Assume Frank's 2000 inch TV is measured on the diagonal. It's HD of course. It's elevated on a base equal to half its height. You're sitting on the roof of a house. How far away can the house be so that you can still see the bottom of the screen without being blocked by the Earth's curvature? State assumptions clearly.
This ought to be, what, the equivalent of a 9th grade geometry problem? (10th grade if you're in Kentucky.)
OK, HDTVs are 16x9, so the height::diagonal is 9::25 (by Pythagorean theoreom). Therefore, the height of the screen is 2000/25*9 = 60 ft. The bottom of the screen is half that height off the surface of the Earth. The viewer is on top of a house, which we'll assume is 15 ft. tall.
Now draw a triangle between the viewer, the bottom of the TV screen, and the center of the Earth. One of the sides is the distance we're interested in (x); the other two sides are r+15 ft. and r+30 ft. Assume a radius of 3,960 miles = 20,908,800 ft.
To find x, we note that the side of length x is tangent to a circle of radius r centered at the opposite vertex. (That's where the surface begins to cut off view.) Draw a radius from the opposite vertex to the point of tangency, dividing the triangle into two right triangles. (A radius is perpendicular to a tangent line.)
This divides x into two parts, x′ and x′′. x′ is on the triangle with r+15. We now have two sides of each right triangle, and the calculation becomes trivial: x′ = sqrt((r+15)^2 - r^2); x′′ = sqrt((r+30)^2-r^2); x = x′ + x′′.
Applying our mighty calculator, we get x=60,464 ft, or 11.5 miles. You can watch Frank's 2000 inch TV from, as the song says, 30 blocks away.
For comparison, a man of normal height (i.e. at least a couple inches about six foot) can see only about 3.3 miles to the horizon (cite).
I'm too lazy to check your math, but I still want to make some sort of statement about how you're wrong about something, so here goes:
While it's true that you can watch Frank's TV from 30 blocks away, it's only true in the same sense that you can watch it from one block away. A standard city block is between 1/8 and 1/12 mile, depending on the city. I think we have to assume the TV is in a city because otherwise "30 blocks" wouldn't be a reasonable way to describe a distance.
Probably it's in Los Angeles: LA is where you tend to get that sort of thing, and it is one of the few spots where someone named Frank is likely to coexist with 30 consecutive blocks that have no obstructions high enough to block the view (because they have all been leveled by earthquakes).
In my Texas public school Geometry was an 11th grade class (one which I never took), so bite me, Mr. "Oh, Look at me! I can do math! I'm so special!" (No, I'm not hostile about math? Why do you ask?)
11th? Shee-oot. My school's math progression was 9th algebra, 10th geometry, 11th algebra II/trig, 12th pre-calc. When I got to ND, I learned that that progression was about a year behind other schools.
TX public schools are in pretty sad shape if they're a year behind KY schools.
I was also in Texas; we were one year ahead of your KY progression, with calculus in 12th. That was at Clear Creek HS, which was where many of the NASA Johnson Space Center employee's kids went, so it was wealthy (actually it was also a magnet school for vocational agriculture, so it was very segregated with the white farmers in one set of buildings and the white rich kids in another set.)
My math progress in Montana was identical to Greg's in Kentucky.
Connecticut was 9th Geometry, 10th algebra , 11th trig/pre-calc, 12th Calc. Advanced kiddos did Geometry in 8th, bumped everything down a level, and took Calc II in 12th.
When I came to NJ at the start of 9th grade, they convinced themselves that my 8th grade geometry couldn't possibly be satisfactory, so I frigging well had to take it over. I am never more miserable than taking a course for which I know all the material, have learned it in a more challenging environment, and don't respect the teacher. All of those things were true in my class. Also, I was tracked with kids my age who were on the regular pre-college track, not the AP track, so I was smarter than all my classmates in addition to already knowing the materials.
Have I mentioned that HS sucked?
Back home in Alabam-my, I had to go to the special math and science school for my 11th and 12th grade years to get anything beyond pre-calc, but I did manage to finesse my hometown school enough to take pre-calc in 10th grade.
It sure was nice taking differential equations and linear algebra classes for the second time as an undergrad; I would have struggle a lot more my first two years at university if not for my higher math exposure in high school.
And balancing my checkbook or doing my taxes still kicks my ass.
Also bored and anal-retentive to boot...(for the record, MN followed the same math/grade progression that Greg's home state did, or at least it did when I went to HS - note as well that math and science were closely related in MN schools in my heyday, which might explain some of my more anal moments below)
"Assume Frank's 2000 inch TV is measured on the diagonal. It's HD of course. It's elevated on a base equal to half its height. You're sitting on the roof of a house. How far away can the house be so that you can still see the bottom of the screen without being blocked by the Earth's curvature? State assumptions clearly."
OK, seems simple enough.
"OK, HDTVs are 16x9, so the height::diagonal is 9::25 (by Pythagorean theoreom)."
Properties of the 30-60-90 right triangle will give you this solution as well without resorting to Pythagoras. With you so far.
"Therefore, the height of the screen is 2000/25*9 = 60 ft."
Whoa, whoa, wait - where did we change from inches to feet? And from the looks of it, it's not clear what order the operations are taking place in. Let's see...
(2000 in.)* (9/25) = 720 in. (multiplying by the ratio of height to diagonal)
720 in. * (1 ft./12 in.) = 60 ft. (converting inches to feet)
Okay, I can see it works out. Good so far.
"The bottom of the screen is half that height off the surface of the Earth."
There's no border? You know, like the case of the TV? After all, this is basically a 60-foot tall piece of...well, whatever HDTVs are made out of.
OK, no problem, we'll give you a pass on this one, since it's mostly a thought exercise.
"The viewer is on top of a house, which we'll assume is 15 ft. tall."
15 feet? For a HOUSE? What is this, some post-WWII neighborhood where they put up a bunch of drywall huts for the returning soldiers to buy with their VA benefits? Anybody who can afford a 2000 inch HDTV is not going to be living in a 60-year old house, otherwise his priorities are majorly messed up.
Okay, okay, it's technically not the guy who owns the TV who's sitting on the house watching it. Besides, it's another part of that whole 'thought-experiment' thing. I get it. Carry on.
It's still weird that the guy's TV stand is twice as tall as the other guy's house...
"Now draw a triangle between the viewer, the bottom of the TV screen, and the center of the Earth. One of the sides is the distance we're interested in (x); the other two sides are r+15 ft. and r+30 ft. Assume a radius of 3,960 miles = 20,908,800 ft."
Okay, I'm with you.
"To find x, we note that the side of length x is tangent to a circle of radius r centered at the opposite vertex. (That's where the surface begins to cut off view.)"
Huh? Why does that line in my triangle show as a chord? Oh, wait, that's right; you're supposed to be able to see the TV from the top of the house, and we're assuming there's no tunnel in the Earth that would let us peer though it. Still, it might have been better, instead of 'draw a triangle', to say something like:
Draw a curve between the bottom of the TV stand and the bottom of the viewer's house representing the curvature of the Earth in this region. Draw a line between the bottom of the TV and the position of the viewer such that this line touches the curved line representing the Earth's curvature at only one point (or in other words, is tangent to the arc of the Earth's curvature), then form a triangle by drawing lines from each end of the sight line to meet at a point representing the center of the Earth. (Note: it's recommended that you do not attempt to draw these lines to scale.)
Okay, back with you.
"Draw a radius from the opposite vertex to the point of tangency, dividing the triangle into two right triangles. (A radius is perpendicular to a tangent line.)"
Have we covered that proof yet? No? Then I'll just take your word for it.
"x′ is on the triangle with r+15. We now have two sides of each right triangle, and the calculation becomes trivial: x′ = sqrt((r+15)^2 - r^2); x′′ = sqrt((r+30)^2-r^2); x = x′ + x′′."
Wait, wait, wait. My calculator only goes up to eight digits, so I can't square an eight-digit number. Let me go back to eighth grade and bust out my algebra and see if I can't simplify this sucker:
x' = sqrt((r+15)^2)-r^2)
x' = sqrt(r^2+30r+225-r^2)
x' = sqrt(30r+225)
x'' = sqrt((r+30)^2)-r^2)
x'' = sqrt(r^2+60r+900-r^2)
x'' = sqrt(60r+900)
So r is 20,908,800? Figures, I still can't force the numbers into the calculator.
Okay, time to fast-forward to physics: Significant Digits!
Compared to a radius of 2.09x10^8 feet, distances of 1.5x10^1 and 3.0x10^1 feet are trivial and can be ignored. Which means we can treat both triangles as equivalent, and all we need to do is calculate either x' or x'' and double the result.
x' = r^2 - r^2 = 0
x = x'*2 = 0*2 = 0
So, despite the size, you still can't see the TV unless you're right up next to it. QED.
Next time your wife asks you 'why do you always have to sit right in front of the TV?', feel free to bust this proof out on her.
heh, cool problem. i might use it in my actual geometry class. i think the kids (9th grade, for the record, but a private school) would like it.
David:
The 9::16::25 right triangle isn't 30-60-90.
Since multiplication is commutative, order of operations doesn't matter when everything is multiplication or division (here, the 2000*9/25/12 operation to figure out the height of the TV.)
A fifteen foot house is a one-story ranch with a low peak to the roof, such as you find in any 1960s suburb. (Keep in mind that the viewer's eyes are probably a foot or two above the actual peak of the roof, and it's actually the height of the eyes we care about.) If you think it's weird that the house is smaller than the TV, just keep in mind we're talking about a TV 90 feet high.
I used the calculator included with Windows to work the numbers, and it appears to have arbitrary precision.
Tim:
I've probably told this story, but it amuses me. Fourth semester of calculus at Notre Dame, we covered linear algebra, which, as you know, largely consists of extracting eigenvalues and eigenvectors from matrices.
However, the physics major track I was on meant that I had already gotten eigenvalues in not one but two other classes, before getting to them in Calculus.
When the matrix test came in calculus, it was easy--things like 3x3 arrays with 6 non-zero values, etc. In several cases, I just wrote down the answers.
I got my test back and I had gotten dinged. I went to the prof. "Why'd I lose points?"
"You didn't show your work."
"I didn't do any work!" He didn't relent.
I didn't go back to class except for tests for the rest of the semester and got an A.
Every bit of math I used as a physics major at Notre Dame, I learned in a physics course before it came up in the ordinary math sequence.
Greg:
You're right about the 30-60-90; I was misremembering. (For the record, the 30-60-90 triangle is where the hypotenuse is exactly double the length of the short side, with the long side equal to the short side times sqrt(3).)
I'm not sure the one-story house can be marked down to *any* 1960s suburb - perhaps my experience is atypical, but I was born in 1967 and never lived in a one-story building until I was in my mid-20s. Single-story 'ramblers' (as we call them here) aren't terribly popular in MN.
And it wasn't so much that the TV was bigger than the house (though by the terms of the problem, the TV is only 60 feet high, not 90 feet high), but that the TV *stand* was bigger than the house. Throw some stairs in that puppy, and it'd make a decent dwelling (as long as your neighbors didn't insist on watching bowling shows on the TV at 2am).
Out of curiosity, what would the purchase DC for d20 Modern be on a 60 foot high HDTV?
Ramblers are dirt common down here in the Southwest. In MN, you're also going to have highly peaked roofs.