Tower....to the moon!March 22, 2007 6:02 PM   Subscribe

Math Question Kinda: I realize it's not possible. But if it was....how tall would the structure need to be if it were to be seen from both US coasts. Or is it impossible because of the curvature of the earth or whatever else I'm not thinking about it.

Let's suggest we're erecting it in Kansas...pretty damn near the center of the country.

Further, how wide would it have to be....? But I guess that's a bit of a curve-ball so let's not attempt that one just yet. Or let's. It's up to you, really.

Cause I got nothin'.

Thanks.
posted by ryecatcher to Technology (26 answers total) 4 users marked this as a favorite

I was an awful student in math, but i believe someone sharper than I would be able to cull some numbers from this article.
posted by ronmexico at 6:15 PM on March 22, 2007

I get roughly 2080 miles from coast to coast at the shortest distance. Using the formula on the horizon page, that gives 135 miles (215 km) for the height of the tower, and it would be in Texas.

Note that low Earth orbit for satellites starts at 200 km, so that's pretty high.
posted by smackfu at 6:26 PM on March 22, 2007

I don't have the numbers on hand, but this is how you find it.

Assume the earth is perfectly smooth, no mountains or anything to get in the way. Assume the structure is exactly midway between the two points.

Let r = radius of the earth.
Let x = the angle between a line drawn from the center of the earth to the first point, and a line drawn from the center of the earth to the structure.
(x can be found with d = distance (along the curve) between structure and one of the points. x in radians = d/r. x in degrees = (d/r)*(180/pi))

Let h = the height of the structure.

cos(x) = (r/(r+h))

The way to visualize this is to draw a circle (the earth) and three points on it, one halfway between the other two. Draw lines from the center of the earth to the coastal points, and one through the in-between point. The angles are x and x. Now, draw tangent lines to the circle that touch at the coastal points. The point at which they meet (which also intersects the line through the in-between point) is the top of the structure. The tangent lines represent line-of sight (since obviously you can't see through the earth). The tangent lines are also perpendicular to the lines through the coastal points, so you now have two identical right triangles with r as one of the sides and r+h as the hypotenuse. The cosine of an angle is the length of the adjacent side over the length of the hypotenuse, and that's where you get the formula above. (You do have to do a little more work to get h by itself, but that's not difficult.)

Incidently, for points on opposite sides of the globe, there is no structure that would be tall enough, because the line-of-sight lines are parallel. (Discounting mountains and other such things that let you 'cheat'.)

As for the width, I think that would depend on what you're looking at it with. With a good enough telescope you might be able to see something the width of a radio tower.
posted by Many bubbles at 6:32 PM on March 22, 2007

--Well, okay, so standing up is, in a sense, the same as being on a mountain in that your line of sight isn't perfectly tangent with the ground you're standing on. Whatever.
posted by Many bubbles at 6:35 PM on March 22, 2007

From the curved distance along the planet's (sea-level) surface to the horizon, your dc = 3362 km / 2, which equals 1681 km from each edge of the coast to your hypothetical building.

Assuming the height of the observer from sea level being zero (the observer's own height being neglible), solving for h you get 228.118 km = 141 miles. Pretty tall building.

I don't know if that formula takes into account light refraction, so in reality there's probably a little give and take in the numbers.
posted by Blazecock Pileon at 6:37 PM on March 22, 2007

SmackFu, your coast-to-coast distance is much too great. The shortest distance would be between San Diego CA and Corpus Christi TX, 1250 miles, and the half-way point would be a little ways south of El Paso, in northern Mexico. (Pick pick pick)

Discussion of how wide such a tower would have to be is meaningless and unanswerable, because it's impossible to build a free-standing structure that tall with the materials and construction techniques available to us now.
posted by Steven C. Den Beste at 6:38 PM on March 22, 2007

...Interesting. I'm getting 617 miles as the height of the tower, using 2080 miles coast-to-coast and 3959 miles as the earth's radius. Anyone want to double-check?
posted by Many bubbles at 6:42 PM on March 22, 2007

... the arc length distance s along the curved surface of the Earth to the bottom of object:

\cos\frac{s}{R}=\frac{R}{R+h}.

Say the continental US is about 4000 km wide, so s is 2000 km, and R is 6400 km; so, unless I screwed up the algebra, the height of the object (in km) is

h = R ( 1/cos(s/R) - 1 )

Or about 325 km (around 200 miles).

Now, what do you mean by "how wide does it have to be?" You mean how big the object needs to be so you can see it from around 200 miles away? Well, the average human eye can see a 1m object from around 1km away, so this satellite/structure/object needs to be around 350m (call it 1000 feet).
posted by phliar at 6:43 PM on March 22, 2007

Bah, nevermind. Forgot what d was supposed to be. I'm getting 140 miles now.
posted by Many bubbles at 6:44 PM on March 22, 2007

Oh--and if the structure isn't exactly halfway between the two points, you just use the larger angle for x. That makes sure it's visible from the farther location, so it'll naturally be visible from the closer one.
posted by Many bubbles at 6:57 PM on March 22, 2007

I'm not going to recompute the figures, but I would like to add that the beanstalk will be visible from coast to coast on whatever continent we build it (except Eurasia, which is a bad choice anyway). You're essentially talking about a tether from the ground to LEO. You'll need a telescope, but you'll see it.
posted by Netzapper at 7:16 PM on March 22, 2007

@Many bubbles -

"Incidently, for points on opposite sides of the globe, there is no structure that would be tall enough, because the line-of-sight lines are parallel. (Discounting mountains and other such things that let you 'cheat'.)"

I'm imagining our observers at the North and South poles with a truly frikkin' massive tower jutting out from the equator. Given a long enough length, and the telescopes that everyone is mentioning, it seems like it would be visible after some length x. Am I missing something? (Just mentally imagining, it seems like a length in the ballpark of half the earth's diameter, some 4000 odd miles, would work.) Are we disqualifying this as being 'visible' because of the elevation of the viewing angle?

@ryecatcher -
Truly great question, I salute you.
posted by mysterious1der at 8:40 PM on March 22, 2007

Response by poster: I have no idea what most of you are saying. But it's an awesome sight to witness.

(The 'how wide' question was relating to what philar answered, about how wide it would have to be to actually see it. Instead of, say, just dangling a thread down 200 miles).
posted by ryecatcher at 8:49 PM on March 22, 2007

Well, the average human eye can see a 1m object from around 1km away, so this satellite/structure/object needs to be around 350m (call it 1000 feet).

Eh... it doesn't matter how wide it is. You just need to make it bright enough. The average human eye can see objects several light years away.
posted by logicpunk at 8:50 PM on March 22, 2007

The household mathematician has tried his hand at it, and this is what he came up with:

Say the radius of the earth is 4000 miles. (a slight overestimate according to wikipedia, but what the hell)

Say the two points we want the tower to be visible from are A and B. Say that A and B are 60 degrees apart. We draw a radius from each point into the center of the earth, we get an angle of 60 degrees at the center of the earth.

The lowest point that A can see, looking east to B, is found by drawing a tangent line (a line at 90 degrees to the radius) at A. Vice versa for point B. So we'll draw two tangent lines.

Now we have drawn a kite or diamond shape. (But different from that linked picture, the right and left angles should be 90 degree angles!) The bottom two lines are the radii from the center of earth (let's call it C) to the surface of earth. The bottom point is the center of the earth. The leftmost point is A, the rightmost point is B. The top to lines are the sight lines from A and B to the tower (to the lowest point on the tower that they can both see -- let's call this point D).
You can draw two more lines, just to make the diagram clearer: a curved line from A to B, representing the surface of the earth between the two cities. (The curve is given by the radius of the earth; visualize drawing the rest of the circle or earth and you'll see where the curve should be.) In the middle of this curve, you can draw a vertical line up to the top point -- this vertical is the tower we're building. Maybe draw a dashed line all the way down to C.

Ok - now we have two 30- 60 -90 triangles: ACD and BCD. Let's just look at the righthand side one: Angle DCB is 30 degrees (half of the 60 degrees we agreed on at the start of the problem for angle ACB). Angle CBD is 90 degrees, because we drew a tangent line (ie a line at 90 degrees to the radius) from point B to point D. This means that angle BDC is 60 degrees.

Geometry tells us that the sides across from the 30 - 60 - 90 angles will be in ratios of 1 - square root of 3 - 2. The "1" side is the distance from B to D. The "square root of 3" side is 4000 miles (the radius of the earth). The distance we're looking for is the height from the surface of the earth to D -- so it's part of the "2" side. The "2" side is the distance from C to D -- so we need to find the length of that side, and then subtract 4000 miles (the underground part), leaving only the part from the surface to D.

So:
(1) 4000/ square root of 3 = x/2
(2) 4000 x 2 = 8000
(3) 8000 / square root of 3 = 4618.8
(4) 4618.8 - 4000 = 618.8 miles

So if A and B are 60 degrees apart, the tower has to be 618.8 miles tall to be visible to both.

Oops:
Well, he just checked the atlas and it's not 60 degrees from coast to coast in the US. It's more like a 50 degree angle between Seattle and Boston for example. So the geometry isn't quite as simple with 30 - 60 - 90 triangle, but trig can help:

Instead it would be 4000/(cos (50/2)) = 4413 miles
4413 miles - 4000 = 413

So for Boston and Seattle to both be able to see the tower, assuming no mountains or optical effects, the tower should be 413 miles tall.
posted by LobsterMitten at 9:41 PM on March 22, 2007

Many Bubbles: The sun rising and setting is visible from both sides of the earth at once. (see??) So if something were 1 AU tall and on the equator it would be visible from both poles or whatever.

Atmospheric refraction is going to play a role in whatever the real answer is, by the way. So this would be a great time for everyone to just give up! :D
posted by aubilenon at 9:59 PM on March 22, 2007

Response by poster: So even if one built a 413 mile tall building, one would have to build it on a level playing field, huh?

I couldn't stand in California, look over into the sky and see this thing stretching its way up from Texas?

Nah, I guess not.

And lo, another dream dies.
posted by ryecatcher at 11:01 PM on March 22, 2007

? ryecatcher, I don't know what you mean.
posted by LobsterMitten at 11:11 PM on March 22, 2007

Response by poster: LobsterMitten --

"So for Boston and Seattle to both be able to see the tower, assuming no mountains or optical effects, the tower should be 413 miles tall."

I don't know what the highest mountain between Seattle and Boston is -I'm certain it ain't 413 miles tall- but even if it's one mile, you're saying that would obstruct my view. Correct?
posted by ryecatcher at 12:47 AM on March 23, 2007

mysterious1der: I'm imagining our observers at the North and South poles with a truly frikkin' massive tower jutting out from the equator. Given a long enough length, and the telescopes that everyone is mentioning, it seems like it would be visible after some length x. Am I missing something? (Just mentally imagining, it seems like a length in the ballpark of half the earth's diameter, some 4000 odd miles, would work.) Are we disqualifying this as being 'visible' because of the elevation of the viewing angle?

Well, as long as you're looking from above ground level, you'll be able to see it at some distance x, because the angle (the one between the line-of-sight and a line through the center of the earth and the object halfway between) will be slightly less than ninety degrees. But at exactly ground level, the tangent lines representing line-of-sight (in a plane, of course) from opposite sides of the earth will be parallel. If they don't intersect, then you can't place something at a point and say "Here's where you start to be able to see it from both places". The object would have to have at least the same diameter as the earth itself, (because then it can intersect both lines-of-sight by itself) and in that case it doesn't matter where you put it, as long as it's equidistant from the two points.

aubilenon: The sun rising and setting is visible from both sides of the earth at once.

This is true. I admit I was only thinking in terms of things smaller than the earth.

Atmospheric refraction is going to play a role in whatever the real answer is, by the way. So this would be a great time for everyone to just give up! :D

Ha! Okay, "assuming the earth is a perfectly smooth sphere, assuming the observer is at exactly ground level, and assuming the earth has no atmosphere".

We're probably getting this thing up there by way of a frictionless pulley, too.
posted by Many bubbles at 1:36 AM on March 23, 2007

In reference to the atmospheric refraction mentioned frequently, light refracts down best on warm days with cold ground (eg. warm winter day with frozen ground / snow) this is because cold air is denser than warm air, which adds to the natural decreasing density gradient from the ground upwards (air higher up is less dense than air on the ground). This gradient is actually large enough to double or even triple the appearance of objects on such days. If you get an opposite effect of a cold day with relatively warm ground, you have an opposite effect, but it is less noticeable as the temperature gradient is affected less. So to find the minimal height of the object, you would need to select a viewing area where warm weather after a significant cold period are common. Probably further north. This could potentially reduce the size of the object to a third of its size projected neglecting atmospheric refraction.
posted by mrw at 1:52 AM on March 23, 2007

The shortest distance would be between San Diego CA and Corpus Christi TX...

IHBT

Discussion of how wide such a tower would have to be is meaningless and unanswerable, because it's impossible to build a free-standing structure that tall with the materials and construction techniques available to us now.

IHBT^2
posted by DU at 4:19 AM on March 23, 2007

This would be... a lot more massive, but there are certain tricks of the light that let you "see" things you shouldn't be able to; i.e. at sunset sometimes you can see Chicago across Lake Michigan but it's really kind of a reflection of the buildings. I don't know how to explain it but my great-great-uncle would if he was still alive.
posted by dagnyscott at 5:25 AM on March 23, 2007

The shortest distance would be between San Diego CA and Corpus Christi TX,

Given that the original question specified "both US coasts," I think it's pretty clear the coasts in question are the East and West, not the Gulf.
posted by Chrysostom at 5:50 AM on March 23, 2007

If you count the Gulf, the shortest distance would be somewhere in Florida...
posted by grateful at 6:34 AM on March 23, 2007

ryecatcher, setting aside the atmospheric issues, the mountains just make the calculation more complicated - they don't make it impossible. Suppose for simplicity that there's one mountain between point A and the tower. Draw a line from A to the tower that passes just over the top of the mountain. Now do the calculation I mentioned above, but using this new CAD triangle.
posted by LobsterMitten at 7:33 AM on March 23, 2007

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