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### Movement on a Sphere, Charted on a Spreadsheet

```Date: 11/20/2010 at 11:21:37
From: Alan
Subject: traversing a sphere given specific inputs

I have a sphere to traverse.

I start at a given latitude and longitude. I face a given angle relative
to north. I travel forward a given distance, measured in a value relative
to revolutions of the sphere.

How do I determine my new position and orientation?

For example, I start at a latitude halfway between the equator and the
north pole, and a longitude of 0. I start facing northeast (pi * 0.25) and
I travel forward half a revolution. I now need to determine my new
latitude, longitude, and angle relative to north given the 4 values
mentioned above.

The radius of the sphere is irrelevant in this case.

I don't know how to construct the equation. I initially used this
approximation:

end_latitude  = start_latitude  + (sin(angle) * distance)
end_longitude = start_longitude + (cos(angle) * distance)

But this only works well near the equator.

It might be different in crucial ways, but my problem seems similar to
this problem:

http://mathforum.org/library/drmath/view/51816.html

With this in mind, I used Doctor Rick's (hopefully better!) equation to
create a speadsheet showing my approximation. The answers seem wrong,
though; for example, traveling no distance results in the end position
being different from the starting position.

```

```
Date: 11/23/2010 at 01:34:08
From: Doctor Vogler
Subject: Re: traversing a sphere given specific inputs

Hi Alan,

Thanks for writing to Dr. Math.

Have you already seen our section on latitude and longitude?

http://mathforum.org/library/drmath/sets/select/dm_lat_long.html

The formula that gives the distance between two points is found at

http://mathforum.org/library/drmath/view/51722.html
http://mathforum.org/library/drmath/view/51879.html

The formula for the bearing can be found at

http://mathforum.org/library/drmath/view/55417.html

The result is this. We use these variables:

lat1 = starting latitude
lon1 = starting longitude
lat2 = latitude at the destination
lon2 = longitude at the destination
d = distance divided by the radius of the sphere
b = bearing (angle from north at the starting point)

Then the distance (haversine) formula is

cos(d) = cos(lat1)*cos(lat2)*cos(lon1 - lon2) + sin(lat1)*sin(lat2)

The bearing formula is

sin(lat2) - sin(lat1)*cos(d)
cos(b) = ----------------------------
sin(d)*cos(lat1)

So you can solve the second equation for sin(lat2) ...

sin(lat2) = cos(b)*sin(d)*cos(lat1) + sin(lat1)*cos(d)

... and then solve the first equation for cos(lon1 - lon2):

cos(d) - sin(lat1)*sin(lat2)
cos(lon1 - lon2) = ----------------------------
cos(lat1)*cos(lat2)

Take an inverse sine to determine lat2, and then take an inverse cosine to
determine lon1 - lon2, and subtract from lon1 to get lon2.

In your example, you have

lat1 = 45 degrees = pi/4 radians
lon1 = 0
b = 45 degrees = pi/4 radians
d = pi radians (half a revolution, or pi*radius)

Then

cos(b) = sin(b) = cos(lat1) = sin(lat1) = sqrt(2)/2
sin(d) = 0
cos(d) = -1

So

sin(lat2) = cos(b)*sin(d)*cos(lat1) + sin(lat1)*cos(d)
= -sqrt(2)/2

This means that lat2 = -pi/4 (since latitudes are always between -pi and
+pi, the inverse sine is unambiguous).

Then we compute

cos(d) - sin(lat1)*sin(lat2)
cos(lon1 - lon2) = ---------------------------- = -1
cos(lat1)*cos(lat2)

This means that lon1 - lon2 = pi, and since lon1 = 0, that means that
lon2 = pi. In this case, since +pi and -pi are the same angle, the inverse
cosine is also unambiguous. More often, though, you won't be able to tell
the sign of the angle (lon1 - lon2) from this formula, because the
distance is indifferent to the sign of (lon1 - lon2). I'll leave it to you
to figure out how to use the bearing to determine the sign of
(lon1 - lon2). Basically, you have to list a few cases when (lon1 - lon2)
is between either -pi and 0, or 0 and +pi.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/23/2010 at 20:18:41
From: Alan
Subject: traversing a sphere given specific inputs

Thanks. This mostly looks great. Your help so far is greatly appreciated.

I have an Excel file now that shows your equation, the inital values, the
final values, and some intermediate ones to indicate the path. They are
plotted on a graph. A large dot marks the start of the path.

If latitudes are between -pi and pi, then how is halfway between the
equator and north pole pi/4? Shouldn't it be pi/2? Trying to start at pi/2
latitude produces strange results, so I used pi/4 in my example.

Longitude appears to range between -pi and 0. Is this correct?

It's not possible to go east at all as setting any easterly course seems
to cause it to reflect. For example, 0.5 goes in the same direction as
-0.5.

Traveling one rotation causes the path to leave one side of the
longitudes; but instead of returning onto the other, it reflects and
returns to the start the wrong way.

```

```
Date: 11/23/2010 at 22:42:11
From: Doctor Vogler
Subject: Re: traversing a sphere given specific inputs

Hi Alan,

Thanks for writing to Dr. Math.

You are right: I should have said that latitudes range from -pi/2 (at the
south pole) to +pi/2 (at the north pole), not from -pi to +pi.

To you to figure out when (lon1 - lon2) is between -pi and 0, and when
it's between 0 and +pi, start by taking (lon1 - lon2) to be between -pi
and 0 if both d and b are between 0 and +pi or neither of them. Take
(lon1 - lon2) to be between 0 and +pi if one of d and b is between 0 and
+pi, and the other is not.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/24/2010 at 12:01:46
From: Doctor Vogler
Subject: Re: traversing a sphere given specific inputs

Hi Alan,

I should add to my previous response that, when I say that d (or b) is
between 0 and pi, what I mean is that its sine is positive. Similarly,
when I said "not between 0 and pi," it would have been more accurate to
say that its sine is negative. When the sine is zero, then you have some
more special cases.

When d = pi, then you'll always get cos(lon1 - lon2) = -1, which is
unambiguous. When b = pi or b = 0, then you have this issue where the
longitude doesn't change until you go around the north/south pole. So

sin(lat2) = sin(lat1 - d),

And while you might take this to mean that lat2 = lat1 - d, this will
sometimes break the rule that -pi/2 <= lat2 <= pi/2. It does so precisely
when you have to change lon2 from lon1 to lon1 + pi.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/24/2010 at 12:34:13
From: Alan
Subject: traversing a sphere given specific inputs

I forgot to mention before that, where you said d is "distance divided by
the radius of the sphere," I used 1 as the radius of the sphere, as I am
working in revolutions rather than units of surface distance, which would
change with sphere size. I was unsure about this, however it seems to have
worked fine.

I implemented in my spreadsheet the part where if both d and b are between
0 and +pi, or neither of them, etc. I didn't mean to miss out this bit
earlier. The maths in this is quite heavy going for me. I had to use
Wolfram|Alpha to rearange the earlier equations as I wouldn't have known
where to start in doing so myself with such complicated equations. The d
and b fix seemed to work well and shows the correct results, fixing both
problems.

It was still buggy where the inital longitude was other than 0, but I
figured out how to correct that myself. The correction is probably not the
neatest, but it's not bad -- and it works.

Provided I haven't missed something major, this all seems to be working.

The last piece of information that I need is the final heading (the
heading afterwards). How is this calculated?

I am happy to provide the latest version of the spreadsheet to anyone
else with a similar problem. This is the end of a long line of trying to
find a solution, and I wouldn't want it to take anyone else so long.

```

```
Date: 11/24/2010 at 21:22:56
From: Doctor Vogler
Subject: Re: traversing a sphere given specific inputs

Hi Alan,

If I understand you correctly, you are asking what the bearing has become
after moving the distance d in a "straight" line (i.e., along a great
circle), since in general the bearing changes as you move. For that
question, you just consider going backwards.

If you switch (lon1, lat1) with (lon2, lat2), then you are talking about
going in the opposite direction, with the bearing calculated at the other
point. So if you add (or subtract) pi to that bearing, then you will get
the bearing moving away from your starting point.

In other words, you have

lat1 = starting latitude
lon1 = starting longitude
lat2 = latitude at the destination
lon2 = longitude at the destination
d = distance divided by the radius of the sphere
b = bearing (angle from north at the starting point)
b' = bearing at end (looking away from start, at ending point)

The formulas I already gave you remain

cos(d) = cos(lat1)*cos(lat2)*cos(lon1 - lon2) + sin(lat1)*sin(lat2)
sin(lat2) = cos(b)*sin(d)*cos(lat1) + sin(lat1)*cos(d)

The new one, gotten by swapping the two points and adding pi to the bearing, is

sin(lat1) = -cos(b')*sin(d)*cos(lat2) + sin(lat2)*cos(d)

Solving for b',

sin(lat2)*cos(d) - sin(lat1)
cos(b') = ----------------------------
sin(d)*cos(lat2)

Then you take the inverse cosine.

Of course, now you have the same issues about b' as before. The cosine is
unaffected by adding pi to b', so you have to figure out whether b' is
between 0 and +pi, or between -pi and 0. It seems to me that b' will be
between 0 and pi (i.e., sin(b) > 0) if and only if b is, since it's just a
question of whether you are walking east or walking west. (Of course, if
you are going due north or due south, then you have special cases, as
before.)

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/25/2010 at 08:48:17
From: Alan
Subject: traversing a sphere given specific inputs

I changed the part about d or b being between 0 and pi to instead ask if
its sine (as in sine wave) is positive. (At first, I wasn't sure if you
meant "sign" as in + and - or "sine" as in the trigonometric function.)
This looks neater and performs the same.

I also implemented the equation you gave me for the final bearing and
included the compensation for eastwards and westwards traveling. From what
I can tell, the values that produce errors are to do with performing
arccos on certain numbers, and occur when the distance to travel is a
multiple of pi. I will probably solve this by adding a very small number
where the distance is a multiple of pi.

Moving north or south doesn't work well for any output but is very simple,
so when I come to code it soon, I will include a check and a simpler
algorithm to use in those cases.

Otherwise, results seem correct!

This should all be correct now? If so, thanks again for the help!
```
Associated Topics:
High School Higher-Dimensional Geometry
High School Trigonometry

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