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Distance of horizon from leaning tower
By Brad Rodgers (P1930) on Sunday, October 22, 2000 - 05:06 am:

At what angle would a tower of height h have to lie (relative to a tangency of the earth) to have maximum viewing distance around the earth?

Thanks,

Brad

By Dan Goodman (Dfmg2) on Sunday, October 22, 2000 - 01:40 pm:

A person at the top of the tower can see everything that a person lower down can see, draw a diagram. Basically, this is because if you draw any line segment going through the middle of the tower, and connect the ends up to the top of the tower, then assuming that the line segment is not obstructed, then the triangle formed will not intersect with the earth, therefore the person at the top can see everything that the person at the middle can see. Therefore, to maximise the viewing area of the tower, you need to maximise the viewing area of a point, which just means maximising the distance from the surface of the earth, so the tower should point straight upwards (along the normal vector at the point).

I hope I haven't made a mistake...

By Brad Rodgers (P1930) on Sunday, October 22, 2000 - 04:32 pm:

That would be my first, intuitive response, but if you draw a picture, you'll see that a very slight tilt allows one to be at a tangency coming from a point farther away than the one seen by a tower straight up.

Actually, when I posted this, I was very tired and thus not able to think real well. But in the past hour I've been able to come up with a solution. It, however, takes probably three lines to write out, you must first change the height of the tower to another term, and it involves about ten uses of sines. If anyone wants me to post this, I probably could though. Or, if anyone thinks that there is a simpler solution that would probably be easier to use.

Thanks,

Brad

By Dan Goodman (Dfmg2) on Sunday, October 22, 2000 - 05:56 pm:

Here's a picture:

Tower problem

What I'm arguing is that everything that anyone at any point on the tower (e.g. at A or B) can see, can also be seen by someone at the top of the tower, T. This is because the vision cones for A and B are contained in the one for T. In other words, in considering the problem, you only need to consider the top of the tower T, as an isolated point. Let d be the distance from the top of the tower to the closest point on the ground (drop a perpendicular). d<=h, but the area seen increases as d increases, therefore the maximum value will be when d=h, i.e. when the tower is normal to the surface of the Earth.

If you have a straight tower and a leaning tower, the leaning tower will be able to see stuff that the straight tower can't, but the straight tower will also be able to see stuff the leaning tower can't, and moreover it will be able to see more stuff that the leaning tower can't.

Nice diagram eh?

By Brad Rodgers (P1930) on Sunday, October 22, 2000 - 09:23 pm:

Yeah, I do like the diagram. How did you have it made? You are right, of course, and you have answered what I asked. But what I really meant to say (reference to said tiredness) was at what angle will the person be able to see out considering only one direction of viewing.

I still have phrased this rather akwardly, so hopefully someone can understand it and write it better; if not, I'll have another go at writing it again.

Brad

By Dan Goodman (Dfmg2) on Sunday, October 22, 2000 - 10:49 pm:

I drew it in Adobe Photoshop (you can download LE for free - limited edition - from www.adobe.com).

I'll get back to you about your question tomorrow.

By Arvan Pritchard (T708) on Monday, October 30, 2000 - 10:44 am:

You can work in a plane containing the centre of the earth, the foot of the tower and the direction you want to look...

Now the top of the tower is on a circle of radius h with centre on the earth's surface (a circle of radius r). The best view is when the line of sight is the common tangent...

I think that if t is the angle from vertical then

sin(t) = (r-h)/r

By Brad Rodgers (P1930) on Monday, October 30, 2000 - 08:08 pm:

My answer is certainly more complicated than that. How did you formulate that though. I'll go ahead and post my answer, but be wary...

First, let B be an angle formed by a tower (this will have radius h) straight up in reference to the line of tangency it is seeing along (that's the best I could do). Now, let

tan(t)=y/x

Where t is the deviation(for tower of radius r) from a tangency to the earth.

Through differentials, we can conclude that (if the radius of the earth is 1) ,

x=1/(1+1/(sin(90-B)(sin(90-B)h+1

Where h can be derived from

r=(x2+(-x/((sin(90-B)(h+1))+h)2)1/2

We can also know that

y=-1/((sin(90-B)(h+1))x+h

From this, we can deduce t, albeit only after alot of work.

As I said, I'm sure that a more elegant solution works (most probably the one posted above me, it's just that I don't undertsand it well), so I won't take the time to post the explanation behind all this. If anyone wants to see my method, I'd be glad to post it; I will try to post a picture to explain this better.

Brad

By Dan Goodman (Dfmg2) on Monday, October 30, 2000 - 10:03 pm:

OK, here's another picture:

Tower Problem mark 2

I don't think I've come up with the neatest method, so I've included lots of information on the diagram so others can refer to it if they wish.

Basically, the two angles marked q are the same because the triangles are similar. We can also use this to show that (h+x)/h = (R+h+x)/R (the ratio EC/CF is the same as EA/AJ by similarity). This gives us x, so we can now write cosq = CF/EC = h/(h+x) = (r-h)/r. Therefore q = cos-1((r-h)/r).

By Dan Goodman (Dfmg2) on Monday, October 30, 2000 - 10:08 pm:

The last answer should (really) be q=min( cos-1((r-h)/r), p/2 ), because you can't build a tower at an angle of greater than p/2 to the vertical. However, you'd have to have a pretty massive tower for this to make a difference.

By Brad Rodgers (P1930) on Tuesday, October 31, 2000 - 12:34 am:

That is right (and a lot easier than my method). I've yet to see if my method will simplify to the above answer, but it would take a lot of work to find out.

Thanks,

Brad