OOPs! I tried to recall the radius of the earth from memory instead of looking it up. It's closer to 4,000 miles instead of 12,000, which is about 1/2 the circumference. That changes the numbers. For an object 400 ft tall, the distance beyond which it can't be viewed is 24.6 miles. Viewed from an obversion point 200 ft above the earth, it's an additional 17.4 miles, or 42.0 miles. Sorry about that -- measure twice, cut once! Roger On 21 Mar 2010, at 7:00 PM, Roger Eichhorn wrote: > These are windmills, essentially, not water turbines. The issue is how much of their height one can see from a distance. I assumed that the height of the turbine was measured from the surface of the earth (lake in this case) to the tip of a blade at its topmost point. I don't know where Walter's numbers came from, but they have to be an approximation of some sort. I took the earth to be a circle of radius 12,000 miles. A distance along that surface is given by the radius times the subtended angle from the center of the earth, measured in radians (2pi radians = 360 degrees). A right triangle with one leg equalling the radius and the hypotenuse equalling the radius plus the height above the surface yields the formula "cosine(angle) = radius/(radius + hight). Solving for the angle and multiplying it by the radius of the earth gives the distance along the surface. One needs a calculator to solve the equation as the height above the surface is such a small number relative to the radius (or one has to expand the cosine in a series and truncate it). > > Other factors that must be considered include refraction along the line of sight caused by temperature and/or humidity variations, as well as surface reflexion -- the common mirage effect. > > Unfortunately, the plain text rules of this forum prevent me from using formulas and diagrams to illustrate the calculations. > > Roger > > On 21 Mar 2010, at 5:34 PM, Walter Mayes wrote: > >> Been following this thread and am not sure what the person is asking. The simple formula for the distance to the horizon is: Take the Square Root of your height in feet and multiply that by 1.23 and that is the number of miles to the horizon. Square Root of 400 feet is 20, times 1.23 = 24.6 miles. If the turbines are "in" the lakes, how much is submerged? What is the size (diameter) of the turbine blades? I'm confused. :-) >> >> Walter Mayes >> >> >> Subject: Re: How tall should the turbines be? >> >> >>> Correct. From the same formula, one can calculate that an observer at 200 ft above the surface would see the "base" of the turbine at a distance of 30.2 miles, therefore all of it. The tip would disappear at a additional distance of 42.6 miles, or 72.8 miles in total. >>> >>> Roger >>> >>> On 21 Mar 2010, at 2:35 PM, Rich Mason wrote: >>> >>>> Don't forget to factor in the height of the observer's position. It's one thing to be at lake level, another to be 200 feet up on a bluff. >>>> >>>> Rich >>>> >>>> On Mar 21, 2010, at 12:38 PM, Roger Eichhorn wrote: >>>> >>>>> Andy, >>>>> >>>>> The tip of the turbine should disappear from view when it is located a distance, along the circumference of the earth, c (away from the observer), equal to the radius of the the earth, r, times the arcos of the ratio of the radius of the earth divided by the radius plus the height of the turbine, where the angle implied is measured in radians. Taking r = 12,000 miles, this gives c = 42.6 miles for an object 400 ft tall. The apparent height should be approximately equal to the actual distance away divided by c times 400, in feet. So, a turbine located 21.3 miles away should appear to be 200 ft tall. I haven't worked out how this translates to image size, focal length, etc. >>>>> >>>>> Roger >>>> >>> >>> Roger Eichhorn >>> Professor Emeritus >>> University of Houston >>> eichhorn@xxxxxx >>> >>> >>> >> > > Roger Eichhorn > Professor Emeritus > University of Houston > eichhorn@xxxxxx > > > Roger Eichhorn Professor Emeritus University of Houston eichhorn@xxxxxx
