Computing height vs distance to horizon?
969415
It's a geometry problem.  Suppose you're high up in a building on the beach overlooking the ocean.  You're at height h above sea level.\n\nNow draw a circle which represents the earth, with radius r.  Your distance from the center of the earth is r+h.  The horizon is some distance away, x.  Your line of sight to the horizon is <tangent> to the radius of the earth, forming a right triangle, where the hypoteneuse is r+h.\n\nUsing the Pythagorean Theorem,\n\nr^2 + x^2 = (r+h)^2 = r^2 + 2rh + h^2\nx^2 = 2rh + h^2\nx = sqrt(2rh + h^2)\n\nThat's the exact answer,  But for practical purposes, 2rh is <much, much> bigger than h^2.  To see this, suppose you're 500 feet in the air.  That's about a tenth of a mile, and using miles, h^2 = 0.1^2 = 0.01.  On the other hand, the earth's radius is about 4000 miles, so 2rh = 2 x 4000 x 0.1 = 800.  This is 80,000 times bigger than h^2.\n\nSo we can safely ignore the h^2, and say x = sqrt(2rh).  That can be broken down further.  Again using miles (you can also do this in kilometers),\n\nx = sqrt(2rh) = sqrt(2r) sqrt h = sqrt(8000) sqrt h = 90 sqrt h\n\n(rounding off, since sqrt 8000 = 89.44; also, h must be in miles).\n\nUsing the formula x = 90 sqrt h, suppose you're in an airplane 4 miles up in the air (21,000 feet).  Then the horizon is 90 sqrt 4 = 180 miles away.\n\nOkay, now let's work it out for h in feet instead of miles.  We'll go back to the formula x = sqrt(2rh), and use r = 3960 miles (a more accurate figure).  Then\n\nx = sqrt(2 x 3960 x h / 5280) = sqrt(3h/2) = 1.225 sqrt h\n\nwhere h is in feet and x is in miles.\n\nTwo examples:  (1) For h = 6 feet, use x = sqrt(3h/2) = 3 miles.  (2) For x = 100 feet, use x = 1.225 sqrt h = 12.25 miles.\n\nTo recap:  For h in miles, use x = 90 sqrt h.  For h in feet, either use x = sqrt(3h/2) or x = 1.225 sqrt h.\n\nIn all cases, x will be in miles.  And of course, all of this can be done using meters and kilometers.
523034
The first answer *edit* was modified so now what I say here makes no sense - but her website confirms that I am right (5km = ~3 miles)*. If we want to know the distance to the horizon from a vantage point on a sphere, the only thing that matters is the radius of the sphere and the height of the vantage point.\n\nSo, assume you are standing with your eyes at 5'5" looking out at a calm ocean. You will typically be able to view the sea surface at 3 arcminutes, or roughly 5 kilometers, out to sea. Beyond that you would still be able to see ships and whatever else sticking above the water quite easily, but upon close inspection realise they appear deeper in the water than they actually are.\n\nAlso note that light bends around mass, so you might be able to "see" farther than your actual vantage point, but I'm not doing that calculation.\n\nI assume that people who have values higher than mine are thinking about higher vantage points than mine, because I think my estimations are close and I'm positive that my logic is right. Correct me if my numbers are off, but I'd like to know why.
880249
Surprisingly for an average height person standing at sea level, the horizon is only about six miles away.  You could see disturbances at that distance.
918805
Time And Relative Dimension In Space
49698
your variable m stays for "miles".\nYou thus just have to put 295 instead of m and do the exact reckoning: 0.05*295 + 30  14.75 + 30 = 44.75\n\nNow choose from the options the one which is closest to your number: 45\n\nIn this way you cannot do wrong
891584
Well, yeah, I suppose this is true, but the difference is quite small.  If we are 1,000 miles apart looking at something 240,000 miles away, the ratio is about the same as if we were 1 foot apart looking at something 80 yards away.  So, if we are standing on the goal line of a footall field looking toward the goal posts at the opposite end (110 yards away) and we are about 17 inches apart, how much different do the goal posts look?
257670
Depends on visibility.\n\nYes, the Earth curves out of sight.\n\nhttp://www.boatsafe.com/kids/distance.htm\n\nhttp://science.howstuffworks.com/question198.htm\n\nHere is a site that will do the math for you:\n\nhttp://www.boatsafe.com/tools/horizon.htm
490274
What several answerers are failing to remember is that St. Louis is not at sea level.  The St. Louis Gateway Arch is 192 meters (630ft) high.  But it is sitting on land that is 132 meters (433ft) above sea level.  This comes to a total height above sea level of 324 meters (1066ft).\n\nI'll assume you are trying to figure out the straight line distance to the horizon (the point where a viewing line is tangent with the curve of the earth).  The curved distance along the surface of the earth is almost the same when close to the earth, so it really isn't worth noting, but if you were in a satellite, there would be notable difference.\n\nAnyway, using triangles, you are trying to find the leg of a right-triangle with sides r, r+h and d.\n\nTo derive a formula figure out that d^2 + r^2 = (r+h)^2.\n\nr = radius of the earth (6378.14km or 6,378,000m)\nh = height above sea level (324m)\n\nAnyway, plug this all in and you get that the distance to the horizon is 64.29km or about 40 miles.\n\n40 miles is the correct answer
1001818
see a minor distance creat such a difrnc, 10/10
406596
no idea...but I think you use the formula (Gravitational constant x mass 1 x mass 2) / distance squared
168445
first find a vicinity plan. \nget a straw(in sodas) and a triangle\nstand on a specific location with the distance of X from the building.\nmake sure to know your height.\ncreate a transit by putting the straw along the hypoteneous of your 45x45 triangle.\nlook through the straw (try to see the zenith of the building) and measure the angle of depression.\nthen, simply apply the law of sines and cosines etc etc.\nthis will not give you exact building height but an instant approximation. this can also be used in finding approximate height of towers, canyons, mountains, and volcanoes.
882346
Laser range finder. Although you could use plum bob, trigonometry,  altitude of both peaks and an observation point between them to figure it out. You need to observe the angle you have to look upwards from a point between the two. There is a lovely diagram sitting on my desk that is just a little to complex to ASCII.
886238
Remember the third law:  P^2=ka^3, where k simply is telling you to use consistent units (e.g. Earth years and a.u.)  So the square of 5[years]=cube of distance in a.u., or cube root of 25 is the distance you want, expressed in astronomical units.  I'll let you finish this, but as a hint, it falls in the outer part of the asteroid belt (Ceres has a period of 2.72 years).
912921
altitude = height = 552km\nAngle of dep = 39.7\n\nTangent of Angle = Alt/ Radius\n\nRadius = Altitude/ Tan(39.7) = 552/tan(39.7) I think... p.s not really sure
1157531
H0: mu = 10 inches\nHA: mu < 10 inches\n\nThe null always has the equal sign in it.  The alternative has what you wish to show is true.
10335
If we assume that visibility is unlimited, then the horizon is always the same distance away from a specific point. How far is the line of the horizon depends on the altitude of the point - the higher the point, the further away the horizon. By the same token, you will be able to see high mountains that are close to, but beyond the low-lying horizon line.\n\nOn other planets/moons the distance to the horizon is a function of their diameter - the bigger the body, the further away the horizon is from a given point.
411926
The height of a triangle is defined as the perpendicular distance from an angle to a side.  Now, since this is a right triangle, and depending on how it is situated, the height could be 10" or 12" if you observe from the two acute angles.  But from the right angle, the height goes from the angle to AC (your hypotenuse).  Finding this one will be tricky but can be done using the Pythagorean Theorem.\n\nIf you drop a height from the right angle to the hypotenuse, it breaks it into two parts.  The length of one part is x (since we don't know and since the height doesn't perfectly bisect the hypotenuse), and the length of the other part is 15.62-x (since the total length is 15.62").  It really doesn't matter which side is x, but for my explanation, I'll use x as the side closer to the leg that is 10".  This height creates two new right triangles, and you can use the Pythagorean Theorem to find out both the height and the missing side x.\n\nOne formula will end up being 10^2 = h^2 + x^2, and the other formula will end up being 12^2 = h^2 + (15.62-x)^2.  You can solve the first equation to find that h^2 = 100 - x^2 and substitute that in to the second equation:\n\n12^2 = h^2 + (15.62-x)^2\n144 = 100 - x^2 + 243.9844 - 31.24x + x^2\n144 = 343.9844 - 31.24x\n-199.9844 = -31.24x\nx = 6.4 (rounded)\n\nKnowing this, we can find that missing height:\nh^2 = 100 - x^2\nh^2 = 100 - (6.4)^2\nh^2 = 100 - 40.98\nh^2 = 59.02\nh = 7.68"\n\nSo it all depends on the specific height you're looking for - it's either 10", 12", or 7.68".\n\nAs for your trough, surface area is the sum of all the areas.  Break your trough into parts that you can find areas of.  If you're just looking for the outer surface area, for example, you have two triangles for the ends and two rectangles for the sides (assuming that it's an open trough).  The area of the triangles is A = 0.5bh, and the area of the rectangle sides is A = bh (b in this case will be a side of the trianlge, 2, and h will be the length of the trough, 16).  So just add up all the surfaces that will comprise your trough and you should have your surface area.  (I'm not sure if you need to include the inside as well, but if so you need 2 triangles and 2 rectangles for the outside, and 2 triangles and 2 rectangles for the inside as well.)