78:1?
1.66:1 = x of 1.78:1
1.66:1 = 1.78:1x
1.66 = 1.78x (the ones can be canceled)
1.66/1.78 = 1.78x/1.78
1.66/1.78 = x

~.93258 = x
~.933 = x
x = 93.3%
1.66:1 is 9.3.3% of 1.78:1
1.66:1 is 6.7% smaller than 1.78:1 in screen size

Eureka. Many thanks, Adam. Now that I have the right process in mind, I can work out the loss or gain in screen size for any film. Actual screen area values (inches, centimeters) would be determined by finding height and width measurements, multiplying them, and then subtracting the larger from the smaller between two screen sizes, of course (the specific number of units of screen size lost or gained would then be known), but for a percentage, the above makes perfect sense.

I can rest easy. Hopefully this brain twisting discussion was of some use to Alan, as well.

 

[Gordon McMurphy]

A-ha ha ha!

Great stuff, Joseph!

You wouldn't happen to know why I failed geography, would you? All the girls were pretty plain in that class! That class was on the third floor - musta been the altitude or something. :b


Gordy

 

[Dan Rudolph]

Gordon, how was the teacher?