At this time of year, my thoughts often turn to….

…the sinusoid?

The sinusoid is a fancy name for a mathematical relationship that is periodic – like the tides or length of day. Sinusoids are also useful in modeling alternating electrical current, light waves, and sound waves.

Below is graph of function y=sin(x)

I became interested in calendrics when I investigated the occurrence of Friday the 13th. Keeping track of time is an almost universal cultural concern.

The Mayan civilization produced a sophisticated mathematical system that was used for both astronomy/astrology and calendrics (which are obviously closely related).

Moon phases and eclipses and the transits of Venus are all contained in the Dresden Codex, one of the few pieces of Mayan knowledge that survived the destruction of their libraries put in motion by Bishop Diego de Landa in 1562.

I studied the Mayan calendar briefly a few years ago because of its complexity.

I was intrigued by the combination of a 365-day solar calendar with a 260-day ceremonial calendar.

The Mayan 365-day solar calendar was comprised of 18 20-day months and 1 short 5-day month. The short 5-day month (Wayeb or Uayeb) may have been used as an intercalary period to make up for the fact that the solar year is not exactly 365 days.

The structure of this solar calendar puzzled me.

Why would the Mayans choose a solar calendar in which there is almost no difference in the calendar from one month to the next?

Granted, they had the ceremonial calendar to denote important ceremonial days. But, for some reason, I was bothered by the structure of the Mayan solar calendar.

I decided (quite without evidence of any kind) that this calendar was a result of the fact that the Mayan homeland on the Yucatan peninsula is situated in the tropics at about 20° north latitude. Consequently, the length of day is fairly constant, never varying more than an hour or so from the 12 hour equinoctal day and night.

Above about 40°N latitude, the change in the length of day from winter to summer solstice is quite noticeable and above about 45°N, the extended sunlight of the summer solstice and darkness of the winter solstice has an undeniable effect on human beings living in or visiting these regions.

I decided that a good calendar for people living above 40°N latitude would acknowledge the importance of the length of day.

The sinusoid is a good function to model the change in the length of day throughout the year. If we make certain adjustments to the original formula y=sin(x), the result models the length of day for a particular region quite nicely.

The equation for the graph below is

y=12+3.5sin((2*PI/365)*(x-80))

In this graph, the x values represent the number of days since January 1st for a particular year and the y values represent the number of hours of sunlight for that day.

The first thing that is important about this relationship is that it is NON-LINEAR. This means that the rate of change of the length of day is not constant.

In this particular application this means that we gain more light from March 1 to March 7 than we do from June 1 to June 7.

Around the winter and summer solstice, we are hardly gaining or losing any light at all. At the Spring and Fall equinoxes, we gain and lose the most light we will gain or lose all year.

Based on this information, I started to look at where the changes in amount of light gained or lost are happening.

If we start with the Winter Solstice, we see that for two weeks on either side of the solstice, there is almost no change in the length of day. After this period there are four weeks of gradually increasing daylight followed by another four weeks of slightly greater increasing daylight.

This brings us to the end of February on the standard Gregorian calendar, which corresponds to an x value of about 59(=31JAN+28FEB) on the graph above.

Notice that, at this point on the graph, the curve nearly straightens out for about 40 days. This is the six week period that lasts from early March to mid-April. Three weeks previous to the equinox and three weeks following the equinox, we gain the most light that we will gain all year.

After this six week period, we continue to gain daylight, but at a slightly reduced rate. Four weeks of gradually reduced gains followed by another four weeks of even smaller gains in length of day bring us to the end of the first week in June.

Two weeks before the summer solstice.

For these two weeks prior to the summer solstice and for two weeks following the solstice, we hardly gain or lose any daylight at all. Although these are the longest days of the year, we are hardly gaining or losing any light at all.

Two weeks after the solstice, the loss of daylight begins to slowly pick up speed for four weeks, and then we lose even more daylight in the following four weeks. This brings us to the end of August, three weeks before the Autumnal Equinox.

For three weeks before the equinox and three weeks after, we will lose the most daylight that we lose all year.

Then, four weeks of gradually declining loss of daylight, followed by four weeks of of even smaller losses of daylight bring us back to the end of the first week in December, which is two weeks before the winter solstice.

Right back where we started.

SO –

Two weeks*Winter Solstice*Two weeks

Four weeks*Four weeks

Three weeks*Spring Equinox*Three weeks

Four weeks*Four weeks

Two weeks*Summer Solstice*Two weeks

Four weeks*Four weeks

Three weeks*Autumnal Equinox*Three weeks

Four weeks*Four weeks

Two weeks*Winter Solstice*Two weeks….

If we add all this up, it comes out to 364 days. Like all calendars, this division of time requires intercalation, or a resetting based on astronomical observation.

This can happen at either solstice. Once the solstice is reached, begin counting the two weeks. This will keep the calendar aligned with the actual procession of the seasons.

Earlier, I mentioned that we gain significantly more daylight from March 1 – March 7 than we do from June 1 – June 7.

Based on this formula, the gain in length of day from March 1 – March 7 is about 21 minutes whereas the the gain in length of day from June 1 – June 7 is about 6 minutes.

This formula I’ve used

y=12+3.5sin((2*PI/365)*(x-80))

is an idealized representation based on a shortest day of 8.5 hours and a longest day of 15.5 hours. The actual lengths of day vary with latitude and, in Portland, Oregon, are closer to 8 hrs. 40 min. for the shortest day and 15 hrs. 40 min. for the longest.

on December 27, 2011 at 12:55 PM |JimI have a hard time grokking why the daylight hours follow a sine curve. Here are my attempts.

What helps me the most is to visualize the circles of latitude as hoops tilting into and out of the earth’s shadow as the earth orbits. Inscribe a square in the hoop so that its edges are square with the earth’s axis and with lines crossing earth’s orbit at the equinoxes and solstices. Consider the line connecting the solstice positions to be parallel to the y-axis of a Cartesian plane. Between the mid-Winter and mid-Spring, and again between mid-Summer and mid-Fall, the shadow line separating the sunlit from the darkened halves of the earth will travel along the y-axis of the squared circle more rapidly than at other times of the year, bringing the hemispheres into and out of darkness at a more rapid rate. (To help see this, imagine a bug crawling from 0 to π/2 on a unit circle with an inscribed square’s corners touching at π/4. Note that a line dropped from π/4 on the orbit to the x-axis is 71/29 the length of a line drawn from that x-axis point outward along the x-axis to the circle. His y-axis change of position will be 71/29 of his x-axis change of position over the course of his walk from 0 to π/4, and then the reverse for the rest of the journey to π/2. This is analogous to the earth’s shadow line’s migration along the portion of the hoop subtended by the y-axis side of the inscribed square).

The daylight hours follow a sine curve because the y-component of the hemisphere surface’s orbital travel is the component that takes its pole into the light in the Spring and out of the light in the Fall. Assuming a complete orbit is 2π and the northern Spring equinox orbital position is 0, the y-component of the travel from 0 to π/4 (mid-Spring) is .71 radians, 2.4 times greater than the .29 y-vector radians from π/4 to the Summer solstice at π/2.

The y-component of the Pole’s travel is in the direction of the sun. The path into and out of the sunlight is a linear path along the y-axis, although it is effected by orbital motion. The y-axis travel is greatest at the equinoxes, and tails off from them until the solstices. Thus daylight hours increase and decrease more rapidly as a function (sin) of their proximity to 0 and π radians (the equinoxes).

on March 7, 2012 at 4:10 PM |richbeveridgeNice – I haven’t worked this quite through, but it makes sense.

I was just thinking about this today as we’re coming into the six weeks around the equinox where we’ll gain the most light all year…

I’m not clear on where the 79/21 comes in – I may not be visualizing this correctly. The way I have it drawn is a square inside the quarter circle between 0 and pi/2, with sides of length 1/sqrt2 and diagonal running from the center to pi/4 (also a radius) =1, which makes their ratio (sqrt2)+1. If you get a chance, let me know where I’m going wrong here.

Thanks for the input!

on March 7, 2012 at 8:08 PM |bojimboThanks for your response. Regarding 71/29: If the orbit were a unit circle and the Spring equinox occurs at theta = 0 on the unit circle, then at theta = pi/4, the sine is 1/2 sqrt(2), approx. .71. The remainder of the y-dimension distance to the Summer solstice (pi/2) is sin(pi/2) – .71, i.e. 1 – .71 = .29. (My square touches the orbit at all 4 corners). So there’s more y-dimension travel in the eighths of a year preceding and following the equinoxes. The hard part for me to visualize is why that correlates with more rapid change of daylight hours per day. If I consider that the solstices (max/mins) are on the y-axis, then I guess it makes sense that y-dimension travel correlates with delta(daylight).

on March 8, 2012 at 10:41 AM |bojimboOk, I see my last post isn’t up yet, but your response got me thinking about this again and I think I’ve clarified, for myself anyway.

Why do the daylight hours follow a sinous curve (number of daylight hours is a sinous function of time)?

From the point of view of a stationary earth, the effect of its orbit is that the sun orbits the earth. Changing that point of view to one where both the earth and the sun are stationary, the earth’s axis can be seen as wobbling such that the poles follow circular paths into and then back out of the sunlight. The poles’ change of position in the dimension of the axis between the circle’s two solstice positions would be the movement into and out of the sunlight. The shadow line would fall along the axis between the two equinox positions. The rate of change of daylight hours would be greater between mid-winter and mid-spring, and again between mid-summer and mid-autumn. If we think of the circle as a unit circle with the pi/2 position being full tilt toward the sun (summer solstice) and the shadow falling from 0 to pi (the x-axis), we see that the poles’ (and hence the hemispheres’) distance from the shadow line (the x-axis) is a sinous function of its movement along the circular path.

on March 8, 2012 at 11:13 AM |bojimboDelete “(and hence the hemispheres’)” from the last sentence above. The exposure of the hemispheres to the sun is a function of the poles’ distance from the shadow line, but not the same thing.