REFERENDUM 2009 RESULTS: IRELAND ON LISBON TREATY

A note by Eamon Henry. 8 October 2009

This short discussion offers a statistical (regression) analysis of the Irish (2 October 2009) Referendum results, of voting for the second time on the Lisbon Treaty. For a turnout of 58.0 percent, there was a YES vote of 67.1 percent of turnout, implying a NO vote of 32.9 percent. For results of the 12 June 2008 Referendum, see the Henry article on http://www.asits.blogspot.com/ , of title “Referendum Results, Actual and Possible”.

Readers not familiar with statistical analysis are asked to accept “on faith” the results presented below via “Simple Regression” analysis of YES percentage as depending on TURNOUT percentage across all 43 constituencies.

The next paragraph gives results in technical format.

There follows a non-technical discussion.

The actual Yes and Turnout percentages, by constituency, are given as an appendix table below, having been extracted by me from the Irish Times of 5 October 2009. The “Data Desk” software package has been used by me to give the regression results and a graph.
Of all 43 constituencies, only two gave a NO percentage result above 50, namely Donegal North East and Donegal South West.

The regression results are as follows and rather technical. For a fairly loose-fitting “Rsquared adjusted” of 15.8 percent, a highly significant (99.5 percent probable) positive coefficient emerged, namely an average 0.758 percent Yes extra for every 1 percent extra turnout. This coefficient has a “standard error” of 0.254. The regression constant has value 22.58 (not significant) with its standard error of value 14.78.
We can estimate a YES percent value on the line of regression as follows. For a turnout of 50 percent, the regression estimate of YES percentage is given by constant plus 50 times the regression coefficient, namely 22.58 + 50X0.758 = 22.58 + 37.90 = 60.48 percent YES. A further 8 percent turnout, to match the actual, would add a further 8X0.758 = 6.06 percent Yes, thus giving a regression estimate of 66.54 percent Yes, close to the actual 67.1 percent quoted in the first paragraph above.

It is of interest to quote corresponding results of the Referendum of 12 June 2008. For a turnout of 53.1 percent of the electorate, there was a Yes vote of only 46.6 percent. Similar regression analysis gave an “Rsquared adjusted” of 13.0 percent, a highly significant (99 percent probable) positive coefficient 0.779 percent extra YES for every 1 percent extra Turnout. This coefficient had a “standard error” of value 0.289. The regression constant had value 4.66 (not significant) with standard error of value 15.39. We may note that the coefficient here is close to the October 2009 value 0.758 quoted above, but the constant here is much smaller (some 18 percentage points) than that of October 2009.
Here again we can estimate a Yes percentage on the line of regression. For a turnout of 50 percent, the regression estimate of YES percentage is given by constant plus 50 times the regression coefficient, namely 4.66 + 50X0.779 = 4.66 + 38.95 = 43.61 percent YES. A further 3.1 percent turnout, to match the actual, would add a further 3.1X0.779 =2.41 percent Yes, thus giving a regression estimate of 46.02 percent Yes, close to the actual 46.6 percent Yes quoted in the last paragraph above.

Appendix Table of Constituency Results, 2 October 2009

Constituency ---------Percent Turnout ------------Percent YES
Carlow-Kilkenny ---------------57.7---------------------------- 70.5

Cavan-Monaghan-------------- 58.1 ----------------------------62.0
Clare ---------------------------56.1---------------------------- 72.3
Cork East ----------------------56.6---------------------------- 66.1
Cork North Central -------------59.6 ----------------------------55.8
Cork North West--------------- 59.4---------------------------- 69.5
Cork South Central------------- 60.3---------------------------- 66.8
Cork South West ----------------59.5 ----------------------------67.2
Donegal North East -------------50.1 ----------------------------48.5
Donegal South West -------------51.1 ---------------------------49.7
Dublin Central ------------------53.3 ---------------------------61.9
Dublin Mid West ----------------55.3 ---------------------------61.5
Dublin North --------------------60.9--------------------------- 72.7
Dublin North Central------------ 65.6 ----------------------------71.1
Dublin North East ---------------63.4 ---------------------------63.5
Dublin North West --------------57.6---------------------------- 55.0
Dublin South --------------------57.6 ----------------------------81.7
Dublin South West --------------57.2 ----------------------------58.9
Dublin South Central ------------55.5 ----------------------------58.0
Dublin South East ---------------54.7---------------------------- 78.7
Dublin West ---------------------59.1 ----------------------------68.5
Dun Laoghaire------------------- 70.7 ----------------------------81.2
Galway East ---------------------54.8 ----------------------------68.1
Galway West --------------------53.1 -----------------------------66.3
Kerry North ---------------------55.4 ----------------------------63.6
Kerry South--------------------- 57.9---------------------------- 66.4
Kildare North --------------------56.6 ----------------------------76.2
Kildare South-------------------- 54.3 -----------------------------69.7
Laois-Offaly ---------------------57.5 -----------------------------73.2
Limerick East --------------------56.5 ----------------------------67.4
Limerick West -------------------56.2---------------------------- 69.3
Longford-Westmeath ------------55.4---------------------------- 65.6
Louth ----------------------------58.2 ----------------------------61.0
Mayo ----------------------------56.2 ----------------------------61.7
Meath East -----------------------54.9 ----------------------------72.3
Meath West---------------------- 60.5 ----------------------------64.9
Roscommon-SthLeitrim--------- 65.0 -----------------------------66.0
Sligo-NorthLeitrim --------------57.7------------------------------- 64.5
Tipperary North -----------------63.5 -----------------------------70.4
Tipperary South -----------------59.6 -----------------------------68.4
Waterford -----------------------59.6 -----------------------------68.5
Wexford -------------------------57.3 ------------------------------65.2
Wicklow -------------------------64.3 -------------------------------70.7

MODELLING THE SUN’S ALTITUDE FROM SUNRISE TO NOON

A note by Eamon Henry, 5 August 2009


Foreword:This note follows the Henry note of 27 July 2009 “How warm is the summer sun!”, now available on http://www.asits.blogspot.com/. The trigonometric model which gives results presented below uses two formulae appearing in a 14-page document “Spherical Trigonometry” available on the web at www.krysstal.com/sphertrig.html. This interesting article even gives 2 pages of (in effect) a “do it you yourself kit” for building your own sundial (just what we all need!!).The Henry Excel 97-2003 worksheet “Sunrise to Noon Sun Altitudes­_rev1”(3 July 2009) gives model results for midsummer day and midwinter day, at 15-minute intervals, computed by him. At present the ASITS blogspot is not able to carry an Excel worksheet as such. Table 1 following gives summary results for 30-minute (0.5- hour) intervals.
Results:
Table 1: Model estimates of Sun’s altitude at Dublin from Sunrise to Noon, Midsummer and Midwinter
Time, local, hours
Midsummer altitude, degrees
Midwinter altitude, degrees
12.00
60.0
13.0
11.30
59.8
12.7
11.00
59.3
11.8
10.30
58.5
10.2
10.00
57.3
8.0
9.30
55.7
5.1
9.00
53.6
1.5
8.48

0.0
8.30
51.1

8.00
48.0

7.30
44.3

7.00
40.0

6.30
35.1

6.00
29.7

5.30
24.0

5.00
18.3

4.30
12.7

4.00
7.4

3.30
2.6

3.12
0.0

“Local time” means calibration so that noon (Sun due south) is 12.00 hours, and thus for Dublin some 25 minutes later than Greenwich Mean Time (GMT) noon, because Dublin is about 6.25 degrees West (longitude); Sun due east will be at 6.00 hours local time, for Earth’s rotation 15 degrees per hour. Thus, in “Azimuth” terms, direction East is 90 degrees away from South, and direction West is 90 degrees away in the opposite direction, as for “Azimuth” angles to be used for model applications described below. “AZIMUTH is the angular distance from the north or south point of the horizon to the intersection with the horizon of a vertical circle passing through a given celestial body” (page 97 of the Oxford Encyclopedic English Dictionary, 1991 edition).

How the Model differs from the Real-World Situation:The model gives 21 June Sunrise as occurring at 3h 12m a.m. The actual Dublin sunrise time is about 3h31m local time, some 19 minutes later. For 21 December, the model gives Sunrise as at 8h48m local time, compared with actual 8h15m local time, some 33 minutes earlier. On page 11 of “Spherical Trigonometry” a calculation is made for Sunrise times at London. The azimuth angles are 39.62 degrees north of east (3h21.2m am local time) and 39.62 degrees south of east (8h38.5m am local time). My Diary 2009 gives London 20 June sunrise as 3h43m, about 22 minutes later, and London 19 December sunrise as 8h03m a.m., some 35 minutes earlier. So the model is wrong in a consistent way, it seems.
One reason for these differences is that the model implicitly assumes that the Earth’s orbit is a circle about the Sun, and thus completely regular. But in fact the orbit is an ellipse with the Sun at one focus, and our distance from the Sun ranges from 147 million kilometers in December to 152 million in June (page 9 of Patrick Moore: “Exploring the Night Sky with Binoculars”). However, the model gives no discrepancy for the Spring and Autumn equinoxes, 21 March and 23 September, with sunrise at 6.00 hours local time everywhere. Altitudes are not calculated for one of these, in the attached Excel sheet, but can be readily derived by changing a few parameters of the model, as will be indicated below.
Note that the model gives exactly 12 hours as the sum of the times from sunrise to noon, 8h48m midsummer plus 3h12m midwinter. The actual sum is about 12h15m, meaning that the Northern Hemisphere does in fact get more Sun-time than the Southern Hemisphere.
Because of these model discrepancies, I am re-calculating (in the Postscript section below) the relative per-hour intensities of the Sunshine for midsummer versus midwinter, as given in my previous note of 27 July 2009. For the summer sunrise situation, one might estimate actual Sun’s altitudes (angles of elevation) by going to the model altitude at least 15 minutes earlier: take the 5.45 hours altitude 26.9 degrees as estimating actual altitude at 6.00 hours, and take model 8.45 hours altitude 52.4 degrees as estimating actual altitude at 9.00 hours. My own estimates (for these times) used in my note of 27 July were 23 degrees and 47 degrees, respectively.Of course, both model and actual altitudes coincide at local noon, so they must gradually come closer as the morning progresses.
Technical description of the Model:The main equation I use is on page 8 of “Spherical Trigonometry” as follows:
sin(D) = sin(L)*cos(Z) – cos(L)*sin(Z)*cos(A)………………………………….. (1)
The sign * means “multiplied by”.I use latin letters where the text uses greek letters. D is the angle of declination (latitude on the Celestial Sphere) of some star, in our case the Sun, at 0 degrees for the equinoxes, 23.5 for summer solstice and -23.5 degrees for winter solstice. L is the Latitude of the point of observation, for Dublin about 53.5 degrees North, for Oxford about 51.6 degrees North. Z is the “Zenith distance”, given as 90 degrees less the Altitude(angle of elevation) of a star (in our case the Sun). A is the Azimuth, measured in degrees from direction South (e.g. direction East has azimuth value “minus 90 degrees”). Here a comment on the Celestial Sphere is in order. At Greenwich at noon on 21st March (spring equinox), the Sun is at location (0,0) on the Celestial Sphere, whose equator coincides with the imaginary circle where the plane of Earth’s equator cuts this visual sphere of stars etc. around the Earth. Longitude is called “Right Ascension” and latitude is called “Declination”. The north pole of the Celestial Sphere is directly above Earth’s north pole, its axis of rotation coinciding with that of Earth.
Let us get the Sunrise equation from equation (1) above. For altitude zero degrees, Z has value 90 degrees, Cos (Z) is zero and sin (Z) has value 1. Thus equation (1) reduces to:
Sin(D) = -cos(L)* cos(A), which can be written cos(A) = - sin(D) / cos(L),………..(2)
And we can write it as cos (A)= sin(D)/cos (L), if we measure A from North direction clockwise, i.e. East is 90 degrees. So, for D and L given we have A, measured from North direction. A gives the Sunrise azimuth angle, as each 15 degrees equal to one hour of time, measured from midnight onwards. For example, find the summer sunrise direction and time for Dublin:
cos(A) =sin 23.5/cos 53.5 (in degrees) = 0.398749/0.594823 =0.670366.A has value 47.90 degrees X 4 = 191.62 minutes of time after midnight (North direction) =3h11.6m a.m.
To use equation (1), see the column headings and function expressions on the Excel worksheet. D has value 23.5 for midsummer (Dublin). All trig calculations must have radian measure of angles, hence the columns of radian values. L has value 53.5 degrees. Angle A (azimuth) is progressed backwards in degrees negative from noon of value 0 degrees. Find Z from this equation for each specified value of azimuth A. In column I you put in estimated altitude values from noon known value 60 degrees back to 0 at time of sunrise. (At equinoxes, Sun’s noon altitude is given by 90-latitude 53.5=36.5; add 23.5 more for Sun coming up by 23.5 degrees (declination) so as to have noon altitude 60 degrees at midsummer).
Now calculate value of right-hand-side (RHS) of (1), and subtract it from left-hand-side to give error. Calculate value of Z-differential of RHS, which is –sin(L)*sin(z) – cos (L)*cos(Z)*cos (A), and divide the error by the latter, to give a Z correction, added on to first Z estimate. (This approach is called Newton’s Method of solving equations.) Re-calculate the RHS of (1) and again find error. These resulting errors are so small (see column R) that a second iteration (to find a further Z increment) is not warranted. So the amended Z is our solution in each row; subtract it (in degrees) from 90, and this is the required altitude (angle of elevation) of the Sun given by the model for that time of morning (as A expressed in time before noon) and shown in column X.
Obviously, Sun’s altitude calculations could be made for other times of year, for any specified Sun declination value D. The Sun’s declination throughout the year is shown on the “Ecliptic” curve on star maps. Just now (early August) it has an approximate value of 12.5 degrees, well below the 23.5 degrees of 21 June.
Postscript (21 August 2009)The right-hand columns Z to AI of the Henry Excel worksheet give detailed calculations of the average per-hour Sun intensity, midsummer and midwinter. In the note “How warm is the summer sun!” rough estimates of per-hour average square- metre areas were given as 3.39 for midsummer and 11.35 for midwinter. Revised figures are 3.419 and 8.587, respectively, as shown on the Excel worksheet. The midwinter area is considerably reduced, due to the Sun’s apparent curve (as given by the model) being well above the previous assumed straight-line movement from sunrise to noon, thus giving larger sine values with matching smaller 1/sine values and thus smaller areas.
These new results are calibrated to match 15-minute intervals from sunrise to noon, but again omitting first 1 degree altitude (elevation) of the Sun after sunrise. The time-factor multiplier is taken to be (for each 15-minute interval) the average time per 1 extra degree of the Sun’s altitude, based on the model results. The relative per-hour midsummer/midwinter intensity is now 2.512, given by 8.587/3.419.

How warm is the Summer Sun

HOW WARM IS THE SUMMER SUN!

A note by Eamon Henry; 27 July 2009

Foreword: This brief note compares the radiation intensity of the midsummer Sun with that of the midwinter Sun. We need first to define that we mean, and next describe how to do it. Results are then given, first a mid-day (noon) comparison, and then an average per hour comparison. A technical appendix gives the geometric (trigonometric) measurement background.

The General Background: We consider Dublin as being about 53.5 degrees (Latitude) north of the Equator. Some facts (without proof) are as follows. At the Spring and Autumn equinoxes, at Dublin the Sun at noon has about 36.5 degrees elevation above the south horizon, given by (90 less 53.5) degrees. At midwinter the Sun will be only about 13 degrees above the south horizon at noon, midway through a day about 7.5 hours long. At midsummer, the Sun at noon will be about 60 degrees above the south horizon, midway through a day about 17 hours long.
These differences in the Sun’s mid-day elevation are due to the fact that on its orbit around the Sun, the Earth’s axis points the northern hemisphere 23.5 degrees lower (than at the Equinox) at midwinter, and 23.5 degrees higher at midsummer, as measured against the plane of the Earth’s orbit.

EXPERIMENT DESCRIPTION:

We imagine a clear sky, and a square window of edge 1 metre long, and thus of area 1 square metre, always pointing directly at the Sun, and thus tilting back and rotating as the Sun rises and gains height. Directly beneath the lower edge of the window we imagine a horizontal floor on which the Sun shines through the window. As the Sun climbs, the sunlit floor area gets gradually smaller, implying more intense radiation. This sunlit area, during time, and of dimension area multiplied by time, is our measure of the Sun’s intensity. Since the window is 1 metre wide, this sunlit rectangle is measured by the length of its edge on the floor.
Because the area is very large for sunrise and shortly after, I omit effects from sunrise to 1 degree elevation of the Sun above the horizon. In treating a day, I assume symmetry, in the sense that effects from sunrise to noon are mirrored (in reverse) from noon to sunset. I also assume that the Sun’s radiation power as such is the same on both days, namely midwinter and midsummer.

RESULTS:The easiest result is to compare the floor area sunlit at noon on midwinter day (21 December) with that on midsummer day (21 June). The midwinter area is 4.445 square metres, as compared with midsummer area 1.155 square metres, giving a ratio 3.85. This indicates that this midsummer radiation intensity is 3.85 times that of corresponding midwinter intensity.
The average per hour areas for the same two days are as follows: midwinter 11.35 square metres , as against midsummer 3.39 square metres, giving a ratio 3.35.
In other words,t he noon intensity at midsummer is 3.85 times that of midwinter, while the per hour average intensity at midsummer is 3.35 times that of midwinter.

TECHNICAL APPENDIX:The vertical cross-section form of the supposed experiment is a right-angled triangle of hypotenuse the length of the sunlit floor, with one side of the 90 degree angle the vertical cross-section of the window pointing directly at the Sun, of length 1 metre, and the other side the line from the window top to the far edge of the sunlit floor. This latter side makes with the floor-line the Sun’s angle of elevation, say A degrees. The sunlit floor length is 1/sine(A), same as cosecant (A). For a constant width of 1 metre, cosec (A) also measures the sunlit floor area.
So, we need to measure cosec(A) multiplied by time, assuming for midwinter a steady increase of elevation from 0 to 13 degrees during 3.75 hours, while omitting cosec(A) values for A between 0 and 1 degrees. However, the noon comparison as such does not involve time , but is simply cosec (13 degrees)/ cosec (60 degrees). For those who know Calculus (Differential and Integral), the integral function of cosec(A) is log (cosec(A) – cotan (A)). The logarithm is to base e, of value 2.718282 to 6 decimal places. For midwinter day, the integral function will be used as the definite integral for A in the range 1-13 degrees. Please note that this assumed same length of time for the Sun elevating through each degree of the stated range is only approximate, implying a straight-line movement, whereas the apparent path of the Sun is a bending curve.

For Midwinter Day:The definite integral value for 1-13 degrees is 2.5744, needing scaling up by 57.2987 (the value of cosec(1 degree)), to give result 147.5098. This has an average midwinter time of 0.28846 hours per degree, obtained by 3.75/13. The area X time product is 42.55, which divided by 3.75 hours gives the average of 11.35 square metres per hour.

For Midsummer Day:I take three separate measures for the 8.5 hours period from sunrise to noon:1. Sunrise to due East during 2.5 hours, for elevation 0 to about 23 degrees:2. Due East to SouthEast 23 to 47 degrees, during 3 hours;3. SouthEast to South 47 to 60 degrees during a further 3 hours to noon.
These approximate angles have been calculated by me, by using a vertical foot-ruler and its shadow length on the floor, to give a tangent value of angle A, at local times 6.00 and 9.00 hours, some 25 minutes later than GMT, for Dublin taken as 6.25 degrees West longitude. The related three separate integrals for cosec (A) give a better estimate than a single average (per degree elevation) calculation.

These three scaled-up integrals (scaling factor being same 57.2987) have values 180.594 for 1-23 degrees, with per degree multiplier 0.1087 hours; 43.533for 23-47 degrees with per degree multiplier 0.125 hours; and 16.238 for 47-60 degrees, with multiplier 0.2308 hours. The combined area X time result is 28.82 for 8.5 hours, giving an average 3.39 square metres per hour.

Without Calculus, the cosecant-based areas can be calculated by using detailed tables such as those for cosecant shown on pages 12-13 of F. Castle’s book “Five Figure Logarithmic and other tables” (Macmillan,1969). For values shown at 0.1 degree intervals, the sum of 11 cosecant values for range 1 to 2 degrees is 440.552, giving an average 40.047. Similarly, the average cosecant value for range 2 to 3 degrees is 23.303, and so on. The sum of these averages for the range 1 to 13 degrees is 147.62, which agrees closely with the Integral value 147.5098 quoted above.

BELOVED HERRING
By Eamon Henry. 3 March 2009

Introduction: This piece of Irish-language verse comprises number 81 in the 1974 anthology EIGSE, compiled by Breandan O Conaire, with publishers Mac Goill agus Macmillan.
The title of the piece is “Mo-chean do theacht, a scadain”, and its author is unknown. The translation into English is my own.
It can be dated to within the 17th century. From its content-matter, its author was probably a student for the priesthood at some Irish College abroad, such as Louvain or Bordeaux. It comprises a mock-serious poem of praise of a salted herring which the author is about to eat, this being the Lenten season. The herring is addressed as a noble young human, whose friendship is highly valued by the poet. No more need be said by way of description here. The translation follows, and renders meaning and sense rather than a tight literal version.

Beloved Herring:
Beloved herring, come at last, draw near, you noble youth. Good health to you a hundred times, you’re welcome here, in truth! Though salmon of the Boyne be good, for you this ode I pen To sing your praise, on oath I swear, upon my soul! Amen!
Oh you of faultless body smooth, sincere your friendship bond!A friend like you I never had, may my response be fond. Did Ireland’s nobles judge your worth, among three kinds of fish, O’er pike and salmon you’d be king, and reign upon a dish!
For Conan of old story lore, among the coastland ground From here to Greece a better fish than herring never found. Oh merry gentle herring friend, of Lent the patron saint, Son of my friend who came last year, your absence left me faint!Though many of your family did last year reach my plate, As clergy’s friend you must forbear to seek revenge or hate!
Oh herring, merry salted one, always of cheerful mien, Your coming gives me purest joy, I love this happy scene! From start of penitential Lent ‘til Easter dawn be here I drink your health in water plain and thank you for good cheer.

I want to be like Mommy

Submitted by: Eamon Henry

Why Parents should always check their children's homework before they hand it in:...Submitted by a 5 year old girl for a homework assignment ...sheer class!!

After it was graded and the child brought it home, she returned to school the next day with the following note:

Dear Ms. Davis,
I want to be very clear on my child's illustration. It is NOT of me on a dance pole on a stage in a strip joint.
I work at B&Q and had commented to my daughter how much money we made in the recent snowstorm.
This photo is of me selling a shovel.
Mrs. Harrington

ASITS Photo Competition winner

"Steps"

Congratulations to ASITS member Sophie Shaw who won our recent Digital Photo competition with her lovely picture "Steps". Sophie won first prize of a Digital Photo frame which was kindly donated by Peats world of Electronics.

Thanks to everyone who took the time to take part we were thrilled with the response and the fantastic pictures that were sent in by members, well done Sophie.