Big Tower Puzzle
July 9th, 2012 at 10:29:23 AM
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How high would you need to build a tower to see 1/3 of the earth's surface?
Assume a perfectly spherical earth with a circumference of 40,075 km.
Assume a perfectly spherical earth with a circumference of 40,075 km.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 9th, 2012 at 11:10:16 AM
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Quote: WizardHow high would you need to build a tower to see 1/3 of the earth's surface?
Assume a perfectly spherical earth with a circumference of 40,075 km.
I get 12,756 km. This is at a distance equal to the earth's diameter above the earth's surface, which is a neat result.
July 9th, 2012 at 11:17:52 AM
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Quote: ChesterDogI get 12,756 km. This is at a distance equal to the earth's diameter above the earth's surface, which is a neat result.
I'll wait until some answers come in before I comment.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 9th, 2012 at 11:36:51 AM
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Not sure where to start on this one so I may not attempt an answer, but good question..... hmmmm.....
Vote for Nobody 2020!
July 9th, 2012 at 11:51:40 AM
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Quote: WizardHow high would you need to build a tower to see 1/3 of the earth's surface?
Assume a perfectly spherical earth with a circumference of 40,075 km.
384,000 km should do just fine. Or, did you want the minimum height necessary?
July 9th, 2012 at 12:04:25 PM
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I get 6,378 km.
July 9th, 2012 at 12:13:57 PM
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Is there an "exact" answer for this? I get something where i have to solve for X in:
X sqrt(R2 - X2) + R2 sin-1 (X/R) = 2R2 / 3
Note that X is not the solution; it is the distance from the center of the earth to the plane that separates the surface into surface areas of 1/3 of the total and 2/3 of the total.
EDIT: Having solved that, I get the height = 12,381 km
X sqrt(R2 - X2) + R2 sin-1 (X/R) = 2R2 / 3
Note that X is not the solution; it is the distance from the center of the earth to the plane that separates the surface into surface areas of 1/3 of the total and 2/3 of the total.
EDIT: Having solved that, I get the height = 12,381 km
Let the radius be 1, and the tower be atop the north pole.
Let H be the distance from the center of the planet to the plane containing the horizon latitude.
Since the surface area of the top half of the planet = 2 PI, the surface area between the equator and the horizon plane = 2/3 PI.
INTEGRAL(0, h) [2 PI sqrt(1 - x^2) dx] = 2/3 PI
INTEGRAL(0, h) [sqrt(1 - x^2) dx] = 1/3
h/2 sqrt(1 - h^2) + 1/2 arcsin h = 1/3
h sqrt(1 - h^2) + arcsin h = 2/3
Solve using Newton-Raphson: h = 0.34
Let O be the center of the planet
T be the top of the tower
S be the point on OT that intersects the surface
P be a horizon point
Q be the point on OT that is on the horizon plane
OQ = h = 0.34
QS = 1 - 0.34 = 0.66
PQ = sqrt(1 - 0.34^2) = 0.940425
TPO is a right angle, so TPO is similar to TQP
TQ / QP = PQ / OQ
TQ = (PQ)^2 / OP = 0.940425^2 / 0.34 = 2.601174
The height of the tower ST = TQ - QS = 2.601174 - 0.66 = 1.941174
Since the actual radius is 6378.13 km, the height of the tower is 1.941174 times this, or about 12,381 km.
Let H be the distance from the center of the planet to the plane containing the horizon latitude.
Since the surface area of the top half of the planet = 2 PI, the surface area between the equator and the horizon plane = 2/3 PI.
INTEGRAL(0, h) [2 PI sqrt(1 - x^2) dx] = 2/3 PI
INTEGRAL(0, h) [sqrt(1 - x^2) dx] = 1/3
h/2 sqrt(1 - h^2) + 1/2 arcsin h = 1/3
h sqrt(1 - h^2) + arcsin h = 2/3
Solve using Newton-Raphson: h = 0.34
Let O be the center of the planet
T be the top of the tower
S be the point on OT that intersects the surface
P be a horizon point
Q be the point on OT that is on the horizon plane
OQ = h = 0.34
QS = 1 - 0.34 = 0.66
PQ = sqrt(1 - 0.34^2) = 0.940425
TPO is a right angle, so TPO is similar to TQP
TQ / QP = PQ / OQ
TQ = (PQ)^2 / OP = 0.940425^2 / 0.34 = 2.601174
The height of the tower ST = TQ - QS = 2.601174 - 0.66 = 1.941174
Since the actual radius is 6378.13 km, the height of the tower is 1.941174 times this, or about 12,381 km.
July 9th, 2012 at 3:49:46 PM
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I got 6,876 km. But i allowed for atmospheric interference.
July 9th, 2012 at 4:00:48 PM
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Quote: WizardDing ding ding! You [meaning jc2286] win the prize for the first correct answer. Perhaps the rest would like to see your solution, but please put it in spoiler tags. If you don't have the time; I can do it.
Here is my solution:
From the top of the tower you see an area of the earth called a "zone." The area of a zone is 2*pi*R*h, where h is the height of the zone and R is the radius of the sphere. (The height of the zone is the distance between the base of the tower and the plane of the horizon.) The zone's area is equal to 1/3 the sphere's area of 4*pi*R^2. So, 2*pi*R*h = 4*pi*R^2/3, so then h = 2R/3.
Now, draw a diagram to see two similar triangles. The bigger triangle has vertices of the earth's center, the tower's top, and a point on the horizon; and it has an hypotenuse of R + x, where x is the height of the tower, and a shorter leg with a length of R. The smaller triangle has vertices of the earth's center, the point on the horizon, and the zone's base's center; and it has an hypotenuse of R and a shorter leg of R - h, or R/3. So, (R+x)/R = R/(R/3). Then R+x = 3R, or x = 2R, or 12,756 km.
Now, draw a diagram to see two similar triangles. The bigger triangle has vertices of the earth's center, the tower's top, and a point on the horizon; and it has an hypotenuse of R + x, where x is the height of the tower, and a shorter leg with a length of R. The smaller triangle has vertices of the earth's center, the point on the horizon, and the zone's base's center; and it has an hypotenuse of R and a shorter leg of R - h, or R/3. So, (R+x)/R = R/(R/3). Then R+x = 3R, or x = 2R, or 12,756 km.
July 9th, 2012 at 4:04:25 PM
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I disagree with this answer . For clarification, where would you locate this tower. Please be specific within 100 miles !!
July 9th, 2012 at 4:16:28 PM
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Quote: buzzpaffI disagree with this answer . For clarification, where would you locate this tower. Please be specific within 100 miles !!
It depends what you want it for. But the best place would be on the Equator. Another good location would be either pole.
But for a tower to be of any use, it would need to be about 35,786 kilometers tall or so, likely more... Can you spare some carbon nanotubes?
Donald Trump is a fucking criminal
July 9th, 2012 at 4:23:15 PM
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Like most mathematical questions, practical use is not in the equation. That is why there are sometimes referred to
as mathematical *****.
as mathematical *****.
July 9th, 2012 at 4:39:09 PM
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How tall would the tower have to be to see 50% of the Earth's surface? (Same parameters).
Simplicity is the ultimate sophistication - Leonardo da Vinci
July 9th, 2012 at 4:43:09 PM
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Quote: AyecarumbaHow tall would the tower have to be to see 50% of the Earth's surface? (Same parameters).
That's a much easier question than the Wizard's question.
infinity
July 9th, 2012 at 4:43:35 PM
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duplicate post
July 9th, 2012 at 4:46:58 PM
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Hee hee... is there an ARCSIN function in Excel?
Simplicity is the ultimate sophistication - Leonardo da Vinci
July 9th, 2012 at 5:11:59 PM
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Quote: ChesterDogHere is my solution:
That is more intuative than the integral calculus I went through. Here is my solution.
1. Know that the surface area of a sphere is 4*pi*r^2, where r is the radius.
2. Assume r=1, where the answer will be expressed relative to the radius of the earth. So the surface area of the earth = 4*pi
3. Assume the tower is on the north pole.
4. Consider the angle x = point on the equator -- center of the earth -- point on the horizon (as seen from the tower)
5. The surface area seen from the tower is (4/3)*pi.
6. The area above the equator, but unseen from the tower is (2/3)*pi.
7. Solve for x such that the sum of the circumference rings from the equation to the line of longitute at x = (2/3)*pi.
(2/3)*pi=integral from 0 to x of 2*pi*cos(a) over the range of integration of 0 to x.
(2/3)*pi=2*pi * integral from 0 to x of cos(a) over the range of integration of 0 to x.
1/3 = sin(a) from x to 0.
1/3 = (sin(a)-0)
sin(a)=1/3
a=.339837 radians or 19.47118 degrees.
8. Consider the triangle formed by the center of the earth, a point on the horizon, and the top of the tower.
9. We know angle formed at the top of the tower is x, or 19.47118 degrees.
10. From simple trig, we know the ratio of distance from the top of the tower to the center of the earth, divided by the radius of the earth is 3.
11. Subtracting the radius of the earth, the height of the tower is 2 times the radius of the earth, or the diameter.
12. We know the circumference of the earth is 40,075 km., so the diameter is 40,075/pi = 12756 km., which is also the height of the tower.
Note: corrections made about 9:00 PM PST 7/9/12 and again 9:30 AM 7/10/12
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 9th, 2012 at 6:51:28 PM
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Quote: AyecarumbaHee hee... is there an ARCSIN function in Excel?
Yes - I think it's ASIN(number); the result is in radians
July 9th, 2012 at 7:05:40 PM
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Quote: WizardThat is more intuative than the integral calculus I went through. Here is my solution.
2/3 = cos(x)
x=30 degrees.
The last time I looked, if x = 30 degrees, cos x = sqrt(3)/2, not 2/3.
The arccosine of 2/3 is about 53.5 degrees.
The arccosine of 2/3 is about 53.5 degrees.
July 9th, 2012 at 7:09:12 PM
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The real question is, assuming normal construction materials; how much area would said tower cover, if it were actually built so someone could climb to the top to see 1/3 of the earths surface? ;-)
July 9th, 2012 at 7:40:20 PM
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Quote: ThatDonGuyEr
I made some mistakes in my solution, but it still came out to the right answer somehow. I just fixed it up.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 9th, 2012 at 9:07:22 PM
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Quote: WizardI made some mistakes in my solution, but it still came out to the right answer somehow. I just fixed it up.
1. Know that the surface area of a sphere is 4*pi*r^2, where r is the radius.
2. Assume r=1, where the answer will be expressed relative to the radius of the earth. So the surface area of the earth = 4*pi
3. Assume the tower is on the north pole.
4. Consider the angle x = point on the equator -- center of the earth -- point on the horizon (as seen from the tower)
5. The surface area seen from the tower is (4/3)*pi.
6. Solve for x:
(4/3)*pi=integral from x to pi/2 of 2*pi*sin(r) dr
If x is the angle (point on the equator - center of the earth - point on the horizon), then you want sin (90-x), as the value you are trying to integrate is a side of the triangle (north pole - center of the earth - point on the horizon); sin (90-x) = cos x, so it should be:
(4/3)*pi = integral from x to pi/2 of 2*pi*cos(r) dr
(4/3)*pi = 2 * pi *integral from x to pi/2 of cos(r) dr
2/3 = integral from x to pi/2 of cos r dr = sin(r) from x to pi/2 = (sin pi/2 - sin x) = 1 - sin x
sin x = 1/3 = cos (90-x)
7. Draw a triangle with points at the center of the earth, hoizon, and top of the tower.
8. The angle top of the tower - center of the earth - horizon is (90-x), so its cosine is 1/3, which is opposite side/hypotenuse;
1/3 = 1 / (1 + h), where h is the distance from the surface of the earth to the top of the tower
3 = 1 + h, so h = 2
2. Assume r=1, where the answer will be expressed relative to the radius of the earth. So the surface area of the earth = 4*pi
3. Assume the tower is on the north pole.
4. Consider the angle x = point on the equator -- center of the earth -- point on the horizon (as seen from the tower)
5. The surface area seen from the tower is (4/3)*pi.
6. Solve for x:
(4/3)*pi=integral from x to pi/2 of 2*pi*sin(r) dr
If x is the angle (point on the equator - center of the earth - point on the horizon), then you want sin (90-x), as the value you are trying to integrate is a side of the triangle (north pole - center of the earth - point on the horizon); sin (90-x) = cos x, so it should be:
(4/3)*pi = integral from x to pi/2 of 2*pi*cos(r) dr
(4/3)*pi = 2 * pi *integral from x to pi/2 of cos(r) dr
2/3 = integral from x to pi/2 of cos r dr = sin(r) from x to pi/2 = (sin pi/2 - sin x) = 1 - sin x
sin x = 1/3 = cos (90-x)
7. Draw a triangle with points at the center of the earth, hoizon, and top of the tower.
8. The angle top of the tower - center of the earth - horizon is (90-x), so its cosine is 1/3, which is opposite side/hypotenuse;
1/3 = 1 / (1 + h), where h is the distance from the surface of the earth to the top of the tower
3 = 1 + h, so h = 2
July 9th, 2012 at 11:07:56 PM
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All joking and sarcasm aside, I got to thinking about the question and answer and started wondering... This 1/3 of the surface we're seeing on top of this tower.. Is it just staring in 1 direction, or is it a 360 degree view and the total of what you're seeing equals 1/3 of said surface? I'm not a math guy by any stretch of the imagination, so please dont tell me to look at the math in previous answers, cuz I'll just get a bigger headache. :)
July 10th, 2012 at 6:06:37 AM
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Quote: WizardDing ding ding! You win the prize for the first correct answer. Perhaps the rest would like to see your solution, but please put it in spoiler tags. If you don't have the time; I can do it.
and fear no man
I did this 2-dimensionally.
Draw a circle (earth).
Draw a point P anywhere outside the circle. This represents the top of the tower.
Draw a line from P to the center of the circle (point C).
Draw a line from P that is tangential to the circle. We'll call that tangential point T.
We now have a triangle PCT, where angle T is 90 degrees by definition.
The length of one side is the radius (r). The length of the hypotenuse is r+x, where x is the height of the tower.
There's an arclength inside the triangle. Because we are looking for 1/3 the circumference, and line PC bisects that, that arclength is 1/6 the circumference (cir/6).
There's a pie wedge at point C, with sides of length r and arclength cir/6.
r is cir/(2pi).
Using the formula theta=s/r, that means that angle C is pi/3 radians, or 60 degrees.
Since angle T is 90 degrees, that means angle P is 30 degrees. We have a 30-60-90 triangle.
In a 30-60-90 triangle, the length of the side opposite 30 degree angle is 1/2 the hypotenuse (you could determine this with trig functions if you didn't know the 30-60-90 properties).
So, r = (r+x)/2, or x=r.
The height of the tower is equal to the radius of the earth, which is cir/(2pi), 400075/(2pi), or ~6,378 km.
Originally I got the same answer as ChesterDog, which is 2x the correct answer. I had taken the entire hypotenuse instead of just the portion that was outside the circle (above the earth's surface). I posted that answer but caught it moments later and edited it.
July 10th, 2012 at 7:05:57 AM
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Quote: jc2286
I did this 2-dimensionally.
Draw a circle (earth).
Draw a point P anywhere outside the circle. This represents the top of the tower.
Draw a line from P to the center of the circle (point C).
Draw a line from P that is tangential to the circle. We'll call that tangential point T.
We now have a triangle PCT, where angle T is 90 degrees by definition.
The length of one side is the radius (r). The length of the hypotenuse is r+x, where x is the height of the tower.
There's an arclength inside the triangle. Because we are looking for 1/3 the circumference, and line PC bisects that, that arclength is 1/6 the circumference (cir/6).
There's a pie wedge at point C, with sides of length r and arclength cir/6.
r is cir/(2pi).
Using the formula theta=s/r, that means that angle C is pi/3 radians, or 60 degrees.
Since angle T is 90 degrees, that means angle P is 30 degrees. We have a 30-60-90 triangle.
In a 30-60-90 triangle, the length of the side opposite 30 degree angle is 1/2 the hypotenuse (you could determine this with trig functions if you didn't know the 30-60-90 properties).
So, r = (r+x)/2, or x=r.
The height of the tower is equal to the radius of the earth, which is cir/(2pi), 400075/(2pi), or ~6,378 km.
Originally I got the same answer as ChesterDog, which is 2x the correct answer. I had taken the entire hypotenuse instead of just the portion that was outside the circle (above the earth's surface). I posted that answer but caught it moments later and edited it.
How do you get 1/3 of the surface area from "We are looking for 1/3 the circumference"?
Also, Wizard, your solution (even in my "corrected" version) has a problem; we both assumed that the integral using polar coordinates is calculated the same way as one using regular coordinates. I think there's a factor of the radius missing somewhere, but I can't figure out where. (This is why my original solution was integrated in a straight line along the line from the center of earth to the tower.)
Let's look at trying to calculate the surface of half of a planet of radius R using polar coordinates and "normal" integration:
it would be the integral from 0 to 90 degrees of (2 PI R sin x dx)
= 2 PI R * the integral from 0 to 90 degrees of (sin x dx)
= 2 PI R * ((- cos 90) - (- cos 0))
= 2 PI R * (0 - (-1)) = 2 PI R.
However, the answer is 2 PI R2.
July 10th, 2012 at 7:15:18 AM
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Quote: ThatDonGuyQuote: jc2286
I did this 2-dimensionally.
Draw a circle (earth).
Draw a point P anywhere outside the circle. This represents the top of the tower.
Draw a line from P to the center of the circle (point C).
Draw a line from P that is tangential to the circle. We'll call that tangential point T.
We now have a triangle PCT, where angle T is 90 degrees by definition.
The length of one side is the radius (r). The length of the hypotenuse is r+x, where x is the height of the tower.
There's an arclength inside the triangle. Because we are looking for 1/3 the circumference, and line PC bisects that, that arclength is 1/6 the circumference (cir/6).
There's a pie wedge at point C, with sides of length r and arclength cir/6.
r is cir/(2pi).
Using the formula theta=s/r, that means that angle C is pi/3 radians, or 60 degrees.
Since angle T is 90 degrees, that means angle P is 30 degrees. We have a 30-60-90 triangle.
In a 30-60-90 triangle, the length of the side opposite 30 degree angle is 1/2 the hypotenuse (you could determine this with trig functions if you didn't know the 30-60-90 properties).
So, r = (r+x)/2, or x=r.
The height of the tower is equal to the radius of the earth, which is cir/(2pi), 400075/(2pi), or ~6,378 km.
Originally I got the same answer as ChesterDog, which is 2x the correct answer. I had taken the entire hypotenuse instead of just the portion that was outside the circle (above the earth's surface). I posted that answer but caught it moments later and edited it.
How do you get 1/3 of the surface area from "We are looking for 1/3 the circumference"?
Also, Wizard, your solution (even in my "corrected" version) has a problem; we both assumed that the integral using polar coordinates is calculated the same way as one using regular coordinates. I think there's a factor of the radius missing somewhere, but I can't figure out where. (This is why my original solution was integrated in a straight line along the line from the center of earth to the tower.)
Let's look at trying to calculate the surface of half of a planet of radius R using polar coordinates and "normal" integration:
it would be the integral from 0 to 90 degrees of (2 PI R sin x dx)
= 2 PI R * the integral from 0 to 90 degrees of (sin x dx)
= 2 PI R * ((- cos 90) - (- cos 0))
= 2 PI R * (0 - (-1)) = 2 PI R.
However, the answer is 2 PI R2.
I flattened it to 2 dimensions to make it easier to solve. The answer is the same.
In your example, yes 2*pi*r^2 would be half the surface. If you flatten it, you divide by 2r, and you get pi*r, which is half the circumference. If we were determining 1/3 the surface area of earth, then we would need to multiply 1/3 circumference by 2r. But we're not... the tower height is 1-dimensional, so it is not affected by my flattening technique.
July 10th, 2012 at 9:44:44 AM
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Quote: ChesterDogI get 12,756 km. This is at a distance equal to the earth's diameter above the earth's surface, which is a neat result.
I admit that I totally blew this one. Sorry for anybody who I led astray. I now admit that ChesterDog is right and as the first one to submit the correct answer wins the prize. Please get it from jc2286, whom I incorrectly gave it to.
Also thanks to ThatDonGuy for his help.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 10th, 2012 at 11:13:59 AM
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I see no fault with his answer. Where did I go wrong? Or is it simply a matter of not being able to flatten to 2D after all?
edit: I think I see. The difference between surface area and circumference length is a factor of 2R. By dropping down a dimension, I eliminate a factor of R, but that still leaves the 2, which I never accounted for. That's why my answer is exactly half the correct answer.
edit: I think I see. The difference between surface area and circumference length is a factor of 2R. By dropping down a dimension, I eliminate a factor of R, but that still leaves the 2, which I never accounted for. That's why my answer is exactly half the correct answer.
July 24th, 2012 at 12:48:39 PM
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Quote: WizardI admit that I totally blew this one. Sorry for anybody who I led astray. I now admit that ChesterDog is right and as the first one to submit the correct answer wins the prize. Please get it from jc2286, whom I incorrectly gave it to.
Also thanks to ThatDonGuy for his help.
Take a look at the first to pic's. They are showing what the earth looks like from almost 8k miles (12,756 meters) high.
You decide if they show 1/3 of the earths surface.
Side note: If you build your tower on the top mount everest you could take almost 9k of the the height of the tower.
http://astrobob.areavoices.com/2012/02/08/earth-from-afar-tales-of-the-blue-marble/
