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"Ron Miller" wrote in message ... "Zarkovic" wrote in message news:ye9ab.8990$CU3.7779@pd7tw3no... Hi guys, here is one question that I can't get for some reason, I just can't. I come close to it, but everytime I am by 300km/hr off. Anyhow here is the question and could someone please tell me how it's done. "Derive the equation from which you could calculate speeds of someone on the surface of the Earth, depending on their latitude. The equation will depend on person's latitude, (Q), the radius of the Earth, (Re), and the period of rotation, (T). ( I tried it, but nothing.) (Hints: the speed of a person on a Northpole is 0 km/hr and speed of a person on the Equator is 1650 km/hr) I don't know what the radius of the earth would have to do with it. Your speed depends on your perpendicular distance from the earth's axis (as though the latitude you are at is the circumference of a wheel that is spinning on the axis). Because that perpendicular distance, is cos(Q)*Re... You either have to have the radius of the Earth, or another way of working out that perpendicular distance. Best Wishes |
#12
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I'm thinking it might be 1650*cos(lat)
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#13
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I'm thinking it might be 1650*cos(lat)
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#14
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"Zarkovic" wrote in message news:XR9ab.9029$CU3.1539@pd7tw3no... Ahhh man, thanks guys. For some reason I turned my sphere around and I used for equator speed cos (90) which equals to 0 , but I should have used cos (0) instead. Sorry for the mix up, and thanks once again to both of you. It is nice to have people around when when the answer and the solution are not in the back of the textbook. ) todd |
#15
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"Zarkovic" wrote in message news:XR9ab.9029$CU3.1539@pd7tw3no... Ahhh man, thanks guys. For some reason I turned my sphere around and I used for equator speed cos (90) which equals to 0 , but I should have used cos (0) instead. Sorry for the mix up, and thanks once again to both of you. It is nice to have people around when when the answer and the solution are not in the back of the textbook. ) todd |
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