Re: Fw: frame grid method
Tom Semadeni ([EMAIL PROTECTED]) wrote: Hi Jim^2,[Sorry, couldn't resist.] Aren't we trying to lay out marks FROM a computer screen or a piece of paper TO the real thing on the ground? So aren't we trying to FIND the radii of the distance circles GIVEN the coordinates of the target point and the location of the second reference point that Ron refers to? That being the case here are my calcs using Ron Anthony's A and B when he first surfaced this two reference point approach on Tue, 25 May 1999 14:17:32 -0700 , after he awoke from his semi-sleep! ;-) You are correct. However, someone (John Carmichael perhaps?) posed the inverse problem, and that is what Jim and I answered. Jim --- -- | Jim Cobb | 540 Arapeen Dr. #100 | [EMAIL PROTECTED] | | Parametric| Salt Lake City, UT | (801)-588-4632 | | Technology Corp. | 84108-1202 | Fax (801)-588-4650 | --- -- If a man's wit be wandering, let him study the mathematics. -- Francis Bacon
Re: frame grid method
Ron: Very cool! Point C is the intersection of two offset circles with centers at A and B, whose radii is known. I wonder what the exact mathematical formula for this is? Wouldn't it come out as an (x,y) coordinate? John Carmichael http://www.azstarnet.com/~pappas All, I'm sorry I was only half awake when this thread started so forgive me if I'm off course. If I had to lay out a large dial (say 100 ft) to a high degree of accuracy (say .1 of an inch) I would plot all the points not as x,y co-ordinates. I would plot them all out as the intersection of two lines from two fixed points. To see what I mean pick 2 points that are well established, e.g., point A where the gnomom meets the dial face, and point B some number of feet due north (in line with the gnomon base) of point A. Every point on the dial face is now at the intersection of two tape measures that start at points A and B. Assuming that the dial face is flat the accuracy would be good as the tape measures used. For the points that are almost inline with the AB line, a third point C could be used as one of the points. Point C could be calculated from points A and B. Of course the computer would have to calculate all of the points for you. As a crude ASCII art: Point X is 30 1 1/4 from point A, and 22 3 7/8 from point B. (A metric tape measure would be a lot handier) B \ \ \ \ \ / X /C / / / / / / A
Re: frame grid method
John, I'm a little weak on the math, (although for my 52nd birthday I took my trigonometry finals). This is how I would attempt it. Pick a point to serve as an origin, say point A. Translate the x,y coordinates of point C to polar. Here's an example written in C. r = the distances from A to C. r = sqrt( sqr(x) + sqr(y) ) For the distance from B to C, x and y would have to be first translated to use B as the origin before translating them to polar: x1 = x -Bx y1 = y-By r1 = sqrt( sqr(x1) + sqr(y1) ) Where: x1,y2 are new coordinates relative to B. Bx,By are the x,y coordinates of point B using A as the origin. For example 0,10 if Point B was 100 meters north of A. ++ron -Original Message- From: John Carmichael [EMAIL PROTECTED] To: Ron Anthony [EMAIL PROTECTED] Cc: sundial@rrz.uni-koeln.de sundial@rrz.uni-koeln.de Date: Wednesday, May 26, 1999 7:48 AM Subject: Re: frame grid method Ron: Very cool! Point C is the intersection of two offset circles with centers at A and B, whose radii is known. I wonder what the exact mathematical formula for this is? Wouldn't it come out as an (x,y) coordinate? John Carmichael http://www.azstarnet.com/~pappas All, I'm sorry I was only half awake when this thread started so forgive me if I'm off course. If I had to lay out a large dial (say 100 ft) to a high degree of accuracy (say .1 of an inch) I would plot all the points not as x,y co-ordinates. I would plot them all out as the intersection of two lines from two fixed points. To see what I mean pick 2 points that are well established, e.g., point A where the gnomom meets the dial face, and point B some number of feet due north (in line with the gnomon base) of point A. Every point on the dial face is now at the intersection of two tape measures that start at points A and B. Assuming that the dial face is flat the accuracy would be good as the tape measures used. For the points that are almost inline with the AB line, a third point C could be used as one of the points. Point C could be calculated from points A and B. Of course the computer would have to calculate all of the points for you. As a crude ASCII art: Point X is 30 1 1/4 from point A, and 22 3 7/8 from point B. (A metric tape measure would be a lot handier) B \ \ \ \ \ / X /C / / / / / / A
Re: frame grid method
OOPS, sorry for the earlier empty post caused by finger trouble caused by brain trouble caused by ... A good approach, Ron, but the complexity lies in: Of course the computer would have to calculate all of the points for you. The complexity can be reduced by following your suggestions for selecting Point A at the centre of the dial face and Point B at the North intersection of the meridian and the dial circle. Then all your fractional and whole hour lines can be easily set by keeping the AC distance constant ( = to the radius of the dial circle or theconstruction layout dial circle) and varying your BC distance to obtain your points, and therefore lines, in either one of two ways: 1) straight line distance from B to C, which requires a complex calculation or, 2) arc length distance from B to C which is simply calculated but is a little more difficult to measure since you are measuring along a curve. (This is an R - THETA, or polar coordinate system, where THETA is determined by arc length rather than by angle directly.) I should have pointed out in my earlier post that this is really easy if you make a template by cutting an arc from a piece of plywood. For Celeste, referred to in my posting, Subject: Re: plotting timelines for giant sundials Date: Mon, 24 May 1999 00:52:52 -0400 To: John Carmichael [EMAIL PROTECTED] CC: Sundial List sundial@rrz.uni-koeln.de, I got over 20 degrees of arc length along the circumference of the dial using an arc cut from an eight foot piece of plywood. And I easily marked it to pencil line width transferring a mark precision of better than a minute (of angle, not time!) to the dial plate. I had four templates from one cut: two concave and two convex, enough to do 45 degrees using the two pieces at once. After reading your post I think that it would also be easier to locate even non-radial figures and lines using polar coordinates than to use rectangular coordinates with the continual right angle problem at layout time. Simply establish Theta using arc length on the template and measure R out on the established line, which could be a surveyors tape. ( I have a thin, 1/4, one, 50m long which would work beautifully.) The computer calculation of R-Theta points is straightforward or automatic in most graphics or CAD programs. It would be a simple matter to square off a rectangle on top of the polar plot, if that is what the shape of the perimeter is to be. Now, if we could just come up with a way to generate ASCII graphics so simply! Or better yet if we could just use FranÁois BLATEYRON's Shadows version 1.5.3 (18 April 1999) program on my iMac!: ^ ( Good stuff, Ron. Cheers, Tom Ron Anthony wrote: All, I'm sorry I was only half awake when this thread started so forgive me if I'm off course. If I had to lay out a large dial (say 100 ft) to a high degree of accuracy (say .1 of an inch) I would plot all the points not as x,y co-ordinates. I would plot them all out as the intersection of two lines from two fixed points. To see what I mean pick 2 points that are well established, e.g., point A where the gnomom meets the dial face, and point B some number of feet due north (in line with the gnomon base) of point A. Every point on the dial face is now at the intersection of two tape measures that start at points A and B. Assuming that the dial face is flat the accuracy would be good as the tape measures used. For the points that are almost inline with the AB line, a third point C could be used as one of the points. Point C could be calculated from points A and B. Of course the computer would have to calculate all of the points for you. As a crude ASCII art: Point X is 30 1 1/4 from point A, and 22 3 7/8 from point B. (A metric tape measure would be a lot handier) B \ \ \ \ \ / X /C / / / / / / A -Original Message- From: John Carmichael [EMAIL PROTECTED] To: François BLATEYRON [EMAIL PROTECTED] Cc: sundial@rrz.uni-koeln.de sundial@rrz.uni-koeln.de Date: Tuesday, May 25, 1999 9:45 AM Subject: frame grid method Perhaps a precise solution would be to calculate the intersection of the hour line with the enclosing frame of your sundial. It must be done by a computer but its easy to give a very good precision. The result would be given as a length and a direction (north, east, south, west side of the sundial), the origin could be one of the two opposite corners. The only problem then is to precisely draw your frame, with parallel sides and a good perpendicularity. With one intersection point you can draw the line by joining it to the gnomon foot. I plan to include such kind of data in my Shadows program in a futur version. Bonjour Francois and everyone else: Your frame method for plotting
Re: Fw: frame grid method
Hi Jim^2,[Sorry, couldn't resist.] Aren't we trying to lay out marks FROM a computer screen or a piece of paper TO the real thing on the ground? So aren't we trying to FIND the radii of the distance circles GIVEN the coordinates of the target point and the location of the second reference point that Ron refers to? That being the case here are my calcs using Ron Anthony's A and B when he first surfaced this two reference point approach on Tue, 25 May 1999 14:17:32 -0700 , after he awoke from his semi-sleep!;-) ASSUMPTIONS: Horizontal, flat (usually impractical) dial surface. Cartesian, X-Y, system has origin at A and +Y is North (LET THE AUSSIES DO IT THEIR WAY!! :-)) Polar, R-Theta, system has origin at A and Zero Theta is East. KNOWN: The location (COORDINATES) of each target point on a computer screen or on paper with reference to one origin in either polar (R, THETA) or cartesian coordinates (X, Y). The location of reference point B with respect to reference point A on the ground. Lets follow everyone's simplification and agree that B is N metres North of A and that A is at the centre of the dial circle or, equivalently, at the intersection of the gnomon and the dial surface. REQUIRED:The distance of the target point from each of two points A and B on the ground. Terminology: Let N = the distance, in metres, of Point B, North of Point A on the ground. A = the distance, in metres, of the target point from Point A on the ground. B = the distance, in metres, of the target point from Point B on the ground. (X,Y) = the cartesian coordinates, in metres, of the target point with respect to an origin at A. (R,T) = the polar coordinates, in metres and radians, of the target point with respect to an origin at A. SOLUTIONS: Cartesian: A = SQRT(X^2 +Y^2) B = SQRT(X^2 + (Y-N)^2) Polar: A = R B = SQRT((RCosT)^2 + (RSinT-N)^2) which simplifies a bit to, B = SQRT(R^2 + N^2 - 2RNSinT) These are easily calculated and displayed using a spreadsheet like Excel, the major challenge being, for me, getting the stuff imported into Excel from my design application. Dratted I/O files! Reminds me of FORTRAN! Yecch! (Okay, if you really want to be an Aussie, put a negative value for N in the above. Only take positive roots, since the radii, A B, are both positive. If you end up with any imaginary roots the evening sun has set past the yardarm so it is time to take a swig o' The Medicinal and wait 'til the mornin' sun and use John Carmichael's Solution C.) BTW, John, have we worked this one to death yet? Cheers, t Jim_Cobb wrote: In the special case described earlier the math is even easier. We want two reference points, P with coordinates (0, 0) and Q, due north of P (or south if you're an Aussie) at (0, a) at distance a from P. Then the points at distance r from P and distance s from Q may be found by solving the two quadratic equations: x^2 + y^2 = r^2 and x^2 + (y-a)^2 = s^2 What is nice about this setup is that y has a very simple solution. Subtract the two equations to get (y-a)^2 - y^2 = s^2 - r^2 expand y^2 - 2 a y + a^2 - y^2 = s^2 - r^2 and conclude y = (a^2 - s^2 + r^2) / 2 a Then solve for x x = +- sqrt(r^2 - y^2) choosing the +/- root depending on whether the point is to the East or West of line PQ. If you find yourself attempting to take a square root of a negative number, the circles do not intersect (more specifically, they have no real points of intersection). You will notice that you never make use of r or s. If you just use the squares of their values the arithmetic is simplified even more. Jim --- -- | Jim Cobb | 540 Arapeen Dr. #100 | [EMAIL PROTECTED] | | Parametric| Salt Lake City, UT | (801)-588-4632 | | Technology Corp. | 84108-1202 | Fax (801)-588-4650 | --- -- Science is trained and organized common sense. -- Alfred, Lord Tennyson Jim Morrison [EMAIL PROTECTED] wrote: Here is one way to calculate the intersection points of two circles: Given circle of radius r1 at origin: x^2 + y^2 = r1^2 Circle at (x0,y0) with radius r2: (x-x0)^2 + (y-y0)^2 = r2^2 Intersection points between the circles lie on a line: y = mx + b Where: m = -x0/y0 b = - 1/2y0 (r2^2 - r1^2 - y0^2 - x0^2) x = (-mb +/- SQR(m^2 b^2 - (m^2+1)(b^2-r1^2)) / (m^2 + 1) If the roots of x are imaginary (i.e. the radical is negative) then there is no intersection. Adjust coordinates of intersecting circle relative to base circle. I have this function coded in C, QBASIC and assembler if anyone wants it. Best regards, Jim James E. Morrison Astrolabe web pages at: http://myhouse.com/mc/planet/astrodir/astrolab.htm -- Tom Semadeni O [EMAIL PROTECTED]
Re: frame grid method
I *really* like this scheme. Very clever... Jim --- -- | Jim Cobb | 540 Arapeen Dr. #100 | [EMAIL PROTECTED] | | Parametric| Salt Lake City, UT | (801)-588-4632 | | Technology Corp. | 84108-1202 | Fax (801)-588-4650 | --- -- You know when you're sitting in a chair and you lean back so you're on just 2 legs and you almost fall over and at the last second you catch yourself? I feel like that all the time. -- Steven Wright Ron Anthony [EMAIL PROTECTED] wrote: All, I'm sorry I was only half awake when this thread started so forgive me if I'm off course. If I had to lay out a large dial (say 100 ft) to a high degree of accuracy (say .1 of an inch) I would plot all the points not as x,y co-ordinates. I would plot them all out as the intersection of two lines from two fixed points. To see what I mean pick 2 points that are well established, e.g., point A where the gnomom meets the dial face, and point B some number of feet due north (in line with the gnomon base) of point A. Every point on the dial face is now at the intersection of two tape measures that start at points A and B. Assuming that the dial face is flat the accuracy would be good as the tape measures used. For the points that are almost inline with the AB line, a third point C could be used as one of the points. Point C could be calculated from points A and B. Of course the computer would have to calculate all of the points for you. As a crude ASCII art: Point X is 30 1 1/4 from point A, and 22 3 7/8 from point B. (A metric tape measure would be a lot handier) B \ \ \ \ \ / X /C / / / / / / A
frame grid method
Perhaps a precise solution would be to calculate the intersection of the hour line with the enclosing frame of your sundial. It must be done by a computer but its easy to give a very good precision. The result would be given as a length and a direction (north, east, south, west side of the sundial), the origin could be one of the two opposite corners. The only problem then is to precisely draw your frame, with parallel sides and a good perpendicularity. With one intersection point you can draw the line by joining it to the gnomon foot. I plan to include such kind of data in my Shadows program in a futur version. Bonjour Francois and everyone else: Your frame method for plotting would be a good practical way to implement John Pickard's (x,y) coordinate method. With a squared frame around the sundial work area, one could use it as a reference to easily measure the horizontal and vertical distances to the x,y coordinates. Let us know when you get your new shadows program up and running. Do you think I'd understand it or would I have to get someone to help me? merci Francois, John Carmichael
Re: frame grid method
Further to Roy Anthony's note on laying out a large dial. His method is similar to the one used for the Swenson dial which is 30 by 60 feet. An arm with a scale attached was pivioted at the base of the gnomon. This was the origin of the original x-y set of points defining the dial lines. Another scale was layed horizontaly along the top of the wall. A seconded arm with attached scale was pivioted at a set of predetermined points along the horizontal scale. Each x-y point was converted to a pair of readings on the pivoted arms. A spread sheet is excellent for this. A drill was placed through a hole on plates attached to each arm. This located the point to about 1/16 of an inch. This is much less than the reading accuracy of the shadow (about 0.5 inch) due to the penumbral effects. Hope this helps. Cheers, John Professor John P.G.Shepherd Physics Department University of Wisconsin-River Falls 410 S. 3rd. St. River Falls,WI 54022 Phone (715)-425-3196, eve. (715)-425-6203 Fax (715)-425-0652
Re: frame grid method
-- Tom Semadeni O [EMAIL PROTECTED] o aka I (Ned) Ames . Britthome Bounty * Box 176 Britt ON P0G 1A0 'Phone 705 383 0195 fax 2920 45.768* North 80.600* West
