We have a question that is causing some debate: How high would a person need to go to see both east and west coasts of Florida? There is one camp saying we need to know the curve of the earth and factor that into drawing the triangle to figure out how high we must go. I think if we drew points at sea level on both coasts and then measured the line straight through the ground, it along with the angle of a person’s field of vision we should get our height. What do you think?
Pick a point midway between the coasts and stand there. I’ll assume you eyes are 5’ above the surface of the earth. Eyes = Point A. Look at the horizon (east or west, you choose). That point on the horizon (Point B) is the far corner of a right triangle with you (5’ of you) as one leg, the line of sight as the hypotenuse, and the line from under your toes (Point C) through the earth to that horizon point (B) as the other leg.
Solve for CB using the theta thing below.
So, get the distance (arc length), over the surface of the earth, from you to the beach, Point B’, (oh say… 150 miles). This is an arc on a circle, radius of curvature equal to the radius of the earth. Get the straight line distance between the two points on that arc (A and the beach, B’) through this:
The arc length (150 miles) equals Theta (angle at center of earth with your toes on one leg and the beach on the other) X (Pi/180) X R. We know 150, R, and Pi, so solve for Theta.
The straight line between under your toes© and Point B’ is the opposite leg from angle theta of an isosolese triangle, close leg lengths both equal Radius of Earth.
Now, the relationship between B and B’ will be the same as
the relationship between 5 feet and the height your eyeballs need to be at to see the coast.
Now is the tricky part. Turn around. You’ll see the other coast.
I do not claim to be a geometry whiz, but it looks good to me.
Corrections, please?
If we assume that earth is a perfect sphere, and you wish to determine the exact height necessary to see two points on the surface of the sphere, you must take into account the curvature of the earth. Even with a sphere the size of the earth, it is possible (theoretically) to be too close to the surface to see two points ten feet apart, much less a hundred miles.
If you do not assume that the earth is sphere-like, then lots of things are possible.
Theoretically? That an easy theory to prove. Just go out in a parking lot, mark off two points, and press your eyes directly on the pavement. What do you see?
In order for you to be close enough for the curvature of the earth to prevent you from seeing the points, your iris would have to be less than .0001 millimeters from the asphalt. I suspect that your eye socket would prevent the iris from getting this close to anything flat like a parking lot. In fact, I would imagine that your cornea would be too thick to make this possible as well.
So you’ve tried it then?
As a practical matter, the minimum height is definitely under 40,000 feet, unless I was hallucinating the last time I flew into West Palm Beach.
UncleBill, I don’t quite understand your explanation. Where exactly is point C, is it below the earth, or right at the surface of the earth?
Here’s how I figured it out. I think it’s right. My explanation is difficult to follow without providing a drawing, but what the hell.
According to the a randomly selected website I just looked at, the diameter of the earth is 12,756.3 km. Therefore the radius of the earth, r, is 6378.15 km.
I am guessing that Florida is roughly 150 km wide at the area of contention.
Imagine a circle that represents the cross section of the earth with it’s center at the origin of the graph. The y-axis will represent height from the center of the earth. There are two points on the circle, A and B, which are 150 km apart along the circle and represent either end of the peninsula. Obviously, to determine the smallest possible height, the y-axis must be equidistant from points A and B, so we will only concern ourselves with one of the points, point B, because the calculations for the other side will be identical.
We will define arclength along the circle between where the y-axis is intercepted by the circle and point B as d.
d = 150/2 = 75.
First we must determine theta, the arc between the y-axis and point B:
The circumference of the earth, c, is equal to 2 x Pi x r. So, c = 40075.1.
Theta is a ratio to 360 degrees that is equal to the ratio of d to c.
So theta = d/c * 360 = 75/40075.1 * 360 = 0.673735
Now we must calculate the x and y location of point B. Since we know theta, this is very easy. Utilizing the teachings of Chief Sohcahtoa, we know that:
By = r x cos(theta) = 6377.71
Bx = r x sin(theta) = 74.9983
Let’s define the line from the center of the circle to point B as OB.
The formula for a line is y = mx + b. Where m is equal to the slope of the line, and b is equal to the y intercept of the line. Let’s solve for the slope so we can find the slope for OB.
slope of OB = (y-b)/x = (6377.71 - 0) / 74.9983 = 85.0381
What we want to do is construct a line that is tangent to the circle and touches the circle at point B. We know that OB is perpendicular to the circle at point B, so the tangent line at point B must have a slope perpendicular to OB. We can do this by taking the negative inverse of the slope of OB, or -0.022759.
Solving the line formula for the slope gives us b = y - mx. We know that y is By, x is Bx and that m is the slope perpendicular the the slope of OB, so:
b = y - mx = By - m x Bx = 6377.71 - (-0.022759) x (74.9983) = 6379.42
b is the height from the center of the earth, so we subtract the radius of the earth:
b - r = 6379.42 - 6378.15 = 1.26689km, converted to feet, this is 4,156 feet, or about 3/4 of a mile.
This passes the reasonableness test for me, so I suspect that I haven’t screwed anything up too badly.
Of course not, that’s why I said it was only theoretical.
But then, I used a different method…
The formula for distance to the horizon is given, along with a rudimentary proof, by a NASA website as:
To make it even easier, they simplify the equation to:
D = SQRT(2r) * SQRT(h)
The page gives the radius ®of the earth as 6371 (assuming that the circle that holds both points on the horizon is going to pass through the center of the earth); and continues to simplify:
D = SQRT(12742) * SQRT(h)
D = 112.88 * SQRT(h)
(measuring in km)
This travel website’s page on the physical geography of Florida mentions the average width of the penninsula is 200km.
So, to see both horizons, the distance (D) would be 100km, half the total width of the penninsula:
100 = 112.88 * SQRT(h)
Solving for h:
100/112.88 = SQRT(h)
0.886 = SQRT(h)
(.886)[sup]2[/sup] = (SQRT(h))[sup]2[/sup]
h = .785km
h = 785m
h = 2575 ft.
Ankh_Too Thanks for the new numbers, I’ll also assume that the width of the peninsula is 200 km and that the radius of the earth is 6371 km. By my method, that would calculate to 2,591 feet.
I think the reason for the difference between our numbers is that in the NASA formula, you are using 100 km as the distance between the eye and the horizon, when if fact, 100 km is the distance along the surface of the earth between the point you are hovering over and the horizon.
That said, I’m a little suspicious that subtracting 8 km from the radius of the earth had such a huge effect on the height I previously calculated and this new one. So I may be totally messing something up.
Point C is on the surface of the earth on a line through your eyes normal to the curvature of the earth, presumably under your toes.
I appreaciate the alternatives/corrections, I’ll do the math tomorrow when I’m sober.
shaking head
sigh
Newbies…
The trouble with this is that you don’t have a right triangle. Your points B and C are on the surface of the earth. Line AB is radial to the earth. They can’t be perpendicular (if angle ACB was square, it would intersect the earth at one and only one point, namely C).
Back to the OP, I don’t see how you can define the meaning of “horizon” without regarding the curvature of earth.
So true. Maybe if he leaned forward a little?
I apologize for dissapointing you and the entire SDMB community, mrblue42. I’ll try much harder next time. 