Check http://www.straightdope.com/classics/a1_167b for what Cecil said regarding horizons.
Check http://boards.straightdope.com/sdmb/showthread.php?postid=1022659#post1022659 for what I say regarding Cecil’s error. He failed to consider the tides!
(post edited to fix links)
[Edited by Arnold Winkelried on 01-04-2001 at 09:36 AM]
Eh, both your links give me a “Page Not Found”. 
The column can also be found on pages 167-168 of Cecil Adams’ book «The Straight Dope».
moderator, «Comments on Cecil’s Columns»
By the way, grienspace, you may want to copy your calculations into this thread, to prevent people from having to make the trip to the other forum. This might generate more discussion.
I think I’m confused. I get what Cecil said about the north south because of the shape of the Earth, but the tide part confuses me since the tide affects the water the same all around you I don’t see how it would make a difference.
But then again we could REALLY confuse everything and say that the Earth is really a Geoid (sp) so horizons everywhere would be different!
Well not quite. Only at slack tides. During a flood or ebb, water is flowing and therefore the surface is on a slope.
Then the water would always be on the same slope at any tide at that same place so you’d still have the same horizion at either low tide or high tide or anything in between.
But the slope/curvature of the wavelength of a tide is not constant. The only time you have have symetry east and west is at the crest and the trough of a wave, the slack tides. Even then successive tides are never identical so I should be safe in saying symetry never occurs at all east-west.
The tides cause the equatorial cross-section of the Earth to resemble an ellipse, with the major axis aligned pretty much toward the Moon. The ellipse’s major axis radius is what, about 2 meters more than its minor axis. The tides are higher in some places, but these are local effects due to sloshing, and wouldn’t apply on the open ocean. So if you’re asking what’s the difference in distance to the horizon East and West, the calculations are complicated, but the involve an ellipse with a major-minor axis difference of a couple of meters.
Now along the North-South direction, the cross section of the Earth is also an ellipse, but the polar radius is about 20000 meters smaller than the equatorial radius. Cecil has told us that the horizon difference North-South is only about 4 inches (100 mm), and although I haven’t seen his calculations this seems reasonable.
So as a rough guess, I would say that the difference in the East-West horizons is about one ten-thousandth of that 100 mm, or about 0.01 mm, since the eccentricity of the North-South ellipse is about ten thousand times that in the East-West direction. Certainly a difference this small is meaningless when you’re talking about the distance to a horizon around 5000 meters away, trying to find the tangent line on top of a moving sea.
I already commented on the original post, but now having a vastly different figure for the polar radius i’ll try again.
Cecil’s column states:
Which implies that the northern horizon is closer to the observer, doesnt it?
Using CurtC’s figure that the polar radius is 20, 000m less than the equatorial radius (6378300m) then assuming the earth follows the equation (at least in the first cartisan plane) x^2/6378300^2 + y^2/6358300^2 = 1 we can take the polar radius and divide it by 2 for someone standing at 45 degrees latitude (I assume, correct me if i’m wrong please).
Taking that value as y and putting it to solve for x we have the guy standing at: (3189150*sqrt(3), 3179150).
Calculating the tangent at this point we get m = -(63583/(21261*sqrt(3))
Taking the negative reciprocal for a person standing straight up at this point we get, m = 21261*sqrt(3)/63583.
Lets say the person’s eye level is 1.524m away from sea level (the figure Cecil used), then his eyes are at: (5.5237711517427210^6, 3.179150763795530410^6). Finding a line that is tangent to the ellipse that is also collinear to this point we solve the equation:
6378300Sqrt((6378300)^2 - x^2)/6358300 = 6378300(x - 5.5237711517427210^6)/(6358300 * Sqrt((6358300)^2 - x^2)) + 3.179150763795530410*^6
we get x = 5.5059791578632110^6, 5.52404625477049510^6
and y = 3.18295724804415*10^6, 3.1753412387125096`*10^6 respectively (by plugging it into our original ellipse equation).
Using the distance formula we get: a distance of: 4400.58m north and 4400.56m south.
4 inches away from Cecil’s figure (which seems correct since we probably used slightly different units).
…
Ok, now we take tides into effect, lets say theres a 6 meter tide on the equatorial radius (the figure you used in the previous post). This time we solve the equation:
3.179150763796068310^6 - (0.9968634305096057x
(-5.52377634789483210^6 + x)/Sqrt(40682787429636 - x^2) = 3179150*Sqrt(40682787429636 - x^2)/3189153
At this point Mathematica 4.0 CANT or doesnt want to deal with the precision required, in other words, the difference must be extremelly small and not worth nit picking over.
After taking a different approach in Mathematica to get greater precision there is about 0.507551mm more viewing distance north and 0.48748mm south with the tides.
oops that should be 0.48748mm less viewing distance south with the tides.
OK, I’m simply reposting what I said in the other thread, although I was also thwarted by the fact that the difference was not really within the capabilities of my trusty HP-48GX calculator. One of these days I’ll make it over to one of my school’s computer labs to try Maple or Matlab (actually probably both, Maple for the symbolic math, and Matlab for the number crunching. I believe that Matlab uses a very lardge number of significant figures, which would be nice)
[/quote]
Or, we could always do this the hard way. With the tidal variation of 6.7 meters, the earth’s equator becomes an ellipse 6.7 meters larger in it’s major axis than its minor axis [I’m actually referring to 1/2 of each axis here, which is consistent with how the numbers are used later]. The equation in cartesian coordinates for an ellipse is:
x[sup]2[/sup]/a[sup]2[/sup] + y[sup]2[/sup]/b[sup]2[/sup] = 1
We will use 6378300 m for b and 6378306.7 m for a. This gives us an equation of:
x[sup]2[/sup]/(4.0682796359E13 m[sup]2[/sup]) + y[sup]2[/sup]/(4.068271089E13 m[sup]2[/sup]) = 1
Now we need to choose a point about mid-tide. We will take (for simplictity sake) the point where the ellipse crosses the line x = y. This point is at (4510146.63079 m, 4510146.63079 m).
Now we need to figure out where the person’s eyes are. We will assume that they are 2 meters off the ground. The slope of the tangent line at that point is given by:
m = - b[sup]2[/sup]x[sub]0[/sub]/a[sup]2[/sup]y[sub]0[/sub]
Since x[sub]0[/sub] = y[sub]0[/sub], they cancel, leaving us with, and we get a result of -.999997899137.
Since the atual location of the person’s eyes is two meters perpendicular to this, we need to find the slope for that line. It is the negative reciprocal of the previous slope, or 1.00000210087. A bit of calculation shows the person’s eyes to be at the point (4510148.045 m, 4510148.04501 m). At about this point, I start wishing my calculator kept track of more significant figures.
Now we are simply left with finding the two tangents that intersect this point. If this is a trivial problem, I’m unaware of it, and hence have the long way here:
we know that these lines will be tangent to the ellipse, so we can easily get the slopes:
m[sub]1[/sub] = b[sup]2[/sup]x[sub]1[/sub] / a[sup]2[/sup]y[sub]1[/sub]
m[sub]2[/sub] = b[sup]2[/sup]x[sub]2[/sub] / a[sup]2[/sup]y[sub]2[/sub]
We also know that they pass through the point (x[sub]0[/sub], y[sub]0[/sub]), and that the points (x[sub]1[/sub], y[sub]1[/sub]) and (x[sub]2[/sub], y[sub]2[/sub]) will be on the ellipse, and thus fulfill the basic ellipse equation of x[sup]2[/sup]/a[sup]2[/sup] + y[sup]2[/sup]/b[sup]2[/sup] = 1.
Thus we have two simultaneous equations for each point:
(y[sub]0[/sub]-y[sub]1[/sub]) / (x[sub]0[/sub]-x[sub]1[/sub]) = b[sup]2[/sup]x[sub]1[/sub] / a[sup]2[/sup]y[sub]1[/sub]
and
x[sub]1[/sub][sup]2[/sup]/a[sup]2[/sup] + y[sub]1[/sub][sup]2[/sup]/b[sup]2[/sup] = 1
and the same for (x[sub]2[/sub], y[sub]2[/sub])
After a bit of work, I reduced this to:
y[sub]1[/sub] = sqrt( (b[sup]2[/sup]/a[sup]2[/sup]) ( (a[sup]2[/sup]y[sub]0[/sub] / b[sup]2[/sup]x[sub]0[/sub]) + (a[sup]2[/sup]/x[sub]0[/sub]) )[sup]2[/sup] - b[sup]2[/sup])
Which is all well and good, except that there should be another possible answer for y[sub]2[/sub]. Perhaps someone who actually is good at this shit could find the other equation.
As for this one, we can now substitute:
a[sup]2[/sup] = 4.0682796359E13 m[sup]2[/sup]
b[sup]2[/sup] = 4.068271089E13 m[sup]2[/sup]
y[sub]0[/sub] = 4510148.04501 m
x[sub]0[/sub] = 4560148.045 m
At this point, I realize that my calculator seems unable to deal with the precision adequately. Maybe I’ll get around to figuring out how to solve it in Maple or Matlab or something. In the meantime, I invite others to give it a try.
Looking over Ian Fan’s comments in the other thread, I realized he is right about the slope at the person’s feet. I had the reciprocal of it. Since I’m recalculating, Ill also use the 1.524 m figure for the height of the person’s eyes. This part:
Should read instead:
[/quote]
Since x0 = y0, they cancel, leaving us with, and we get a result of -1.00000210087.
Since the atual location of the person’s eyes is two meters perpendicular to this, we need to find the slope for that line. It is the negative reciprocal of the previous slope, or .999997899137. A bit of calculation shows the person’s eyes to be at the point (4510147.70842 m, 4510147.63079 m). At about this point, I start wishing my calculator kept track of more significant figures.
[/quote]
These new values also take the place of x[sub]0[/sub] and y[sub]0[/sub].
Also, there is a slight problem with the units in my final equation, I’ll update that as soon as I find the mistake. Or perhaps I’ll just steal Ian Fan’s equation for finding tangents.