Van Brummelen Chapter 9: Navigating by the Stars

Table of Contents

1. Pappas Question of the Day
2. Van Brummelen Chapter 9: Navigating by the Stars
3. Exercise 9.5
4. Exercise 9.9
5. This Isn't a Blaugust Post!
6. Conclusion

Pappas Question of the Day

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

Find x.

Almost all of the given information for this problem is in the diagram, and so I have no choice but to go into a long description. The main figure is a quadrilateral. Two of its opposite angles are a degrees while the other two are b degrees. One of its sides is x while another is 15. Finally, from one vertex (between the x and 15 sides) there are two segments drawn to the remaining two sides. The segment closer to the x is labeled 6, while the segment closer to the 15 is labeled 10.

First of all, we notice that the quadrilateral is a parallelogram because opposite angles are congruent (Lesson 7-7, pgram test/Sufficient Conditions for a pgram part b).

Now I assume that Pappas intends the 6 and 10 to be heights of the pgram so that we can use the pgram area formula -- 10 is the height to the base x, and 6 is the height to base 15. In other words, the 10 and 6 form right angles with their respective bases. But to me, the assumption that these are right angles is unwarranted, especially since they don't even look like right angles on the page (more like 95 or 100 degrees the way I see it).

Without the assumption that these are right angles, there is insufficient information to determine x. So we'll just assume that they are, since I know that Pappas wants us to use the formula A = hb.

Anyway, this is a Pappas favorite -- either side of a pgram can be taken as a base, and so there's two ways to find its area. And the area of the pgram taking one side as the base must equal the area taking the other side as the base, since after all, the area is the area:

10x = 6(15)
10x = 90
x = 9

Therefore x = 9 -- and as I explained yesterday, today's post is dated the ninth.

As it turns out, there's a way to find x even if the angles between the so-called "heights" and the bases aren't 90, provided they equal each other. Along with a pair of congruent angles from the pgram, we have two triangles that are similar by AA~. This allows us to form a proportion, which after we cross multiply, becomes the same 10x = 6(15) that we found earlier, and so x = 9 as before.

By the way, it appears that coming up this weekend, I caught another error on the Pappas calendar:

The polynomial long division of 4x^2 + ?x + 13 by 2x +1 has remainder 7/(2x + 1).

This problem is dated the 11th. Instead of finding the answer, let's just prove that 11 is impossible:

Divide 4x^2 + 11x + 13 by 2x + 1:

                          2x         
2x + 1 | 4x^2 + 11x + 13
             4x^2 +  2x
                          9x  + 13

At this point we must divide 9x by 2x -- yet 9 isn't evenly divisible by 2. Notice that polynomial long division is typically an Algebra II problem (although it might appear occasionally in Algebra I). In these high school classes, every single problem is set up to avoid fractions in the quotient -- either the divisor has leading coefficient 1, or else the leading coefficient after each subtraction step is divisible always by the leading coefficient of the divisor.

In fact, suppose we're dividing 2x + a into 4x^2 + bx + c. The first term of the quotient is 2x, so let's assume the quotient is 2x + d and reverse it by multiplying:

(2x + a)(2x + d) = 4x^2 + 2(a + d)x + ad

The middle coefficient b = 2(a + d), so it's even -- which means it can't be 11. This means that it's impossible for the answer to be the odd number 11 -- yet Pappas posted it on the 11th.

The remainder of 7 doesn't matter, since this only affects the constant, not the coefficient of x. In other words, c = ad + 7, but b = 2(a + d) -- therefore it's still even (assuming, of course, that all of the coefficients in the entire problem are integers).

In fact, the correct answer to the Pappas question is 14, not 11:

                          2x  + 6  
2x + 1 | 4x^2 + 14x + 13
             4x^2 +  2x
                        12x  + 13
                        12x +  6
                                   7

Van Brummelen Chapter 9: Navigating by the Stars

Chapter 9 of Glen Van Brummelen's Heavenly Mathematics is called "Navigating by the Stars." Here is how it begins:

"B.M. Brown's complaint in the previous chapter against Cesaro's remarkable approach to spherical trigonometry might have been made by an astronomer or navigator. For the practitioner already in command of the important theorems and looking ahead to their uses in science, a pit stop to examine alternative approaches is a restless, impatient exercise."

Van Brummelen writes that early Renaissance sailors originally used Euclidean trig to navigate -- this was called "dead" reckoning. He tells us that "dead" was short for "deduced," but if they weren't careful, "dead" meant "dead":

"During the Age of Exploration, an error of several miles easily could be difference between a successful passage and death, either by sailing past an island containing needed provisions, or by contending with dangerous rocks off shore."

As the author points out:

"Finding one's terrestrial latitude at sea is relatively easy: measure the altitude of the North Star above the horizon."

But finding one's longitude was much more difficult:

"So the problem of longitude remained vital to western European nations' prosperity and security. Several astronomical approaches were attempted, especially using distances measured from the center of the Moon to the Sun, a planet, or some reference star."

Van Brummelen explains that in the 18th century, the longitude problem was solved. John Harrison, a British clock maker, realized that if he could build a clock that could keep accurate time on a moving ship, then he could determine the time zone -- and thus the longitude -- simply by comparing the time on the clock to the position of the sun in the sky. The author doesn't pursue this further since it has nothing to do with spherical trig, but he does point out that there was an A&E miniseries made about this story (Longitude, which first aired the day after Y2K).

"We conclude our voyage through spherical trigonometry by exploring one of the most common techniques of determining one's position at sea, the Method of Saint Hilaire, which revolutionized navigation in the late 19th century."

Here is Van Brummelen's description of this method:

"It is early in the evening of June 22nd, 2010, and we are sailing our ship eastward to the west coast of North America (figure 9.2). By dead reckoning we have a rough idea of our current position, known today as the assumed position or AP. In our case it is phi = 47 degrees 30 minutes N, lambda = 126 degrees 45 minutes W."

But as the author reminds us, the AP might not be correct, so our navigational skills are required.

"The sun has just set, and Venus is a bright evening star trailing the Sun in the western sky. Meanwhile, just west of south, Spica is shining brightly. So their azimuths (the direction of the object along the horizon measured from the north point; see figure 9.3) differ by about 90 degrees. We check our chronometer set to Greenwich Mean Time; conveniently, it reads exactly 5:00 AM on June 23rd, 2010. Using our handy sextant, we measure the altitudes of our two celestial bodies; for Venus we get h_O = 16 degrees 25.1 minutes, and for Spica h_O = 28 degrees 14.1 minutes.

"Now, since we are very unlikely to be exactly at the AP, our values for h_O will not quite match the altitudes at the AP; it is these differences that will allow us to fix the ship's position. So our next task is to compute the altitudes h_C of Venus and Spica at the AP, as well as their azimuths Z. So we have no choice but to compute Z.

"We appeal to the astronomical triangle, defined by connecting our star, the North Pole P, and the zenith Z (figure 9.4). The sides of this fundamental triangle are all familiar quantities: the complement of our known latitude, the complement of the star's known declination, and the complement of the star's sought altitude."

Van Brummelen uses a bar (vinculum) to denote "complement of." On this blog, I'll use strikethrough so that the three sides of the astronomical triangle are phi, delta, and h_C.

"Two of the angles are useful as well: Z is equal to the star's azimuth, which becomes clear if we extend both of the sides departing from Z down to the horizon; and the angle at P is the star's local hour angle t. (The third angle, called the parallactic angle, will not concern us here.)

"For Venus (as well as the Sun, Moon, and other planets), consider figure 9.5. Place point M at the top of the circle, representing the local meridian, and draw a radius connecting M to the center. Next place Greenwich G on our diagram; since our assumed longitude is lambda = 126d 45' W, Greenwich's meridian is 126d 45' east of ours. We turn next to the Nautical Almanac (see figure 9.6); it tells us that the Greenwich hour angle GHA of Venus at our time is 212d 58.2'."

At this point the author mentions an online equivalent of the Nautical Almanac. As it turns out, the URL has changed since he wrote this book:

http://reednavigation.com/lunars/nadata_v5.html

"So we place Venus 212d 58.2' counter-clockwise from Greenwich. From the diagram, then, we see that the local hour angle is t = 212d 58.2' - 126d 45' = 86d 13.2'. For Spica (or any star) the hour angle process involves an extra step. In figure 9.7, draw M and G as before."

Here the author basically tells us that we must look up the GHA for the vernal equinox (Aries) first, then the displacement of the star from Aries (SHA, sidereal hour angle). He finds all of these using that same Nautical Almanac. The values he finds here are 346d 15.9' from GHA Aries and 158d 33.4' for SHA Spica. So, measured westward from M, Spica's local hour angle t is -126d 45' + 346d 15.9' - 360d = 18d 0.43' -- and now we may continue.

At this point, Van Brummelen goes into a digression. He tells us that we have enough information to use the Law of Cosines:

cos h = cos delta cos phi + sin delta sin phi cos t.

But the problem is that before calculators and computers, this equation is difficult to solve. So instead of cosines, navigators used two functions called "versine" and "haversine" (half-versine):

vers theta = 1 - cos theta = 2sin^2(theta/2)
hav theta = (1/2)(1 - cos theta) = sin^2(theta/2)

The haversine function is never negative, and it increases from (0, 0) to (180, 1). The author points out that even in the calculator era, haversine may be more convenient than cosine:

cos 0.01 = 0.999999984769
hav 0.01 = 7.615 * 10^-9

Rounded to seven significant figures, cos 0.01 = 1, but hav 0.01 is distinct from zero. Thus hav is less prone to rounding error than cos. Actually, this reminds me of an old joke:

https://www.theonion.com/nation-s-math-teachers-introduce-27-new-trig-functions-1819575558

Most of the functions listed here (such as "gamsine") are jokes. But a few days after this Onion article, Evelyn Lamb wrote an article about real "secret trig" function, such as versine and haversine:

https://blogs.scientificamerican.com/roots-of-unity/10-secret-trig-functions-your-math-teachers-never-taught-you/

(I wrote about Evelyn Lamb's math column nearly 12 months ago, in my August 23rd post.)

Now that we have the haversine, Van Brummelen introduces the Method of Saint Hilaire, named for a late 19th century French naval officer. We are going to find h_C using three known values -- the local hour angle t = 86d 13.2', the declination of Venus delta = +19d 32.4' (from the Nautical Almanac), and the local latitude from dead reckoning, phi = +47d 30'.

Derivation:
We start with:

cos h_C = cos delta cos phi + sin delta sin phi cos t.

Applying the formula cos theta = 1 - 2 hav theta to cos h_C and cos t, we get the ungainly:

1 - 2hav h_C = cos delta cos phi + sin delta sin phi - 2sin delta sin phi hav t.

But cos delta cos phi + sin delta sin phi = cos(delta - phi) = cos(phi - delta). If we replace this latter expression with its haversine equivalent and clean up a bit, we arrive at the haversine formula of navigation:

hav h_C = hav(phi - delta) + cos phi cos delta hav t.

In our case, the formula gives us h_C = 16d 46.3' for Venus (compared to h_O = 16d 25.1'), and h_C = 29d 06.9' for Spica (compared to h_O = 28d 14.1'). The author adds, "Of course, the reader following along with one of those rare calculators lacking a haversine button may feel free to use the Law of Cosines instead."

Now Van Brummelen uses the Law of Sines to find the azimuth Z:

sin h/sin t = sin delta/sin Z.

"So for Venus, from sin Z = 0.98214 we deduce that Z = 79d 09.3' west of North, and for Spica, from sin Z = 0.34829 we deduce that Z = 20d 22.9' west of South.

"Now that Z is known, we can imagine moving forward or backward in that direction on the water's surface along the azimuth line (figure 9.9). As we move, only Venus's altitude (not its azimuth) will change; and if we move forward far enough, we will reach Venus's geographical position, or GP.

"At some point in our journey back and forth along the azimuth line, Venus's altitude will match our observed altitude h_O = 16d 25.1' exactly. This point might be our true position. We will assume that our true position is somewhere on the straight line perpendicular to the place on the line of azimuth where Venus's altitude matches h_O. We then draw the line of position, or LP, at right angles to the azimuth line, and we know that we are somewhere on that line. But how far from the AP should we travel to reach the LP?

"The intercept, the distance from the AP to the LP, is where our method derives one of its names, and it is surprisingly easy to find. Figure 9.10 is the cross section of the universe through the center of the Earth that contains Venus. Form a right triangle by drawing a tangent to the circle at the AP and joining it to the line of sight from the Earth's center to Venus. The angles in this triangle will be 90, h_C, and h_C. The angle at the center of the Earth between the assumed and true positions will be h_O - h_C = h_C - h_O. But this angle, measured in minutes of arc, is equal to the distance on the surface measured in nautical miles! So to calculate the intercept, we need only determine 60(h_C - h_O). In Venus's case the intercept is 21.2 nautical miles; for Spica it is 52.8 nautical miles.

"We are now ready to use a plotting chart, a simple version of which is shown in figure 9.11."

It's worth describing this chart to you. It consists of a circle where all multiples of 10 degrees, from 10 to 360 (clockwise from North), are marked. The x- and y-axis are drawn in, as well as horizontal lines drawn tangent to the top and bottom of the circle. The y-axis is labeled in multiples of 10 nautical miles, from 10 to 50 (the radius of the circle is 60 nautical miles). For some reason, no nautical miles are labeled on the x-axis.

In Figure 9.12, the author fills in the chart so that it fits our problem. The x-axis is our AP latitude, phi = 47d 30' N, and the y-axis is the AP longitude, lambda = 126d 45' W. Now 60 nautical miles is one degree of latitude, so 48d 30'N is the top line and 46d 30' N is the bottom line. But 60 nautical miles is not one degree of longitude -- instead, 60cos phi miles is one degree. The author knows that 60cos phi is the x-coordinate if phi is measured from the x-axis (that is, 42d 30' from North), and so he draws vertical lines on either side there and labels them 125d 45' W and 127d 45'.

Van Brummelen continues:

"Earlier we calculated Venus's azimuth to be 79d 09.3' west of North, so we draw the azimuth line onto our chart. The intercept is 21.2 nautical miles, so we must move that distance away from the center of the circle. But in which direction? As seen on figure 9.10, since h_C > h_O then we must move away from Venus, and if h_C < h_O then we must move toward it. Navigators remember this rule by memorize the phrase 'computed greater away.' Now that we have located Venus's intercept (to the right and a little below the center), we draw a perpendicular. This marks Venus's line of position (LP), and we know that our ship is someone along it.

"Of course, one LP isn't enough to pin down our location, but we had the foresight to make two observations," the second from Spica. This trick of using two observations to draw lines and then finding where the lines intersect often appears on Square One TV: Mathnet, where it's usually referred to as "triangulation."

The author summarizes the results:

"In our chart, we find that our ship is actually around 55 nautical miles northeast of the AP, at about phi = 48d 15' N, lambda = 126d W; that position is indicated in figure 9.13. It's a good thing we have a navigator on board."

Van Brummelen closes the chapter by revealing the actual location of the ship at the time he made these observations -- it's indeed 48d 15' N, 126d W:

"The true position is so close to our fix that the thickness of the lines at the intersection of the two LP's covers both locations. We have pinpointed the ship to a distance of less than 1000 feet."

Exercise 9.5

The formula derived in the previous exercise may be used to build a device called the haversine nomogram, capable of solving some triangles visually. Make a scale as in figure E-9.5.1, where the position of each tick mark corresponds to the haversine of that angle. (The more tick marks you can make, the more accurate your result.) Aline three of these scales in a rectangle opened at the top, as in figure E-9.5.2. Imagine that the triangle has sides a = 87, b = 52, and c = 106. Then a - b = 35 and a + b = 139. Draw a diagonal line from 35 on the left scale to 139 on the right scale. Then draw a horizontal line from the 106 point on the right scale and move down to the bottom scale when you reach the diagonal line. The angle at that place, 115, is the value of C.

(a) Solve the triangle of question 3(b) using a haversine nomogram.

As I mentioned yesterday, summer is almost over, and I'm rushing through these last few chapters just so I can say that I finished this book before summer ends. So I randomly choose only a few questions to do today. Exercise 9.5 seems like a poor choice since it refers to both 9.3 and 9.4, but all I need to do is look up values and equations given in those exercises. We don't need to complete 9.3 and 9.4 to be able to do 9.5.

Meanwhile, the haversine nomogram is an interesting device, even if it's not easy to communicate to you on a blog. Van Brummelen describes how to make one above, but since I'm looking at the one he shows us in the book, I can give you a few more tips:

• First of all, Van Brummelen describes it as a "rectangle," and it is -- but the shape is better described as a square. So we can begin with a sheet of paper and cut out a square.
• The haversine of 0 is 0, and the haversine of 180 is 1. So the lower left corner is (0, 0), the lower right corner is (180, 0), the upper left corner is (0, 180), and the last is (180, 180).
• The haversine of 90 is 1. So we can fold the square in half each way, and then this tells us where to label the points (90, 0), (0, 90), and (180, 90).
• What's the haversine of 60? Well, hav 60 = (1/2)vers 60 = (1/2)(1 - cos 60) = (1/2)(1/2) = 1/4, and also hav 120 = 3/4. So we fold the square again and label the points (60, 0) and (120, 0), as well as (0, 60), (0, 120), (180, 60), and (180, 120).
• But hav 30 = (1/2)vers 30 = (1/2)(1 - cos 30) = (1/2)(1 - sqrt(3)/2) = 0.067. And so it's not easy to find hav 30 or hav 150 = 0.933 easily without actually measuring. And we'll have to label many more points if we want to use it to solve triangles.

Now let's look at the triangle that we're supposed to solve:

a = 52, b = 39, c = 44

We'll follow the author's steps from earlier. We have a + b = 91 and a - b = 13. Now 91 isn't that hard to find (very close to 90), but 13 is squeezed down there between 0 and 30. Since I don't know whether you followed my steps to make a haversine nomogram, I'll just find the haversines:

hav 13 = 0.0128

hav 91 = 0.5087

So the first step is to draw a line from (0, 13) to (180, 91) -- or (0, 0.0128) to (1, 0.5087). Notice that the slope of this line is 0.4959, so its equation is y = 0.4959x + 0.0128.

Now the second step is to find c on the y-axis:

hav 44 = 0.1403

Notice what we're actually doing is plugging in y = 0.1403 and then solving for x:

0.1403 = 0.4959x + 0.0128

0.1275 = 0.4959x

x = 0.2571

Now we must find the angle whose haversine is 0.2571. Well, we just double it (to find the versine), subtract it from 1 (to find the cosine), and then take the inverse cosine:

cos C = 1 - 2(0.2571) = 0.4858

C = 61

The instructions asked us to solve the triangle, which means to find all the angles, not just C.

Angle B:

a - c = 8

a + c = 96

hav 8 = 0.0049

hav 96 = 0.5523

Draw line at y = 0.5474x + 0.0049

hav 39 = 0.1114

0.1114 = 0.5474x + 0.0049

0.1065 = 0.5474x

x = 0.1946

cos B = 1 - 2(0.1946) = 0.6108

B = 52

(Actually B is more like 52.3 or 52.4, but I'm assuming that on a real nomogram, we'd be lucky to discern 52 at all, much less tenths of a degree.)

Angle A:

c - b = 5

c + b = 83

hav 5 = 0.0019

hav 83 = 0.4391

Draw line at y = 0.4372x + 0.0019

hav 52 = 0.1922

0.1922 = 0.4372x + 0.0019

0.1903 = 0.4372x

x = 0.4353

cos B = 1 - 2(0.4353) = 0.1294

A = 83

So our solution is A = 83, B = 52, C = 61. Again, I wonder how we'd do this on a real nomogram, where hav 5, hav 8, and hav 13 are all very difficult to locate on the y-axis.

(b) Explain why this method produces the correct answer. (Hint: use the formula of question 4(b), solved for hav C.)

Again, we didn't do that problem, so let's me just give the formula mentioned in that question:

hav c = hav(a - b) + [hav(a + b) - hav(a - b)]hav C.

And Van Brummelen's hint is to solve it for hav C, so let's do so:

hav c = hav(a - b) + [hav(a + b) - hav(a - b)]hav C

hav c - hav(a - b) = [hav(a + b) - hav(a - b)]hav C

[hav c - hav(a - b)]/[hav(a + b) - hav(a - b)] = hav C

Actually, I'd rather just rearrange the original formula:

hav c = [hav(a + b) - hav(a - b)]hav C + hav(a - b)

Notice that this is in the form:

y = mx + b

The line we drew on the nomogram does have slope hav(a + b) - hav(a - b) and y-intercept hav(a - b), so it matches. Then we plugged in y = hav c and found x = hav C. Therefore it all matches up. QED

(c) Devise a method to use a haversine nomogram to find the third side if two sides and their included angle are given. [Nielsen/Vanlonkhuyzen 1944, 120-121]

Well, let's look at this formula again:

hav c = [hav(a + b) - hav(a - b)]hav C + hav(a - b)

This time, we're given a, b, and c and asked to find C. So we can draw the same line again, except this time we find hav C on the x-axis and then move up to the line to find y = hav C.

Exercise 9.9

(Assumes calculus) Find the derivative of the Sun's altitude with respect to local hour angle. Explain from the result why solar observations taken when the Sun is in the East or West were preferred to when the Sun is in the South (near noon). [courtesy of Joel Silverberg]

What makes this question so frustrating is that nowhere on any of the 192 pages of Van Brummelen's text does he ever state a formula giving the Sun's altitude as a function of the local hour angle. There are several pages referring to the altitude of the planets and distant stars, but not one word referring to the altitude of the Sun.

All I can do is make an educated guess. Earlier in the book (that is, in the chapters we covered last summer), we recall the formula:

sin delta = sin lambda sin epsilon

where delta is the Sun's declination, lambda is the Sun's longitude, and epsilon is the ecliptic. We know that epsilon is a constant 23.44 degrees. And lambda is zero at the spring equinox, 90 at the summer solstice, 180 at the fall equinox, and 270 at the winter solstice.

We find that delta = 0 at the equinoxes, 23.44 (that is, epsilon) at the summer solstice, and -23.44 at the winter solstice. This means that at noon, the sun is directly overhead at the Equator on the equinoxes, latitude 23.44N (Tropic of Cancer) at the summer solstice, and latitude 23.44S (Tropic of Capricorn) at the winter solstice. In other words, the altitude of the sun is 90 degrees.

Now here is where my guesswork begins. If the Sun's altitude is 90 on the summer solstice at the Tropic of Cancer (23.44N), then maybe it's 89 one degree away (24.44N and 22.44N), 88 two degrees away (25.44N and 21.44N), 87 three degrees away (26.44N and 20.44N), and so on. Then 90 degrees away (66.56S, the Antarctic Circle) the Sun's altitude would be zero. This makes sense -- when it's summer on the Tropic of Cancer, there are 24 hours of darkness at the Antarctic Circle.

Thus we obtain the following formula for the Sun's altitude at noon:

h at noon = 90 - |phi - delta|

where phi is the local latitude and delta comes from sin delta = sin lambda sin epsilon.

But now we need the Sun's altitude at times other than noon. Van Brummelen uses the variable t to represent the local hour angle, and it appears that t = 0 at noon. Then t = 90 at 6PM, t = 180 at midnight, and t = 270 at 6AM.

Well, since we already used the fact that there are 24 hours of darkness at the Antarctic Circle in June, maybe the face that there's 24 hours of light at the Arctic Circle in June might come into play.

For example, when it's noon at some point on the Tropic of Cancer, then at noon on the Arctic Circle, the Sun's altitude is 46.88 degrees (2epsilon). But on the opposite side of the Arctic Circle (the midnight side), there's a point that is exactly one quadrant away from the point where the sun is currently overhead. Thus the Sun's altitude is zero at midnight -- and since the altitude is never negative, there are 24 hours of light.

In other words, suppose we let S be the point at which the Sun is overhead. Then the Sun's altitude must be 90 at S, -90 at the antipodal point of S, and 0 at the "equator" of S (that is, the great circle of which S is a pole). In other words, if P is any point on the Earth's surface, then:

h = 90 - SP

The coordinates of S are easy to find -- its latitude is delta and its longitude equals the longitude of P plus the local hour t.

So let's set up a triangle in preparation for the Law of Cosines. As usual, our third point will be placed at the North Pole N. The angle at N is t, the side NP is 90 - phi (or phi), side NS is 90 - delta (or delta), and side SP is 90 - h (or h). Then the Law of Cosines (page 157 in the book and alluded to earlier today) gives us:

cos h = cos delta cos phi + sin delta sin phi cos t

Now we can finally follow Van Brummelen's instuctions and take the derivative:

-sin h dh/dt = -sin delta sin phi sin t

And since h = 90 - h, dh/dt = -dh/dt. This allows us to eliminate that annoying negative sign:

sin h dh/dt = -sin delta sin phi sin t
cos h dh/dt = -sin delta sin phi sin t
dh/dt = -sin delta sin phi sin t/cos h

Notice that here we're treating phi and delta as constants. Of course, phi (our local latitude) is a constant, but delta changes every day, every hour, every second. But delta changes by much less than one degree per day, and so it hardly changes at all during the course of a day. So to keep things simple, we treat delta as a constant. And in fact, to remind us that phi and delta are constants, let's write sin delta sin phi by the positive constant k:

dh/dt = -k sin t/cos h

Finally, let's answer Van Brummelen's question. At sunrise, t is around -90 and -k sin t is positive, while at sunset, t is around 90 and k sin t is negative, and at noon t = 0 and k sin t is 0. This means that the sun is moving the fastest at sunrise and sunset and the slowest at noon. When the sun is moving faster, we can estimate the time t more accurately (since an error of a few minutes is much more noticeable). Therefore observations near sunrise and sunset are preferred.

I'm a bit worried about that cos h term. Near noon, cos h is tiny and so 1/cos h is huge. It might be enough to make the sun move faster near noon, even though sin t is approaching 0. I fear that all of this work is incorrect.

I was considering attempting a third exercise in tonight's post. But I'd randomly selected Exercise 9.7, which requires us to make up our own navigation problem. We're supposed to locate objects in the sky, measure their altitudes h_O, and then follow the Method of Saint Hilaire to determine our actual true location.

But I live in Southern California -- in an urban area where it's hard to identify celestial objects. I think I see one object in the southern sky, and as I mentioned last week, Mars is especially bright tonight --but the object I see is just as likely to be an airplane as the Red Planet. Or it might even be a meteor, since the Perseid meteor shower is always visible the second week in August.

So two questions are enough for tonight's post. I found these questions difficult -- particularly those in Chapters 8 and 9, since I rushed through those chapters during the final week of vacation.

After the exercises, Van Brummelen concludes his book:

"Our tour through the world of spherical trigonometry has ended, but there are countless journeys that may be taken from here."

And after listing several books for further reading, he writes:

"If you care to linger a while in these dusty old textbooks, you will find that the playground of spherical trigonometry contains many more forgotten delights."

It reminds me of why I read Van Brummelen's book and do his exercises in the first place. I enjoy learning about new mathematical concepts -- but it also makes me feel like a student again. Just as I struggle to learn spherical trig, many of our students struggle to learn plane trig. By learning an advanced topic, I can empathize with my students as they learn topics that they find advanced. I find haversines confusing, and our students find cosines confusing. Empathy on my part will be helpful -- that is, if and when I find myself with my own math classroom once more.

This Isn't a Blaugust Post!

As I explained yesterday, I'm not signing my blog up on Shelli's Blaugust list. But I still can't help but contemplate about some of the topics on her list.

Since today's post is dated the ninth, let's look at the ninth topic from Shelli's list:

• Share your #MTBoS Photo Challenge photos (#MathPhoto18)

I don't take classroom photos that often, and I didn't have very many photos to post even back when I was in the classroom two years ago. Thus there's no way I'd be able to respond to this prompt.

But as I wrote yesterday, I regret not signing up for Twitter two years ago when Shelli, the leader of the Blaugust challenge, invited me to join. I notice that the most popular MTBoS Twitter teachers (recall that the T in MTBoS really does mean "Twitter") embed classroom photos in their tweets. So unless I did the same, most of my tweets would have gone unnoticed.

As part of piloting the Illinois State text, we were required to submit photos of students working on projects to Illinois State every two weeks -- in other words, I really did have classroom photos. But most of these photos contain students' faces and so were inappropriate for posting. Submitting student photos to Illinois State is one thing, but I really shouldn't post them to Blogger or Twitter without parent permission.

Notice that for the first project of the year, I posted photos of the mousetrap cars to this blog. I took them after school, so no students were in the photos. All other photos contained students' faces.

I admit that this blog doesn't have very many readers. Most posts only get a few dozen views, but the most popular MTBoS blogs draw hundreds or even thousands of views. I could have created my own Twitter account, posted the few mousetrap car pics that I had, added the #MTBoS tag, and even linked back to my own blog -- then maybe I'd have more followers on both Blogger and Twitter. But I felt guilty about doing so -- this all seemed liked "fishing for page views" (posting photos for the sole purpose of drawing followers). If I can't draw readers to my blog naturally without "fishing," then I'd just accept that this blog isn't worth reading.

But Shelli's reason for my joining Twitter was so that I could find help in improving my class. And so if I had to "fish for page views" to find someone to help me out, then it's worth it. Once again, my first year might have been more successful if I had joined Twitter, fished for views by posting pics, and conversed with more experienced teachers who could help me. And maybe I'd still be in the classroom today, preparing for my third year of teaching.

The mousetrap car photo could have been the first pic I tweeted. During the week of the mousetrap car project, I blogged on Monday, Tuesday, and Thursday, but not Wednesday. And so Wednesday would become a Twitter day -- the day I'd tweet the mousetrap car pics.

One way to obtain tweetable photos is to ask my students to take them. For example, the no cellphone rule could have been something like "no phones unless you're taking a math photo for me." Then if a student took out her phone, I'd punish her unless she took a math photo. I'd remind her that no student faces are allowed, because the photos will be posted to Blogger and/or Twitter. If the student fails to send me the tweetable photo, then she'd be punished for having phones out in class.

Twitter is a two-way street. I can't expect anyone to help me unless I help others. And so on each Twitter day, I must also leave at least one comment for another MTBoS teacher.

But in the end, I didn't create a Twitter account, and I never received the help I needed. Once again, Blaugust reminds me of all the mistakes I made and all the bad choices I regret.

Conclusion

Yes, there are plenty of fires occurring right here in Southern California. One of them is close enough to my districts that outdoor P.E. would be cancelled had it been past the first day of school.

Today is officially my August 9th post. There is a Google Doodle for August 9th, and since it's related to STEM, I do wish to highlight it on this blog. The Doodle features Mary G. Ross, who is considered to be the first Native American woman to work as an engineer.

Ross worked for Lockheed Martin in Northern California and studied math and engineering at my alma mater, UCLA. Her interest was in planning flights to Mars and Venus. Therefore I wonder whether she ever studied any of the spherical trig that we learned about in Van Brummelen's book.

Meanwhile, back in my November 13th post, I wrote about another group of famous mathematicians, namely the Fields medalists. I wrote that the medals were awarded every four years and mentioned that there was an awards ceremony in August 2006.

This means that August 2018 -- this month -- should be another quadrennial award ceremony. And indeed, the ceremony was held on the first of this month in Rio de Janeiro, Brazil.

https://www.nytimes.com/2018/08/01/science/fields-medals-mathematics.html
https://www.nytimes.com/2018/08/02/world/europe/fields-medal-theft-caucher-birkar.html

The four Fields medalists this year are:

Caucher Birkar, 40, England
Akshay Venkatesh, 36, USA (Princeton and Stanford University)
Alessio Figalli, 34, Switzerland
Peter Scholze, 30, Germany

Caucher Birkar is of Kurdish descent. (Did he read a conjecture of Erdos in Kurdish?) But a tragedy has occurred -- as soon as he received the medal, it was stolen. Since then it's been replaced -- I've heard nothing about whether the original medal has been found or not.

Scholze is one of the youngest mathematicians ever to earn a Fields medal, while Birkar is at the maximum age to receive the honor, 40. Right now I'm 37, so in four years when the next Fields medals are awarded, I'd be 41, one year too old to earn one -- bummer!

Oh well, I guess I'll never be a Fields medalist -- or a great space engineer like Mary Ross. But one goal is still possible, I hope -- to become a great teacher, simply to become worthy of MTBoS and Shelli's list of great teachers participating in the Blaugust challenge.

The work to become a great math teacher -- even if it's just as a sub -- starts up again on August 15th, the first day of school in my district and the date of my next post.

And don't worry -- I'm done with most of the extra side topics I wrote about this summer. Spherical geometry is over, unless I stumble upon one of those books suggested by Van Brummelen, or I find some reason to sneak spherical geometry into the high school curriculum. And music, the topic of so many posts dating back to May, is also going away from my blog posts. The only reasons for me to write about music are for holiday music posts, or when I sub in a music class (or I suddenly get my own math class where I can sing math songs).

From this point on, it's back to Geometry. Like last year, we will start out with Michael Serra's Discovering Geometry for Chapter 0, then return to the U of Chicago Geometry text.

Enjoy what's left of your summer!