tag:blogger.com,1999:blog-4946003991896518492017-08-16T09:06:43.418-07:00httprover's 2nd blogJim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.comBlogger859125tag:blogger.com,1999:blog-494600399189651849.post-4848586857698379462017-08-13T17:55:00.001-07:002017-08-13T19:13:29.529-07:00The Sun's Apparent Motion in the Plane of the Ecliptic<br /> I redid the series of fits for the Sun's apparent position this time in the plane of the Ecliptic. The primary motion is of course a Keplerian ellipse. The horizontal and vertical axes are the major and minor axes and at the beginning of the year the Sun in near perigee on the right and moves upwards. The units are AU.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-lSsWQdqjjgk/WZDtt-CB23I/AAAAAAAAHLg/glxI-JonUoojU0WlzifDICJjjsqqAYiEgCLcBGAs/s1600/Sun%2B3d%2Bfit%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="321" data-original-width="349" src="https://1.bp.blogspot.com/-lSsWQdqjjgk/WZDtt-CB23I/AAAAAAAAHLg/glxI-JonUoojU0WlzifDICJjjsqqAYiEgCLcBGAs/s1600/Sun%2B3d%2Bfit%2B01.PNG" /></a></div><br />The Keplerian elements for the fit are as follows.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-0wVvYpQ98pk/WZDveTV_k3I/AAAAAAAAHLs/qbngyiem8GgEmL5Fgt42IZwWwA0Tds-SQCLcBGAs/s1600/Sun%2B3d%2Bfit%2B01a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="109" data-original-width="258" src="https://4.bp.blogspot.com/-0wVvYpQ98pk/WZDveTV_k3I/AAAAAAAAHLs/qbngyiem8GgEmL5Fgt42IZwWwA0Tds-SQCLcBGAs/s1600/Sun%2B3d%2Bfit%2B01a.PNG" /></a></div><br />The residuals of this fit form a <a href="http://en.wikipedia.org/wiki/Rose_(mathematics)">rose curve</a> which appears to be due to solar pulls and <a href="http://en.wikipedia.org/wiki/Torque">torques</a> acting on the Moon's orbit. Again the units are AU.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-u-dQch4codk/WZDxPMuEY2I/AAAAAAAAHL4/sWHvTztqR3YOfAZFcVxZ6K0BBeWDKJ1JwCLcBGAs/s1600/Sun%2B3d%2Bfit%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="350" data-original-width="371" src="https://4.bp.blogspot.com/-u-dQch4codk/WZDxPMuEY2I/AAAAAAAAHL4/sWHvTztqR3YOfAZFcVxZ6K0BBeWDKJ1JwCLcBGAs/s1600/Sun%2B3d%2Bfit%2B02.PNG" /></a></div><br />The residuals of the second fit are down to μAUs and more random in appearance. There's an odd step in the direction of increasing perigee at the end of the year. Could the Earth be slowing down and spending more time at perihelion? What effect would that have on global warming?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-usBjg11EZcE/WZDyPsaqaBI/AAAAAAAAHME/1bxESzpXtLsGcy78rg1-gVTxEiE7PB63ACLcBGAs/s1600/Sun%2B3d%2Bfit%2B03.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="290" data-original-width="482" src="https://1.bp.blogspot.com/-usBjg11EZcE/WZDyPsaqaBI/AAAAAAAAHME/1bxESzpXtLsGcy78rg1-gVTxEiE7PB63ACLcBGAs/s1600/Sun%2B3d%2Bfit%2B03.PNG" /></a></div><br />Fits can produce some deviations when all the error isn't accounted for.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-68957379765378698712017-08-09T01:23:00.002-07:002017-08-09T01:51:15.586-07:00Found an Odd Error for a Fit of the Sun's Position<br /> I was preparing for the solar eclipse later this month, doing a fit of the Sun's relative position from the center of the Earth, and the fit didn't turn out as I expected. A linear least squares fit of the Sun's position for the functions indicated resulted in the following relative differences between the fit and the calculated positions.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-4wgS4xnBXQ4/WYq_IMyTcfI/AAAAAAAAHK0/1UQwPgBEROQFVsmeGX5CLxp-6F0E5QGmACLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="302" data-original-width="405" src="https://2.bp.blogspot.com/-4wgS4xnBXQ4/WYq_IMyTcfI/AAAAAAAAHK0/1UQwPgBEROQFVsmeGX5CLxp-6F0E5QGmACLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B01.PNG" /></a></div><br />The sinusoidal function is mainly due to the Moon's pull on the Earth but there appears to be another component present. How can one explain the displacement of mean error from zero? It turns out there are some <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomials#First_kind">Chebyshev polynomials</a> present.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-GO20zX95Zps/WYrAcl-96xI/AAAAAAAAHLA/8ndb8nM-KuE9KLDQ1oyfOE3wxX64apXLACLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="210" data-original-width="155" src="https://4.bp.blogspot.com/-GO20zX95Zps/WYrAcl-96xI/AAAAAAAAHLA/8ndb8nM-KuE9KLDQ1oyfOE3wxX64apXLACLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B02.PNG" /></a></div><br />This polynomial series generates a displacement of the following form given in AU.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-qrWf2ABV8F4/WYrBQhXwuJI/AAAAAAAAHLI/TEm5dGpZwLMsEzCPwhdVtGE98ePQ6u2RgCLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B03.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="261" data-original-width="420" src="https://2.bp.blogspot.com/-qrWf2ABV8F4/WYrBQhXwuJI/AAAAAAAAHLI/TEm5dGpZwLMsEzCPwhdVtGE98ePQ6u2RgCLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B03.PNG" /></a></div><br />After subtracting this and the displacement due to the pull of the Moon the following error remains.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-iLOtLvYPf5Y/WYrCGWRgH4I/AAAAAAAAHLQ/FjBdOjsjCA4j44AD6bP23dwxpVrcBwdAQCLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B04.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="263" data-original-width="541" src="https://2.bp.blogspot.com/-iLOtLvYPf5Y/WYrCGWRgH4I/AAAAAAAAHLQ/FjBdOjsjCA4j44AD6bP23dwxpVrcBwdAQCLcBGAs/s1600/Sun%2Berror%2Bcomponent%2B04.PNG" /></a></div><br />The error can be measured in micro AU (μAU). For comparison the Earth moves about the Sun at a mean rate of 2π/365.24=0.0172AU/day=12μAU/min. It probably wouldn't hurt to get accurate measurements of the start and end eclipse times.<br /><br />Essentially the same error or perhaps correction is present in both MICA and HORIZONS data for the Sun's position relative to the Earth.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-39459761710984633842017-08-02T16:56:00.002-07:002017-08-02T16:57:48.437-07:00With Light Sail Acceleration in the Direction of Sunlight Preferred<br /> One can study various modes of light sail operation to see how it will perform. The parasitic reduction of perigee seems to be associated with acceleration in the direction of sunlight and requires more energy for insertion into a higher circular orbit. The following mode of operation prefers acceleration in the direction of sunlight. The attitude of the sail is directed as follows.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-LvgG4Ec0rL0/WYJd26Z9zJI/AAAAAAAAHKM/VW9KD1Wa4c8allqY5L1gXKgVGd0_5F7dACLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="116" data-original-width="161" src="https://3.bp.blogspot.com/-LvgG4Ec0rL0/WYJd26Z9zJI/AAAAAAAAHKM/VW9KD1Wa4c8allqY5L1gXKgVGd0_5F7dACLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B01.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-_MPAaI1oxw0/WYJd5ipFjxI/AAAAAAAAHKQ/PXovlnaZrl4B56H6d13MtDvE1ncKq0F0QCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="217" data-original-width="489" src="https://1.bp.blogspot.com/-_MPAaI1oxw0/WYJd5ipFjxI/AAAAAAAAHKQ/PXovlnaZrl4B56H6d13MtDvE1ncKq0F0QCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B02.PNG" /></a></div><br />The relative change in the apogee of the sail increases with time but the perigee also decreases even if a small 10 second step is used in the calculation.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-ota-pbqqB5Y/WYJedVXdAyI/AAAAAAAAHKU/ym9b8SzU3ychI5tZmL1KWbU2etv9NT3ZgCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B03.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="267" data-original-width="404" src="https://3.bp.blogspot.com/-ota-pbqqB5Y/WYJedVXdAyI/AAAAAAAAHKU/ym9b8SzU3ychI5tZmL1KWbU2etv9NT3ZgCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B03.PNG" /></a></div><br />Again, more Δv is required to put the light sail into a circular orbit than for a ballistic transfer orbit. More time would be required to compensate for the loses when the Δv is in the proper direction orbit insertion.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-Bgnm7DRoBBk/WYJel6lQuVI/AAAAAAAAHKY/7Z1FJmwSj3gU6HCJKwERqu-ZssWOUV6FQCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B04.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="289" data-original-width="407" src="https://4.bp.blogspot.com/-Bgnm7DRoBBk/WYJel6lQuVI/AAAAAAAAHKY/7Z1FJmwSj3gU6HCJKwERqu-ZssWOUV6FQCLcBGAs/s1600/solar%2Bsail%2Boutwards%2Bprefered%2B04.PNG" /></a></div><br />If nature favors some modes of light sail operation over others the preferred mode would make the desired orbital changes in the least possible time. It may require a little R & D to determine the optimal solution. The rate of the gain in altitude at apogee is greatest for α=0 when the light sail moves in the direction of sunlight. If the drop in perigee isn't compensated for light sail won't climb out of the Earth's gravity well before it encounters atmospheric drag. So one will have to do the climb stepping from one circular orbit to a higher one.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-26937163223566098272017-08-01T02:16:00.000-07:002017-08-01T23:35:38.139-07:00The Effect of Changing the Step Size on the Light Sail's Orbits<br /> Since the light sail's orbit appeared to show signs of a deteriorating perigee I tried playing with the attitude of the sail relative to the direction of sunlight to get relatively more angular acceleration. In the figure below the direction on sunlight is to the left. The light sail's acceleration is along its normal n which is at an angle α relative to the horizontal axis. The changes to the equations of motion are also shown.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-yxbk1n_7qyA/WYA-BjCa2pI/AAAAAAAAHJU/XBBLFRe0YLY9ibXVvCDteqhttA5-TC6VACLcBGAs/s1600/solar%2Bsail%2Bcap%2B17.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="217" data-original-width="216" src="https://4.bp.blogspot.com/-yxbk1n_7qyA/WYA-BjCa2pI/AAAAAAAAHJU/XBBLFRe0YLY9ibXVvCDteqhttA5-TC6VACLcBGAs/s1600/solar%2Bsail%2Bcap%2B17.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/--ftPLA4Az8I/WYA-E18t6tI/AAAAAAAAHJY/_xJwKaCbAaI4Cavx3wrNsCJk2gImm3U1QCLcBGAs/s1600/solar%2Bsail%2Bcap%2B18.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="68" data-original-width="227" src="https://3.bp.blogspot.com/--ftPLA4Az8I/WYA-E18t6tI/AAAAAAAAHJY/_xJwKaCbAaI4Cavx3wrNsCJk2gImm3U1QCLcBGAs/s1600/solar%2Bsail%2Bcap%2B18.PNG" /></a></div><br />The angle of the sail was set to α=π/2-θ when it was above the x-axis and moving away from the Sun and α=π/2 when it was below in order to restrict the acceleration some. The behavior was sensitive to the size of the step so Δt was set to 10 seconds. This gave better results for the behavior of Δr=r-r<sub>0</sub>.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-NRdfaBTdrJ4/WYBCE5mEzuI/AAAAAAAAHJg/Cfz_fo-0LU8rqMAtP3gBU_GxUruZoazSgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B14.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="267" data-original-width="462" src="https://2.bp.blogspot.com/-NRdfaBTdrJ4/WYBCE5mEzuI/AAAAAAAAHJg/Cfz_fo-0LU8rqMAtP3gBU_GxUruZoazSgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B14.PNG" /></a></div><br />It still drifts away from a ballistic object in the original circular orbit due to changes in the eccentricity and the period of the orbit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/--0MTEFHE2Cc/WYBCi02sddI/AAAAAAAAHJk/cDuL3d4INL0Pq65BdTefQbvkYxLgxBazgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B15.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="256" data-original-width="405" src="https://1.bp.blogspot.com/--0MTEFHE2Cc/WYBCi02sddI/AAAAAAAAHJk/cDuL3d4INL0Pq65BdTefQbvkYxLgxBazgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B15.PNG" /></a></div><br />Changing the step size resulted in a closer match of the light sail's Δv's with the transfer orbit Δv's for insertion into the higher circular orbit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-bPu6huXZrZA/WYBDHB2-gSI/AAAAAAAAHJs/SxZyYKVW-zoibV-kaZIj7iRfcPNUnwM9QCLcBGAs/s1600/solar%2Bsail%2Bcap%2B16.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="315" data-original-width="407" src="https://4.bp.blogspot.com/-bPu6huXZrZA/WYBDHB2-gSI/AAAAAAAAHJs/SxZyYKVW-zoibV-kaZIj7iRfcPNUnwM9QCLcBGAs/s1600/solar%2Bsail%2Bcap%2B16.PNG" /></a></div><br />So it appears that a light sail can make changes to a transfer orbit. Unfortunate, the apogee is on the sunward half of the orbit and one can't use sunlight to accelerate towards the sun to go into a higher circular orbit. This is where an auxiliary propulsion system would be needed if one wanted to do orbit changes. A flyby mission would not require orbit insertion.<br /><br />Supplemental (Aug 1): After the passage of half a year the sunlight will be in the opposite direction (to the right in the first figure) and the Δv will be directed properly for insertion into the higher circular orbit. It may be possible for a light sail to slowly work its way out of a gravity well.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-35096458708903755112017-07-31T18:02:00.000-07:002017-08-01T01:33:43.356-07:00LightSail 2 Needs to Compensate for the Decrease in Perigee Periodically<br /> One can compare the energy change in LightSail 2's orbit that with that required to go into a <a href="http://en.wikipedia.org/wiki/Hohmann_transfer_orbit">transfer orbit</a>. After a number of revolutions however the perigee of the light sail starts to drop. This loss of energy has to be compensated for if one desires to go into a circular orbit at the light sail's apogee.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-kux2a8niKiw/WX_Ogq5HN4I/AAAAAAAAHJE/ZRfq_khsy7g484rZ93KP74cxJvdtvt9ggCLcBGAs/s1600/solar%2Bsail%2Bcap%2B12a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="316" data-original-width="476" src="https://1.bp.blogspot.com/-kux2a8niKiw/WX_Ogq5HN4I/AAAAAAAAHJE/ZRfq_khsy7g484rZ93KP74cxJvdtvt9ggCLcBGAs/s1600/solar%2Bsail%2Bcap%2B12a.PNG" /></a></div><br />In the plot above the blue curve indicates the velocity change Δv required to go from a point of the light sail's orbit into a circular orbit. The solid red line is the total Δv required to go from the original circular orbit into a circular orbit at the apogee of the transfer orbit. The dashed red line indicates just the Δv required to go from the apogee of the transfer orbit into a circular orbit. Initially the orbit injection Δv's of the light sail match up with the Δv's for the transfer orbit. After about six revolutions more Δv is required to compensate for the drop in perigee. The implication appears to be that the orbit of the light sail needs to be periodically corrected to a circular orbit to keep the light sail from becoming parasitic. The major deficiency of the light sail is that the direction of its thrust is limited. It's lack of angular acceleration requires an auxiliary propulsion system for some orbit changes.<br /><br />Supplemental (Aug 1): The deviations in the Δv's for the light sail when going to the higher circular orbit are affected by the step size. Here the step size was Δt=60 seconds. Compare next blog.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-89371500839745788532017-07-30T03:25:00.001-07:002017-07-30T03:41:07.123-07:00How Long Would It Take LightSail 2 To Escape From the Earth?<br /> One can compute LightSail 2's gain in energy over time and estimate the average rate at which it will gain energy.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-0JK_PmFdg8Q/WX2yitibZ-I/AAAAAAAAHI4/LIAo_KO4xlEmAmW6xFBYL_lqb9LoH_CKwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B11.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="290" data-original-width="538" src="https://1.bp.blogspot.com/-0JK_PmFdg8Q/WX2yitibZ-I/AAAAAAAAHI4/LIAo_KO4xlEmAmW6xFBYL_lqb9LoH_CKwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B11.PNG" /></a></div><br />This average rate appears to be linear and allows one to estimate how long it would take for the light sail to acquire enough energy to escape from the Earth's gravity well. The answer is 6.4 years.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-pm8b0_tl3po/WX2zKaOAbRI/AAAAAAAAHI8/PUXm1dCffrsADJOSiUI06bG-wEOvDlZfQCLcBGAs/s1600/solar%2Bsail%2Bcap%2B12.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="162" data-original-width="194" src="https://2.bp.blogspot.com/-pm8b0_tl3po/WX2zKaOAbRI/AAAAAAAAHI8/PUXm1dCffrsADJOSiUI06bG-wEOvDlZfQCLcBGAs/s1600/solar%2Bsail%2Bcap%2B12.PNG" /></a></div><br />The model used was overly simplified neglecting the Earth's shadow and didn't take into account the need to raise the height of perigee to avoid atmospheric drag. As the Earth moves about the Sun in its orbit the direction of sunlight will slowly change.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-79823531105153969432017-07-28T01:01:00.001-07:002017-07-28T14:21:30.631-07:00LightSail 2's Behavior in Orbit<br /> The Planetary Society's <a href="http://sail.planetary.org/">LightSail 2</a> is to go into orbit aboard SpaceX's <a href="http://www.spacex.com/falcon-heavy">Falcon Heavy</a> later <a href="http://www.businessinsider.com/falcon-heavy-first-launch-date-2017-6">this year</a> if all goes well. The initial orbit will be a circular orbit 720 km above the Earth's surface. The light sail will face the Sun as it moves away from it and the plane of the sail will lie along the direction of a line from the Sun as it moves toward it. The effect of switching back and forth between its boost and cruise phases will be a slow but steady increase in the satellite's orbital energy.<br /><br />The following figure defines the plane of the orbit with the x-axis pointing towards the Sun. The unit vector n indicates the direction of the acceleration caused by the light pressure on the sail during the boost phase. The equations below indicate the specific radial and angular forces acting on light sail as it orbits the Earth. μ is constant GM for the Earth.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-xu09Y_Suvf8/WXrjY2-e2CI/AAAAAAAAHGY/zZ1Ej9z8n-oZ2xuztK66URspugXWkjLpwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="307" data-original-width="232" src="https://2.bp.blogspot.com/-xu09Y_Suvf8/WXrjY2-e2CI/AAAAAAAAHGY/zZ1Ej9z8n-oZ2xuztK66URspugXWkjLpwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B01.PNG" /></a></div><br />The set of equations above are difficult to solve analytically but it's not too difficult to do a numerical calculation. The following calculations used 1 second steps in time. While in cruise phase the satellite coasts in a Keplerian orbit. The following table gives values for the selected points of the first orbit. The units for time, angle and radius are seconds, radians and meters. The highest and lowest points of the orbit are the apogee r<sub>a</sub> and perigee r<sub>p</sub>. The cruise phase was allowed to continue past the switching point at θ=2π in order to get the values for the perigee. Interpolation was used to get a better estimate of the values.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-3TpMYZtzDyQ/WXrpYEIVeaI/AAAAAAAAHGw/H4JknYY3rbIrg-x2xE8OMUlKuzqu7NxUgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B02b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="170" data-original-width="449" src="https://1.bp.blogspot.com/-3TpMYZtzDyQ/WXrpYEIVeaI/AAAAAAAAHGw/H4JknYY3rbIrg-x2xE8OMUlKuzqu7NxUgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B02b.PNG" /></a></div><br />From this table we can compute some of the orbital elements for the Keplerian orbit of the cruise phase. There is a gain in energy during the boost phase.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-ZmUNp8YLiU4/WXrsGQoNnsI/AAAAAAAAHG8/s46BWR56qSIUWDv8Rj9Tr45MHOk-DOxLACLcBGAs/s1600/solar%2Bsail%2Bcap%2B03a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="214" data-original-width="286" src="https://4.bp.blogspot.com/-ZmUNp8YLiU4/WXrsGQoNnsI/AAAAAAAAHG8/s46BWR56qSIUWDv8Rj9Tr45MHOk-DOxLACLcBGAs/s1600/solar%2Bsail%2Bcap%2B03a.PNG" /></a></div><br />The light sail will spiral away from a ballistic object in the same initial circular orbit since their angular separation will increase over time in addition to the radial separation.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-440dUL0NsFI/WXr9Nz5b4yI/AAAAAAAAHHg/KppPwbeCucArwA2DIBXQTFTkm2nWRfIZwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B06.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="324" data-original-width="350" src="https://2.bp.blogspot.com/-440dUL0NsFI/WXr9Nz5b4yI/AAAAAAAAHHg/KppPwbeCucArwA2DIBXQTFTkm2nWRfIZwCLcBGAs/s1600/solar%2Bsail%2Bcap%2B06.PNG" /></a></div><br />The separation viewed from the perspective of revolutions shows the alternating crossing of the axes over time.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-a4mHHvOm3rM/WXr-jESjCfI/AAAAAAAAHHo/zq1-uQNOOd80kbQR2fLC5mPFdgxZRdzcgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B07.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="327" data-original-width="499" src="https://2.bp.blogspot.com/-a4mHHvOm3rM/WXr-jESjCfI/AAAAAAAAHHo/zq1-uQNOOd80kbQR2fLC5mPFdgxZRdzcgCLcBGAs/s1600/solar%2Bsail%2Bcap%2B07.PNG" /></a></div><br />Supplemental (Jul 28): Two additional plots to clarify relative positions. The first is Δr=r-r<sub>0</sub> where r<sub>0</sub> is the radius of the circular orbit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-BhVGVf67Uys/WXsRaFxLO8I/AAAAAAAAHH8/DCE-vXOhWmADoQEZnPpXmC7aaeBGdAtiACLcBGAs/s1600/solar%2Bsail%2Bcap%2B08.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="290" data-original-width="480" src="https://3.bp.blogspot.com/-BhVGVf67Uys/WXsRaFxLO8I/AAAAAAAAHH8/DCE-vXOhWmADoQEZnPpXmC7aaeBGdAtiACLcBGAs/s1600/solar%2Bsail%2Bcap%2B08.PNG" /></a></div><br />The second is the angular separation of the light sail from a ballistic object in the original circular orbit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-cBZi49ZlhL0/WXunuD4McFI/AAAAAAAAHIU/7BV1SdGU-KgTAFZxLPD8YSVk5rGvu2zsACLcBGAs/s1600/solar%2Bsail%2Bcap%2B10.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="292" data-original-width="482" src="https://2.bp.blogspot.com/-cBZi49ZlhL0/WXunuD4McFI/AAAAAAAAHIU/7BV1SdGU-KgTAFZxLPD8YSVk5rGvu2zsACLcBGAs/s1600/solar%2Bsail%2Bcap%2B10.PNG" /></a></div><br />Edit (Jul 28): Found an error in the first two plots and removed them.<br /><br />Edit (Jul 28): Found an error in the Δθ plot. It was a dumb mistake. You can't subtract radians from degrees. Replaced plot. Found a minor error in the spiral plot calculation which didn't affect the result much. Left plot as it was. I used 1 minute steps for the longer time period used in all the plots so there is a little additional error present.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-51174069135629230702017-07-13T13:32:00.000-07:002017-07-13T13:32:46.647-07:00Refractive Index of Air as a Function of Wavelength<br /> In 1908 Rentschler published a study on the <a href="http://books.google.com/books?id=VOwEAAAAYAAJ&pg=PA357#v=onepage&q&f=false">refractive index of a gas</a> for different wavelengths in the Astrophysical Journal. I used his data for air in Table II to find the coefficients of a Cauchy formula to fit the data. One can also use the <a href="http://emtoolbox.nist.gov/Wavelength/Edlen.asp">NIST calculator</a> to get data for air and do another fit.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-hDBPf2Pn8gs/WWfVYEX2-rI/AAAAAAAAHGI/ma1VzxPZuywMGMMQ6ygUMtODi7dsnkx2QCLcBGAs/s1600/n-1%2Bfor%2Bair%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="481" data-original-width="441" src="https://3.bp.blogspot.com/-hDBPf2Pn8gs/WWfVYEX2-rI/AAAAAAAAHGI/ma1VzxPZuywMGMMQ6ygUMtODi7dsnkx2QCLcBGAs/s1600/n-1%2Bfor%2Bair%2B02.PNG" /></a></div><br />The Rentschler data appears to be for dry air. If one computes the index of refraction for the CRC Handbook data using the given wavelengths for air and computed wavelengths for the Balmer series for hydrogen the plot appears to be displaced somewhat. Note that the humidity of the air also affects the index of refraction. I used the ratio of the wavelength in air to 5000Å for the formula so it would be easier to substitute wavelengths given in nanometers (nm) using 500/λ with the same coefficients.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-85314630239733040992017-07-10T21:30:00.002-07:002017-07-10T21:30:19.710-07:00Balmer Series Fit Using More Accurate CRC Data<br /> The hydrogen wavelengths in air found in the CRC Handbook of Chemistry & Physics are lines of the Balmer series and I got better results for a fit of these lines.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-6qk1e8ttzjQ/WWRQr5X1XaI/AAAAAAAAHF0/RPTiLlr9g3Ukl12vEQIKkJTgbuczeTd4QCLcBGAs/s1600/CRC%2Bdata%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="254" data-original-width="358" src="https://4.bp.blogspot.com/-6qk1e8ttzjQ/WWRQr5X1XaI/AAAAAAAAHF0/RPTiLlr9g3Ukl12vEQIKkJTgbuczeTd4QCLcBGAs/s1600/CRC%2Bdata%2B01.PNG" /></a></div><br />One has to use the reduced mass of the electron to get R<sub>H</sub>=R<sub>∞</sub>/(1+m<sub>e</sub>/m<sub>p</sub>) where m<sub>e</sub> and m<sub>p</sub> are the electron and proton masses to compute the limit of the series in air and divide that by the index of refraction for air to get the limit of the series for air. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-u8o5IZyqp00/WWRREq3NiuI/AAAAAAAAHF4/8VvN5NRieFIAVkR734jMxeJCLMTH-Vg-ACLcBGAs/s1600/CRC%2Bdata%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="143" data-original-width="394" src="https://4.bp.blogspot.com/-u8o5IZyqp00/WWRREq3NiuI/AAAAAAAAHF4/8VvN5NRieFIAVkR734jMxeJCLMTH-Vg-ACLcBGAs/s1600/CRC%2Bdata%2B02.PNG" /></a></div><br />The index of refraction for air is a function of wavelength and that also affects the observed values and so the fit has an error that varies with the number of the series.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-36555426720788625322017-07-09T19:26:00.002-07:002017-07-09T19:33:54.393-07:00Scaling Factors That Could Potentially Affect Huggins' Hydrogen Lines<br /> Relative to the minimum deflection wavelength for Huggins' spectroscope there appears to be a progressive shift to higher wavenumbers which is a blue shift. But relative to the limit of the series, λ<sub>∞</sub>, there is a redshift. There are a number of factors which might cause a scale change in the observed wavelengths and need to be taken into consideration.<br /><br /><u>Calibration of the Spectroscope</u><br /><a href="http://books.google.com/books?id=7RgepdyskJAC&pg=PA70#v=onepage&q=calibration&f=false">minimum deflection</a><br /><a href="http://books.google.com/books?id=bSUyAQAAMAAJ&pg=PA92#v=onepage&q&f=false">calibration curve</a><br /><br /><u>Air Wavelengths vs Vacuum Wavelengths</u><br />Definition of <a href="http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure">standard conditions</a><br /><a href="http://en.wikipedia.org/wiki/Snell%27s_law">Snell's Law</a> (measurement in air results in a blue shift relative to a vacuum )<br /><a href="http://en.wikipedia.org/wiki/Rydberg_constant">Rydberg Constant</a> (R<sub>∞</sub> is for a vacuum)<br /><br /><u>Light nucleus</u><br />R<sub>M</sub> needed for <a href="http://en.wikipedia.org/wiki/Rydberg_constant#Occurrence_in_Bohr_model">light nucleus</a> (R<sub>∞</sub> is for a heavy nucleus)<br /><a href="http://en.wikipedia.org/wiki/Reduced_mass">reduced mass</a><br /><a href="http://en.wikipedia.org/wiki/Electron_rest_mass">electron mass</a><br /><br /><a href="http://en.wikipedia.org/wiki/Doppler_effect">Doppler shift</a><br /><br /><a href="http://en.wikipedia.org/wiki/Gravitational_redshift">Gravitational redshift</a><br /><br />All of the above contribute to a change in scale. What did Huggins miss?<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-85300302411577768772017-07-07T19:45:00.000-07:002017-07-08T11:58:47.881-07:00A More General LS Formula for Estimating a Common Factor<br /> One can generalize the formula in the last post to find the best estimate of a common factor.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-hHgM2ss1U_Y/WWA_kRVnkgI/AAAAAAAAHFM/zBZoTbAicZQVWVEw-PA3z5n3RLvPZnzkgCEwYBhgL/s1600/least%2Bsquares%2Bderivation%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="219" data-original-width="295" src="https://3.bp.blogspot.com/-hHgM2ss1U_Y/WWA_kRVnkgI/AAAAAAAAHFM/zBZoTbAicZQVWVEw-PA3z5n3RLvPZnzkgCEwYBhgL/s1600/least%2Bsquares%2Bderivation%2B02.PNG" /></a></div><br />Using this formula gives a better estimate for the wavenumber limit of the Balmer series as the following calculation shows. The formula gives 2744.8 while the average for the individual estimates is 2744.6 and the rms errors are 1.14 and 1.13 respectively.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-oubqXlvdeKs/WWA_chDu2aI/AAAAAAAAHFM/A2Ky1kQBx-gv7NnZpliaBNfO7R6F4FLIACEwYBhgL/s1600/Balmer%2B07.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="310" data-original-width="435" src="https://3.bp.blogspot.com/-oubqXlvdeKs/WWA_chDu2aI/AAAAAAAAHFM/A2Ky1kQBx-gv7NnZpliaBNfO7R6F4FLIACEwYBhgL/s1600/Balmer%2B07.PNG" /></a></div><br />The improvement over an average is marginal and doesn't appear to compensate for a scaling error.<br /><br />Supplemental (Jul 8): Huggins adjusted his spectroscope for minimum deviation for the H line corresponding to 4340 Å or n=5. This looks like a better fit for the wavenumbers with an rms err of 0.46 which could result from rounding to the nearest Ångström.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-96StJk9lP1I/WWCNq_3gzWI/AAAAAAAAHFk/l_xqVa_sK2Qdl0cDy22Zali1XK9T9SaWgCLcBGAs/s1600/Balmer%2B09a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="285" data-original-width="295" src="https://1.bp.blogspot.com/-96StJk9lP1I/WWCNq_3gzWI/AAAAAAAAHFk/l_xqVa_sK2Qdl0cDy22Zali1XK9T9SaWgCLcBGAs/s1600/Balmer%2B09a.PNG" /></a></div><br />This rescaling brings the estimate for the limit of the wavelengths to within a quarter of an Ångström of the currently accepted value.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-B8TSOQHIXhw/WWCM8HwdEvI/AAAAAAAAHFc/zAu7Kb7roCc0sX7XI9-GJ9kWEEtamNqhgCLcBGAs/s1600/Balmer%2B10.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="73" data-original-width="221" src="https://3.bp.blogspot.com/-B8TSOQHIXhw/WWCM8HwdEvI/AAAAAAAAHFc/zAu7Kb7roCc0sX7XI9-GJ9kWEEtamNqhgCLcBGAs/s1600/Balmer%2B10.PNG" /></a></div><br />Supplemental (Jul 8): The deflection of light by a prism depends on the <a href="http://en.wikipedia.org/wiki/Prism">index of refraction</a> n and the <a href="http://en.wikipedia.org/wiki/Sellmeier_equation">wavelength</a> λ. Shorter wavelengths are bent more than longer ones so blue light is bent more than red. The deviation from linearity also increases as the wavelength decreases so this might explain the need to rescale the spectroscope readings. The changes are in the right direction but I haven't compare the observed errors with calculated errors.<br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-25405659653835874242017-07-07T19:31:00.000-07:002017-07-07T19:31:38.956-07:00An Error in Huggins' Hydrogen Wavelengths?<br /> There appears to be a systematic error in the hydrogen wavelengths in Huggins' paper when one compares them with the fit for the Balmer series. Notice that there is an progressive increase in the value of the error, Δν. If one uses the first line as the reference we can set that error equal to zero and compute the relative errors Δν'.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-5h8cQH2LGxU/WWA9rUeLfVI/AAAAAAAAHEs/d0gxRhgq210TX0CH1JsLPJiyNkNP54ZhACLcBGAs/s1600/Balmer%2B04.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="325" data-original-width="419" src="https://2.bp.blogspot.com/-5h8cQH2LGxU/WWA9rUeLfVI/AAAAAAAAHEs/d0gxRhgq210TX0CH1JsLPJiyNkNP54ZhACLcBGAs/s1600/Balmer%2B04.PNG" /></a></div><br />Trying to fit a straight line to this data is a little tricky but least squares allows us to derive a formula for the best fit for a line going through the origin.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-a0IoJhEiwFg/WWBAPPNsgTI/AAAAAAAAHFA/OMZZWv6NMg4V3odCF6t6eu0CGtgdiLOoACLcBGAs/s1600/Balmer%2B05.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="329" data-original-width="249" src="https://2.bp.blogspot.com/-a0IoJhEiwFg/WWBAPPNsgTI/AAAAAAAAHFA/OMZZWv6NMg4V3odCF6t6eu0CGtgdiLOoACLcBGAs/s1600/Balmer%2B05.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-2-VsIbZRLSI/WWBAX9mPs2I/AAAAAAAAHFE/g8KYbPFcZ6kE-L9cNuKzizsuUpiJUIeyQCLcBGAs/s1600/Balmer%2B06.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="287" data-original-width="298" src="https://4.bp.blogspot.com/-2-VsIbZRLSI/WWBAX9mPs2I/AAAAAAAAHFE/g8KYbPFcZ6kE-L9cNuKzizsuUpiJUIeyQCLcBGAs/s1600/Balmer%2B06.PNG" /></a></div><br />The adjusted lines give a closer estimate for the limit of the lines of the Balmer series.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-GA94jVgMKwI/WWBBqn_GXjI/AAAAAAAAHFI/wq4Q6FB9ngQ2ohDtmOoJtbmL0D693FZWwCLcBGAs/s1600/Balmer%2B08.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="355" data-original-width="386" height="294" src="https://4.bp.blogspot.com/-GA94jVgMKwI/WWBBqn_GXjI/AAAAAAAAHFI/wq4Q6FB9ngQ2ohDtmOoJtbmL0D693FZWwCLcBGAs/s320/Balmer%2B08.PNG" width="320" /></a></div><br />It's possible that the Iceland spar prism that Huggins used was slightly nonlinear in its dispersion of light. Dispersion is dependent on frequency. This may be why rescaling gives a better result. The least squares derivation of the formula above is fairly simple. The quantities on the right of the formula are the column vectors in the second table.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-51VOIuy75Ho/WWA_3estj1I/AAAAAAAAHFM/qGJcslTzW_sIPBrgpV2AxXLxsHQTCysigCEwYBhgL/s1600/least%2Bsquares%2Bderivation%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="195" data-original-width="303" src="https://4.bp.blogspot.com/-51VOIuy75Ho/WWA_3estj1I/AAAAAAAAHFM/qGJcslTzW_sIPBrgpV2AxXLxsHQTCysigCEwYBhgL/s1600/least%2Bsquares%2Bderivation%2B01.PNG" /></a></div><br /><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag:blogger.com,1999:blog-494600399189651849.post-9552420097668566832017-07-05T17:04:00.000-07:002017-07-05T23:36:44.550-07:00The Balmer Series<br /> It's customary on the 4th of July for Americans view local <a href="http://en.wikipedia.org/wiki/Fireworks">fireworks</a> displays. What the experts can do with patterns, colors and sound is quite amazing. History tells us that fireworks originated in ancient China and the ancient alchemists experimented with colors. Knowledge of the colors that can be produced can be acquired through the use of <a href="http://en.wikipedia.org/wiki/Flame_test">flame tests</a>. This tool is still used in modern chemistry.<br /><br /> The scientific study of colored light was advanced in the late 17th Century by <a href="http://en.wikipedia.org/wiki/Isaac_Newton">Newton's</a> work on the prism. At the beginning of the 19th Century the introduction of the spectroscope allowed scientists to study the Sun's spectrum and discover the dark lines known as <a href="http://books.google.com/books?id=gWNZ64zhzggC&pg=PT1#v=onepage&q&f=false">Fraunhofer lines</a>. At the same time flame tests of various elements allowed scientists to connect these lines to <a href="http://books.google.com/books?id=EqLLh-PmRfAC&pg=PA89#v=onepage&q&f=false">chemical elements</a> present. In 1868 <a href="http://books.google.com/books?id=vqURAAAAYAAJ&pg=PA31#v=onepage&q&f=false">Ångström</a> published accurate values for the lines of the solar spectrum with the elements associated with them.<br /><br />In 1885 Balmer published a <a href="http://books.google.com/books?id=fzs-AQAAMAAJ&pg=PA80#v=onepage&q&f=false">notice</a> giving the <a href="http://books.google.com/books?id=fzs-AQAAMAAJ&pg=PA86#v=onepage&q&f=false">formula</a> for a series of hydrogen lines. How might he have accomplished this? His data came from a notice by <a href="http://books.google.com/books?id=uNdKAQAAMAAJ&lpg=PA415&pg=PA680#v=onepage&q&f=false">Huggins</a> on the hydrogen lines present in the spectra of certain stars. The data is included in a footnote referring to a note he received from <a href="http://books.google.com/books?id=uNdKAQAAMAAJ&lpg=PA415&pg=PA678#v=onepage&q&f=false">Johnstone Stoney</a>, a fellow of the Royal Society, who states that the lines might belong to a series.<br /><br />What happens if we try to do an empirical fit for the data? Notice that Stoney also includes the wave numbers, ν=1/λ, and we can try to fit these. The lines appear to converge in one direction so we might first try to fit a formula that is quadratic in 1/n (fit1). The results are quite good with an rms err of 0.4. Using n=3 for the first line gives the best fit. The value for B is relatively quite small when compared with the others and the ratio of C to A is very close to 4. Redoing the fit for just two terms makes the ratio even closer to 4 so we can just try to get a value for A by computing a value for each line and taking the average.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-XydETK1I-Go/WV17kPg4P3I/AAAAAAAAHD0/35F3Uqu1WbMJwG6mSmr_XI_yGvTOQE4IgCLcBGAs/s1600/Balmer%2B02.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="185" data-original-width="350" src="https://3.bp.blogspot.com/-XydETK1I-Go/WV17kPg4P3I/AAAAAAAAHD0/35F3Uqu1WbMJwG6mSmr_XI_yGvTOQE4IgCLcBGAs/s1600/Balmer%2B02.PNG" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-ctT_aYqPlNc/WV17d3dtAVI/AAAAAAAAHDw/YHgHNOp-cpw8Ar0E9LysSSV0pSsflOv9gCEwYBhgL/s1600/Balmer%2B01.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="311" data-original-width="496" src="https://4.bp.blogspot.com/-ctT_aYqPlNc/WV17d3dtAVI/AAAAAAAAHDw/YHgHNOp-cpw8Ar0E9LysSSV0pSsflOv9gCEwYBhgL/s1600/Balmer%2B01.PNG" /></a></div><br />Although the rms error is greater, we still get a fairly good fit to the data.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-pYcEac15kaE/WV192vh89UI/AAAAAAAAHEE/oj-oPZxKMeUr1s0tYF1CIlkxcfPhLcBoACLcBGAs/s1600/Balmer%2B03.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="342" data-original-width="322" src="https://1.bp.blogspot.com/-pYcEac15kaE/WV192vh89UI/AAAAAAAAHEE/oj-oPZxKMeUr1s0tYF1CIlkxcfPhLcBoACLcBGAs/s1600/Balmer%2B03.PNG" /></a></div><br />Notice that Stoney includes a curve passing through the data points. Did he know the formula for the series of lines? His table suggests he used a difference formula to fit the data.<br /><br />Supplemental (Jul 5): Fraunhofer's lines:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-OmkSo7FLtqA/WV3X7FfYKnI/AAAAAAAAHEY/JDRhyW8cdsQTCvFASEMEc88nxIgpVupwgCLcBGAs/s1600/Fraunhofer%2B-%2Bspectrum%2Blines.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="307" data-original-width="572" src="https://1.bp.blogspot.com/-OmkSo7FLtqA/WV3X7FfYKnI/AAAAAAAAHEY/JDRhyW8cdsQTCvFASEMEc88nxIgpVupwgCLcBGAs/s1600/Fraunhofer%2B-%2Bspectrum%2Blines.jpg" /></a></div><br />Huggins Plate 33 showing the line spectra of a number of stars. The second row appears to be <a href="http://books.google.com/books?id=AL1AAQAAMAAJ&pg=PA73#v=onepage&q&f=false">α Lyræ</a> (Vega) containing the first twelve lines of the data above:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-qiGrgq0JLbc/WV3YOHMs3xI/AAAAAAAAHEc/bPiBL10GuW4vsXcvJuHe9OOWF1VY0CUNQCLcBGAs/s1600/Huggins%2B-%2Bplate%2B33.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="252" data-original-width="573" src="https://1.bp.blogspot.com/-qiGrgq0JLbc/WV3YOHMs3xI/AAAAAAAAHEc/bPiBL10GuW4vsXcvJuHe9OOWF1VY0CUNQCLcBGAs/s1600/Huggins%2B-%2Bplate%2B33.jpg" /></a></div><br />Jim Bergquisthttp://www.blogger.com/profile/10032393950433918867noreply@blogger.com0tag: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.com0