No, if the line of the guy’s spine moves away from the center of the Earth, then his head will move towards the northern horizon by the same factor, since these are small angles, except removing the square root of two factor. That’s h times e[sup]2[/sup] then, and that’s a difference of 2.04 cm, or .4 inches, both ways–0.8 inches total, which is nearly as large as your answers.
The slope of the ellipse at each point is easy to describe, algebraically, F(x,y), or even f(x). Set it equal to -1 to find the 45 degree latitude point, then add h times root two to each of x and y. That’s your viewpoint, P. For each horizon point X=(x,y) on the ellipse, the slope between P and X is equal to the slope of the ellipse at X, since the view to the horizon is tangent to the horizon at the horizon. Set that slope equal to f(x), and solve for x. That’s fairly easy to plug into Excel and let it solve for it–and I got that the north horizon was farther by 2.15 cm, or .848 inches. If the person’s feet were pointing towards the center of the Earth, it’d be 1.6 inches–artificially inflating the distance to twice. Hmm…
I checked the spreadsheet with a constant radius of 6367km, and it came up with a distance to horizon of 4405.294 km, which is equal to sqrt(6367001.524^2-6370000^2) so that seems to work.
I’d like to see how Cecil got his 4 inch value. Seems too big by a factor of five.
The problem with what you’re saying here is that when he moves his head, his views of the horizons change as well. So the farthest point on the Northern horizon he can see will move a little north, as will the farthest point he can see on the Southern horizon. This will greatly reduce the effect you’re describing.
The effects of him leaning can be described by an effective change in his height (of order e[sup]4[/sup]), and an effective change in his latitude (of order e[sup]2[/sup] (h/a) - not e[sup]2[/sup] (h/a)[sup]2[/sup], which is what I had earlier).
Great. Now let’s write up a report and put it in one of those clear plastic covers that has a green plastic spine that you slip over the outside that you see at the school supplies section of WalMart, and send it off to Cecil! I’ll make up the cover sheet! I’m putting our usernames in alphabetical order.
Well, it’s been two months and the column is still the same, and I don’t think anyone has come up with any idea of where four inches may have come from. Any new insights?
For what it’s worth, about a month ago I was bored and did the problem again, using a method which was essentially similar to this one:
The only thing I really did different was solve algebraically for all the variables, and then run the equations through my calculator. This method should be exact to however much precision your calculator has; mine should produce results good to 3-4 decimals.
Here are the variables I used:
a, b, h, and L are the givens: major axis, minor axis, person’s height, and latitude. (x[sub]0[/sub], y[sub]0[/sub]) is where their feet are, and (p, q) is where their head is. (x[sub]N[/sub], y[sub]N[/sub]) is the tangent point of the Northern horizon, and (x[sub]S[/sub], y[sub]S[/sub]) for the Southern horizon. d[sub]N[/sub] and d[sub]S[/sub] for the distances to the horizons. Here are the equations I got:
x[sub]0[/sub] = a(1 + b[sup]2[/sup] tan[sup]2[/sup]L / a[sup]2[/sup])[sup]-1/2[/sup]
y[sub]0[/sub] = b(1 + a[sup]2[/sup] cot[sup]2[/sup]L / b[sup]2[/sup])[sup]-1/2[/sup]
p = x[sub]0[/sub] + h cos(L)
q = y[sub]0[/sub] + h sin(L)
The horizon tangent point x values are the two real roots of the equation:
x[sup]2[/sup] - 2A x + C = 0, where:
A = p a[sup]2[/sup] b[sup]2[/sup] / (p[sup]2[/sup] b[sup]2[/sup] + a[sup]2[/sup] q[sup]2[/sup])
C = a[sup]4[/sup](b[sup]2[/sup] - q[sup]2[/sup]) / (p[sup]2[/sup] b[sup]2[/sup] + a[sup]2[/sup] q[sup]2[/sup])
That is:
x[sub]N[/sub] = A - (A[sup]2[/sup] - C)[sup]1/2[/sup]
x[sub]S[/sub] = A + (A[sup]2[/sup] - C)[sup]1/2[/sup]
y[sub]N[/sub] = b(1 - x[sub]N[/sub][sup]2[/sup] / a[sup]2[/sup])[sup]1/2[/sup], and similarly with y[sub]S[/sub].
d[sub]N[/sub] = ((x[sub]0[/sub] - x[sub]N[/sub])[sup]2[/sup] + (y[sub]0[/sub] - y[sub]N[/sub])[sup]2[/sup])[sup]1/2[/sup], and similarly for d[sub]S[/sub].
Here are the numbers I got, in units of 1000’s of kilometers:
a = 6.3781370
b = 6.3567523142
h = 0.000001524
L = 45°
(x[sub]0[/sub], y[sub]0[/sub]) = (4.517590878865, 4.487348408818)
(p, q) = (4.5175919564957, 4.4873494864487)
A = 4.5175897939961
C = 20.408607842933
(x[sub]N[/sub], y[sub]N[/sub]) = (4.5144746880589, 4.4904624443584)
(x[sub]S[/sub], y[sub]S[/sub]) = (4.5207048999333, 4.484232232494)
d[sub]N[/sub] = 0.0044054355615418
d[sub]S[/sub] = 0.0044054150878294
d[sub]N[/sub] - d[sub]S[/sub] = 0.0000000204737124 = 2.047 cm.
As this agrees with my previous result to better than 1%, of course I believe it.
Obviously, bikertrash, you don’t go to the right sort of boathouses. I’m sure there are plenty of babes who would be impressed by a guy who can do precision analytical geometry calculations. Now if only I could find any of them…
this was simply the observation of an old san diago sailor who , while standing mid watch on the after crane 20 feet above the surface of the water was suposed to be able to see aproxamatly five miles that gave me enough area to wory abouy with out you adding 4 more inches all the way around
