Re: Emergency Sundial
> Edley, Steve, Andrew, et al, > > Approximate 'trig.' from memory is all very well, but > why not simple geometry of equality and bisection? > Because nobody ever pointed it out to me before !! > With a string, twig, or weed-stem part as arbitrary > unit-length b, swing a circular arc. Cut successive > chords of length b, which if connected to the center > form 60° equiangular triangles. Then successively > bisect ... Yep, it's nifty. S.
Does your watch do the same?
Hi everyone! Last Christmas a mate of mine told me this conversation he had with somobody: --- And what's your hobby? --- In my spare time I make sundials. --- Huh, sundials ?! My watch tells the time more accurately and it's much cheaper. What a waste of time and effort! --- But, tell me, does your watch tell you how much time is left for the sunset? --- Well, no, what should I need know it for? --- Maybe you're travelling by car or on a trip through the forest and want to know how much time of light you've left, maybe you like fishing or birdwatching and want to know when the animals leave,... (and so on). And my dial does it in very a simple and elegant way: it can be just a cone lying on the ground with its axis pointing towards the North Star. So it is not so much useless as you thought! Ah, the old italic hours and the splendid conical-gnomon dial coming to our help! Does anybody know more about other kinds of italic hour dials (apart from the flat ones, of course)? Anselmo Perez Serrada [ 41.63 N 4.73 W]
RE: Emergency Sundial
Edley, Steve, Andrew, et al, Approximate 'trig.' from memory is all very well, but why not simple geometry of equality and bisection? With a string, twig, or weed-stem part as arbitrary unit-length b, swing a circular arc. Cut successive chords of length b, which if connected to the center form 60 equiangular triangles. Then successively bisect with any fixed length object longer than the chords of the sub-angles in question, to get 30, 15, 7 1/2, 3 3/4... etc. as wanted. Binary math is also great for graduating lineal scales, as for example, the traditional English inch. ( for Andrew) Sciagraphically, Bill Maddux > Hi Andrew, > > A closer and somewhat easier method would be to continue dividing the > string in half til it had been divided into 16 equal parts. Measure > out 15 of those parts and then at a right angle go 4 parts. As I > mentioned to Steve LeLievre this approximates the tangent of 15 > degrees with better than 1/2 percent accuracy. [ tan(15) roughly is > equal to 0.267949 and 4/15 is about equal to 0.27, arctan(4/15) > is about 14.93 degrees ]. Of course any object marked out in equal > divisions would do as well to mark out the 15 and 4. It just so > happens that my fist is right at 4 inches wide and the length > measured from my elbow to the end of my fist is right at 15 inches, > so, I'm lucky, I carry this measure about with me! > > How many of the rest of you are so lucky? How many find your fist > width to elbow-fist length ratio to be even closer to the tangent of > 15 degrees? > > Are there even better measurements taken from our self dimensions? > > Edley. > > You wrote: > > Steve Lelievre suggested using arc tan 1/4 to find 15 degrees "in > > emergency". > > > > A slightly better approximation is to go via > > sin(15) = 0.2588 and arcsin(0.25) = 14.4775 deg > > cf. tan(15) = 0.2679 and arctan(0.25) = 14.0362 degrees > > > > So, take your string and find a quarter of its length as before by two > > halvings. Start from the centre point and mark a point on a first line at > > the string's length away. From that point draw an arc (or mark a few > > points to indicate it) using the quarter string as radius. Go back to the > > centre and take the tangent to that arc as the second line. > > > > The angle subtended is arcsin 0.25 or about 14 degrees 29 minutes, close > > enough for practical purposes I would think. > > > > If you want to get still closer, then do it both ways and add the > > difference between them to the larger of the two angles to get 14 degrees > > 55 minutes. (And if you dislike estimating the addition to make rather > > than constructing it, then take two of the sine angles one after the other > > i.e. added together and go back one of the tan angles to subtract it.) > > > > Or of course you could just use the same process, with half the string > > length instead of a quarter, to go via arcsin 1/2 and obtain 30 degrees, > > then take the string and mark two points equidistant from the origin on > > the two lines 30 degrees apart, and use doubling the string to find the > > point midway between those two - hence bisecting the angle into two angles > > of exactly 15 degrees with no error at all. But we're getting close to > > terrestrial origami I fear ... > >
RE: Emergency Sundial
Hi Andrew, A closer and somewhat easier method would be to continue dividing the string in half til it had been divided into 16 equal parts. Measure out 15 of those parts and then at a right angle go 4 parts. As I mentioned to Steve LeLievre this approximates the tangent of 15 degrees with better than 1/2 percent accuracy. [ tan(15) roughly is equal to 0.267949 and 4/15 is about equal to 0.27, arctan(4/15) is about 14.93 degrees ]. Of course any object marked out in equal divisions would do as well to mark out the 15 and 4. It just so happens that my fist is right at 4 inches wide and the length measured from my elbow to the end of my fist is right at 15 inches, so, I'm lucky, I carry this measure about with me! How many of the rest of you are so lucky? How many find your fist width to elbow-fist length ratio to be even closer to the tangent of 15 degrees? Are there even better measurements taken from our self dimensions? Edley. You wrote: > Steve Lelievre suggested using arc tan 1/4 to find 15 degrees "in > emergency". > > A slightly better approximation is to go via > sin(15) = 0.2588 and arcsin(0.25) = 14.4775 deg > cf. tan(15) = 0.2679 and arctan(0.25) = 14.0362 degrees > > So, take your string and find a quarter of its length as before by two > halvings. Start from the centre point and mark a point on a first line at > the string's length away. From that point draw an arc (or mark a few > points to indicate it) using the quarter string as radius. Go back to the > centre and take the tangent to that arc as the second line. > > The angle subtended is arcsin 0.25 or about 14 degrees 29 minutes, close > enough for practical purposes I would think. > > If you want to get still closer, then do it both ways and add the > difference between them to the larger of the two angles to get 14 degrees > 55 minutes. (And if you dislike estimating the addition to make rather > than constructing it, then take two of the sine angles one after the other > i.e. added together and go back one of the tan angles to subtract it.) > > Or of course you could just use the same process, with half the string > length instead of a quarter, to go via arcsin 1/2 and obtain 30 degrees, > then take the string and mark two points equidistant from the origin on > the two lines 30 degrees apart, and use doubling the string to find the > point midway between those two - hence bisecting the angle into two angles > of exactly 15 degrees with no error at all. But we're getting close to > terrestrial origami I fear ...
Ceiling Sundial
Dialists, In my mind I wanted to make a try for a polar ceiling dial at latitude 52 degrees, mid Netherlands. In the attached picture you see the mirror M at a distance g from the ceiling. The polar pattern for this configuration is drawn below left and I want to have the hours 8 am - 4 pm on the dial, if possible. ( no longitude correction in this story ) The line BC then is about the limit of the space I need on the ceiling, or with other words, the mirror should be inward the room over a distance AB. Let's now have a look if the sun will shine on the mirror all year at noon. It's obvious that I need to look for the limits at summer solstice. It is seen that during the summer solstice the sun can't shine on the mirror at all. The first change I need to make is to cut of the ceiling at the line PCQ in which C is the intersection point of the summer ray and the ceilng. In the hourline pattern the hours before 10 am and after 2 pm will be cut of too and the first conclusion is that at summer solstice the dial is useable from about 10 am to 2 pm. However, there are more things to concider. In this configuration the mirror will catch the sun at summer solstice only at noon. Shortly before and after noon the mirror is in the shade. To see this we need to imagine a vertical south facing dial through AM with a (pin)gnomon of length AC. That pattern is drawn below at the right side. At noon the edge of the ceiling gives a line of shadow KL and the mirror just catches the sun. Before noon and after noon this line of shadow moves down ( to the right in the drawing ) and the mirror won't catch the sun. It sounds strange, after noon the altitude of the sun decreases and also the shadow on a south facing dial decreases. Yes, this is true and can be seen with the shape of the dateline for the summer solstice. Before and after noon this date line in going down. So at the summer solstice my dial only gives a "flash" at noon and no more... The third problem to concider is the width of the window. Because the mirror is inside the room the azimut of the lines from mirror through the edges of the window also will give limitations For this an analemmatic dial or Oughtred dial could be of use to see what the limitations are. Who will show us the first real polar-mirror-ceiling dial? Who has an inclined ceiling, higher at the south end? Won't that be better? Fer. Fer J. de Vries [EMAIL PROTECTED] http://www.iae.nl/users/ferdv/ Eindhoven, Netherlands lat. 51:30 N long. 5:30 E Attachment converted: Macintosh HD:ceiling4.gif (GIFf/JVWR) (00039C8C)
RE: Emergency Sundial
Steve Lelievre suggested using arc tan 1/4 to find 15 degrees "in emergency". A slightly better approximation is to go via sin(15) = 0.2588 and arcsin(0.25) = 14.4775 deg cf. tan(15) = 0.2679 and arctan(0.25) = 14.0362 degrees So, take your string and find a quarter of its length as before by two halvings. Start from the centre point and mark a point on a first line at the string's length away. From that point draw an arc (or mark a few points to indicate it) using the quarter string as radius. Go back to the centre and take the tangent to that arc as the second line. The angle subtended is arcsin 0.25 or about 14 degrees 29 minutes, close enough for practical purposes I would think. If you want to get still closer, then do it both ways and add the difference between them to the larger of the two angles to get 14 degrees 55 minutes. (And if you dislike estimating the addition to make rather than constructing it, then take two of the sine angles one after the other i.e. added together and go back one of the tan angles to subtract it.) Or of course you could just use the same process, with half the string length instead of a quarter, to go via arcsin 1/2 and obtain 30 degrees, then take the string and mark two points equidistant from the origin on the two lines 30 degrees apart, and use doubling the string to find the point midway between those two - hence bisecting the angle into two angles of exactly 15 degrees with no error at all. But we're getting close to terrestrial origami I fear ... Regards Andrew James 51 04 W 01 18 N
Re: Finding the meridian with a GPS
Fernando, I am not quite sure of exactly what you were trying to determine! You need have no worries re the accuracy of your GPS. By "old GPS", I assume that you mean an 8-satellite one (like my 8 year old brick!). Have no worries about the accuracy of an 8-satellite GPS. I bought my son (an avid explorer of rugged bushland near Sydney, that involved him in a couple of epics when he went into the wrong creek, necessitating nearly 10 abseils (US = rappel) including two of 75 m on 70 m ropes!). I set it up for him before I gave it to him, and out of curiosity, I checked my old 8-satellite Garmin against the new 12-satellite Garmin. The only difference in multiple readings was in the last digit (i.e. the metres digit). This convinces me that now that the CIA (sorry, US government) has removed selective availability, you can use the old GPS with no worries at all. BTW, my staff have also tested 12-channel GPSs on surveyed points in the landscape. Again, the only difference was in the metres digit. Given that the new Garmin Etrex costs AUD $380 (i.e. probably around $US 150), this is pretty acceptable accuracy. In fact, it is damn fine accuracy. Having said all that, the little yellow thing of electronics in your hand doesn't have the same aesthetic attraction as a theodolite! Cheers, John "Far better an approximate answer to the right question which may be difficult to frame, than an exact answer to the wrong question which is always easy to ask" John W Tukey, statistician - Original Message - From: "Fernando Cabral" <[EMAIL PROTECTED]> To: "sundial mailing list" Sent: Tuesday, January 08, 2002 11:29 PM Subject: Finding the meridian with a GPS After buying an old theodolite I merrily travelled to my little farm with the firm purpose of finding the local meridian with high precision. And, as an extra, I would survey the region and have a very precise map. For no purpose at all, but what is a theodolite good for if you don't find ways to use it? They say that if you give a child a hammer she will try to fix the whole world using it. So, if an adult gives himself a theodolite, he MUST find what to do with it, right? Well, theodolite, almanac, calculator, everything in place except... the moon, the sun, the stars. It rained cats and dogs for several days. Starting on December 20 until January 5, we had plenty of water from the skies. As chance uses to have it, if I had to go to main street to buy some stuff or perhaps visit a friend, that's when we had a brief period of clear sky. But experience had taught me: if I decided to rush back home clouds would immediately fly in and start pouring the water god wanted to send us... After several days I picked up my old GPS and decided to use it to replace the stars. I chose a far away point. A tree about 1000 m from the place I set up the theodolite. I walked to it and took a reading (in fact, several readings). Then I walked back to where the theodolite was and took some readings. According to the GPS distance was a little bellow 1000m. I used the GPS to find the bearing from one point to the other. I set the limbo to 0 at that tree. From there I could find any other bearing. My reasoning was that even if the GPS would give me an error of, say, 20 meters to one side (worst case) I'd still be very close to the real azimuth. In fact, let's suppose (this was what I reasoned) I have a right triangle with two legs of 707,1 m each. The third side would be 999,99 (that is, 1000 m). Now, the angle each of the two other angles would be 45 degrees. Now, if, instead of having a leg of 707,1m I had it with 727,1, the difference in the angles would be arc tan (727,1 / 707,1) = 45,7 so, an error of less than 1 degree. Close enough for most purposes, I guess, And probabily better than the other simple methods. Of course, if I wanted to secure higher precision I could still try to go further away (terrain permiting). With a tower (or tree) 5 km away my error would be of less than 10 minutes for a GPS error of 20 m in the worst direction. I still expect to refine my findings when I have a good star in the proper position and a cloudless sky. But for the time being I am happy that I could put my o'theodolite to use. Now I am a happy (ancient) kid! - fernando -- REDUZIR, REUSAR, RECICLAR -- Dever de todos, amor aos que virão REDUCE, REUSE, RECYCLE -- Everybody's duty, love to those who are to come Fernando CabralPadrao iX Sistemas Abertos mailto:[EMAIL PROTECTED] http://www.pix.com.br Fone Direto: +55 61 329-0206 mailto:[EMAIL PROTECTED] PABX: +55 61 329-0202 Fax: +55 61 326-3082 15º 45' 04.9" S (23 L 0196446/8256520) 47º 49' 58.6" W 19º 37' 57.0" S (23 K 0469898/7829161) 45º 17' 13.6" W
