![[H]ard|Forum](/styles/hardforum/xenforo/logo_dark.png){kind=logo}
This is about LCD panels, not CRTs.
Just the other day I was visiting some offices and noticed some really big monitors. They turned out to be traditional 4:3 20"ers, but I was really surprised at how big they looked even though their bezel was rather narrow. Then I went home and had to admit that my 22"er didn't look as large as before. A week before I was invited to dinner by some friends who have a 52" SD 4:3 projection TV. Next day I went to Circuit City to see their 52" monitors and they didn't look that big anymore either. Of course, the picture was HD and some of the Samsungs and Panasonics looked quite astounding. Yet, the question of their and the monitors' dimensions kept bugging me and wandering whether I was beset with optical illusion. Therefore, here are the results of my musings. Do not be put off by the mathematics, it's simple algebra and some trigonometric function manipulation.
For the very beginning, when we trace the monitor's diagonal the result is the obvious two right triangles. This is why we calculate the area of a triangle as that of a rectangle devided by two. OK, so let's take on one of the triangles, either one since the two triangles are equal. Remember, the ratio between the two adjacent sides to the right angle is 4 (long side) : 3 (short side). The opposite side to the right angle is the hypotenuse. Now it turns out that the relationship with this hypotenuse in a 4:3 triangle is 5. Therefore the relationship that determines all 3 sides is 4:3:5. Since we're talking about 22" diagonals in such a triangle, finding the length of the adjacent legs is easy. We need some kind of a token unit length which multiplied by the numbers in the ratio relationship will give us the lengths.
We know the hypotenuse, so the length of this token unit will be 22"/5 which is 4.4".
Then the other sides will be 3x4.4"=13.2" and 4x4.4"=17.6".
These numbers are for the sides of the rectangle as well. Now we're interested in how much real estate (surface area) the traditional 4:3 monitor has. This is given by the area of the rectangle A = 13.2" x 17.6" = 232. 32" sq.
All well and good so far. However, wide monitors are not of 4:3:5 but 16:10 ratio and the hypotenuse is not in the same relationship with the other sides as in the former setting. Therefore, we can resort to two methods. The first would be the simplest, physically measuring the monitor IF we have one. I did mine and came up with 11.7" x 18.5" (both approximations). This means that area is ~ 216.45" sq. The difference with the traditional 4:3 is therefore 232.32"sq - 216.45" sq = 15.87"sq. How is this possible, though in reality when the difference is calculated rigorously it turns out to be smaller (14.32"sq), nevertheless even so it's quite a difference..
A second method, more painful, but much more useful and precise is the usage of trigonometric functions. This, if we don't have a monitor. Don't be put off, just follow the steps. The trig functions will also come in very handy when we want to find out by how much bigger should the diagonal be in order to get the same real estate for a wide 22" monitor as from the traditional 4:3 monitor.
Here we go. Again, the hypotenuse is 22" long. Again, remember, we do not use the hypotenuse for the trig purposes, except for illustrating the ratio relationships in the triangle and later in the algebra part of the operations. In order to find the length of the adjacent legs to the right angle, we'll need to work with one of the non-right angles. For this we'll use the very ratio that's given to us, 16:10 between the two adjacent legs (sides) of the right angle. I am going to turn the ratio upside down, which won't change anything, it will just save some steps (the cotangent). Let's label our triangle: If the leg with back slashes (c) is not in the right position it's because I don't know how to place these characters on the page such that they form the AB leg correctly. Please get a piece of paper and draw the triangle for yourself, uniting A with B the right way. The low case letters (a, b, c, e, e, f) stand as monikers for the sides.
A
| \
| \
| \
| \
| \
| b \ c
| \
| \
| \
| \
|__________\ B
C a
Take angle A. We call the Tangent of A the opposite side divided by the adjacent side which in our case would be a/b. In our case the tangent is given: 10:16 or 10/16 which is 0.625. The angle will be the inverse tangent of A which is 32 degrees. Now, the sine of A will be given by the ratio between the opposite side and the hypotenuse, that is a/c. But we know the angle, so from a scientific calculator sin32 = 0.53. Therefore the short side (a) will be gotten from the following equation:
a/c = 0.53 so a = 0.53c = 0.53 x 22" = 11.66" and the long side will be gotten from the cosine (adjacent side divided by the hypotenuse) function. From the calculator we get the cosine of A which is 0.85.
and for b (long side): b/c = 0.85 so b = 0.85c = 0.85 x 22" = 18.7"
These numbers pretty much correspond with the physical measurements above. And the monitor's display area is 11.66" x 18.7" = 218" sq.
The real difference with the traditional monitor therefore is: 232.32" sq - 218" sq = 14.32" sq, still a marked one.
Now, a more arcane question arises, namely, by how much should the diagonal be longer in order to get the same real estate (display area) on the wide monitor as the traditional 4:3 monitor? Well, this can be solved by setting up a system of equations. The data that we have is sufficient for calculations. We just have to visualize that the result will be a bigger triangle, longer legs and hypotenuse. We have the area of the 4:3 monitor (the area we're looking for to emulate for the wide monitor), which is given by d x e (the adjacent sides, short and long, to the right angle in the bigger triangle). So the first equation will be
|\
| \
| \
| \
| \
| \
| \
| e \ f
| \
| \
| \
| \
| ___________\
d
(1) d (short) x e (long) = 232.32" sq.
and the second is
(2) e/d = 1.6 (the ratio of the monitor) so d = e / 1.6 and by substituting this into equation (1) we get (e / 1.6) x e = 232. 32" sq.
Next, e sq / 1.6 = 232.32" then e sq = 1.6 x 232. 32" sq and e = sq root of 371.2 which yields 19.28", the length of the long side of the bigger triangle. And, since the ratio between the two sides is 1.6, that is e/d = 1.6, the short side will be e / 1.6 = d = 19.28/1.6 = 12.04".
Now, by the Pythagorean theorem, the hypotenuse (diagonal of our new, bigger triangle) will be f = square root of (d sq + e sq) = sq root of 515.33 sq in = 22.7"
Therefore, for a 22" diagonal wide screen to get the same useful display area as a traditional 4:3 monitor, the diagonal will have to be only .7" longer, that is 22.7", quite close to the 23" monitors which are priced double or triple. Well these 23" inchers are PVAs or IPSs.
All mistakes and bad assumptions are those of the author, so please feel free to comment and/or correct.
Just the other day I was visiting some offices and noticed some really big monitors. They turned out to be traditional 4:3 20"ers, but I was really surprised at how big they looked even though their bezel was rather narrow. Then I went home and had to admit that my 22"er didn't look as large as before. A week before I was invited to dinner by some friends who have a 52" SD 4:3 projection TV. Next day I went to Circuit City to see their 52" monitors and they didn't look that big anymore either. Of course, the picture was HD and some of the Samsungs and Panasonics looked quite astounding. Yet, the question of their and the monitors' dimensions kept bugging me and wandering whether I was beset with optical illusion. Therefore, here are the results of my musings. Do not be put off by the mathematics, it's simple algebra and some trigonometric function manipulation.
For the very beginning, when we trace the monitor's diagonal the result is the obvious two right triangles. This is why we calculate the area of a triangle as that of a rectangle devided by two. OK, so let's take on one of the triangles, either one since the two triangles are equal. Remember, the ratio between the two adjacent sides to the right angle is 4 (long side) : 3 (short side). The opposite side to the right angle is the hypotenuse. Now it turns out that the relationship with this hypotenuse in a 4:3 triangle is 5. Therefore the relationship that determines all 3 sides is 4:3:5. Since we're talking about 22" diagonals in such a triangle, finding the length of the adjacent legs is easy. We need some kind of a token unit length which multiplied by the numbers in the ratio relationship will give us the lengths.
We know the hypotenuse, so the length of this token unit will be 22"/5 which is 4.4".
Then the other sides will be 3x4.4"=13.2" and 4x4.4"=17.6".
These numbers are for the sides of the rectangle as well. Now we're interested in how much real estate (surface area) the traditional 4:3 monitor has. This is given by the area of the rectangle A = 13.2" x 17.6" = 232. 32" sq.
All well and good so far. However, wide monitors are not of 4:3:5 but 16:10 ratio and the hypotenuse is not in the same relationship with the other sides as in the former setting. Therefore, we can resort to two methods. The first would be the simplest, physically measuring the monitor IF we have one. I did mine and came up with 11.7" x 18.5" (both approximations). This means that area is ~ 216.45" sq. The difference with the traditional 4:3 is therefore 232.32"sq - 216.45" sq = 15.87"sq. How is this possible, though in reality when the difference is calculated rigorously it turns out to be smaller (14.32"sq), nevertheless even so it's quite a difference..
A second method, more painful, but much more useful and precise is the usage of trigonometric functions. This, if we don't have a monitor. Don't be put off, just follow the steps. The trig functions will also come in very handy when we want to find out by how much bigger should the diagonal be in order to get the same real estate for a wide 22" monitor as from the traditional 4:3 monitor.
Here we go. Again, the hypotenuse is 22" long. Again, remember, we do not use the hypotenuse for the trig purposes, except for illustrating the ratio relationships in the triangle and later in the algebra part of the operations. In order to find the length of the adjacent legs to the right angle, we'll need to work with one of the non-right angles. For this we'll use the very ratio that's given to us, 16:10 between the two adjacent legs (sides) of the right angle. I am going to turn the ratio upside down, which won't change anything, it will just save some steps (the cotangent). Let's label our triangle: If the leg with back slashes (c) is not in the right position it's because I don't know how to place these characters on the page such that they form the AB leg correctly. Please get a piece of paper and draw the triangle for yourself, uniting A with B the right way. The low case letters (a, b, c, e, e, f) stand as monikers for the sides.
A
| \
| \
| \
| \
| \
| b \ c
| \
| \
| \
| \
|__________\ B
C a
Take angle A. We call the Tangent of A the opposite side divided by the adjacent side which in our case would be a/b. In our case the tangent is given: 10:16 or 10/16 which is 0.625. The angle will be the inverse tangent of A which is 32 degrees. Now, the sine of A will be given by the ratio between the opposite side and the hypotenuse, that is a/c. But we know the angle, so from a scientific calculator sin32 = 0.53. Therefore the short side (a) will be gotten from the following equation:
a/c = 0.53 so a = 0.53c = 0.53 x 22" = 11.66" and the long side will be gotten from the cosine (adjacent side divided by the hypotenuse) function. From the calculator we get the cosine of A which is 0.85.
and for b (long side): b/c = 0.85 so b = 0.85c = 0.85 x 22" = 18.7"
These numbers pretty much correspond with the physical measurements above. And the monitor's display area is 11.66" x 18.7" = 218" sq.
The real difference with the traditional monitor therefore is: 232.32" sq - 218" sq = 14.32" sq, still a marked one.
Now, a more arcane question arises, namely, by how much should the diagonal be longer in order to get the same real estate (display area) on the wide monitor as the traditional 4:3 monitor? Well, this can be solved by setting up a system of equations. The data that we have is sufficient for calculations. We just have to visualize that the result will be a bigger triangle, longer legs and hypotenuse. We have the area of the 4:3 monitor (the area we're looking for to emulate for the wide monitor), which is given by d x e (the adjacent sides, short and long, to the right angle in the bigger triangle). So the first equation will be
|\
| \
| \
| \
| \
| \
| \
| e \ f
| \
| \
| \
| \
| ___________\
d
(1) d (short) x e (long) = 232.32" sq.
and the second is
(2) e/d = 1.6 (the ratio of the monitor) so d = e / 1.6 and by substituting this into equation (1) we get (e / 1.6) x e = 232. 32" sq.
Next, e sq / 1.6 = 232.32" then e sq = 1.6 x 232. 32" sq and e = sq root of 371.2 which yields 19.28", the length of the long side of the bigger triangle. And, since the ratio between the two sides is 1.6, that is e/d = 1.6, the short side will be e / 1.6 = d = 19.28/1.6 = 12.04".
Now, by the Pythagorean theorem, the hypotenuse (diagonal of our new, bigger triangle) will be f = square root of (d sq + e sq) = sq root of 515.33 sq in = 22.7"
Therefore, for a 22" diagonal wide screen to get the same useful display area as a traditional 4:3 monitor, the diagonal will have to be only .7" longer, that is 22.7", quite close to the 23" monitors which are priced double or triple. Well these 23" inchers are PVAs or IPSs.
All mistakes and bad assumptions are those of the author, so please feel free to comment and/or correct.