Longitude and time problems

[Puma Lion]

I am certified to teach English and History but I am trying to develop a cross-curriculum Navigation unit that would integrate math and science with the subject matter that is more familiar to me. I found online a 1901 mathematics text for middle or early high school students ("seventh year") with five problem sets involving latitude and longitude but it does not have an answer key. I think I calculated correct answers for the earlier lessons (although I might ask for advice on rounding numbers or other modern practices). But the last problem set was too complicated for me! Too many variables? I am hoping I can provide some examples of problems (1,2,6 and 9, below) here and you all can tell me how to set up equations or solve a sample problem to show me how to approach the others in the book. Thank you!

Some instructions are included in the previous, slightly simpler lessons:

Lesson 18
Since the earth turns upon its axis from west to east once in 24 hr., the sun appears to revolve from east to west around the earth in the same time. Therefore a circumference (360°) is described by the apparent revolution of the sun around the earth in 24 hr.
Since the sun appears to travel through 360° of longitude in 24 hr., in 1 hr. it appears to travel through 1/24 of 360°, or 15° of longitude; in 1 min. through 1/60 of 15°, or 15’; and in 1 sec. through 1/60 of 15’, or 15” of longitude.
15° of longitude correspond to 1 hr. of time,
15’ of longitude correspond to 1 min. of time,
15” of longitude correspond to 1 sec. or time.
All places east of a certain point have later time, all places west, earlier time.
For example, when it is 10 o’clock a.m. at Philadelphia, it is 11 o’clock a.m. at a point 15° east of Philadelphia; 12 o’clock 30° east; 1 o’clock p.m. 45° east, etc. Again, when it is 10 o’clock a.m. at Philadelphia, it is 9 o’clock 15° west; 8 o’clock 30° west; 7 o’clock 45° west, etc.

Lesson 19 (sample problem provided in the problem set)
1. The difference in longitude between two places is 35° 4’. What is the difference in time?
Since a difference of 15° longitude corresponds to a difference of 1 hr. in time, 15’ of longitude to 1 min. of time, and 15” of longitude to 1 sec. of time, it is evident that 1/15 of the difference in longitude will give a corresponding difference in time. By dividing the difference in longitude by 15, the difference in time is found to be 2 hr. 20 min. 16 sec.

Examples of problems with which I could use help:

Lesson 20
1. The difference in time between two places is 45 min. 30 sec.; the longitude of the one having the faster time is 75° 10’ west. What is the longitude of the other place?
2. The longitude of New York is 74° 3” west. When it is 4 o’clock p.m. at New York, it is about 3:18 p.m. at Cincinnati. What is the longitude of Cincinnati?

6. When it is 4:30 p.m. at Berlin, it is 52 min. 9-1/5 sec. past 10 a.m. at Boston, longitude 71° 3’ 58” west. What is the longitude of Berlin?

9. Chicago is 87° 34’ 9” west. When it is 10:30 p.m. July 24, at Chicago, what time is it at Paris, longitude 2° 20’ 9” east?

Thank you!

 

[Dr.Peterson]

I don't have time at the moment to dig deeply, but here are a couple thoughts before actually solving anything:

First, it's essential to realize that this deals with local time (that is, where the sun will actually be at a specific place), and ignores time zones. I assume you are aware of that; are the students?

Second, for modern usage, longitudes would probably be found using decimal degrees rather than minutes and seconds, in my own experience; for example, if I Google "longitude of Chicago", I am told "41.8781° N, 87.6298° W", though a the first link given shows "41.881832, -87.623177", though I can also see that its coordinates are "41° 52' 54.5952'' N and 87° 37' 23.4372'' W". These are three different ways to express it (and, of course, the differences beyond format are due to choosing different points within the city).

In particular, going all the way to seconds is a little excessive!

Part of my point is that whatever part you find difficult here may go away if you stick with more modern formats.

Can you show us where you having trouble? Is it in converting minutes and seconds, or something else?

 

[Puma Lion]

Thank you for your reply!

It had dawned on me that the longitude format was archaic. One of the lessons in the larger unit will explore the history of the chronometer and the ways that time and astronomy were "competing" to provide solutions for the longitude puzzle. I was thinking that it was good for the students to understand the original notation and methods for calculation before making the connection to the notation used for modern GPS coordinates. (The history also mentions that it could take hours for a navigator to calculate longitude before reliable chronometers were invented. I thought maybe the students should have a taste of that complexity!)

As you point out, though, maybe it is too much to expect to work out equations down to seconds. The 1901 textbook was definitely written for a different generation. It may be that lesson 20 would be too hard for all but the most eager students of mathematics. Maybe I should just rely on lessons 16 through 19. On the other hand, I would like it if a range of teachers could adapt the material for different age and ability groups.

I set this aside a long time ago and took it out again to work on during the virus "stay at home" down time. I should take a closer look at my notes to try to figure out what stumped me. I remember thinking that there were too many steps required to find the answers.

Actually, looking at my notes now, apparently I did make an attempt at some of the problems. Here's what I had for question 1.

1. The difference in time between two places is 45 min. 30 sec.; the longitude of the one having the faster time is 75° 10’ west. What is the longitude of the other place?

multiply time by 15
1 hour = 15 °
.75 hour = 11.25°
11° 15'
1 min = 15"
.5 min = 7.5"
Then add
75° 10'
11' 15"
7' 30"
75° 28' 45"
subtract?
74° 51' 15"

I can see my attempt isn't correct. It looks as if I calculated the value of the 30 seconds in the time difference in terms of seconds of longitude but then mistakenly turned it into minutes instead of sticking with seconds when I was finding the difference between the two points.

I was also confused by the term "faster time" in the question and wasn't sure whether to add or subtract. If I were using this question in the lesson, I would probably reword that question. It should probably say something like "... the longitude of the eastern place is 75° 10' west. What is the longitude of the western place?"

I may be ok with my calculations for question 2. Here's what I had:

2. The longitude of New York is 74° 3” west. When it is 4 o’clock p.m. at New York, it is about 3:18 p.m. at Cincinnati. What is the longitude of Cincinnati?

42 min. x 15 = 630
630 / 60
74° 0' 3" New York longitude add
10° 30' (42 min. time difference)
84° 30' 3" Cincinnati longitude

I'll go back to see if I can figure out whether I was on the right track with some of the other questions.

 

[Dr.Peterson]

I hadn't focused on the fact that your main topic is navigation, so that of course you are talking specifically about apparent local time, and you do want to show some idea of precision. It's also a good idea to have hard stuff as at least an optional unit. I should point out, though, that the difficulty of finding latitude before accurate clocks was an entirely different thing than the difficulty here, which is largely a matter of unit conversion. Before clocks, depending on what method that statement referred to, it might be something like astronomical observations of the moons of Jupiter! What you're asking about is not really difficult math; it is just more work than we find necessary today.

Let's look at problem 1. First, "faster" time presumably means that their clock is "faster" in the sense of being ahead of the other. So that location is farther east. So we'll be adding, not subtracting (I think). Second, your work is otherwise correct, except for one major silly error: you somehow changed 11° 15' to 11' 15" before adding.

But I would have done it quite differently. My instinct is to convert complex units into single units before doing math on them, and then to convert the resulting single unit to a complex one. So I would first convert 45 min 30 sec to hours:

30 sec / 60 = 0.5 min
45 min + 0.5 min = 45.5 min
45.5 min / 60 = 0.758333... hr

Then find the longitude difference:

0.758333... hr * 15° .hr = 11.375°

Then convert to mixed units (or I could convert the given longitude to decimal, and add in decimal form instead):

0.375° * 60 = 22.5'
0.5' * 60 = 30"

So the longitude difference is 11° 22' 30"

Add: W75° 10’ + 11° 22' 30" = W86° 32' 30"

But I did it that way because I could use a calculator, and decimals were not a problem. In the old days, working separately with degrees, minutes, and seconds would feel safer. The methods the book teaches are probably meant to avoid decimals and resulting approximations.

For #2, I probably would similarly have converted time to hours first; but what you did may be better practically.

I do find that it helps to think of the work in chunks: the whole process is just "convert, multiply, convert", in three main steps, and the individual calculations you did are just parts of those steps. So it doesn't feel like so many steps when they are organized in your mind.

 

[Puma Lion]

Thank you! This is very helpful. I want the kids to use methods that draw on their regular ways of approaching math problems, so the single units simplification is probably a good instinct. Maybe I will do a few problems showing both types of work and let the students choose the approach they like best.

Thanks again!