Hi John, I agree with that enthusiasme.
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: <[email protected]> Sent: Monday, May 27, 2002 5:40 AM Subject: Re: Shadow Velocity Calculation? > Hi Fer > > You wrote: > >But what do you want to do with it? > > I tried to find out what you may do with these average values but can't > find > > a good reason. > > There's just one good reason to know the average shadow velocity of a giant > sundial like this: It's damn impressive! 25 feet per hour! (And there are > places on the dial face where it's moving even faster) I think non-dialists > love to hear this stuff. Especially if you say it with enthusiasm. > > > 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> > ----- Original Message ----- > From: "fer j. de vries" <[EMAIL PROTECTED]> > To: <[email protected]> > Sent: Sunday, May 26, 2002 1:04 PM > Subject: Re: Shadow Velocity Calculation? > > > > 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> > > > > > > - > > > > > > > - > > > > - > -
