RE: Question about analemmic sundials
Troy [EMAIL PROTECTED] asked < I'm curious about analemmic sundials. Is it possible to move the gnomon of the dial (along an analemma course marked with the proper days) to achieve the same effect? > If I understand Troy correctly, to be asking about simple dials with "figure eight" analemmas on the hour lines, I think the answer is "no" as what is wanted must affect all the hour lines equally, which could be achieved by a rotation of the dial about the gnomon axis, and I believe no displacement of the gnomon will achieve this. I'm open to correction, though! Andrew James N 51 04 W 01 18
Re: Question about analemmic sundials
Hello Troy: Are you talking about an "analemmic" dial or an "analemmatic" dial? John Carmichael >I'm curious about analemmic sundials. Is it possible to move the gnomon of >the dial (along an analemma course marked with the proper days) to achieve >the same effect? I wonder since the analemma would likely be confusing to >laymen viewing the dial. > >Thanks, >Troy > >
Determining hour-angle from azimuth
I'm trying to convert an azimuth value into an hour-angle, and would appreciate some help with the formula and with my algebra. Using dcl=declination, lat=latitude, azi=azimuth,ha=hour-angle, I started with a standard formula for azimuth given hour-angle: tan(azi) = sin(ha)/(sin(lat)*cos(ha) + tan(dcl)*cos(lat)) After multiplying out and rearranging, I got: sin(lat)*tan(azi)*cos(ha) - sin(ha) = -tan(azi)*tan(dcl)*cos(lat) Substituting into a standard fromula from my old notes from school, this solves as: ha = arctan(-1 / sin(lat)*tan(azi)) + - arccos( -tan(azi)*tan(dcl)*cos(lat) / sqrt( sin(lat)^2 * tan(azi)^2 +1 )) where + - means plus or minus Questions: Is this the correct solution? If so, how should I interpret the plus or minus element of the solution. Surely for any given azimuth-latitude-declination there is only one possible hour-angle, not the two implied by the plus or minus option? Many thanks, Steve
Re: Question about analemmic sundials
Thanks, that answers my question. In a message dated 12/12/00 10:40:34 AM Eastern Standard Time, [EMAIL PROTECTED] writes: << Troy, In issue 5-1 of the NASS Compendium there is an article by Yvone Masse about such a dial. The dial face is a straight line with hour marks. The gnomon is a string (or movable rod) that is attached to a pole at one end, The other end is set to a point on a figure 8 according to the date thus giving the correction for the EOT. ++ron >>
Re: Determining hour-angle from azimuth
Thanks to Jorge Ramalho for pointing out a sign error in my last message. It
should have read as below. I still have the same questions, tho'
Steve
I'm trying to convert an azimuth value into an hour-angle, and would
appreciate some help with the formula and with my algebra. Using
dcl=declination, lat=latitude, azi=azimuth,ha=hour-angle, I started with a
standard formula for azimuth given hour-angle:
tan(azi) = sin(ha)/(sin(lat)*cos(ha) - tan(dcl)*cos(lat))
After multiplying out and rearranging, I got:
sin(lat)*tan(azi)*cos(ha) - sin(ha) = tan(azi)*tan(dcl)*cos(lat)
Substituting into a standard formula from my old notes from school
{solutions of a*cos(x)+b*sin(x)=c}, this means my solution is:
ha = arctan(-1 / sin(lat)*tan(azi)) + - arccos( tan(azi)*tan(dcl)*cos(lat)
/ sqrt( sin(lat)^2 * tan(azi)^2 +1 ))
where + - means plus or minus
Questions:
Is this the correct solution?
If so, how should I interpret the plus or minus element of the solution.
Surely for any given azimuth-latitude-declination there is only one possible
hour-angle, not the two implied by the plus or minus option?
Many thanks, Steve
I think there can be 2 hour angle solutions for a single azimuth
Steve L, I think there can be 2 hour angle solutions for a single azimuth. I think one h.a. will be less than 90 degrees, and the other greater than 90. See attached DeltaCad file. I wish I could comment on the derivation of your math, but I am unfamiliar with your arctan, arc cos solution for ha. I hate to say it, but this is where spherical trig will yeild the answers. Any takers? Bill G. Content-Type: application/octet-stream; name="STEVE AZIMUTH.DC" Content-Disposition: attachment; filename="STEVE AZIMUTH.DC" Attachment converted: Macintosh HD:STEVE AZIMUTH.DC (/) (0001FCE2)
