Hi John, Your calculation only is valuable at the summer solstice. At winter solstice or any other day you have to choose the daylenghth of that day and the lenght of the hourscale for that day. So you get an average for each day.
But what do you want to do with it? Sometimes the length between 2 hour marks on your scale is longer and at another part it is smaller. So also for each hour you get different average values. I tried to find out what you may do with these average values but can't find a good reason. In any case, the velocity is smallest where the scale is shortest between a certain time space. Best wishes, Fer. Fer J. de Vries mailto:[EMAIL PROTECTED] http://www.iae.nl/users/ferdv/ Eindhoven, Netherlands lat. 51:30 N long. 5:30 E ----- Original Message ----- From: "John Carmichael" <[EMAIL PROTECTED]> To: "Sundial List" <[email protected]> Sent: Saturday, May 25, 2002 5:42 PM Subject: Shadow Velocity Calculation? > Hello List: > > Would this be a true statement? > > (In this case I'm refering to the Kitt Peak Telescope Sundial, but I think > it might be a true statement for any sundial): > > The average velocity of the shadow (m/hr) on the sundial face as it moves > along a line connecting the time point markers is equal to the total length > of these lines (m) divided by the longest possible daylength (in hours). > > In this case that would be 118 meters /14.17 hours = 8.327 m/hr (or > 27ft.hr). > > John > > John L. Carmichael Jr. > Sundial Sculptures > 925 E. Foothills Dr. > Tucson Arizona 85718 > USA > > Tel: 520-696-1709 > Email: [EMAIL PROTECTED] > Website: <http://www.sundialsculptures.com> > > - > -
