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| uk.sci.weather (UK Weather) (uk.sci.weather) For the discussion of daily weather events, chiefly affecting the UK and adjacent parts of Europe, both past and predicted. The discussion is open to all, but contributions on a practical scientific level are encouraged. |
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#31
May 21st 05, 10:00 PM
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Furthest you can see a CB cloud?
"Joe Hunt" wrote in message ... Surely when talking of relatively short distances, the effects of the Earth's curvature are minimal ? Joe Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger |
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#32
May 21st 05, 10:35 PM
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Furthest you can see a CB cloud?
"Roger Smith" wrote in message news  Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe |
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#33
May 21st 05, 10:35 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
"Roger Smith" wrote in message news Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe |
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#34
May 21st 05, 10:35 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
"Roger Smith" wrote in message news Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe |
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#35
May 21st 05, 10:59 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
On Sat, 21 May 2005 21:47:12 +0100, Joe Hunt wrote:
Surely when talking of relatively short distances, the effects of the Earth's curvature are minimal ? I think you underestimate how curved the surface really is. Humber bridge the support towers are 36mm further apart at the top than the bottom. They are only 155m + a bit high but the best part of mile apart. http://www.humberbridge.co.uk/interest.htm http://www.humberbridge.co.uk/techspec.htm With a cloud top at 20,000' the horizon is 190 miles away, not a short distance... -- Cheers Dave. pam is missing e-mail |
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#36
May 21st 05, 10:59 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
On Sat, 21 May 2005 21:47:12 +0100, Joe Hunt wrote:
Surely when talking of relatively short distances, the effects of the Earth's curvature are minimal ? I think you underestimate how curved the surface really is. Humber bridge the support towers are 36mm further apart at the top than the bottom. They are only 155m + a bit high but the best part of mile apart. http://www.humberbridge.co.uk/interest.htm http://www.humberbridge.co.uk/techspec.htm With a cloud top at 20,000' the horizon is 190 miles away, not a short distance... -- Cheers Dave. pam is missing e-mail |
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#37
May 21st 05, 10:59 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
On Sat, 21 May 2005 21:47:12 +0100, Joe Hunt wrote:
Surely when talking of relatively short distances, the effects of the Earth's curvature are minimal ? I think you underestimate how curved the surface really is. Humber bridge the support towers are 36mm further apart at the top than the bottom. They are only 155m + a bit high but the best part of mile apart. http://www.humberbridge.co.uk/interest.htm http://www.humberbridge.co.uk/techspec.htm With a cloud top at 20,000' the horizon is 190 miles away, not a short distance... -- Cheers Dave. pam is missing e-mail |
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#38
May 21st 05, 11:03 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
"Joe Hunt" wrote in message ... "Roger Smith" wrote in message news Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe Joe, taking a diametral section through the Earth as being circular, you can use very simple geometry to determine the approximate distance of a point on a tangent from the nearest point to it on the circle as a function of the distance of that point from where the tangent is in contact with the circle. A simple but fairly good approximation is h=d^2/D where D is the diameter of the Earth h is the distance of a point on the Earth's surface below the observer's horizon plane at a distance d from the observer. For determining the visibility of distant objects this formula ignores the effects of refraction. Regards, Roger |
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#39
May 21st 05, 11:03 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
"Joe Hunt" wrote in message ... "Roger Smith" wrote in message news Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe Joe, taking a diametral section through the Earth as being circular, you can use very simple geometry to determine the approximate distance of a point on a tangent from the nearest point to it on the circle as a function of the distance of that point from where the tangent is in contact with the circle. A simple but fairly good approximation is h=d^2/D where D is the diameter of the Earth h is the distance of a point on the Earth's surface below the observer's horizon plane at a distance d from the observer. For determining the visibility of distant objects this formula ignores the effects of refraction. Regards, Roger |
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#40
May 21st 05, 11:03 PM
posted to uk.sci.weather
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Furthest you can see a CB cloud?
"Joe Hunt" wrote in message ... "Roger Smith" wrote in message news Joe "On a clear day you can see for ever" Do you really think so? The Earth is NOT flat. Work it out. Regards, Roger Roger, I fully understand the Earth is not flat, however, what equations are used by websites such as the one quoted earlier in the post ? Joe Joe, taking a diametral section through the Earth as being circular, you can use very simple geometry to determine the approximate distance of a point on a tangent from the nearest point to it on the circle as a function of the distance of that point from where the tangent is in contact with the circle. A simple but fairly good approximation is h=d^2/D where D is the diameter of the Earth h is the distance of a point on the Earth's surface below the observer's horizon plane at a distance d from the observer. For determining the visibility of distant objects this formula ignores the effects of refraction. Regards, Roger |
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