Forgot how much fun math is. I flunked out of engineering science classes and switched majors but actually was only one math class away from a math minor. Now I suddenly have the urge to be a graphic designer with a math minor.
@@The_TylerDurden_Narrator I feel you bro. 4th year final in about a week, then undergrad thesis defence. Prolly jobless after that. Me CS major. What about you?
@@oop1761 yeah more like applied mathematics. but we do have to do some hardcore electrical courses. good for me, I opted out of those courses (my uni offers that option).
@@minhazulislam4682if you dont mind me asking, why do you think you will be jobless? Im a first year CS major. Is the job market bad right now? I was told that if you have a good resume you shouldn’t have trouble finding a job.
@@Alborzhakimi7010 I was more of an academic geek with high cg. I do not have a presentable github right now. My initial goal of being a faculty at a uni is at a halt for personal reasons and that's why I have to look for opportunities elsewhere. I may take an online course (has great support) to break into the industry while dabbling in teaching O and A level stem subjects just to earn in the meanwhile. and yes, the people said correct who told you if you have a good resume, it would be easy. my advice would be to learn whatever programming language you like and stick to it for at least 2 years. Learn some frameworks of that language and just build stuff. You don't need to invent something new in the process (takes off the burden), just replicate existing stuff at a primary level and improve gradually. If you have a solid understanding of any framework, then you know how projects basically work. You can take those experience and use it in a completely new framework because essentially you would be doing similar stuff, but the tools are different. It's all about knowing your tool better. Anyway, if you find any of my rambling random words helpful, you are welcome.
I could calculate this, but I could also take a look at the limits. If one side of the rectangle approaches to the length of the diagonal, the other side becomes zero and thus the area. This is symmetric in both directions. So, if you plot a graph of the side length against the area, you get a horizontally flipped parabola with a maximum at a square. And since 4:3 is closer to 1:1 than 16:9, the screen has the larger area.
does this mean that the closer a rectangle is to a square the larger its area would be compared to its diagonal? is a square the largest area to diagonal ratio rectangle possible?
@AndyMath That's quite interesting! I immediately went to grab a piece of paper and a pen to try and prove it after writing my comment haha, but I wasn't quite sure if my proof was correct. Thanks for confirming it! :D
@@mhaskams i looked at the thumbnail and pretty quickly came to the conclusion that the old tv would be bigger. i just imagined the new tv getting longer and longer until it was almost a 27 inch line, and the area would be tiny, and as you move from the flat line to the square-ish tv, the area increases.
@@HugoStuffI assume mostly math people watch this channel and finding the rectangle (ratios of sides or diagonal) with the largest area is a very common calculus problem in highschool, so they probably already know it.
If you want to take it to the next level, the older tech tube (CRT) TV's like in the 90s were curved outwards in the Z direction, which I believe was counted in the 27" calculation. It wasn't a ruler you needed to measure, you needed a measuring tape. I'm not really sure what the curvature was, you'd have to google it to find out. Now it's not a simple triangle calculation, but you'd need to take into account the 3D z-axis & arc lengths. Would like to see another math video tackling that problem. ;) There of course were flat CRTs (which were "amazing" new tech, but that's not really until the early 2000s).
Literally none of that is necessary. Squares already have more area than rectangles with the same perimeters by definition so adding a Z axis just compounds on the advantage Squares already have
@@jdrmanmusiqking Yeah but the point is if you wanted to find the exact area of the older TV you would need to do that. The result of "which is larger" is the same but the "by how much" would be different
cause it does. because yes. its a very long derivation, but its clearly logical because the more uniform the shape is the more area it will cover. eg 5*5>6*4>3*7>2*8.@@hvglaser
@@hvglaser if we have a fixed value for a diagonal "d" in a rectangle of base "b" and height equal to "b" times some ratio "r" the area "A" of the rectangle is "b*(r*b) = r*b²". By pythagoras "d² = b² + (b*r)²", if we isolate the base squared we get that "b² = d² / (1 + r²)", substituting the "b²" term in the area we get that "A = d² * r/(1+r²)". Now what value of "r" would give the maximum area for a constant "d"? We can use calculus for that: the derivative of "A" in respect to "r" is "(d²)*(1-r²)/(1+r²)²", if this first derivative is zero it means that "r" should be equal to 1 or -1, but "r" can't be negative so what remains is the number 1. The second derivative of "A" in respect to "r" will let us finish this, it is equal to "2*r*(r² - 3)/(r² + 1)³", such that when "r" equals 1 it becomes negative. And what all that means is that when the ratio between the sides of a rectangle is equal to 1 we get the maximum area of the rectangle for a fixed diagonal.
I really love your work sir I haven't had this much fun since I never excelled at math but watching your video made me realise the joys of math. I'm currently in the process of practicing and solving different equation because I was picked to compete in a math quiz bee and wouldn't have done it without these videos
If you have two rectangles with the same diagonal length but different side length and height, the shape that more closely resembles a square will always have the larger area of the two shapes. In other words, when comparing the ratio of the side length and height of each shape, the shape that produces a ratio that is closer to one will have the larger area.
You can also do the same thing by drawing inside a semi circle, R=27, x and y parameters, and see for yourself by drawing dotted lines, and telling visibly! The closer to 1:1 you are you maximize your area
I also though of something similiar! I found the area as a function of the ratio, which is A(x) = d² · x / (1 + x²). d is the diameter of the semi circle and x is the ratio.
I learned from maximization / minimization problems in calculus that the larger area will always be the more square or whichever is closest to 1:1 (assuming same diagonal length)
this always bothered me when the phone market moved from 16:9 to 18:9 and then even 19.5 and beyond... they started marketing phones as if their screen area had gone up as much as their diagonal, when if you did the math, the actual area was in many cases smaller. Realizing this I was always on the lookout for phones with small bezels that were still in the 16:9 aspect ratio. I ended up with a Huawei Mate 10 (non pro) that I had to import via aliexpress. After that they stopped getting produced almost entirely. The marketing won.
But that's why it's more handy to use you can just buy smarthone that has approximately same width. Let's do some maths widht is the same but height is larger that's why you can see more with the same size. That's not correct for tablets, tv, monitors because you use them in portrait mode. P.s my English is not perfect because it's not my first language.
@@MikleSokoloff I'm actually more annoyed by tall aspect ratios in phones for a few reasons: 1. I usually have more trouble reaching the top than the left half 2. the scaling of icons and browser content is always relative to the screen width, so extra width is what gets you a real major UX benefit as opposed to just minor scroll space increase 3. most of the "action" in apps happens in the central 16:9 space even if there is more space available so I would turn your statement on its head: I think "why couldn't the 19.5:9 phone be a few millimeters wider", not "why couldn't the 16:9 phone be taller"
One caveat. When they counted the size of the tube TV it actually is the size of the tube not of the image. Typically the edges of the tuber hidden underneath the bezels of the screen. In an LCD panel it is edge to edge. So you physically have to measure with a ruler the diagonal for the tube TV. Sometimes what you have is almost exact same area in inches squared. But the LCD panel will be a flat screen instead of a curved surface.
In fact, if we seek to maximize area, it turns out a square is ideal. This can be proven by considering the unit circle, wherin A(α) = sin(α)cos(α), where α is the angle of the diagonal and A is the area of the rectangle. to maximize area, we take the derivative ( (fg)' = f'g+g'f ) A'(α) = cos^2(α)-sin^2(α) = 0 => cos(α) = sin(α) => sin(π/4-α) = sin(α) which has solutions α=π/4 +2πn, n=1,2,..., aka 45° + 360°n. And therefore, a square. Since the unit circle can be scaled up without changing the math, this applies regardless of the size of the diagonal.
If you plot the area as a function of the angle of the diagonal, 0 and 90 degrees gives zero, and its obviously symmetrical, so the maximum should be at the middle 45 degrees. The 4:3 tv is closer to that so it has the larger area.
One shouldn't just assume that the maximum is perfectly in between the zeroes in things like this. Like, it actually is in this case but it won't be for every arbitrary relationship ever.
@@droopy_eyes New TVs have pixels and are flat, however old TVs have no pixels and are rounded so the "size" or area will be more than actually calculated due to the curvature and either way the video did not include the edges which I think should be considered in terms of sizing the two TVs in regard to the question answered. You can think of an old TV as a reverse of a camera, the actual glass display doesn't mean much for the mechanism if you don't care about the quality of visuals displayed.
A 27" CRT TV had a viewable area of maybe 25". On modern flat panels the viewable area is the size, or close to it. My 32" LG is actually 80 cm which is close to 31.5".
I think you are mixing up the computer CRT with the TV CRT measuring methods/facts. Shortly after the round TV picture tubes went out of style the confusion was people used to measure the TV screen by measuring from top to bottom. Now we were told it is corner to corner for the more rectangular TV screen shape. Computer screens were less due to the fact you needed a more square corner and the CRT screen type of the time was rounded so to avoid your print to be rolled off in the corner they put a cover around the front to square up the corners, making it the "viewable area".
I always just remember that for any given perimeter of a rectangle, a square has the largest area. The closer it is to a square, the larger it is. Same thing with ellipses. The largest ellipse for a given circumference is a circle and the longer and skinner it is, the smaller the area becomes.
the algorithm decided to show me your circle/square area problem and I subbed immediately. this is one of the coolest channels around. i feel like i walk away knowing something new with every video. please keep it up!
This was cool but simply knowing squares have larger areas than rectangles in general is enough. Just visualize sliding the down hypotenuse. As C² approches the X-Axis, the Y becomes smaller as the space inside intuitively decreases
Take 16/9 vs 4/3 and put them into commonly stated decimal format, 1.78 vs 1.33. Subtract each from square root of 27 (5.19) then use those sums to determine your percentages. Yes, it works, and is very close to your answer but slightly different due to rounding off. You can use square root of any same diagonal TV measures and you will get the same result of course.
I ended up coming out with the older tv (4:3 tv) was bigger than the modern one by about 30in² (as in the area was bigger by about 30in²). Then I thought about how that makes sense that the closer a rectangle is to a square, the bigger the area it'll have as long as the diagonal length stayed constant. Overall I quite enjoyed it
I mean, the 4:3 screen obviously. A perfect square has the highest possible area for a given diagonal, the more extreme the difference in sides the smaller the area, but beside that it's also much thicker so if we think in terms of volume (since it's a three dimensional object) it wins by an enormous landslide, it should also likely win in terms of mass.
I used Thales' Theorem to arrive at the solution: If we take the TV diagonal as the diameter of a circle, the right-angled vertex of the yellow triangle also falls on the circle. We also know from the triangle area formula that a larger altitude yields a larger area; the largest altitude length occurs when the right-angle vertex is closest to the midpoint of the semicircle (ie the triangle is closest to isosceles). Since 4:3 is closer to 1:1 than 16:9, the 4:3 TV is larger
Don't forget that on old CRT monitors/televisions, you'd lose about an inch of viewable area on all four sides. So, a television that's marketed as being a 27-inch really only has 25 to 25.5 inches of diagonal viewable area.
i'd love to see a visual example of a 16:9 rectangle inside a 4:3 rectangle with equal diagonals, just to make it more clear at a glance how much the difference is
4:3 is the same as 16:12 so 16:9 fits into it with black bars top & bottom that fill the remaining 3/12ths aka 1/4th. IOW, 16:9 is 75% of the height of 4:3. Though if you keep the diagonal the same it gets wider and wouldn't fit inside but stick out and also no longer be 75% of the height
If you draw the 2 rectangles and overlap by one 27" diagonal, you will see clearly that the non overlapping triangles, 2 (of 4:3 rectangle) triangles (congruent) are considerably bigger than and similar to the other 2 congruent triangles (of 16:9 rectangle).
In CRT TVs it is a measurement of the outer edge to edge of the image tube whereas on flat sreens like LCD, Plasma and OLED it is a meassure image area. So the diagonal of a 17” 4:3 CRT will be approximately the same size as a 13-15” 4:3 LCD. I remember being disappointed and thinking I got fooled when I realized that my 14” CRT was actually only 11,5” when I measured it.
The tradeoff is of course, the viewing area. While the physical area is lower, the way the picture is formatted, you *gain* 33% viewing area which is important for video games and maintaining the correct aspect ratio for movies and television
Without doing any math I just think about the square cube law and would assume that you get more space per unit on the square TV because a square is the most most mathematical efficient way of combining things into a surface area. It hardly matters though because the visual information you can display is greater on a widescreen, since very little of human perception is based around vertical information.
Very nicely explained. But it only works if you have measured the viewable area of the CRT monitor to 27 inches. In the old days when you bought a “27 inch” monitor it was the measurement including casing, the viewable area was smaller.
.......uuhh no. It was always measured by the viewable area. Otherwise you could have a TV with an extremely large bezel around it and say "check out my 32 inch tv" when it was in fact 27 or something. You're wrong, and it's not too late to delete your comment BTW.
It didn't include all the casing, but you're making a valid point. A 27" screen would have had about 26" viewable, and looked more disingenuous when newer LCD panels gave their measurements correctly. I'm not sure why manufacturers were able to get away with it.
@@droopy_eyes I am not lying, it might not have been the entire casing and it might also have changed over the years to be less significant deviation. But I do remember measuring my own CRT monitor many years back and being disappointed.
The question of "which is larger?" Can also be solved with basic logic. The more regular a shape is, i.e., the closer it gets to a circle, the greater the area.
The ratio of areas between TV 1 with aspect ratio R_1 and TV 2 with aspect ratio R_2 if they have the same diagonal size is: A_1 / A_2 = (R_2 + 1/R_2) / (R_1 + 1/R_1). So they are inversely proportional to R + 1/R, which has a global minimum at R = 1, so a 1:1 aspect ratio gives the maximum area for a given diagonal size, anything smaller or higher has a smaller area.
Or for a simpler way, consider the extremes, a straight line and a square. For a straight line, the diagonal is the length and the area is 0. For a square it is the best possible case, with the area being the diagonal squared on 2, or in this case 27^2/2 = 364.5 Then realise 4:3 is closer to a square than 16:9 (as 4:3 is equivalent to 16:12), so it must be larger.
And for a more rigorous proof, recognise that the diagonal can trace out a circle, with the area of the screen bring proportional to the area of the right angle triangle made from that, or more specifically cos(t)*sin(t). This is equivalent to sin(2t)/2, which has a maximum at t=pi/4, corresponding to a square. You can do this graphically by noting that for a tiny change in the angle from the line, you will add more area than you take away; and a tiny change from the square, you will take away more than you add.
The 1995 TV is bigger, because it ain't flat. Edit and before anyone asks, the screens were rounded on old TVs so they weren't flat at all, which added surface area to the screen. Oh and let's not forget that flat screen TVs are flat while the old CRT TVs were big bulky and heavy. So they always win in largest.
There's an important fact missing from this video. When cathode ray tube TVs were measured diagonally, the measurement included the thickness of the bezel around the TV screen. In contrast, when modern flat panel screens are measured diagonally, only the visible screen is included in the measurement. This means that all the calculations and the comparisons between the calculations depicted in this video are not 100% correct. All of this being said a square is geometrically more efficient than a rectangle, and the TV screen will still have a larger viewable area than a rectangle even when taken into account the bezel thickness.
If the question is only to determine which more boundary cases can be taken. The square and strip are 27 inches long and 1mm high. The narrower the monitor, the smaller it becomes.
Interesting! I decided to solve it for myself before looking at the rest of the video, and I got exactly the same numbers, but my mind immediately went towards calculating the length of one of the sides of a TV using the angle of atan(3/4), and then getting the length of the vertical side using that angle and sine. Ain't it fun? There's so many different ways to approach one and the same problem, and still arrive at the correct answer!
It's super easy, what you need is a cheap ruler, then just measure the height and width of both TV. Then use the formula to find area of each TV (height * width), last, just compare the result to find out which one larger. Everything is simple when you have a cheap ruler.. :)
Take a line of length 27 as a diameter of a circle. Since the corners of a tv screen are right angles, we know regardless of the ratio, the corner will be on the circle (i forgot the theorem name). This triangle has half the area of the tv. Now consider the edge cases. A triangle with one infinitesimally short side would have no area, while a triangle with two equal sides would fill the most of the circle. In other words, the closer the ratio to 1:1 the bigger the area. So the 4:3 tv is bigger than the 16:9
I am glad you just proven that my 18.5: 9 phone gets even worse from that 😅, now i am itching to find out if my old phone had bigger display, than my current one😆
Of course the old one is larger, it's ratio is closer to a square, which will have the most area given a diagonal length compared to any other rectangle
what's funny is if instead it was measured as more of a taxicab distance, this effect is still present By taxicab I mean the width plus the height is taken into account, I learned that term once and I can't be bothered to search it again
Problem is that a 4:3 will -always- usually crop to 16:9 or whatever other widescreen ratio when watching movies. The actual useful area of a 4:3 is smaller than the 16:9's because of this. 16:9 screens either don't crop at all, or have very little cropping of the image for other ratios. It's still good for other things, as most programs adapt to the monitor's aspect ratio
Often it can be zoomed as if it was something closer to 4:3, which can be preferable on a 4:3 depending on the actual size of the screen. Older TV series were also made for 4:3, but are cropped/botched to 16:9 these days in ways that are worse than zooming-in a 16:9 movie.
Thats not the problem, the problem was the pricing of older tvs which were more expensive and luxurious, now adays a 27 inch tv is a monitor, not a tv, tvs are much larger, you can get a 27 inch monitor for 200 dollars and the old tv same size would cost 10 times that at least
Or you could scale down the 16:9 ratio by a factor of 4 to make the horizontal sides of both televisions equal. We're left with 9/4 for the vertical side of the 16:9 television, and 3 for the vertical side of the 4:3 television. And since 9/4 < 3, the 4:3 television would be bigger.
I answered it without doing math just by realizing that a square is the highest area rectangle, and the area of any rectangle with the same diagonal but a higher aspect ratio has to be progressively lower as you approach a 27" line segment. It is notable that a 27" TV in 1995 was maybe about mid-sized, and now it is considered quite small.
Well, there was a reason for that. A 27" CRT television in 1995 weighed about 95lbs/43 kilos. A 32" LCD TV, which is pretty much the smallest TV for home use these days, weighs 7.5lbs/3.4 kg
Well its pretty easy to figure out cause the closer to a square a rectangle is the more surface area it got for its diameter. The only math I gotta do is scaling up 4:3 to 16:12 and I know it got a larger surface area than 16:9. Aint got precise numbers but it answers the question.
Old TVs and monitors would usually measure the size of the tube, and not what you visibly get. Whatever part of the tube was behind plastic, was wasted. It was common for review sites to include both measures, but not stores. This became a non-issue with digital displays as there was no part of the screen concealed by plastic. So your 27" 4:3 TV, could have easily been 26.5". Of course you could have gotten 4:3 on an LCD display but that wasn't the norm. A lot of those first LCDs were also 5:4.
You know, for whatever reason, the entire video I was thinking the ratios were equivalent. I just could not comprehend why they would be different sizes.
I'm sure you know this but to optimize the area of a quadrilateral, we want to make it square so it would make sense without even computing that the 4:3 has a larger area
This is comparing the area of the screen, which is reasonable for a math problem. However, when comparing how big an image actually appears on the screen, it's even worse. For example, imagine a man standing in the middle of the screen taking up the full height of the frame. In the older TV, that man would be displayed at 16.2" tall, while in the widescreen he would only appear 13.2" tall. This means that the image on the older TV actually appears 22.7% larger than newer TVs. This is because while widescreens increase the area as they get wider, they are just adding more of the scene that wasn't visible before, but aren't actually making the image appear larger when compared to an old TV of the same height. I think it would make more sense to measure TVs by their height instead of by the diagonal, because the image would appear the same size on both, but you just would see more of the background on the wider TV, as expected.
I approached this challenge in a different way, by looking at the limits of the shapes. if the angle between the diagonal and , lets say, lower length of the rectangle goes towards zero, the area of the rectangle also goes to zero. vice versa, when the angle increases the area increases until the shape is a square and then the situation reverses again and the are decreases again to zero, hence the square with side ratio 1 is the largest possible area.
When a object has 4 corners the biggest area you can achieve is in the form of a square. The closer you are to a square the closer you are to the “max” area. So by knowing that fact the question becomes obvious. As an example you have a square with the sides of 5. The area is 5*5 which is 25. If we now to a rectangle with the sides of 7 and 3 (note that both the squares sides used and the rectangle sides used adds to 10) the are becomes 7*3 which is 21. And with the sides 9 and 1 we get 9 area. With rectangle with the sides 5.1 and 4.9 we get 24.99 area, as close as the max which is a square of the area 25. Peace out
The 16:9 is a flat panel, however the 4:3 is not. You forgot to account for the convex surface of the CRT television which would increase its surface area.
Reminds me of my logic where you can call something technically larger as being smaller. Say that you are at a Starbucks, and you get yourself a large coffee. The large coffee that you get can be more large than a gigantic, because the large is more in the large range. That's it.
Our eyesight is wide angle so it makes more sense that wide angle will appear larger and give us a more significant view. In general there is not as much real estate to be seen vertically
Forgot how much fun math is. I flunked out of engineering science classes and switched majors but actually was only one math class away from a math minor. Now I suddenly have the urge to be a graphic designer with a math minor.
@@The_TylerDurden_Narrator I feel you bro. 4th year final in about a week, then undergrad thesis defence. Prolly jobless after that. Me CS major. What about you?
@@minhazulislam4682 cs isnt engineering
@@oop1761 yeah more like applied mathematics. but we do have to do some hardcore electrical courses. good for me, I opted out of those courses (my uni offers that option).
@@minhazulislam4682if you dont mind me asking, why do you think you will be jobless? Im a first year CS major. Is the job market bad right now? I was told that if you have a good resume you shouldn’t have trouble finding a job.
@@Alborzhakimi7010 I was more of an academic geek with high cg. I do not have a presentable github right now. My initial goal of being a faculty at a uni is at a halt for personal reasons and that's why I have to look for opportunities elsewhere. I may take an online course (has great support) to break into the industry while dabbling in teaching O and A level stem subjects just to earn in the meanwhile.
and yes, the people said correct who told you if you have a good resume, it would be easy. my advice would be to learn whatever programming language you like and stick to it for at least 2 years. Learn some frameworks of that language and just build stuff. You don't need to invent something new in the process (takes off the burden), just replicate existing stuff at a primary level and improve gradually. If you have a solid understanding of any framework, then you know how projects basically work. You can take those experience and use it in a completely new framework because essentially you would be doing similar stuff, but the tools are different. It's all about knowing your tool better.
Anyway, if you find any of my rambling random words helpful, you are welcome.
I could calculate this, but I could also take a look at the limits. If one side of the rectangle approaches to the length of the diagonal, the other side becomes zero and thus the area. This is symmetric in both directions. So, if you plot a graph of the side length against the area, you get a horizontally flipped parabola with a maximum at a square. And since 4:3 is closer to 1:1 than 16:9, the screen has the larger area.
I "solved" this problem just by glance with this method you said seeing the video preview.
Thought the same thing, you immediately know 4:3 is larger, not sure by how much tho
It is nice to remember the optimal shapes. Square, circle, cube, sphere/ball.
@@herrjemeneh368 12.something%
This was exactly what I did. 4:3 is closer to being a square, ergo it has a larger area
does this mean that the closer a rectangle is to a square the larger its area would be compared to its diagonal? is a square the largest area to diagonal ratio rectangle possible?
Yes! A square has the largest area for a given diagonal length. That would've been cool if I mentioned that the 4:3 is closer to a square than a 16:9.
@@AndyMathonly the cool people that read the comments will get this fun info
@AndyMath That's quite interesting! I immediately went to grab a piece of paper and a pen to try and prove it after writing my comment haha, but I wasn't quite sure if my proof was correct. Thanks for confirming it! :D
@@mhaskams i looked at the thumbnail and pretty quickly came to the conclusion that the old tv would be bigger. i just imagined the new tv getting longer and longer until it was almost a 27 inch line, and the area would be tiny, and as you move from the flat line to the square-ish tv, the area increases.
@@HugoStuffI assume mostly math people watch this channel and finding the rectangle (ratios of sides or diagonal) with the largest area is a very common calculus problem in highschool, so they probably already know it.
If you want to take it to the next level, the older tech tube (CRT) TV's like in the 90s were curved outwards in the Z direction, which I believe was counted in the 27" calculation. It wasn't a ruler you needed to measure, you needed a measuring tape. I'm not really sure what the curvature was, you'd have to google it to find out. Now it's not a simple triangle calculation, but you'd need to take into account the 3D z-axis & arc lengths. Would like to see another math video tackling that problem. ;)
There of course were flat CRTs (which were "amazing" new tech, but that's not really until the early 2000s).
A few models even came with HDMI. Extremely rare though.
Literally none of that is necessary. Squares already have more area than rectangles with the same perimeters by definition so adding a Z axis just compounds on the advantage Squares already have
@@jdrmanmusiqking Yeah but the point is if you wanted to find the exact area of the older TV you would need to do that. The result of "which is larger" is the same but the "by how much" would be different
@@jdrmanmusiqking it is necessary because the given information is the diagonal, not the side lengths.
I've been on the fence for so long but this video helped me decide to go ahead with the 1995 27" TV. Thanks Andy.
As the shape becomes more square, its area increases.👍👏
While using less perimeter also*
why
cause it does. because yes. its a very long derivation, but its clearly logical because the more uniform the shape is the more area it will cover. eg 5*5>6*4>3*7>2*8.@@hvglaser
@mohammadfaaiz1050wdym
@@hvglaser if we have a fixed value for a diagonal "d" in a rectangle of base "b" and height equal to "b" times some ratio "r" the area "A" of the rectangle is "b*(r*b) = r*b²".
By pythagoras "d² = b² + (b*r)²", if we isolate the base squared we get that "b² = d² / (1 + r²)", substituting the "b²" term in the area we get that "A = d² * r/(1+r²)".
Now what value of "r" would give the maximum area for a constant "d"? We can use calculus for that: the derivative of "A" in respect to "r" is "(d²)*(1-r²)/(1+r²)²", if this first derivative is zero it means that "r" should be equal to 1 or -1, but "r" can't be negative so what remains is the number 1.
The second derivative of "A" in respect to "r" will let us finish this, it is equal to "2*r*(r² - 3)/(r² + 1)³", such that when "r" equals 1 it becomes negative. And what all that means is that when the ratio between the sides of a rectangle is equal to 1 we get the maximum area of the rectangle for a fixed diagonal.
The old TV is about 100 times heavier too
I learned a long time ago that the most "optimal area" is a square. The closer to a square, the large your area will be with the same constraints.
No, the most optimal area is a circle
I really love your work sir I haven't had this much fun since I never excelled at math but watching your video made me realise the joys of math. I'm currently in the process of practicing and solving different equation because I was picked to compete in a math quiz bee and wouldn't have done it without these videos
This is 8th grade (14yo) level math lol
@@erixccjc2143💀
@@erixccjc2143 and what? it's fun!
@@erixccjc2143 how u know he isnt in 8th grade
@@erixccjc2143 so?
If you have two rectangles with the same diagonal length but different side length and height, the shape that more closely resembles a square will always have the larger area of the two shapes. In other words, when comparing the ratio of the side length and height of each shape, the shape that produces a ratio that is closer to one will have the larger area.
Yeah exactly, this is what I was looking for
You can also do the same thing by drawing inside a semi circle, R=27, x and y parameters, and see for yourself by drawing dotted lines, and telling visibly! The closer to 1:1 you are you maximize your area
I also though of something similiar! I found the area as a function of the ratio, which is A(x) = d² · x / (1 + x²). d is the diameter of the semi circle and x is the ratio.
I just figured 4:3 is equal to 16:12, the gap is smaller so it’s side lengths are closer together so it is bigger. Like 3x3 is bigger than 2x4 or 5x1
Math truly is the well that will never run out
Maths can tell if it looks big it ain’t big 😂😂
27'' TV in 1995: the height of luxury
27'' TV in 2023: bargain bin
I learned from maximization / minimization problems in calculus that the larger area will always be the more square or whichever is closest to 1:1 (assuming same diagonal length)
this always bothered me when the phone market moved from 16:9 to 18:9 and then even 19.5 and beyond... they started marketing phones as if their screen area had gone up as much as their diagonal, when if you did the math, the actual area was in many cases smaller. Realizing this I was always on the lookout for phones with small bezels that were still in the 16:9 aspect ratio. I ended up with a Huawei Mate 10 (non pro) that I had to import via aliexpress. After that they stopped getting produced almost entirely. The marketing won.
But that's why it's more handy to use you can just buy smarthone that has approximately same width. Let's do some maths widht is the same but height is larger that's why you can see more with the same size. That's not correct for tablets, tv, monitors because you use them in portrait mode. P.s my English is not perfect because it's not my first language.
@@MikleSokoloff I'm actually more annoyed by tall aspect ratios in phones for a few reasons:
1. I usually have more trouble reaching the top than the left half
2. the scaling of icons and browser content is always relative to the screen width, so extra width is what gets you a real major UX benefit as opposed to just minor scroll space increase
3. most of the "action" in apps happens in the central 16:9 space even if there is more space available
so I would turn your statement on its head: I think "why couldn't the 19.5:9 phone be a few millimeters wider", not "why couldn't the 16:9 phone be taller"
@AB-ho1cr yeah iPhone and new Samsung phones.
Yeah, my phone is 20.5:9 and i don't really like the aspect ratio but other than that, my current phone is better than my previous phone
@@LCTeslaisn't that more because of the size of your phone and / or your hands ?
Went in rooting for 4:3, went out and realised “Oh I learnt more math” 🤔
Cool video :)
One caveat. When they counted the size of the tube TV it actually is the size of the tube not of the image. Typically the edges of the tuber hidden underneath the bezels of the screen. In an LCD panel it is edge to edge. So you physically have to measure with a ruler the diagonal for the tube TV.
Sometimes what you have is almost exact same area in inches squared. But the LCD panel will be a flat screen instead of a curved surface.
Some LCD screens were measured like that as well. That's why you sometimes see the "viewable" measurement in the specs.
In fact, if we seek to maximize area, it turns out a square is ideal. This can be proven by considering the unit circle, wherin
A(α) = sin(α)cos(α), where α is the angle of the diagonal and A is the area of the rectangle.
to maximize area, we take the derivative ( (fg)' = f'g+g'f )
A'(α) = cos^2(α)-sin^2(α) = 0 => cos(α) = sin(α) => sin(π/4-α) = sin(α)
which has solutions α=π/4 +2πn, n=1,2,..., aka 45° + 360°n. And therefore, a square. Since the unit circle can be scaled up without changing the math, this applies regardless of the size of the diagonal.
If you plot the area as a function of the angle of the diagonal, 0 and 90 degrees gives zero, and its obviously symmetrical, so the maximum should be at the middle 45 degrees. The 4:3 tv is closer to that so it has the larger area.
One shouldn't just assume that the maximum is perfectly in between the zeroes in things like this.
Like, it actually is in this case but it won't be for every arbitrary relationship ever.
@@natchu96 yes but this is also obviously monotonous until the extrema
Just one small issue, display ratio does mean actual ratio; we use display ratio to match the image displayed and not the dimensions of the TV itself.
Displays are made to specific ratio dimensions. Most common for TVs is 16:9.
@@droopy_eyes New TVs have pixels and are flat, however old TVs have no pixels and are rounded so the "size" or area will be more than actually calculated due to the curvature and either way the video did not include the edges which I think should be considered in terms of sizing the two TVs in regard to the question answered. You can think of an old TV as a reverse of a camera, the actual glass display doesn't mean much for the mechanism if you don't care about the quality of visuals displayed.
A 27" CRT TV had a viewable area of maybe 25". On modern flat panels the viewable area is the size, or close to it. My 32" LG is actually 80 cm which is close to 31.5".
Indeed, this is a significant consideration.
I think you are mixing up the computer CRT with the TV CRT measuring methods/facts. Shortly after the round TV picture tubes went out of style the confusion was people used to measure the TV screen by measuring from top to bottom. Now we were told it is corner to corner for the more rectangular TV screen shape. Computer screens were less due to the fact you needed a more square corner and the CRT screen type of the time was rounded so to avoid your print to be rolled off in the corner they put a cover around the front to square up the corners, making it the "viewable area".
I always just remember that for any given perimeter of a rectangle, a square has the largest area. The closer it is to a square, the larger it is. Same thing with ellipses. The largest ellipse for a given circumference is a circle and the longer and skinner it is, the smaller the area becomes.
the algorithm decided to show me your circle/square area problem and I subbed immediately. this is one of the coolest channels around. i feel like i walk away knowing something new with every video. please keep it up!
This was cool but simply knowing squares have larger areas than rectangles in general is enough.
Just visualize sliding the down hypotenuse. As C² approches the X-Axis, the Y becomes smaller as the space inside intuitively decreases
Take 16/9 vs 4/3 and put them into commonly stated decimal format, 1.78 vs 1.33. Subtract each from square root of 27 (5.19) then use those sums to determine your percentages. Yes, it works, and is very close to your answer but slightly different due to rounding off. You can use square root of any same diagonal TV measures and you will get the same result of course.
I ended up coming out with the older tv (4:3 tv) was bigger than the modern one by about 30in² (as in the area was bigger by about 30in²). Then I thought about how that makes sense that the closer a rectangle is to a square, the bigger the area it'll have as long as the diagonal length stayed constant. Overall I quite enjoyed it
Fun fact - displays were 4:3 but the pixels weren't square with ratio 5:4 - that's why retro tv games played on modern displays looks squeezed
Glad this Channel was recommended. Love this!
I mean, the 4:3 screen obviously. A perfect square has the highest possible area for a given diagonal, the more extreme the difference in sides the smaller the area, but beside that it's also much thicker so if we think in terms of volume (since it's a three dimensional object) it wins by an enormous landslide, it should also likely win in terms of mass.
Fun fact: by 2030, TVs and phones will both be an infinitely long line with no area.
next question: given (a,b) coordinates on the unit circle (edit: in the first quadrant, ie 0
I used Thales' Theorem to arrive at the solution: If we take the TV diagonal as the diameter of a circle, the right-angled vertex of the yellow triangle also falls on the circle. We also know from the triangle area formula that a larger altitude yields a larger area; the largest altitude length occurs when the right-angle vertex is closest to the midpoint of the semicircle (ie the triangle is closest to isosceles). Since 4:3 is closer to 1:1 than 16:9, the 4:3 TV is larger
I like your deadpan delivery of 'How exciting.'
But the 2023 TV is sooooo much lighter. Those 1995 TV were heavy!
So true, Travis.
Don't forget that on old CRT monitors/televisions, you'd lose about an inch of viewable area on all four sides. So, a television that's marketed as being a 27-inch really only has 25 to 25.5 inches of diagonal viewable area.
The closer you get to a circle the more surface something has for it's size. Pretty straightforward.
A small correction at the end it's 11.23% larger
It took me way too long to see if anyone else noticed
i'd love to see a visual example of a 16:9 rectangle inside a 4:3 rectangle with equal diagonals, just to make it more clear at a glance how much the difference is
4:3 is the same as 16:12 so 16:9 fits into it with black bars top & bottom that fill the remaining 3/12ths aka 1/4th. IOW, 16:9 is 75% of the height of 4:3. Though if you keep the diagonal the same it gets wider and wouldn't fit inside but stick out and also no longer be 75% of the height
cool method of going straight to area from a. was wondering why you were cooking initially
I have never seen it done that way before. I just thought of it and it seemed smoother than solving for a and plugging it back in.
Finally, I solved a question from your channel orally.
you somehow proved we get more size out of size by just rewinding
I love simple math like this that solves real life problem that we conjure up during shower.
It's the 4:3 cause rectangle is optimal for area enlargment
If you draw the 2 rectangles and overlap by one 27" diagonal, you will see clearly that the non overlapping triangles, 2 (of 4:3 rectangle) triangles (congruent) are considerably bigger than and similar to the other 2 congruent triangles (of 16:9 rectangle).
In CRT TVs it is a measurement of the outer edge to edge of the image tube whereas on flat sreens like LCD, Plasma and OLED it is a meassure image area.
So the diagonal of a 17” 4:3 CRT will be approximately the same size as a 13-15” 4:3 LCD.
I remember being disappointed and thinking I got fooled when I realized that my 14” CRT was actually only 11,5” when I measured it.
The tradeoff is of course, the viewing area. While the physical area is lower, the way the picture is formatted, you *gain* 33% viewing area which is important for video games and maintaining the correct aspect ratio for movies and television
I knew the older TV was bigger because anytime a newer version of something is made a corner or 2 gets cut off.
Without doing any math I just think about the square cube law and would assume that you get more space per unit on the square TV because a square is the most most mathematical efficient way of combining things into a surface area.
It hardly matters though because the visual information you can display is greater on a widescreen, since very little of human perception is based around vertical information.
Very nicely explained. But it only works if you have measured the viewable area of the CRT monitor to 27 inches. In the old days when you bought a “27 inch” monitor it was the measurement including casing, the viewable area was smaller.
.......uuhh no. It was always measured by the viewable area. Otherwise you could have a TV with an extremely large bezel around it and say "check out my 32 inch tv" when it was in fact 27 or something.
You're wrong, and it's not too late to delete your comment BTW.
@@MF-rtard89 Not casing, but diagonal of the tube itself. Viewable area + thickness of glass at the edges
@@igorbondarev5226 negligible. The point of the video still stands
It didn't include all the casing, but you're making a valid point. A 27" screen would have had about 26" viewable, and looked more disingenuous when newer LCD panels gave their measurements correctly. I'm not sure why manufacturers were able to get away with it.
@@droopy_eyes I am not lying, it might not have been the entire casing and it might also have changed over the years to be less significant deviation. But I do remember measuring my own CRT monitor many years back and being disappointed.
The question of "which is larger?" Can also be solved with basic logic. The more regular a shape is, i.e., the closer it gets to a circle, the greater the area.
The most efficiently sized TV will have equal sides. The more you move away from a perfect square, the less area you get with the same sized diagonal.
The ratio of areas between TV 1 with aspect ratio R_1 and TV 2 with aspect ratio R_2 if they have the same diagonal size is: A_1 / A_2 = (R_2 + 1/R_2) / (R_1 + 1/R_1). So they are inversely proportional to R + 1/R, which has a global minimum at R = 1, so a 1:1 aspect ratio gives the maximum area for a given diagonal size, anything smaller or higher has a smaller area.
Or for a simpler way, consider the extremes, a straight line and a square.
For a straight line, the diagonal is the length and the area is 0.
For a square it is the best possible case, with the area being the diagonal squared on 2, or in this case 27^2/2 = 364.5
Then realise 4:3 is closer to a square than 16:9 (as 4:3 is equivalent to 16:12), so it must be larger.
And for a more rigorous proof, recognise that the diagonal can trace out a circle, with the area of the screen bring proportional to the area of the right angle triangle made from that, or more specifically cos(t)*sin(t).
This is equivalent to sin(2t)/2, which has a maximum at t=pi/4, corresponding to a square.
You can do this graphically by noting that for a tiny change in the angle from the line, you will add more area than you take away; and a tiny change from the square, you will take away more than you add.
The 1995 TV is bigger, because it ain't flat.
Edit and before anyone asks, the screens were rounded on old TVs so they weren't flat at all, which added surface area to the screen.
Oh and let's not forget that flat screen TVs are flat while the old CRT TVs were big bulky and heavy. So they always win in largest.
There's an important fact missing from this video.
When cathode ray tube TVs were measured diagonally, the measurement included the thickness of the bezel around the TV screen.
In contrast, when modern flat panel screens are measured diagonally, only the visible screen is included in the measurement.
This means that all the calculations and the comparisons between the calculations depicted in this video are not 100% correct.
All of this being said a square is geometrically more efficient than a rectangle, and the TV screen will still have a larger viewable area than a rectangle even when taken into account the bezel thickness.
If the question is only to determine which more boundary cases can be taken. The square and strip are 27 inches long and 1mm high. The narrower the monitor, the smaller it becomes.
The problem is, the width of the space allows you to see more of a scene, so 4:3 looks smaller by virtue of not being as wide relative to its height
YESSS! This is the first Andy Math problem I’ve solved before watching this video.
Cost cutting shrinks everything.
I made similar mathematical arguments when widescreen LCD monitors were competing with 4:3 monitors. I lost that battle.
This is the type of video that will be recommended to someone in a decade at 3:00am
Interesting! I decided to solve it for myself before looking at the rest of the video, and I got exactly the same numbers, but my mind immediately went towards calculating the length of one of the sides of a TV using the angle of atan(3/4), and then getting the length of the vertical side using that angle and sine.
Ain't it fun? There's so many different ways to approach one and the same problem, and still arrive at the correct answer!
Just look at them. As the TV gets more rectangular, the hight shrinks. The limit is a line with no area.
Instructions unclear; threw out flat screen TV for a CRT.
It's super easy, what you need is a cheap ruler, then just measure the height and width of both TV. Then use the formula to find area of each TV (height * width), last, just compare the result to find out which one larger. Everything is simple when you have a cheap ruler.. :)
Perfect example of why Pythagorean Theory is daily relevant.
The '95 TVs also have a more rotund, bulging screen as opposed to the basically flat screen of a modern 16:9 TV
Take a line of length 27 as a diameter of a circle. Since the corners of a tv screen are right angles, we know regardless of the ratio, the corner will be on the circle (i forgot the theorem name). This triangle has half the area of the tv.
Now consider the edge cases. A triangle with one infinitesimally short side would have no area, while a triangle with two equal sides would fill the most of the circle. In other words, the closer the ratio to 1:1 the bigger the area. So the 4:3 tv is bigger than the 16:9
I am glad you just proven that my 18.5: 9 phone gets even worse from that 😅, now i am itching to find out if my old phone had bigger display, than my current one😆
The control it takes to not say, "your mom".
To expected, as time goes on everything gets smaller, but more powerful with better tech
Of course the old one is larger, it's ratio is closer to a square, which will have the most area given a diagonal length compared to any other rectangle
what's funny is if instead it was measured as more of a taxicab distance, this effect is still present
By taxicab I mean the width plus the height is taken into account, I learned that term once and I can't be bothered to search it again
Problem is that a 4:3 will -always- usually crop to 16:9 or whatever other widescreen ratio when watching movies. The actual useful area of a 4:3 is smaller than the 16:9's because of this. 16:9 screens either don't crop at all, or have very little cropping of the image for other ratios.
It's still good for other things, as most programs adapt to the monitor's aspect ratio
Often it can be zoomed as if it was something closer to 4:3, which can be preferable on a 4:3 depending on the actual size of the screen. Older TV series were also made for 4:3, but are cropped/botched to 16:9 these days in ways that are worse than zooming-in a 16:9 movie.
@@petitio_principii generally speaking, we watch more content in 16:9 than in 4:3, which in my opinion makes this irrelevant
Thats not the problem, the problem was the pricing of older tvs which were more expensive and luxurious, now adays a 27 inch tv is a monitor, not a tv, tvs are much larger, you can get a 27 inch monitor for 200 dollars and the old tv same size would cost 10 times that at least
@@tbqhwyfit’s not irrelevant because you said always, and you’re completely wrong as a result. It’s not always that way.
Someone is too young to remember pan & scan.
Or you could scale down the 16:9 ratio by a factor of 4 to make the horizontal sides of both televisions equal. We're left with 9/4 for the vertical side of the 16:9 television, and 3 for the vertical side of the 4:3 television. And since 9/4 < 3, the 4:3 television would be bigger.
Not only a larger area but also a larger volume if you remember how thick those rhings were, almost a cube.
I wonder what a 55 inch old tv looks like
The largest CRT to enter production was a 43" PVM; any need for larger displays was fulfilled by either projectors or multiple tubes in a grid.
yep, since 4:3 is much closer to 1:1 as compared to 16:9 it has more area :)
It’s probably even a few percentage points larger still because a modern tv is flatscreen, whereas the old analog tv is slightly bulbous
If you know that those propositions are the length of sides of the rectangle, you can just multiply the left one by "4". 4×4=16, 3×4=12
16×12>16×9
That way both diagonals won't be 27 inch.
Both diagonals have to be 27" for the comparison.
I answered it without doing math just by realizing that a square is the highest area rectangle, and the area of any rectangle with the same diagonal but a higher aspect ratio has to be progressively lower as you approach a 27" line segment. It is notable that a 27" TV in 1995 was maybe about mid-sized, and now it is considered quite small.
Well, there was a reason for that. A 27" CRT television in 1995 weighed about 95lbs/43 kilos. A 32" LCD TV, which is pretty much the smallest TV for home use these days, weighs 7.5lbs/3.4 kg
Well its pretty easy to figure out cause the closer to a square a rectangle is the more surface area it got for its diameter. The only math I gotta do is scaling up 4:3 to 16:12 and I know it got a larger surface area than 16:9. Aint got precise numbers but it answers the question.
In both cases, the diagonal has to be 27".
@@Mash4096 exactly
But yeah basically, the closer to a square, the bigger the surface area, with the same diagonal.
Old TVs and monitors would usually measure the size of the tube, and not what you visibly get. Whatever part of the tube was behind plastic, was wasted. It was common for review sites to include both measures, but not stores. This became a non-issue with digital displays as there was no part of the screen concealed by plastic. So your 27" 4:3 TV, could have easily been 26.5". Of course you could have gotten 4:3 on an LCD display but that wasn't the norm. A lot of those first LCDs were also 5:4.
You know, for whatever reason, the entire video I was thinking the ratios were equivalent.
I just could not comprehend why they would be different sizes.
I'm sure you know this but to optimize the area of a quadrilateral, we want to make it square so it would make sense without even computing that the 4:3 has a larger area
its definetly larger in the third axis too not to mention its mass old tvs are fun XD
never thought i'd be watching this on winter break
Never forget what they took from us.
Basically a 16:9 aspect ratio doesnt give you more width it gives you less height. At least, in terms of area.
Now consider how much of that surface area would be used if you're watching something in 16:9 format (with black bars at top and bottom).
The closer a rectangle is to a square, the more area it will have
And unlike the flat panel OLEDs that wouldn't last even during the warranty.. the ones fro. 1990s may still actually work
This is comparing the area of the screen, which is reasonable for a math problem. However, when comparing how big an image actually appears on the screen, it's even worse. For example, imagine a man standing in the middle of the screen taking up the full height of the frame. In the older TV, that man would be displayed at 16.2" tall, while in the widescreen he would only appear 13.2" tall. This means that the image on the older TV actually appears 22.7% larger than newer TVs.
This is because while widescreens increase the area as they get wider, they are just adding more of the scene that wasn't visible before, but aren't actually making the image appear larger when compared to an old TV of the same height. I think it would make more sense to measure TVs by their height instead of by the diagonal, because the image would appear the same size on both, but you just would see more of the background on the wider TV, as expected.
I approached this challenge in a different way, by looking at the limits of the shapes. if the angle between the diagonal and , lets say, lower length of the rectangle goes towards zero, the area of the rectangle also goes to zero. vice versa, when the angle increases the area increases until the shape is a square and then the situation reverses again and the are decreases again to zero, hence the square with side ratio 1 is the largest possible area.
Also the old tv's screen is more convex, so its screen must be even bigger
Not always. They made 4:3 flat screens as well.
When a object has 4 corners the biggest area you can achieve is in the form of a square. The closer you are to a square the closer you are to the “max” area. So by knowing that fact the question becomes obvious.
As an example you have a square with the sides of 5. The area is 5*5 which is 25. If we now to a rectangle with the sides of 7 and 3 (note that both the squares sides used and the rectangle sides used adds to 10) the are becomes 7*3 which is 21. And with the sides 9 and 1 we get 9 area. With rectangle with the sides 5.1 and 4.9 we get 24.99 area, as close as the max which is a square of the area 25.
Peace out
The 16:9 is a flat panel, however the 4:3 is not. You forgot to account for the convex surface of the CRT television which would increase its surface area.
Mind you thos is assuming the old tv is flat. Old tv screens usually bulge out a bit
The old TV would also take up less floor space. If you ignore the thickness 😅
Reminds me of my logic where you can call something technically larger as being smaller. Say that you are at a Starbucks, and you get yourself a large coffee. The large coffee that you get can be more large than a gigantic, because the large is more in the large range. That's it.
Our eyesight is wide angle so it makes more sense that wide angle will appear larger and give us a more significant view. In general there is not as much real estate to be seen vertically