tag:blogger.com,1999:blog-4946003991896518492017-06-04T16:51:06.751-07:00httprover's 2nd blogJim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.comBlogger846125tag:blogger.com,1999:blog-494600399189651849.post-34121710474064700372017-06-01T03:36:00.000-07:002017-06-04T16:51:06.780-07:00Newton's Temperature Scale<br /> The <a href="http://en.wikipedia.org/wiki/Thermoscope">thermoscope</a>, a bulb containing air with a long tube that was immersed in water, was developed by Galileo and others to measure temperature during the first half of the 17th Century. Boyle studied similar "weather-glasses" and introduced the <a href="http://books.google.com/books?id=GUxTAAAAcAAJ&pg=PA489#v=onepage&q&f=false">hermetically sealed thermometer</a> in England by 1665. In 1701 Newton anonymously published an article, <a href="http://books.google.com/books?id=x8NeAAAAcAAJ&pg=PA824#v=onepage&q&f=false">Scala graduum caloris</a>, which described a temperature scale ranging from the freezing point of water to that of a fire hot enough to make iron glow. An English translation of Newton's article can be found in Magie, A Source Book in Physics, p. 225.<br /><br />Newton's temperature scale has a geometric series and an arithmetic series associated with it. The geometric series corresponds to the temperatures and the arithmetic series is associated with cooling times.<br /><br /> "This table was constructed by the help of a thermometer and of heated iron. With the thermometer I found the measure of all the heats up to that at which lead melts and by the hot iron I found the measure of the other heats. For the heat which the hot iron communicates in a given time to cold bodies which are near it, that is, the heat which the iron loses in a given time, is proportional to the whole heat of the iron. And so, if the times of cooling are taken equal, the heats will be in a geometrical progression and consequently can easily be found with a table of logarithms."<br /><br />After finding a number of temperatures with the aid of a thermometer, Newton describes how the hot iron was used.<br /><br />"...I heated a large enough block of iron until it was glowing and taking it from the fire with a forceps while it was glowing I placed it at once in a cold place where the wind was constantly blowing; and placing on it little pieces of various metals and other liquefiable bodies, I noted the times of cooling until all these bodies lost their fluidity and hardened, and until the heat of the iron became equal to the heat of the human body. Then by assuming that the excess of the heat of the iron and of the hardening bodies above the heat of the atmosphere, found by the thermometer, were in geometrical progression when the times were in arithmetical progression, all heats were determined."<br /><br />Newton's temperature scale can be constructed mathematically as follows where I've noted some corresponding temperatures on the Fahrenheit temperature scale for comparison.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-idwF2VraWXk/WS_lfEAPiJI/AAAAAAAAHCY/Hju_Cwr-hD8aAhZH0mkmWBN4RL-RAytFACLcB/s1600/Newton%2BTemp%2BScale%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="373" data-original-width="335" src="https://3.bp.blogspot.com/-idwF2VraWXk/WS_lfEAPiJI/AAAAAAAAHCY/Hju_Cwr-hD8aAhZH0mkmWBN4RL-RAytFACLcB/s1600/Newton%2BTemp%2BScale%2B01b.PNG" /></a></div><br />The temperature point between the melting point of wax and the boiling point of water is an average. I used the geometric average which works best. One can put together a table as follows to compare the Fahrenheit temperatures with the index number, k, above.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-hIO6DFRLKxM/WS_nlE-JhEI/AAAAAAAAHCs/L66qV6dYq3EV_gEcOKcRBe7YHcx8QduyQCLcB/s1600/Newton%2BTemp%2BScale%2B01c1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="152" data-original-width="239" src="https://1.bp.blogspot.com/-hIO6DFRLKxM/WS_nlE-JhEI/AAAAAAAAHCs/L66qV6dYq3EV_gEcOKcRBe7YHcx8QduyQCLcB/s1600/Newton%2BTemp%2BScale%2B01c1.PNG" /></a></div><br />A graphical comparison shows that the logs are fairly linear. Using 66°F for the temperature difference gave the best fit for human body temperature at the lower left of the plot.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-z5rq3dEQfRo/WS_nftE8IsI/AAAAAAAAHCo/YpEbjTZ41UgWkma0JWuZUgx_HUKYZs_aACLcB/s1600/Newton%2BTemp%2BScale%2B01e.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="259" data-original-width="334" height="248" src="https://1.bp.blogspot.com/-z5rq3dEQfRo/WS_nftE8IsI/AAAAAAAAHCo/YpEbjTZ41UgWkma0JWuZUgx_HUKYZs_aACLcB/s320/Newton%2BTemp%2BScale%2B01e.PNG" width="320" /></a></div><br />The slope of the fitted line can be used to convert Farenheit temperatures to points on Newton's scale.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-jvIo6NP8ESw/WTSck3axlAI/AAAAAAAAHDg/5nZ5cFD8D0oWwN5IFOgKsw6lUTYl0zIBwCLcB/s1600/Newton%2BTemp%2BScale%2B01h.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="49" data-original-width="182" src="https://1.bp.blogspot.com/-jvIo6NP8ESw/WTSck3axlAI/AAAAAAAAHDg/5nZ5cFD8D0oWwN5IFOgKsw6lUTYl0zIBwCLcB/s1600/Newton%2BTemp%2BScale%2B01h.PNG" /></a></div><a href="http://en.wikipedia.org/wiki/Newton%27s_law_of_cooling"><br /></a> <a href="http://en.wikipedia.org/wiki/Newton%27s_law_of_cooling">Newton's law of cooling</a> can be in be expressed as the difference between the temperature of an object at some time and the ambient temperature being proportional to an exponential term involving time. This can to shown to be equivalent to the differential form of the law.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-V9JxWxuoaE0/WS_slGvoH1I/AAAAAAAAHDQ/iwRXN1oe1hogcPpnddiAyB6pO6ZEOJXFQCLcB/s1600/Newton%2BTemp%2BScale%2B01f.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="177" data-original-width="174" src="https://1.bp.blogspot.com/-V9JxWxuoaE0/WS_slGvoH1I/AAAAAAAAHDQ/iwRXN1oe1hogcPpnddiAyB6pO6ZEOJXFQCLcB/s1600/Newton%2BTemp%2BScale%2B01f.PNG" /></a></div><br />Supplemental (Jun 1): <a href="http://books.google.com/books?id=VsY5AAAAcAAJ&pg=PA101#v=onepage&q&f=false">Leurechon Thermometer</a> (1627)<br /><br />Supplemental (Jun 2): 65°F gives a better fit for body temperature. Was this the ambient temperature at which the experiments were done? It's doubtful there was a standard temperature yet in Newton's time. For more on the history of early thermometers see <a href="http://books.google.com/books?id=-386AAAAMAAJ&pg=PA1#v=onepage&q&f=false">Bolton, Evolution of the Thermometer, 1592-1743</a>.<br /><br />Supplemental (Jun 2): The average of the freezing point of water and body temperature is (32+98.6)/2= 65.3. Did this originate with <a href="http://books.google.com/books?id=-386AAAAMAAJ&pg=PA38#v=onepage&q&f=false">Accademia del Cimento</a>?<br /><br />Supplemental (Jun 4): Corrected conversion formula for k.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-55920104633379988742017-05-22T20:13:00.001-07:002017-05-22T20:14:05.963-07:00Fermat's Problem in Three Dimensions<br /> Verified the Newton's method works in three dimensions. I choose the four vertices of a tetrahedron as the given points. The Fermat point which makes the sum of the distances from the given point a minimum turned out to be mean of the given points.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-xuYDw80yEYE/WSOnlkfaXOI/AAAAAAAAHCE/k62IYSalsTARSIVXWM37CGmYpcIxyYW-ACLcB/s1600/tetrahedron%2BFermat%2Bpoint%2Banaglyph%2B03a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-xuYDw80yEYE/WSOnlkfaXOI/AAAAAAAAHCE/k62IYSalsTARSIVXWM37CGmYpcIxyYW-ACLcB/s1600/tetrahedron%2BFermat%2Bpoint%2Banaglyph%2B03a.PNG" /></a></div><br />I used Excel to create an anaglyph. You will need red-cyan glasses to view it properly.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-ZR8Vty15FG0/WSOoDVOdwTI/AAAAAAAAHCI/7oCV69F-uzAbZcwvy0-_R5r3LOui3okTgCLcB/s1600/tetrahedron%2BFermat%2Bpoint%2Banaglyph%2B03b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-ZR8Vty15FG0/WSOoDVOdwTI/AAAAAAAAHCI/7oCV69F-uzAbZcwvy0-_R5r3LOui3okTgCLcB/s1600/tetrahedron%2BFermat%2Bpoint%2Banaglyph%2B03b.PNG" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-21739718867612446992017-05-20T18:56:00.002-07:002017-05-21T17:01:49.721-07:00Is the Minimum for the Four Point Fermat Problem Where We Thought?<br /> I've been trying to convince myself that minimum in the four point Fermat problem of the last blog is not slightly displaced from the point c. The plot below shows changes in the sum L=Σℓ<sub>i</sub> of the lengths of the links from the known points to the unknown point x for changes along two lines, u and v through the point (0.700,0.700) in the plane of x.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-hkt8jcWXp3Y/WSDsObZDeuI/AAAAAAAAHB0/L9x3dtUuj5UrS83EF3wuKyxyIIVj0b-HgCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01d.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-hkt8jcWXp3Y/WSDsObZDeuI/AAAAAAAAHB0/L9x3dtUuj5UrS83EF3wuKyxyIIVj0b-HgCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01d.PNG" /></a></div><br />Below 0.700 both lines decrease and increase above this value. The slopes are fairly linear on each side. But it's difficult to be certain that point c is the actual minimum just going by the data because of the discontinuity in the slope. Notice that the angle from horizontal is not the same for both lines. It may be possible for the slope on the right to be decrease also but at a lower rate. But under the circumstances it does look like c is the actual point of intersection for the two line segments.<br /><br />Supplemental (May 21): Obviously we can't use an extension of Newton's method to solve this type of minimum problem since the gradients are not zero at the minimum. Fermat's theory of <a href="http://en.wikipedia.org/wiki/Maxima_and_minima">maxima and minima</a> is not a general theory. When doing searches for curve fits one often encounters local minima that appear to be line segments. This might happen if the minimum is paraboloidal in shape and the contour lines are elliptical.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-52147079016624215732017-05-19T13:46:00.001-07:002017-05-19T13:46:54.238-07:00An Insoluble Fermat Problem for the Method<br /> There's a four point Fermat problem that can't be solved by linearizing the function for the sum of the distances of the unknown point.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-ZtWIbbmkW3s/WR9VRXtfPNI/AAAAAAAAHBE/gsSBKAvNtUMH6PtfVGNDuoW-vzPNvuY7ACLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-ZtWIbbmkW3s/WR9VRXtfPNI/AAAAAAAAHBE/gsSBKAvNtUMH6PtfVGNDuoW-vzPNvuY7ACLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01.PNG" /></a></div><br />If one tries one ends up with division by zero. The gradient of L at point c is not continuous.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-C3EV8X0_og0/WR9WKurStxI/AAAAAAAAHBM/9ZEW-ujqQocRrcBpSDrdQfKSvfMXLjYsACLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-C3EV8X0_og0/WR9WKurStxI/AAAAAAAAHBM/9ZEW-ujqQocRrcBpSDrdQfKSvfMXLjYsACLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01a.PNG" /></a></div><br />The distance function near a point is cone shaped.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-n0DAicLCCn0/WR9W7st3OrI/AAAAAAAAHBU/F1meRyYVGMQ4OXy9NrN9G3fki1BKpCqSQCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-n0DAicLCCn0/WR9W7st3OrI/AAAAAAAAHBU/F1meRyYVGMQ4OXy9NrN9G3fki1BKpCqSQCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01b.PNG" /></a></div><br />The individual gradients are not well behaved near a given point as this plot shows.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-k4TLWF0NkT4/WR9XsYhbcOI/AAAAAAAAHBc/U5RRrCN74JwQVkzRI4jtuY-T-MO1nrCxwCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-k4TLWF0NkT4/WR9XsYhbcOI/AAAAAAAAHBc/U5RRrCN74JwQVkzRI4jtuY-T-MO1nrCxwCLcB/s1600/Insoluble%2BFermat%2Bproblem%2B01c.PNG" /></a></div><br />For a problem like this one can compute the gradient function for two points displaced from the minimum and try to find where two lines in their directions through the chosen points intersect to get a better estimate of the minimum.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-82796267140618369762017-05-18T03:22:00.002-07:002017-05-18T03:30:48.672-07:00An Oversight on the Fermat Point Solution<br /> I just noticed an error in my the solution for the Fermat Point in the last blog but it didn't affect the results. One can solve the f correction equations for dx directly and the normal equations are not needed.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-eWVYgFAnROk/WR10To-nFUI/AAAAAAAAHAg/-Snpwmtp35kO8li88z-ppKWUbaWgjy5wACLcB/s1600/Fermat%2Bpoint%2B01d.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-eWVYgFAnROk/WR10To-nFUI/AAAAAAAAHAg/-Snpwmtp35kO8li88z-ppKWUbaWgjy5wACLcB/s1600/Fermat%2Bpoint%2B01d.PNG" /></a></div><br />One can read f|<sub>x</sub>≠0 as "f evaluated at x is not equal to zero." The normal equations are useful when one has more equations than unknowns which often occurs when one is doing least squares fits. This method might be considered the equivalent of <a href="http://en.wikipedia.org/wiki/Newton%27s_method">Newton's method</a> for finding the zero of an equation in higher dimensions.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-42639610443027544072017-05-17T01:31:00.002-07:002017-05-17T01:35:13.135-07:00Finding the Fermat Point Given Three Arbitrary Points<br /> At the end of a <a href="http://archive.org/stream/oeuvresdefermat01ferm#page/152/mode/2up">letter to Mersenne</a> in about 1640 concerned with finding maxima and minima Fermat proposed this problem:<br /><br /> "Datis tribus punctis, quartum reperire, a quo si ducantur tres rectæ ad data puncta, summa trium harum rectarum sit minima quantitas."<br /><br /> "Given three points, the forth to be found, from which you draw three lines to the given points, the sum of these three lines is to be a minimum quantity."<br /><br />So, given three arbitrary points, and using the method in the previous blogs, we can find the Fermat point as follows. The distances are the ℓ<sub>i</sub> whose sum is to be minimized. Taking the derivative we find for an assumed value of x that dL=f<sup>T</sup>dx where f is the sum of three unit vectors pointing to x. For the position of x for the minimum value of L the change dL has to be zero for arbitrary changes in position, dx, and the only way that this can happen is if f is equal to zero too. But the value of f at the assumed point is not necessarily zero so we look at changes in f with position and see find the value of dx for which f+df=f+Mdx=0. These are the correction equations for f. The matrix M is found by extracting the derivative of the vector function f(x). Using the method of least squares one can show that dx is a solution of the normal equations.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-gYz2xcSLDGk/WRwB92dQxXI/AAAAAAAAG_w/5ccd617E8nEydIL7O3kSZ-f482CK0fOZgCLcB/s1600/Fermat%2Bpoint%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-gYz2xcSLDGk/WRwB92dQxXI/AAAAAAAAG_w/5ccd617E8nEydIL7O3kSZ-f482CK0fOZgCLcB/s1600/Fermat%2Bpoint%2B01.PNG" /></a></div><br />Using the above equations in Excel and repeatedly correcting the value for x we arrive at the Fermat point after just a few iterations.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-6Wx2W_YsOyI/WRwGgzRHgMI/AAAAAAAAG_8/riQUXXgZzqwsIFsl7XWUKry6w8LxznUbQCLcB/s1600/Fermat%2Bpoint%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-6Wx2W_YsOyI/WRwGgzRHgMI/AAAAAAAAG_8/riQUXXgZzqwsIFsl7XWUKry6w8LxznUbQCLcB/s1600/Fermat%2Bpoint%2B01a.PNG" /></a></div><br />Checking the angles between the lines from x to the given points we find they are all 120° which was deduced from the minimum condition.<br /><br />The correction equations for Fermat's problem are simpler than the reflection problem since we do not have a constraint on the change for dx.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-2568891248136460812017-05-16T14:06:00.001-07:002017-05-16T14:30:56.774-07:00Reflection as an Example of the Shortest Path for Light<br /> There's a simpler version of the Steiner Tree Problem and that is Hero's problem of finding the <a href="http://books.google.com/books?id=ry3gAAAAMAAJ&pg=PA325#v=onepage&q&f=false">shortest path</a> for a reflected ray of light. Again, for the general problem, we have the "gradient" equal to the sum of two unit vectors pointing to the unknown point, x. An additional complication is the constraint of the motion of x along a line so that d<b>x=î</b>dx'.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-TR65Vm6SjWU/WRtj9MLrSwI/AAAAAAAAG_U/O4pARms34EsYYsGUf_Q-SV-mRWP_EUmYQCLcB/s1600/Hero%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-TR65Vm6SjWU/WRtj9MLrSwI/AAAAAAAAG_U/O4pARms34EsYYsGUf_Q-SV-mRWP_EUmYQCLcB/s1600/Hero%2B01a.PNG" /></a></div><br />A solution for the reduced normal equations verifies that the angle of incidence equals the angle of reflection.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-wAnZdfCT_oI/WRtlgHqE5oI/AAAAAAAAG_c/mvMQuoAcbCYlzkf5qlabcgEi8bhDNkWgwCLcB/s1600/Hero%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-wAnZdfCT_oI/WRtlgHqE5oI/AAAAAAAAG_c/mvMQuoAcbCYlzkf5qlabcgEi8bhDNkWgwCLcB/s1600/Hero%2B01.PNG" /></a></div><br />Reflecting the second point above the line illustrates Euclid's <a href="http://books.google.com/books?id=CIEAAAAAMAAJ&pg=PA22#v=onepage&q&f=false">Prop. XX</a> in his Elements Bk 1 asserting the sum of any two sides of a triangle is greater than the third or, equivalently, a straight line is the shortest distance between two points.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/--W5FEfefnwI/WRtmbDOdBVI/AAAAAAAAG_g/fcnXyr7hzfUR9PaMIK4zY2T5I3PKLMCVwCLcB/s1600/Hero%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/--W5FEfefnwI/WRtmbDOdBVI/AAAAAAAAG_g/fcnXyr7hzfUR9PaMIK4zY2T5I3PKLMCVwCLcB/s1600/Hero%2B01b.PNG" /></a></div><br />One can see that the triangles in the two problems are similar and the math works out the same.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-29375042313759863992017-05-13T17:23:00.000-07:002017-05-13T17:23:40.236-07:00Solving a Steiner Tree Problem in Excel<br /> Solving a <a href="http://en.wikipedia.org/wiki/Steiner_tree_problem">Steiner tree problem</a> can be challenging but I managed to get Excel to do this using an iterative process for correcting the positions of the unknown points. The problem seeks to find a set of links between a number of points that has a minimal sum for the lengths. One can derive the minimum conditions as follows. The required condition is that the sum of a set of unit vectors toward or away from the unknown points x and y is equal to zero.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-I-doXQieqEM/WRefP-fGh-I/AAAAAAAAG_E/i9cTUr15-d41R1iKkZxr3Z7IOjEB8MDrwCLcB/s1600/Steiner%2Bminimum%2Bcondition%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" src="https://4.bp.blogspot.com/-I-doXQieqEM/WRefP-fGh-I/AAAAAAAAG_E/i9cTUr15-d41R1iKkZxr3Z7IOjEB8MDrwCLcB/s320/Steiner%2Bminimum%2Bcondition%2B01a.PNG" width="320" /></a></div><br />Note that the conditions imply that these unit vectors can be arranged to form the sides of an <a href="http://en.wikipedia.org/wiki/Equilateral_triangle">equilateral triangle</a> making the angles at the points x and y equal to 120°.<br /><br />The conditions for minimum are not linear making them difficult to solve for x and y but we can linearize them by assuming values for x and y and seeking corrections dx and dy for which the above functions are zero.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-Vc7LRnzurUQ/WRec3RonHSI/AAAAAAAAG-w/ZVcqNcMWV7QcKY5Py8kOR7oGgc50w-JXwCLcB/s1600/Steiner%2Btree%2Bsolution%2B02a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-Vc7LRnzurUQ/WRec3RonHSI/AAAAAAAAG-w/ZVcqNcMWV7QcKY5Py8kOR7oGgc50w-JXwCLcB/s1600/Steiner%2Btree%2Bsolution%2B02a.PNG" /></a></div><br />The values for x and y allow one to find the corrections dx and dy which produce the more accurate solutions x' and y'. One nice thing about Excel is that one can use a macro to replace the original values of x and y with the new ones, x' and y', and rapidly recompute them using a shortcut key such as Ctrl-Shift-R.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-76567460030085007832017-05-09T12:48:00.000-07:002017-05-10T17:42:29.668-07:00Binomial Distribution Fit Curvature<br /> If one takes the natural log of the probabilities for the binomial distribution and the fit in the last blog one gets the curves below. The 2nd differences which are a measure of the curvature of the curves are also given. The 2nd differences for the fit are constant as expected.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-lEIpVwl38Mc/WRIcWbXE8oI/AAAAAAAAG-g/hbfXfqM7b8ghJXBTm8hVMOQJz5t4pbd4QCLcB/s1600/binomial%2Bfit%2B01e.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-lEIpVwl38Mc/WRIcWbXE8oI/AAAAAAAAG-g/hbfXfqM7b8ghJXBTm8hVMOQJz5t4pbd4QCLcB/s1600/binomial%2Bfit%2B01e.PNG" /></a></div><br />The relatively large differences at the ends are less critical since they correspond to relatively small values for the probabilities.<br /><br />Supplemental (May 9): The 2nd differences for the normal distribution function are also uniform and equal to -0.04 or 1/λ exactly.<br /><br />Supplemental (May 10): Technically, curvature depends on changes in the tangent of a curve with path length but I think it's fair to say that deviation from a straight line is a form of curvature even if it is not constant. For a parabolic arc the rate of change of the slope with a change in the "horizontal" distance is constant. I got Excel to find a center for the circular arc of the lower curve of the second plot above and radius of curvature turned out to be a little over 25,000. We don't have to worry about units here since both axes are just real numbers.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-56649342186369069152017-05-08T17:58:00.001-07:002017-05-08T19:47:37.026-07:00The Normal Dist. as an Empirical Fit to the Binomial Dist.<br /> The <a href="http://en.wikipedia.org/wiki/Binomial_distribution">binomial distribution</a> is a rather complicated function and the factorials are difficult to deal with so one might be tempted to seek a simpler function that approximates it. The binomial distribution is fairly symmetric about the mean value, μ, and a logarithmic plot reveals an approximate quadratic function so we might try to fit a function of the form,<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-9dOXA2F2PQU/WREMebKw-vI/AAAAAAAAG9g/jOOw50ilDyUxmgq0gG2eX2RGFFU-OFPGACLcB/s1600/binomial%2Bfit%2B01c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-9dOXA2F2PQU/WREMebKw-vI/AAAAAAAAG9g/jOOw50ilDyUxmgq0gG2eX2RGFFU-OFPGACLcB/s1600/binomial%2Bfit%2B01c.PNG" /></a></div><br />This is a discrete probability function and its sum over all values of k is equal to 1 so A is actually a function of β too. We could try a search for β that minimizes the mean square error or find an approximate solution and try to improve on it. The second method gave the following fit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-tNPfl7AW83A/WREO0YNt6oI/AAAAAAAAG9w/xfMBNfAxqNk0ughjGHiFmtvJ4TSEWj24gCLcB/s1600/binomial%2Bfit%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-tNPfl7AW83A/WREO0YNt6oI/AAAAAAAAG9w/xfMBNfAxqNk0ughjGHiFmtvJ4TSEWj24gCLcB/s1600/binomial%2Bfit%2B01.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-SPAKQDZoo8Y/WREO-JzJEUI/AAAAAAAAG90/v2t137LNQ84U0FzexjpZ2M7KNS478VHxwCLcB/s1600/binomial%2Bfit%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-SPAKQDZoo8Y/WREO-JzJEUI/AAAAAAAAG90/v2t137LNQ84U0FzexjpZ2M7KNS478VHxwCLcB/s1600/binomial%2Bfit%2B01a.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-z617XJpEz-Q/WREPGM8thxI/AAAAAAAAG94/bOLUmU0QWFMRPBXBbfRe5oQbDrmwjqpgwCLcB/s1600/binomial%2Bfit%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-z617XJpEz-Q/WREPGM8thxI/AAAAAAAAG94/bOLUmU0QWFMRPBXBbfRe5oQbDrmwjqpgwCLcB/s1600/binomial%2Bfit%2B01b.PNG" /></a></div><br />The first plot compares the fit with the binomial distribution which is quite good in this example. The red points in the second plot show the deviation of the fit from the binomial distribution and the blue points give the same for the normal distribution formula using the expected values <k> for μ and <k<sup>2</sup>> for λ. For most values of p the two error "curves" are nearly equal but for p near 1/2 the least squares fit has lower bounds on the error.<br /><br />Supplemental (May 8): Recomputed the normal distribution error evaluating the erf function at k±0.5 to get:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-bWFihDIL7R8/WRErclzl_II/AAAAAAAAG-Q/xy_kcYeVGB8BjEECA-GMR7fXuzqDBlQugCLcB/s1600/binomial%2Bfit%2B01d1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-bWFihDIL7R8/WRErclzl_II/AAAAAAAAG-Q/xy_kcYeVGB8BjEECA-GMR7fXuzqDBlQugCLcB/s1600/binomial%2Bfit%2B01d1.PNG" /></a></div><br />The curvature of the binomial distribution near the peak and wings may be responsible for the deviations of the two approximation functions.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-73459671958895997432017-04-28T02:58:00.000-07:002017-04-28T02:58:44.764-07:00Interpreting Error Bounds for the Trapazoidal Distribution Fit<br /> I've got a better handle on the error bounds for the fit of a trapazoidal distribution to the deviations of the time of the Equinox. Here's a review and a corrected plot with error bounds showing the expected deviation from expected value for δt.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-P4rv_sZMVHg/WQMESkl_XYI/AAAAAAAAG8w/wplZKGAw_GMIjdZBRWBGeL_qiIjGSUiqQCLcB/s1600/interpreting%2Berr%2Bbnds%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-P4rv_sZMVHg/WQMESkl_XYI/AAAAAAAAG8w/wplZKGAw_GMIjdZBRWBGeL_qiIjGSUiqQCLcB/s1600/interpreting%2Berr%2Bbnds%2B01.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-YdLyBbPGUsw/WQMEXysi0yI/AAAAAAAAG80/T2VYAnb2xmoh2GVojoPpjEwcs1uS-5JtgCLcB/s1600/interpreting%2Berr%2Bbnds%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-YdLyBbPGUsw/WQMEXysi0yI/AAAAAAAAG80/T2VYAnb2xmoh2GVojoPpjEwcs1uS-5JtgCLcB/s1600/interpreting%2Berr%2Bbnds%2B02.PNG" /></a></div><br />The error bounds used the values of the estimate of the deviation, δf*, for the probability densities, obs_f*, for the intervals in the table above. One can get a better understanding of what the error bounds mean by looking at the expected relative frequency, f<sub>i</sub>=n<sub>i</sub>/n, for the intervals chosen. The x values indicate the center of the interval. Using the values of a and b for the fitted trapazoidal distribution we can compare the observed counts with the expected counts, k=nf and their expected rms deviation of the counts, δk=√[nf(1-f)]. The expected variation in the relative frequency will then be δf=√[f(1-f)/n]. But what does all this tell us about the observations themselves? One can look at the terms of the <a href="http://en.wikipedia.org/wiki/Binomial_distribution">binomial distribution</a> with p=f and determine the probability of observing exactly k counts in each interval. Then we can add up the probabilities for those values of k which are within a distance of δk from the expected value for k. The last column on the right shows the probability of this occurring for each interval. A calculation shows the odds aren't uniform for the intervals but equal to 0.6252 ± 0.0375. The probabilities associated with the error bounds are less than those for a normal distribution and fluctuate a little because we are dealing with a discrete probability distribution and taking sum of those values of k between <k>-δk and <k>+δk. One can show that probabilities associated with bounds for a given number of standard deviations in a normal distribution is equal to P(k)=erf(k/√2). So we would expect slightly more observations to be outside the error bounds for the binomial distribution than would be the case for a normal distribution<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-0neQiuHAuQ8/WQMGg4C4MNI/AAAAAAAAG9A/FupLtdYsJYogsRNyg-VfRxa3EqUz3GZHQCLcB/s1600/interpreting%2Berr%2Bbnds%2B03.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-0neQiuHAuQ8/WQMGg4C4MNI/AAAAAAAAG9A/FupLtdYsJYogsRNyg-VfRxa3EqUz3GZHQCLcB/s1600/interpreting%2Berr%2Bbnds%2B03.PNG" /></a></div><br />The binomial distributions for each of the intervals can be plotted together for comparison.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-hlg8ya-m2Yk/WQMQurAmRmI/AAAAAAAAG9Q/Pr1MbP3Il_kItXAIlUJ7wI_S8NrGsa8AwCLcB/s1600/interpreting%2Berr%2Bbnds%2B04.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-hlg8ya-m2Yk/WQMQurAmRmI/AAAAAAAAG9Q/Pr1MbP3Il_kItXAIlUJ7wI_S8NrGsa8AwCLcB/s1600/interpreting%2Berr%2Bbnds%2B04.PNG" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-88557887165306997062017-04-24T01:56:00.000-07:002017-04-25T14:13:52.103-07:00A Fit of the Equinox Deviations Using Expected Values<br /> I got a little bogged down with some technical details associated with copying formulas from one Excel worksheet to another. It's a little annoying when Excel crashes, restarts and you have to redo the stuff you haven't saved. It may have been how an error crept into one of my previous pages. You literally lose track of what you are doing. One needs to constantly check one's formulas and trace dependencies when transferring material from one page to another.<br /><br />I used the expected value formulas to do the fit for the deviations of the Equinox times. The results were similar. The frequencies that I've been using were relative frequencies defined in as the ratio of counts for an interval to the total count. One can also define the function f as a probability density or probability per unit interval. I had to use this definition to get the fit to work properly for the Equinox times. The value of f here is the previous value divided by the width of the interval dx.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-z6gN3bOQUfY/WP23mqfF_9I/AAAAAAAAG7s/BFciQteeqCQzowiDgVrT5fraSDNyJrgbQCLcB/s1600/equinox%2Bexp%2Bvalue%2Bfit%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-z6gN3bOQUfY/WP23mqfF_9I/AAAAAAAAG7s/BFciQteeqCQzowiDgVrT5fraSDNyJrgbQCLcB/s1600/equinox%2Bexp%2Bvalue%2Bfit%2B01.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-u01tEDzMtI4/WP234cYcAPI/AAAAAAAAG7w/gSCmDY1QkCY15JbVmbapjUwt2WlmzOtyQCLcB/s1600/equinox%2Bexp%2Bvalue%2Bfit%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-u01tEDzMtI4/WP234cYcAPI/AAAAAAAAG7w/gSCmDY1QkCY15JbVmbapjUwt2WlmzOtyQCLcB/s1600/equinox%2Bexp%2Bvalue%2Bfit%2B01a.PNG" /></a></div><br />The error bounds are nominal in the sense that they are typical of the observed variations for a trapazoidal distribution. The fit values for the trapazoidal distribution are a=4.470 min and b=15.089.<br /><br />Supplemental (Apr 24): The trapazoidal distribution has an interesting series for formulas for its expected values. The pattern holds for higher powers of x. Technically this might be called a folded trapazoidal distribution since the probabilities for the positive and negative values of x are combined.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-UpQfqqFR3vA/WP3Ap-uVMEI/AAAAAAAAG8A/OW6hA_sPPgksIMxBKv7Gi4mGIY_RMVutwCLcB/s1600/trapazoidal%2Bexpected%2Bvalue%2Bformulas%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-UpQfqqFR3vA/WP3Ap-uVMEI/AAAAAAAAG8A/OW6hA_sPPgksIMxBKv7Gi4mGIY_RMVutwCLcB/s1600/trapazoidal%2Bexpected%2Bvalue%2Bformulas%2B01.PNG" /></a></div><br />Supplemental (Apr 25): The variations in the relative frequencies are scaled down versions of the expected variation in the counts for an interval as this derivation shows.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-oM0Lqdu8YUw/WP-pqHI_lmI/AAAAAAAAG8Q/JGPOtdg_4z0hJ7RymakjmzR0eu0PpjsVwCLcB/s1600/frequency%2Bvariation%2Bformula%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-oM0Lqdu8YUw/WP-pqHI_lmI/AAAAAAAAG8Q/JGPOtdg_4z0hJ7RymakjmzR0eu0PpjsVwCLcB/s1600/frequency%2Bvariation%2Bformula%2B01.PNG" /></a></div><br />In evaluating f and δf in the table above I used the observed values for the interval's density, obs_f (=n<sub>i</sub>/n/dx), as an approximation. It was intended as a check of the trapazoidal density formula whose maximum value is 2/(a+b)=0.1023. Using the same letter for the relative frequencies of the counts and the probability density formula may have be a little too confusing. So the error bounds in the plot are a little too large. Using f* for the density the correct formula for the expected rms error in the density would be as follows with Δx=dx.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-0JLDVicmAQc/WP-57KwFqGI/AAAAAAAAG8g/gTn-wxuamyU4MARbtfb6s5TOmT-XjatjgCLcB/s1600/frequency%2Bvariation%2Bformula%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-0JLDVicmAQc/WP-57KwFqGI/AAAAAAAAG8g/gTn-wxuamyU4MARbtfb6s5TOmT-XjatjgCLcB/s1600/frequency%2Bvariation%2Bformula%2B02.PNG" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-78738170036742725432017-04-21T00:34:00.002-07:002017-04-21T00:37:24.763-07:00Using Estimated Expected Values to Fit a Trapazoidal Distribution<br /> From the definition of the trapazoidal distribution and its integral one can obtain formulas for the expected values ⟨x⟩ and ⟨x<sup>2</sup>⟩.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-tYxPy0jbOSE/WPmu_yV4atI/AAAAAAAAG7I/_JXLSAnBTp0pFC7bkDibPoGS9as7RsfpQCLcB/s1600/trapazoidal%2Bfit%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-tYxPy0jbOSE/WPmu_yV4atI/AAAAAAAAG7I/_JXLSAnBTp0pFC7bkDibPoGS9as7RsfpQCLcB/s1600/trapazoidal%2Bfit%2B01.PNG" /></a></div><br />Then given a set of random numbers from a trapazoidal generator one can analyze the set by counting the number of values that fall within chosen intervals and then estimate the distribution for the intervals and the expected values ⟨x⟩ and ⟨x<sup>2</sup>⟩.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-hn1CSayJ4wY/WPmwn1qHBuI/AAAAAAAAG7U/MRPJLXQU6dUGflbjpzsDnUhW-mNsReqDACLcB/s1600/trapazoidal%2Bfit%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-hn1CSayJ4wY/WPmwn1qHBuI/AAAAAAAAG7U/MRPJLXQU6dUGflbjpzsDnUhW-mNsReqDACLcB/s1600/trapazoidal%2Bfit%2B01a.PNG" /></a></div><br />We now have two equations which can be solved for a and b which can be used to fit the observations and compare the results with the original values of a and b. In the example below the original values were a=0.5 and b=1.0 and the fit values were a=0.538 and b=0.993.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-IWz2e19_feI/WPmxb3w77zI/AAAAAAAAG7c/D4YCgNDvmCIHRluvm_ejnemnebuCQhNzQCLcB/s1600/trapazoidal%2Bfit%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-IWz2e19_feI/WPmxb3w77zI/AAAAAAAAG7c/D4YCgNDvmCIHRluvm_ejnemnebuCQhNzQCLcB/s1600/trapazoidal%2Bfit%2B01b.PNG" /></a></div><br />This was a lot easier to do since the equation for ⟨x⟩ can be transformed into a quadratic function of ρ=a/b whose solution can be found if a set of values are assumed for b. The set of values for ρ can then be used to find a zero of the expression for ⟨x<sup>2</sup>⟩.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-46163358650982240262017-04-19T12:31:00.000-07:002017-04-19T12:31:09.102-07:00Why the Least Squares Fit Failed<br /> The least squares fit failed primarily because it favored the majority at the expense of a minority. The histogram cells closer to the mean time had the highest probabilities while those beyond the value of b had zero probability. The result was that the data for the last cell on the right could be ignored when computing the rms error of a trapazoidal distribution if the b value was too small.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-7VyCMd2gwvI/WPe3OJappTI/AAAAAAAAG6s/wxwTLO4fEdMf5NvWbgzbZqtpNuVbZxa2QCLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03_.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-7VyCMd2gwvI/WPe3OJappTI/AAAAAAAAG6s/wxwTLO4fEdMf5NvWbgzbZqtpNuVbZxa2QCLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03_.PNG" /></a></div><br />The additional constraint for the minimum maximum magnitude of the z-scores for the probability distribution assured that an unlikely situation would not occur. The probability of a histogram interval was found by using the difference of the integral of the trapazoidal distribution of its upper and lower bounds.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-IdIf5wuuxF0/WPe5NWrKrxI/AAAAAAAAG64/fqFNZJX9Qsg_xBalwJoGzfgrTKMkHX0yACLcB/s1600/least%2Bsquares%2Bfailure%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-IdIf5wuuxF0/WPe5NWrKrxI/AAAAAAAAG64/fqFNZJX9Qsg_xBalwJoGzfgrTKMkHX0yACLcB/s1600/least%2Bsquares%2Bfailure%2B01.PNG" /></a></div><br />When setting bounds for curve fits one has to make certain that significant data is not ignored.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-53432654593128654372017-04-19T04:06:00.000-07:002017-04-19T04:06:12.875-07:00A Better Trapazoid Fit For the Equinox Time Deviations<br /> The trapazoidal distribution fit for MICA's Spring Equinox time deviations proved to be a little difficult. The b values were difficult to fit since they favored lower values at the expense of large z-scores for the last interval of the histogram. I tried minimizing the maximum absolute value of the z-scores while minimizing the rms error and got what appears to be a better fit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-qSKU8P_8mC8/WPdC3dOcfUI/AAAAAAAAG6c/cQo5Ly0h_coGXhC7sMQk80kUPIMF5ZweACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03_.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-qSKU8P_8mC8/WPdC3dOcfUI/AAAAAAAAG6c/cQo5Ly0h_coGXhC7sMQk80kUPIMF5ZweACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03_.PNG" /></a></div><br />Here are some statistics for the fit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-CspE3_Jitqg/WPdB4Sa19VI/AAAAAAAAG6Q/_JLC5UOMhpUsjijitqjy9pJCsrckXWCVgCLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-CspE3_Jitqg/WPdB4Sa19VI/AAAAAAAAG6Q/_JLC5UOMhpUsjijitqjy9pJCsrckXWCVgCLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B03a.PNG" /></a></div><br />Judging by the z-scores its a marginal trapazoidal distribution at best.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-4494162499283584472017-04-18T01:21:00.000-07:002017-04-19T13:47:11.724-07:00Trapazoid Fit For the Equinox Time Deviations<br /> Just got through doing a rough trapazoidal fit of the deviations in the time of the Spring Equinox.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-De9nypOgucA/WPXJt6q_WTI/AAAAAAAAG5o/3YoNV-tHh54_tyOv_d1InrW2XOgSZ5XzACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B00.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-De9nypOgucA/WPXJt6q_WTI/AAAAAAAAG5o/3YoNV-tHh54_tyOv_d1InrW2XOgSZ5XzACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B00.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-b9nzfp8b53E/WPXPQRudiXI/AAAAAAAAG58/CgM2p_MTRF0kbOxII7gCTHDJZHx1GLKGACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-b9nzfp8b53E/WPXPQRudiXI/AAAAAAAAG58/CgM2p_MTRF0kbOxII7gCTHDJZHx1GLKGACLcB/s1600/Equinox%2Bdev%2Btrap%2Bfit%2B02.PNG" /></a></div><br />This fit uses the MICA times of the 251 Spring Equinoxes from 1800 through 2050. We probably shouldn't take the trapazoidal distribution too seriously but it may be wise to keep an open mind about the actual shape of the distribution. The minimum error was for a=2.05 min and b=15.15 min. The statistics hint that a slight deviation from the mean time is the most probable situation.<br /><br />Supplemental (Apr 19): This fit ended up somewhat off the mark due to an error in evaluating the probability for the histogram intervals. I found the error in data used last night and corrected it in the next blog. Here too I had problems in choosing a good value for b.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-58722412954707707942017-04-17T19:08:00.000-07:002017-04-17T19:22:28.270-07:00Trapazoid Fit Video<br /> I did a video to show the trapazoid fit process in action. I was able to compensate for Google's processing somewhat but not entirely. Excel recalculates the worksheet if a cell's content is changed to moving a cell about is an easy way to recalculate the worksheet. Each time the selected box is moved the worksheet computes 1000 random trapazoidal numbers, does the data analysis and computes a fit for the data.<br /><br /><div class="separator" style="clear: both; text-align: center;"><object width="320" height="266" class="BLOG_video_class" id="BLOG_video-6932d022189ebf91" classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,40,0"><param name="movie" value="https://www.youtube.com/get_player"><param name="bgcolor" value="#FFFFFF"><param name="allowfullscreen" value="true"><param name="flashvars" value="flvurl=https://redirector.googlevideo.com/videoplayback?id%3D6932d022189ebf91%26itag%3D5%26source%3Dblogger%26requiressl%3Dyes%26app%3Dblogger%26cmo%3Dsecure_transport%3Dyes%26cmo%3Dsensitive_content%3Dyes%26ip%3D0.0.0.0%26ipbits%3D0%26expire%3D1500620587%26sparams%3Dip,ipbits,expire,id,itag,source,requiressl%26signature%3D5BA7B0A34700C6E2ED842EA8F7E0A5A232BD6830.48391A2E34304EFD3D21CCDDFB0C51735AA5CEB0%26key%3Dck2&iurl=http://video.google.com/ThumbnailServer2?app%3Dblogger%26contentid%3D6932d022189ebf91%26offsetms%3D5000%26itag%3Dw160%26sigh%3DQxeqEir585ke6VPeI-fz5Py2yt0&autoplay=0&ps=blogger"><embed src="https://www.youtube.com/get_player" type="application/x-shockwave-flash" width="320" height="266" bgcolor="#FFFFFF" flashvars="flvurl=https://redirector.googlevideo.com/videoplayback?id%3D6932d022189ebf91%26itag%3D5%26source%3Dblogger%26requiressl%3Dyes%26app%3Dblogger%26cmo%3Dsecure_transport%3Dyes%26cmo%3Dsensitive_content%3Dyes%26ip%3D0.0.0.0%26ipbits%3D0%26expire%3D1500620587%26sparams%3Dip,ipbits,expire,id,itag,source,requiressl%26signature%3D5BA7B0A34700C6E2ED842EA8F7E0A5A232BD6830.48391A2E34304EFD3D21CCDDFB0C51735AA5CEB0%26key%3Dck2&iurl=http://video.google.com/ThumbnailServer2?app%3Dblogger%26contentid%3D6932d022189ebf91%26offsetms%3D5000%26itag%3Dw160%26sigh%3DQxeqEir585ke6VPeI-fz5Py2yt0&autoplay=0&ps=blogger" allowFullScreen="true" /></object></div><br />You can pause the video to study a particular fit. It helps to zoom in a little too.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-70499094446341383252017-04-17T12:57:00.000-07:002017-04-17T13:05:54.655-07:00The Trapazoidal Distribution<br /> Over the weekend I've been studying the <a href="http://en.wikipedia.org/wiki/Trapezoidal_distribution">Trapazoidal Distribution</a>. One only needs two numbers a and b to define this distribution. Its height, h, can be found since the area under the curve equals 1.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-jAjSlnc_e6o/WPUX_n6NbEI/AAAAAAAAG40/BqACbrmM65UWqCEtcFHKciXdPXWEnEaVQCLcB/s1600/trapazoidal%2Bdistribution%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-jAjSlnc_e6o/WPUX_n6NbEI/AAAAAAAAG40/BqACbrmM65UWqCEtcFHKciXdPXWEnEaVQCLcB/s1600/trapazoidal%2Bdistribution%2B02.PNG" /></a></div><br />I wrote an Excel user function to generate 1000 random numbers that fit this distribution with a=0.5 and b=1 then determined the observed frequencies for the following set of intervals.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-9jnkp1hvmoo/WPUZ4xnkRbI/AAAAAAAAG5I/Y7rkzkIRG-8pm4bx6KoiCdBtweQePQfpgCLcB/s1600/trapazoidal%2Bdistribution%2B02b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-9jnkp1hvmoo/WPUZ4xnkRbI/AAAAAAAAG5I/Y7rkzkIRG-8pm4bx6KoiCdBtweQePQfpgCLcB/s1600/trapazoidal%2Bdistribution%2B02b.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-RrqoyYCoXf0/WPUZS8fhFNI/AAAAAAAAG5A/GI8fYlp3OcQwTAEa50-PThK2WsMGtnxGgCLcB/s1600/trapazoidal%2Bdistribution%2B02a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-RrqoyYCoXf0/WPUZS8fhFNI/AAAAAAAAG5A/GI8fYlp3OcQwTAEa50-PThK2WsMGtnxGgCLcB/s1600/trapazoidal%2Bdistribution%2B02a.PNG" /></a></div><br />To check the trapazoidal random number generator it was necessary to determine the distribution which best fit the data. To do this a search was necessary to find the values for a and b which minimized the root mean square error for the fit. Since the trapazoidal distribution is symmetric it can be folded horizontally to reduce the number of intervals needed in the calculation.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-UJq3Hre0bRE/WPUbr7RHqFI/AAAAAAAAG5U/phIZRoB4aSg7b1kguwa8VyPkNQaEgL1fwCLcB/s1600/trapazoidal%2Bdistribution%2B02d.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-UJq3Hre0bRE/WPUbr7RHqFI/AAAAAAAAG5U/phIZRoB4aSg7b1kguwa8VyPkNQaEgL1fwCLcB/s1600/trapazoidal%2Bdistribution%2B02d.PNG" /></a></div><br />The fit turned out to be fairly good.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-YP86rf2uZu8/WPUcNMSMhVI/AAAAAAAAG5Y/A7A_yyy4v7IeF205NlIYncK0IqzMwAeggCLcB/s1600/trapazoidal%2Bdistribution%2B02c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-YP86rf2uZu8/WPUcNMSMhVI/AAAAAAAAG5Y/A7A_yyy4v7IeF205NlIYncK0IqzMwAeggCLcB/s1600/trapazoidal%2Bdistribution%2B02c.PNG" /></a></div><br />Above the blue dots represent the observed frequencies and the blue line is the fit.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-35405846509573023342017-04-13T13:37:00.000-07:002017-04-13T13:37:12.239-07:00Getting Control of Excel's Histogram Intervals<br /> I've solved the problem I've been having with setting the intervals on Excel's histograms. I tried adjusting the bounds by setting the Overflow and Underflow bins to ±18.0 but they were ignored. Apparently, these bounds need to be less than the maximum and minimum for the data series used for the histogram. It's debatable whether or not one needs empty columns at the ends of a histogram. These entries set up the histogram in an acceptable manner.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-ZcSGkX_PCjg/WO_bgfY8ncI/AAAAAAAAG38/WaHyJBn94y0CKyLJyJOxJSi8KDfjfUBLACLcB/s1600/Equinox%2Bhistograms%2B01d.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-ZcSGkX_PCjg/WO_bgfY8ncI/AAAAAAAAG38/WaHyJBn94y0CKyLJyJOxJSi8KDfjfUBLACLcB/s1600/Equinox%2Bhistograms%2B01d.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Pdw80IAoQ_U/WO_dBVCWG3I/AAAAAAAAG4I/I1-bQg6GToccdO6-wIc3i-raH3ZJ1n_LACLcB/s1600/Equinox%2Bhistograms%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-Pdw80IAoQ_U/WO_dBVCWG3I/AAAAAAAAG4I/I1-bQg6GToccdO6-wIc3i-raH3ZJ1n_LACLcB/s1600/Equinox%2Bhistograms%2B01a.PNG" /></a></div><br />Again, the observed counts from the histogram can be used for comparison with the statistically expected counts for a normal distribution.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-OiDo0QxFK3M/WO_dq2BGdHI/AAAAAAAAG4Q/EPw-EkWbDGwd2NSPQ_2LWcu7B-hguryGwCLcB/s1600/Equinox%2Bhistograms%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-OiDo0QxFK3M/WO_dq2BGdHI/AAAAAAAAG4Q/EPw-EkWbDGwd2NSPQ_2LWcu7B-hguryGwCLcB/s1600/Equinox%2Bhistograms%2B01b.PNG" /></a></div><br />The distribution above looks similar to a normal distribution and the z-scores for the intervals can't be rejected on statistical grounds but one cannot always get a clear impression of the actual distribution for deviations from a mean value with a single dataset. In another histogram the distribution appears to be more like a <a href="http://en.wikipedia.org/wiki/Trapezoid">trapazoid</a>.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-hKcQP7WFYeY/WO_fuWMFYII/AAAAAAAAG4g/BgEB9jC_xksNGgIalgyqBDLdZNJ2cvAAQCLcB/s1600/Equinox%2Bhistograms%2B01c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-hKcQP7WFYeY/WO_fuWMFYII/AAAAAAAAG4g/BgEB9jC_xksNGgIalgyqBDLdZNJ2cvAAQCLcB/s1600/Equinox%2Bhistograms%2B01c.PNG" /></a></div><br />We would need more data to come to a conclusion about what the actual distribution for the Equinox times is.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-38125551281420467892017-04-10T03:24:00.000-07:002017-04-10T13:36:46.885-07:00Using the Error Function to Evaluate Expected Values<br /> Excel's <a href="http://en.wikipedia.org/wiki/Error_function">error function</a>, ERF(x), can be used to evaluate the probabilities for the intervals in a histogram and their expected values. The probability integral, p, below is simplified by the use of a <a href="http://en.wikipedia.org/wiki/Standard_score">z-score</a> eliminating the standard deviation found in a <a href="http://en.wikipedia.org/wiki/Normal_distribution">normal probability distribution</a>. The error function is simplified even further since the factor of 1/2 is missing from the exponential so its arguments require the z-scores a and b of the interval to be divided by the square root of 2. This can be done for the deviations for the times of the Spring Equinoxes. Formulas for <a href="http://en.wikipedia.org/wiki/Binomial_distribution">binomial distribution</a> can be use to compute the expected value, k=np, for an interval once the probability is known and also the deviation, δk, for the expected value. δk<sup>2</sup>=np(1-p). The z-scores for the observed counts is found in the last column and gives us a measure of how good the counts for the intervals are. One would expect nearly all the observations to be within 3z for the interval as is the case.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-AfNsitZmfs4/WOtX2IbIaNI/AAAAAAAAG3o/68ERd9Yep3ophRdzzlKoayTm-P4WEHCLACLcB/s1600/deviation%2Bstatistics%2B03a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-AfNsitZmfs4/WOtX2IbIaNI/AAAAAAAAG3o/68ERd9Yep3ophRdzzlKoayTm-P4WEHCLACLcB/s1600/deviation%2Bstatistics%2B03a.PNG" /></a></div><br />So even though the Earth's orbit deviates in a complicated spiral motion from the Keplerian ellipse, the observed deviations time of the Equinoxes appears to be random.<br /><br />Supplemental (Apr 10): Another way of looking at the analysis above is that we compared the distribution for the deviations in time with a normal distribution. The z-scores in the last column indicate the central peak is somewhat flattened and the "outliers" at the ends are under represented in our sample. I had a part-time job in the Physics Shop while attending university and one of my teachers used a <a href="http://en.wikipedia.org/wiki/Bean_machine">bean machine</a> to study the statistics. He said that the distributions he was getting were not what was expected. We considered the possibility that the balls might be gaining a little momentum as they fell and were more likely to bounce off the pins and move outwards or the pins may get bent after prolonged use. But rare events are less likely to be represented in a sample and the peak may be more likely to lose its balls than gain them if the total number of balls is small.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-50896958154151902522017-04-07T19:30:00.000-07:002017-04-08T11:18:46.885-07:00Another Look at the Spring Equinox Statistics<br /> I redid the statistics for the deviations from fit formula and got something a little more normal using fewer intervals<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-qdRtsosxcSk/WOhIP-lUwNI/AAAAAAAAG20/CG0AyIp6vg4DYw8ebNOlEarT-z7wr6YpgCLcB/s1600/deviation%2Bstatistics%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-qdRtsosxcSk/WOhIP-lUwNI/AAAAAAAAG20/CG0AyIp6vg4DYw8ebNOlEarT-z7wr6YpgCLcB/s1600/deviation%2Bstatistics%2B02.PNG" /></a></div><br />Microsoft's histograms are a little difficult to do. For this one I set the underflow and overflow values at -16 and 16 respectively and used 6.4 for the interval width. One can compare the observed values with the expected values and their standard deviations.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-tpGyiESBlX0/WOhJDSz8v-I/AAAAAAAAG28/J-Wi6J560csu8G4xsBH7VoSPk1q_ofO7wCLcB/s1600/deviation%2Bstatistics%2B02a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-tpGyiESBlX0/WOhJDSz8v-I/AAAAAAAAG28/J-Wi6J560csu8G4xsBH7VoSPk1q_ofO7wCLcB/s1600/deviation%2Bstatistics%2B02a.PNG" /></a></div><br />One would expect most of the observed values to be within 3 standard deviations of the expected values. Under the circumstances one cannot rule out a normal distribution for the deviations.<br /><br />Supplemental (Apr 8): I used numerical integration to evaluate the probability integrals for the expected values and their deviations. A more accurate calculation shows that for the 16< interval the expected value is 1.6.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-37430815011542853092017-04-06T23:50:00.000-07:002017-04-07T12:18:06.635-07:00The Mean Time of the Spring Equinox<br /> I converted the MICA times for the Spring Equinox between 1800 and 2050 to Julian Dates and got a linear equation for the fit. A plot of the deviations appears to be fairly random with a maximum deviation of about 16 minutes.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-EJRwoqTr5V0/WOc53pQv5NI/AAAAAAAAG2M/qNEJdTHl4e8i9w5nfY8bALn7hl7KfzH3QCLcB/s1600/deviation%2Bfrom%2Bmean%2BEquinox%2B01b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-EJRwoqTr5V0/WOc53pQv5NI/AAAAAAAAG2M/qNEJdTHl4e8i9w5nfY8bALn7hl7KfzH3QCLcB/s1600/deviation%2Bfrom%2Bmean%2BEquinox%2B01b.PNG" /></a></div><br />The "resonance" seems to have disappeared.<br /><br />Supplemental (Apr 7): The probability distribution for the deviations from the linear fit doesn't appear to be normal. Its peak appears to be somewhat flattened.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-0haLGYZvwk8/WOflnoEXEAI/AAAAAAAAG2k/RhJQE7tUp9Ath2UuLjMDMOAwj9XMixL7gCLcB/s1600/deviation%2Bstatistics%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-0haLGYZvwk8/WOflnoEXEAI/AAAAAAAAG2k/RhJQE7tUp9Ath2UuLjMDMOAwj9XMixL7gCLcB/s1600/deviation%2Bstatistics%2B01.PNG" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-1771315560450745002017-04-05T10:52:00.000-07:002017-04-05T10:52:21.400-07:00Keplerian Ellipse Apsides Needed For Calculating Positions<br /> I tried using the apsides from MICA's 2017 positions for the Earth to calculate positions on the Keplerian ellipse but got a 0.02 radian difference between the curves for MICA's radii and the ellipse radii. Using apsides based on the fit brought both curves together.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-SSIwJdiTCAg/WOUtsbVXn1I/AAAAAAAAG1Y/X5iZNSFIXbcLLeLZtrJ3IovkfRBJDzAnQCLcB/s1600/Perihelion%2B%2526%2BAphelion%2B04.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-SSIwJdiTCAg/WOUtsbVXn1I/AAAAAAAAG1Y/X5iZNSFIXbcLLeLZtrJ3IovkfRBJDzAnQCLcB/s1600/Perihelion%2B%2526%2BAphelion%2B04.PNG" /></a></div><br />These angular separation between these two apsides was very close to π as it should be.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-79379847171559539762017-04-05T00:57:00.001-07:002017-04-06T00:21:34.178-07:00Deviations in the Orbital Plane<br /> One can plot the deviations of MICA's Earth positions from the best fitting Keplerian ellipse positions. The result is rather curious. The motion appears to be that of compounded spirals. The smaller cycles are due to the Moon's motion.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-XzUhKgCmDQg/WOSi7G4apLI/AAAAAAAAG1I/PxgMxMlAZbgtCJ9zvllPsk6Bd_0r1uAlgCLcB/s1600/deviation%2Bin%2Borbital%2Bplane%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-XzUhKgCmDQg/WOSi7G4apLI/AAAAAAAAG1I/PxgMxMlAZbgtCJ9zvllPsk6Bd_0r1uAlgCLcB/s1600/deviation%2Bin%2Borbital%2Bplane%2B01a.PNG" /></a></div><br />The blue dot at the end point marks the last data point.<br /><br />Supplemental (Apr 6): A 0.0002 AU distance deviation along the ecliptic path of the Sun is enough to account for the observed 16 min deviation in the time of the Spring Equinox.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-CTvBj_CFiC4/WOXsFSjAPZI/AAAAAAAAG1s/Rya9FdKKFBc8qhbNZNwXHj84xOg2yyTXgCLcB/s1600/time%2Bdiff%2Bcalc%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-CTvBj_CFiC4/WOXsFSjAPZI/AAAAAAAAG1s/Rya9FdKKFBc8qhbNZNwXHj84xOg2yyTXgCLcB/s1600/time%2Bdiff%2Bcalc%2B01.PNG" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-49812253708271258262017-04-04T11:18:00.001-07:002017-04-04T12:26:34.382-07:00More on the Shifted Time of Spring Equinox<br /> The formula to the shift in time of the Spring Equinox could be clarified somewhat. The Sun has a mean position and mean path. The Moon and other bodies can shift the Earth's position a little and this will affect the apparent position of the Sun which travels parallel to and at the same speed as the mean Sun. So for some arbitrary deviation, δ, relative to the mean position we can calculate the extra distance the Sun needs to travel in order to cross the Celestial Equator and the maximum value.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-OxvMJWu7oOs/WOPhJXBNONI/AAAAAAAAG0o/1NMP84XhLL8ppojoquQt_hQw7A569NtggCLcB/s1600/time%2Bdeviation%2B03a.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-OxvMJWu7oOs/WOPhJXBNONI/AAAAAAAAG0o/1NMP84XhLL8ppojoquQt_hQw7A569NtggCLcB/s1600/time%2Bdeviation%2B03a.png" /></a></div><br />Note that the error is magnified by small angles of inclination for the path. As stated previously the maximum error is for θ = 0 or a deviation perpendicular to the plane of the Celestial Equator. Of course when the deviated position is above the Celestial Equator the Sun will cross before the Mean Sun does.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0