Hello! In New Scientist this week (actually next week!), there was a question concerning the curvature at the horizon. The formula is as follows for the distance to the horizon: (2*6373*h)^1/2 km; where h is the height of the individual from the ground. Using the exaple it states a towe 150m high the horizon will be 44km away and displaced down by 0.39 degrees. That's all fine, but the next bit states that by holding a 1m stick 1m in front of you, the ends of the stick will be 0.8mm above the horizon. Intrested to know how this was arrived at. Although I'm not sure whether this is in the right place or not, I'm pretty sure it'd be beneficial for anyne else with trig questions! Thanks in advance!