When you look out at the horizon, how far is the farthest you can see?
The distance to the horizon (in miles) is the square root of your height (in feet) times 9/10. If you’re 25 feet high, your 4.5 miles from the horizon.
Hi and welcome.
The answer can be found at
http://boards.straightdope.com/sdmb/showthread.php?threadid=23213
It all depends how tall you are, but 3 miles is an average distance for most people.
It might depend on which direction you’re facing:
http://www.straightdope.com/classics/a1_167b.html
Arjuna34
Well assuming the Earth is perfectly spherical and the radius of the earth is 6.3*10^6 meters then you can use simple coordinate geometry.
The equation for the “Earth” is x^2 + y^2 = (6.310^6)^2 and say a person is standing at (6.310^6 + h, 0), where h is his eye level relative to the sea level. A point from the eye level of the person drawn at a tangent to the circle by definition is perpendicular to the origin. Then using Pythagoras the equation for how far you can see on a perfect day on an area of the earth that is perfectly spherical is Sqrt((6.310^6 + h)^2 - (6.310^6)^2 + x^2) where h is your eye level relative to the sea level in meters and x is the height of the horizon relative to the sea level.
This is however just the distance calculated using purely math, so it means less in the real world when you consider the things like the the temperature of the air.
Thankyou for the link Arjuna34.** I can now say with utter certitude that I have caught the esteemed Cecil in error.**
Cecil said** “The eastern and western horizons are equidistant from the observer”** Cecil was considering the case of an observer at 45 degrees latitude in the middle of the ocean.
** But Cecil failed to consider the effect of the tides**, primarily an east west phenomenon. I would guess that the more extreme ebb and flood tides would result in significant enough elevation change from 3 miles west to 3 miles east resulting in a difference of horizon distance far greater than 4 inches.
but your certitude is misplaced. Think of it like this. Take a cross-section of the Earth along an East-West direction, from your perspective in the middle of the ocean. That cross-section is basically ellipse-shaped because of the tides, with the difference in the major and minor diameters being maybe ten feet.
This is a similar setup to taking a North-South cross section, which Cecil addressed, except the ellipse in the North-South direction is out-of-round by ten or twenty kilometers, IIRC.
It’s the ellipse shape that causes the difference in distance to the horizon, and the North-South effect due to the equatorial bulge is thousands of times larger than the East-West effect due to the tides.
Okay then CurtC, try and follow me. The maximum tide range was measured 3 miles off shore Australia at 6.7 meters. The circumference of the earth is nominally 40,000 km. The tide range covers 1/4 of the circumference, 10,000 km. The gradient(slope) of the ocean surface in the mid-range therefore is in the order of 6.7/10,000,000 m/m. The difference in elevation between the observer and the horizon (nominal distance is 5,000m or 3 mi.)is 6.7/10,000,000 x 5000m = .00335m or**.011 ft.**
Now Arjuna34 provided a link giving us a formula for horizon distance.
1,23 x sqrt(height of observer) x 5280 ft/mi = distance to horizon in feet.
Case 1. SLACK TIDE observer’s eyes at 6 ft
1.23 x sqrt(6,0) x 5280 = 15908 ft.
Case 2. MID TIDE observers eyes at 6.011 ft wrt to elevation of horizon in the direction of tide flow.
1.23 x sqrt(6.011) x 5280 = 15923 ft.
Now the observation in one direction has increased by 15 feet. In the other direction, it follows that the horizon decreases by 15 feet. The total difference east to west is 30 feet. Now that is 90 times the 4 inch figure that Cecil claims for the north south difference.
I have certainly taken some mathematical liberties here,but I’m certainly in the right order of magnitude. I know that the tide range is not directly east west which will reduce the figure somewhat, and the 6.7 m figure is high for a normal tide range, But I have a lot a room for error.
Perhaps someone else might investigate and provide a more rigorous mathematical proof.
Someone else? Should I not be offended? 
Your error is in assuming that a very slight slope to the surface of the ocean is equivalent to a change in the observer’s elevation, and therefore lets him see farther in one direction and not as far in the other.
If that were the way that it worked, then you could make the same argument for the North-South direction. The gradient of the ocean along the North-South slice is 10,000/10,000,000 (the Earth’s polar radius is about 10 km less than the equatorial radius), and multiply this by the 5000m distance to the horizon, and you get 5 meters, instead of the measley 0.00335 meters along the East-West slice.
But you can’t say that the distance to the North horizon is equivalent that for a 7-meter person (5 plus about 2 for his true height), nor is the distance to the South horizon equivalent to the distance for a (-3) meter person.
So there’s something wrong with your method.
I realize I’m not using a rigorous method, but I would assume that with such a small increment of relative datums such as .011 feet or 1/8th of an inch , that the approximation might be warranted. That is why I’m asking for someone perhaps you to provide me with a mathematical proof of what the difference actually is. I just can’t get the vision of tidal wave increasingly blocking my view of a palm tree in the distance as I stand on the shore of an atoll as the crest approaches.
last sentence should conclude
** out of my head **
OK, here’s a way to point out why you can’t do the calculation that way. Draw on a piece of paper a curved line representing a segment of the surface of the Earth, with a little guy standing in the middle of it. Now imagine it with the curved surface angled to the right a very slight amount. The guy’s head will now be higher above the part of the curve on the right than it was, and lower than the curve to the left. But his distance to the horizon will not change in either direction.
I remember seeing a video of a motivational speaker a couple of years ago. The whole thing was about an adventure he took many years before.
When he was young he drove from England to South Africa.
During part of it he had to drive through the Sahara desert. He said that on the road there were oil drums every so often. They were about 5 (I forget the measurement, but KM I assume, miles are too far). That way when you were driving, the set behind you would vanish at the same time the set in front of you appear.
True, I’ve considered that, the tangent of the curve changes, but so does the degree of curvature. That is why I can’t dismiss the idea of a difference more significant than 4 inches out of hand.
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. 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.
Thanks waterj2
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Shouldn’t you double the 6.7m figure as the high tide point really only represents 1/2 the major axis? Remember, we get two high tides per day.
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I remember a problem once in determining how high off the surface of the earth a string 3 ft longer than the circumference of the earth was. Some simple math reduced the problem as independant of the original circumference and I was able to just divide the difference , 3 ft by 2pi. That solved the problem of significant number of digits. Any chance this might help here?
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I like your approach, even though I think what we have here is a sine wave whose axis is a circle rather than a straight line and the wave length is pi*r. Somehow I think that doesn’t help.
Carry on:)
I have the figures right, the terminology was wrong. a and b represent half of each axis. Also, I think it is an ellipse, as it’s the tidal forces from the moon squishing the water. At least that’s the approximation I’m using. I’ve already got three pages of scrawled notes on this damn thing.
Using strictly the numbers and equation you gave Mathematica returned: 4.2240640319874664 * 10^16, I didn’t bother with units. But looking at the equation it doesn’t seem to get the correct units.
Just assume a b c and d are different values with no units, it simplifies to:
Sqrt( a * (b + c meters)^2 - d meters^2)
equals
Sqrt( ab^2 + 2abc meters + ac^2 meters^2 - d meters^2)
Well for simplicity’s sake lets just say the person is standing at the y=x line on the ellipse, not at 45 degrees latitude (i’d assume you would just take the polar radius and divide it by 2 to find this out but I’m too lazy to do this).
Then the point of intersection of the ellipse and the y=x line in the first cartisan plane for the equation waterj2 gave is (figured out by my TI-89 algebriacally) is (4510141.551, 4510141.551).
Taking the equation x^2/a^2 + y^2/b^2 = 1, solving for y and taking the positive square root then differentiating it yields:
y’ = -6378306.7x/(6378300*Sqrt(6378300^2 - x^2))
Filling it in with the point of intersection gets a slope of m = -1.000002101.
Taking the negative reciprocal to find the equation for the person standing perpendicular to the “Earth” we have: m = 0.9999978991. Assuming his eye level is 2 meters from sea level then the equation for him is: y = 0.9999978991x - 0.0490042. Note: I got that from my calculator’s graph, his eyes are at (4510143.551, 4510143.551).
So now we have to find a line thats tangent to the ellipse that has a point at (4510143.551, 4510143.551).
After a couple pieces of scrap paper the solution is approximately:
6005.88meters of viewing distance when looking north, 6006.377meters when looking south. Which contradicts what Cecil said (that the northern horizon is farther away). If anyone else wants to try these calculations however, be my guest.