Re: Capuchin and Regiomontanus dials
Thanks for the Regiomontanus slide. Then the original designer of that dial must have just checked out the result of that way of setting the bead, by doing the calculation to find out if squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec lat sec dec, as a trial-and error trial that? Or, I don't know, is that a trigonometric fact that would be already known to someone who is really experienced in trig? --- What's the purpose of the lower latitude scale, on the dial shown in that slide? When I described my folded-cardboard portable equatorial-dial, I mis-stated the declination arrangement: Actually, the sliding paper tab (made by making two slits in the bottom of the tab, and fitting that onto an edge of the cardboard) is positioned via date-markngs along that edge. The declination reading, and therefore the azimuth, is correct when the shadow of a certain edge of the tab, perpendicular to the cardboard edge on which it slides, just reaches the hour-scale on the surface that's serving as a quarter of an Disk-Equatorial dial. Actually, that dial was intended as an emergency backup at sea, where there would always be available a horizon by which to vertically orient the dial. The use of a plumb-bob for that purpose was my idea, because, on land there often or usually isn't a visible horizon, due to houses, trees, etc. Maybe, in really flat land, even without an ocean horizon, even a land-horizon could be helpful, but such a horizon isn't usually visible in most places on land. But then, with the plumb-line, it's necessary to keep the vertical surface parallel to the pendulum-string, and keep the pendulum-string along the right degree-mark, while making sure that the declination-reading is right, when reading the time. ...Four things to keep track of at the same time. ...maybe making that the most difficult-to-use portable dial. With the Equinoctical Ring-Dial, the vertical orientation, about both horizontal axes, is automatically achieved by gravity, so only time and declination need be read. And, with a pre-adjustable altitude-dial, only the sun-alignment shadow and the time need to be read. With my compass tablet-dials, one mainly only had to watch the compass and the time-reading. Of course it was necessary to hold the dial horizontal, without a spirit-level, but that didn't keep them from being accurate. Michael Ossipoff On Sun, May 14, 2017 at 5:18 PM, Fred Sawyer wrote: > Michael, > > See the attached slide from my talk. All the various dials work with a > string of this length. They vary simply in where the suspension point is > placed. The pros and cons of the various suspension points were part of my > presentation. > > Fred Sawyer > > > On Sun, May 14, 2017 at 4:40 PM, Michael Ossipoff > wrote: > >> When I said that there isn't an obvious way to measure to make the >> plumb-line length equal to sec lat sec dec, I meant that there' s no >> obvious way to achieve that *with one measurement*. >> >> I was looking for a way to do it with one measurement, because that's how >> the use-instructions say to do it. >> >> In fact, not only is it evidently done with one measurement, but that one >> measurement has the upper end of the plumb-line already fixed to the point >> from which it's going to be used, at the intersection of the appropriate >> latitude-line and declination-line. >> >> That's fortuitous, that it can be done like that, with one measurement, >> and using only one positioning of the top end of the plumb-line. >> >> But of course it's easier, (to find) and there's an obviously and >> naturally-motivated way to do it, with *two* measurements, before fixing >> the top-end of the plumb-line at the point where it will be used. >> >> The line from that right-edge point (from which the first horizontal is >> drawn) to the point where the appropriate latitude-line intersects the >> vertical has a length of sec lat. >> >> So, before fixing the top end of the plumb-line where it will be used >> from, at the intersection of the appropriate lat and dec lines, just place >> the top end of the plumb line at one end of that line mentioned in the >> paragraph before this one, and slide the bead to the other end of that >> line. ...to get a length of thread equal to sec lat. >> >> Then, have a set of declination marks at the right edge, just like the >> ones that are actually on a Regiomontanus dial, except that the lines from >> the intersection of the first horizontal and vertical lines, to the >> declination (date) marks at the right-margins are shown. >> >> Oh, but have that system of lines drawn a bit larger, so that the origin >> of the declination-lines to the right margin is a bit farther to the left >> from the intersection of the first horizontal and the first vertical. >> ...but still on a leftward extension of the first horizontal. >> >> That's so that there will be room for the 2nd measurement, the >> measurement tha
Re: Capuchin and Regiomontanus dials
I asked: "Or, I don't know, is that a trigonometric fact that would be already known to someone who is really experienced in trig?" Well, alternative expressions for the product of two cosines is something that might be basic and frequently-occurring enough to be written down somewhere, where someone could look it up. Maybe someone who'd thoroughly studied trig, and done a lot of it, would know it without looking it up. It might be especially notable that one of the alternative expressions for sec x sec y is the square-root of the sum of two squares, a Pythagorean distance. Michael Ossipoff On Mon, May 15, 2017 at 11:32 AM, Michael Ossipoff wrote: > Thanks for the Regiomontanus slide. > > Then the original designer of that dial must have just checked out the > result of that way of setting the bead, by doing the calculation to find > out if > squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec lat sec dec, > as a trial-and error trial that? > > Or, I don't know, is that a trigonometric fact that would be already known > to someone who is really experienced in trig? > > --- > > What's the purpose of the lower latitude scale, on the dial shown in that > slide? > > > > When I described my folded-cardboard portable equatorial-dial, I > mis-stated the declination arrangement: > > Actually, the sliding paper tab (made by making two slits in the bottom of > the tab, and fitting that onto an edge of the cardboard) is positioned via > date-markngs along that edge. The declination reading, and therefore the > azimuth, is correct when the shadow of a certain edge of the tab, > perpendicular to the cardboard edge on which it slides, just reaches the > hour-scale on the surface that's serving as a quarter of an Disk-Equatorial > dial. > > Actually, that dial was intended as an emergency backup at sea, where > there would always be available a horizon by which to vertically orient the > dial. > > The use of a plumb-bob for that purpose was my idea, because, on land > there often or usually isn't a visible horizon, due to houses, trees, etc. > Maybe, in really flat land, even without an ocean horizon, even a > land-horizon could be helpful, but such a horizon isn't usually visible in > most places on land. > > But then, with the plumb-line, it's necessary to keep the vertical surface > parallel to the pendulum-string, and keep the pendulum-string along the > right degree-mark, while making sure that the declination-reading is right, > when reading the time. > > ...Four things to keep track of at the same time. ...maybe making that > the most difficult-to-use portable dial. > > With the Equinoctical Ring-Dial, the vertical orientation, about both > horizontal axes, is automatically achieved by gravity, so only time and > declination need be read. > > And, with a pre-adjustable altitude-dial, only the sun-alignment shadow > and the time need to be read. > > With my compass tablet-dials, one mainly only had to watch the compass and > the time-reading. Of course it was necessary to hold the dial horizontal, > without a spirit-level, but that didn't keep them from being accurate. > > Michael Ossipoff > > > > On Sun, May 14, 2017 at 5:18 PM, Fred Sawyer wrote: > >> Michael, >> >> See the attached slide from my talk. All the various dials work with a >> string of this length. They vary simply in where the suspension point is >> placed. The pros and cons of the various suspension points were part of my >> presentation. >> >> Fred Sawyer >> >> >> On Sun, May 14, 2017 at 4:40 PM, Michael Ossipoff > > wrote: >> >>> When I said that there isn't an obvious way to measure to make the >>> plumb-line length equal to sec lat sec dec, I meant that there' s no >>> obvious way to achieve that *with one measurement*. >>> >>> I was looking for a way to do it with one measurement, because that's >>> how the use-instructions say to do it. >>> >>> In fact, not only is it evidently done with one measurement, but that >>> one measurement has the upper end of the plumb-line already fixed to the >>> point from which it's going to be used, at the intersection of the >>> appropriate latitude-line and declination-line. >>> >>> That's fortuitous, that it can be done like that, with one measurement, >>> and using only one positioning of the top end of the plumb-line. >>> >>> But of course it's easier, (to find) and there's an obviously and >>> naturally-motivated way to do it, with *two* measurements, before >>> fixing the top-end of the plumb-line at the point where it will be used. >>> >>> The line from that right-edge point (from which the first horizontal is >>> drawn) to the point where the appropriate latitude-line intersects the >>> vertical has a length of sec lat. >>> >>> So, before fixing the top end of the plumb-line where it will be used >>> from, at the intersection of the appropriate lat and dec lines, just place >>> the top end of the plumb line at one end of that line m
Re: Capuchin and Regiomontanus dials
Michael, I seem to recall that sec^2(x)=1+tan^2(x) Therefore sec^2(lat).sec^2(dec)=(1+tan^2(lat)).(1+tan^2(dec)) =1+tan^2(lat)+tan^2(dec)+tan^2(lat).tan^2(dec) =(1+tan dec tan lat)^2 + (tan dec - tan lat)^2 I guess that this relationship, which is just a variant of sin^2+cos^2=1, should have been known to the dial designer. Geoff On 15 May 2017 at 16:32, Michael Ossipoff wrote: > Thanks for the Regiomontanus slide. > > Then the original designer of that dial must have just checked out the > result of that way of setting the bead, by doing the calculation to find > out if > squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec lat sec dec, > as a trial-and error trial that? > > Or, I don't know, is that a trigonometric fact that would be already known > to someone who is really experienced in trig? > > --- > > What's the purpose of the lower latitude scale, on the dial shown in that > slide? > > > > When I described my folded-cardboard portable equatorial-dial, I > mis-stated the declination arrangement: > > Actually, the sliding paper tab (made by making two slits in the bottom of > the tab, and fitting that onto an edge of the cardboard) is positioned via > date-markngs along that edge. The declination reading, and therefore the > azimuth, is correct when the shadow of a certain edge of the tab, > perpendicular to the cardboard edge on which it slides, just reaches the > hour-scale on the surface that's serving as a quarter of an Disk-Equatorial > dial. > > Actually, that dial was intended as an emergency backup at sea, where > there would always be available a horizon by which to vertically orient the > dial. > > The use of a plumb-bob for that purpose was my idea, because, on land > there often or usually isn't a visible horizon, due to houses, trees, etc. > Maybe, in really flat land, even without an ocean horizon, even a > land-horizon could be helpful, but such a horizon isn't usually visible in > most places on land. > > But then, with the plumb-line, it's necessary to keep the vertical surface > parallel to the pendulum-string, and keep the pendulum-string along the > right degree-mark, while making sure that the declination-reading is right, > when reading the time. > > ...Four things to keep track of at the same time. ...maybe making that > the most difficult-to-use portable dial. > > With the Equinoctical Ring-Dial, the vertical orientation, about both > horizontal axes, is automatically achieved by gravity, so only time and > declination need be read. > > And, with a pre-adjustable altitude-dial, only the sun-alignment shadow > and the time need to be read. > > With my compass tablet-dials, one mainly only had to watch the compass and > the time-reading. Of course it was necessary to hold the dial horizontal, > without a spirit-level, but that didn't keep them from being accurate. > > Michael Ossipoff > > > > On Sun, May 14, 2017 at 5:18 PM, Fred Sawyer wrote: > >> Michael, >> >> See the attached slide from my talk. All the various dials work with a >> string of this length. They vary simply in where the suspension point is >> placed. The pros and cons of the various suspension points were part of my >> presentation. >> >> Fred Sawyer >> >> >> On Sun, May 14, 2017 at 4:40 PM, Michael Ossipoff > > wrote: >> >>> When I said that there isn't an obvious way to measure to make the >>> plumb-line length equal to sec lat sec dec, I meant that there' s no >>> obvious way to achieve that *with one measurement*. >>> >>> I was looking for a way to do it with one measurement, because that's >>> how the use-instructions say to do it. >>> >>> In fact, not only is it evidently done with one measurement, but that >>> one measurement has the upper end of the plumb-line already fixed to the >>> point from which it's going to be used, at the intersection of the >>> appropriate latitude-line and declination-line. >>> >>> That's fortuitous, that it can be done like that, with one measurement, >>> and using only one positioning of the top end of the plumb-line. >>> >>> But of course it's easier, (to find) and there's an obviously and >>> naturally-motivated way to do it, with *two* measurements, before >>> fixing the top-end of the plumb-line at the point where it will be used. >>> >>> The line from that right-edge point (from which the first horizontal is >>> drawn) to the point where the appropriate latitude-line intersects the >>> vertical has a length of sec lat. >>> >>> So, before fixing the top end of the plumb-line where it will be used >>> from, at the intersection of the appropriate lat and dec lines, just place >>> the top end of the plumb line at one end of that line mentioned in the >>> paragraph before this one, and slide the bead to the other end of that >>> line. ...to get a length of thread equal to sec lat. >>> >>> Then, have a set of declination marks at the right edge, just like the >>> ones that are actually on a Regiomontanus di
Re: Capuchin and Regiomontanus dials
Wow. What can I say. Your approach makes more sense in every way, than the way that I'd been trying to find how the bead-setting procedure could have been arrived at. I'd wanted to start with various pairs of points, and then find out if any of them are separated by a distance of sec lat sec dec. But of course (now it's obvious) it makes a lot more sense to start with sec lat sec dec, and find out if it can be made into a distance. ...which is of course how you approached the problem. If we expect the distance on the dial to be a diagonal distance, then it will be the sum of two squares, all in a square-root sign. Most likely it will be a diagonal distance, which means the it will be the sum of two squares, all in a square-root sign. Of course it _needn't_ be a diagonal distance. It could be all horizontal or all vertical. But, a lot of distances already on the dial are expressed as tangents, and more could be. So, converting the sec to tan makes sense, for a start. The familiar identity that relates sec and tan involves their squares. That suggests a diagonal distance, adding some confirmation to the initial impression that a diagonal distance might be more likely. So (tan lat + 1) and (tan dec + 1) are multiplied together, resulting in four terms, each of which is a square (...though of course the 1 needn't snecessrily have been gotten by squaring--except that it wasn't gotten by multiplying other numbers. So maybe it should be considered a square). The fact that there are four squares suggests that the two squared expressions are both binomials. ...and that the squares' middle terms cancel eachother out. (Of course maybe the inventor didn't have a way to be sure that sec lat sec dec can be written as a distance on the dial at all. But, if not, he evidently checked out the possibility.) So, if the four squares are the squares of the terms of two binomials, with their middle terms canceling out, there are 3 ways in which the two binomials could be assembled from the square roots of those four squares. In a way, it doesn't matter which way it's done, as long as it results in a distance. But, for that diagonal distance, of course it's necessary that the two squared binomials reapresnt distances in mutually perpendicular directions. Well there's an obvious distance there, among the square-roots of those terms: tan lat tan dec. It's horizontal, and is the distance of the string-hang-point forward (sunward) from the middle vertical. And if the 1 is added to it, that's the horizontal distance of the string-hang-point from the rear edge of the dial-card. Of the square-roots of the other two terms, tan lat is the vertical distance of the string-hang point above the main horizontal, the first horizontal. So that works--a horizontal distance and a vertical distance, which are needed for a diagonal distance. And of course naturally (tan lat - tan dec) would be a vertical distance from the string-hang-point, to the upper-end of a line has been drawn across half the dial-card's 2-unit width, one end on the first horizontal, with the line angled up by the declination-angle. Since the horizontal distance suggested was from the string-hang-point to the rear edge of the dial, and because the string-hang point is tan lat above the first horizontal, then that suggests that the measurement should be from the string-hang point, to a point that is tan dec above the first horizontal, on the rear margin of the dial. ...leading to the Regiomontanus's way of setting the bead. And, in that way, that beat-setting method is naturally arrived at. So thanks for pointing out that natural approach, making choices than make more sense than the approach I was considering. Michael Ossipoff On Mon, May 15, 2017 at 2:54 PM, Geoff Thurston wrote: > Michael, > > I seem to recall that sec^2(x)=1+tan^2(x) > > Therefore sec^2(lat).sec^2(dec)=(1+tan^2(lat)).(1+tan^2(dec)) > > =1+tan^2(lat)+tan^2(dec)+tan^2(lat).tan^2(dec) > > =(1+tan dec tan lat)^2 + (tan dec - tan lat)^2 > > I guess that this relationship, which is just a variant of sin^2+cos^2=1, > should have been known to the dial designer. > > Geoff > > On 15 May 2017 at 16:32, Michael Ossipoff wrote: > >> Thanks for the Regiomontanus slide. >> >> Then the original designer of that dial must have just checked out the >> result of that way of setting the bead, by doing the calculation to find >> out if >> squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec lat sec dec, >> as a trial-and error trial that? >> >> Or, I don't know, is that a trigonometric fact that would be already >> known to someone who is really experienced in trig? >> >> --- >> >> What's the purpose of the lower latitude scale, on the dial shown in that >> slide? >> >> >> >> When I described my folded-cardboard portable equatorial-dial, I >> mis-stated the declination arrangement: >> >> Actually, the sliding paper tab (made by making two s
Re: Capuchin and Regiomontanus dials
Of course, because only the four squared-terms are present, the two binomials have to be chosen so that, when they're both squared, their resulting middle terms cancel eachother out. (tan lat tan dec + 1) and (tan lat - tan dec) meet that requirement. Michael Ossipoff On Mon, May 15, 2017 at 9:25 PM, Michael Ossipoff wrote: > Wow. What can I say. > > Your approach makes more sense in every way, than the way that I'd been > trying to find how the bead-setting procedure could have been arrived at. > > I'd wanted to start with various pairs of points, and then find out if any > of them are separated by a distance of sec lat sec dec. > > But of course (now it's obvious) it makes a lot more sense to start with > sec lat sec dec, and find out if it can be made into a distance. ...which > is of course how you approached the problem. > > If we expect the distance on the dial to be a diagonal distance, then it > will be the sum of two squares, all in a square-root sign. > > Most likely it will be a diagonal distance, which means the it will be the > sum of two squares, all in a square-root sign. > > Of course it _needn't_ be a diagonal distance. It could be all horizontal > or all vertical. But, a lot of distances already on the dial are expressed > as tangents, and more could be. So, converting the sec to tan makes sense, > for a start. > > The familiar identity that relates sec and tan involves their squares. > That suggests a diagonal distance, adding some confirmation to the initial > impression that a diagonal distance might be more likely. > > So (tan lat + 1) and (tan dec + 1) are multiplied together, resulting in > four terms, each of which is a square (...though of course the 1 needn't > snecessrily have been gotten by squaring--except that it wasn't gotten by > multiplying other numbers. So maybe it should be considered a square). > > The fact that there are four squares suggests that the two squared > expressions are both binomials. ...and that the squares' middle terms > cancel eachother out. > > (Of course maybe the inventor didn't have a way to be sure that sec lat > sec dec can be written as a distance on the dial at all. But, if not, he > evidently checked out the possibility.) > > So, if the four squares are the squares of the terms of two binomials, > with their middle terms canceling out, there are 3 ways in which the two > binomials could be assembled from the square roots of those four squares. > > In a way, it doesn't matter which way it's done, as long as it results in > a distance. But, for that diagonal distance, of course it's necessary that > the two squared binomials reapresnt distances in mutually perpendicular > directions. > > Well there's an obvious distance there, among the square-roots of those > terms: tan lat tan dec. It's horizontal, and is the distance of the > string-hang-point forward (sunward) from the middle vertical. And if the 1 > is added to it, that's the horizontal distance of the string-hang-point > from the rear edge of the dial-card. > > Of the square-roots of the other two terms, tan lat is the vertical > distance of the string-hang point above the main horizontal, the first > horizontal. > > So that works--a horizontal distance and a vertical distance, which are > needed for a diagonal distance. And of course naturally (tan lat - tan dec) > would be a vertical distance from the string-hang-point, to the upper-end > of a line has been drawn across half the dial-card's 2-unit width, one end > on the first horizontal, with the line angled up by the declination-angle. > > Since the horizontal distance suggested was from the string-hang-point to > the rear edge of the dial, and because the string-hang point is tan lat > above the first horizontal, then that suggests that the measurement should > be from the string-hang point, to a point that is tan dec above the first > horizontal, on the rear margin of the dial. > > ...leading to the Regiomontanus's way of setting the bead. > > And, in that way, that beat-setting method is naturally arrived at. > > So thanks for pointing out that natural approach, making choices than make > more sense than the approach I was considering. > > Michael Ossipoff > > > > > > > > > On Mon, May 15, 2017 at 2:54 PM, Geoff Thurston > wrote: > >> Michael, >> >> I seem to recall that sec^2(x)=1+tan^2(x) >> >> Therefore sec^2(lat).sec^2(dec)=(1+tan^2(lat)).(1+tan^2(dec)) >> >> =1+tan^2(lat)+tan^2(dec)+tan^2(lat).tan^2(dec) >> >> =(1+tan dec tan lat)^2 + (tan dec - tan lat)^2 >> >> I guess that this relationship, which is just a variant of sin^2+cos^2=1, >> should have been known to the dial designer. >> >> Geoff >> >> On 15 May 2017 at 16:32, Michael Ossipoff wrote: >> >>> Thanks for the Regiomontanus slide. >>> >>> Then the original designer of that dial must have just checked out the >>> result of that way of setting the bead, by doing the calculation to find >>> out if >>> squrt((1+tan dec tan lat)^2 + (tan dec - tan lat)^2)) = sec
S-Town, John B McLemore
Hello Sundial Listers, I was wondering if anyone of us knew John B Mclemore, the Horologist protagonist of the This American Life radio show S-Town. The story includes his fascination with sundials and astrolabes. Was he known to any of us? -Bill --- https://lists.uni-koeln.de/mailman/listinfo/sundial