# How can I find the slope of a line tangent to a small circle on a sphere? [closed]

I draw a circle on the earth, so that it passes through the north pole. I then begin walking around the circle, keeping track of my latitude. How do I tell what direction that I'm facing by knowing only the diameter of the circle and my latitude? It should be understood that the line tangent to the circle lies in the plane formed by the great circle that is also tangent at the same point.

## closed as off-topic by Servaes, kjetil b halvorsen, user91500, Adrian Keister, José Carlos SantosJan 12 at 16:19

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• Does the rotation of the earth matter? – user Jan 11 at 20:44
• Most of those lines of latitude that you’re recording intersect the circle at two points with different facings. – amd Jan 11 at 20:49
• Is the known diameter a straight-line distance, or the arc over the surface of the sphere? – DJohnM Jan 11 at 21:03
• My direction is relative the earth, which doesn't depend on motion of the earth. If I know whether I'm moving clockwise or counterclockwise, and whether I'm moving north or south, I should be able to determine which one of four possibilities is the correct direction that I am facing. – steveupson Jan 11 at 21:04
• The diameter can be either a straight-line distance along the plane of the sphere, or it can be an arc length along the sphere surface. I generally view the problem such that the diameter is expressed as twice the arc length from the pole of the small circle to the edge of the small circle. – steveupson Jan 11 at 21:07

Consider a spherical triangle (composed of portions of great circles), with one vertex on the North Pole, one vertex at the center of the drawn circle, and one vertex at your location

You know your latitude, so you know your co-latitude, so you know the side of the spherical triangle from the Pole to you.

You know that the drawn circle passes through the Pole, and its diameter, so you know the distance from the Pole to the center of the drawn circle, so you know the side of the spherical triangle from the Pole to the center of the drawn circle.

You know that you are one the drawn circle, so you know the side of the spherical triangle from you to the center of the drawn circle.

With side-side-side you can solve the spherical triangle for the required information...

• These are not the correct angles to solve for. I want to find the direction that I'm facing relative to the earth, or in other words, the slope (or direction) of the line that is tangent to the circle at the point where I am standing. – steveupson Jan 11 at 21:44
• Stand at your location, facing the center of the drawn circle; the solution to the spherical triangle gives this info. Then turn 90 degrees. You are facing along the drawn circle... – DJohnM Jan 11 at 22:13
• The plane where I am standing (tangent to the sphere) is not the same plane as the plane of the circle. The line tangent to the circle does not lie in the plane in which I am standing. It must be projected onto the plane by some method. – steveupson Jan 11 at 22:16
• @steveupson "Slope" is a slippery concept in three dimensions. The plane in which the circle lies is completely irrelevant to any of the calculations in this answer--it isn't involved. All the angles described in the answer are already "projected" onto the tangent plane where you are standing, in the sense that the angles all are measured between lines lying in that plane and tangent to the relevant arcs. This is the correct answer. – David K Jan 12 at 3:11
• @steveupson You have calculated your example wrong. The pole of the small circle will not be anywhere near southwest from the place you say it is. It will be very nearly west, 90 degrees from the direction to the pole. The direction of the arc to the pole of a small circle is always 90 degrees from the direction along the circle, without exception. All you need to do to prove this is to draw an alternative set of lines of latitude and longitude using the small circle's pole as the "north" pole. – David K Jan 12 at 3:27

First, I want to say explicitly that I believe we're all assuming the Earth is perfectly spherical, because on an ellipsoidal model of the Earth the only "small circles" that are actually circular are the lines of latitude, and on the actual geoid as far as we know it it's even worse.

As you have said,

The coordinate system is not important. We can convert the answer from one coordinate system to another.

This is true. And if we convert to a coordinate system $$C'$$ in which the center of the small circle becomes the "north pole," then you will always be walking either due east or due west along the small circle (according to that coordinate system), because the small circles centered at the north pole are the lines of latitude.

At the same time, from any point on the small circle the north pole is always due north, $$90$$ degrees away from the direction in which you are walking.

Now we convert back to our original coordinate system $$C$$ and find the compass direction in which we're walking around the small circle and the compass direction toward the pole of the small circle. If we were walking due east (compass heading $$90$$ degrees) according to coordinates $$C',$$ we are probably walking in a different direction according to coordinates $$C.$$ For example, the direction of travel in system $$C$$ might be $$150$$ degrees instead of $$90.$$

In that case, we know that at the point where we're standing right now, the compass rose for coordinate system $$C$$ is rotated $$60$$ degrees relative to the compass rose for coordinate system $$C'.$$ The compass heading to the pole of the small circle is no not due north ($$0$$ degrees) in coordinate system $$C.$$ Instead, it's $$60$$ degrees.

At different places along the small circle you may have different rotations of the compass rose for coordinate system $$C$$ relative to system $$C'.$$ But at any one place there is only one compass rose for each system, and converting from one system to another adds or subtracts the same number of degrees from any of the headings from that point to any other point on the Earth, including the center of the small circle and the point one step forward along the small circle. The angle between those two directions is always $$90$$ degrees.

You can then apply the method in the answer by DJohnM to find the actual angles for any particular small circle and latitude.

Here's another way to think about the direction along the small circle and the direction to the pole of the small circle. Stand at any point on the small circle, choose a direction of travel along the small circle, and face in that direction. Now turn $$180$$ degrees in place. You are now facing in the other direction of travel along the small circle. If the pole of the small circle is on your left when you face one way, it's on your right when you face the other way. But the geometry of the small circle and of your position on it is completely mirror-image symmetric in a plane through the pole of the small circle, your position, and the center of the Earth. If the small circle pole is $$x$$ degrees to your left when you face one way, it's $$x$$ degrees to your right when you face the other way. And the only way the heading to a point can change from $$x$$ degrees left to $$x$$ degrees right when you turn yourself $$180$$ degrees is if $$x = 90.$$

Let's work on the particular example of a small circle of diameter $$178$$ degrees. We can use the navigation calculator at http://edwilliams.org/gccalc.htm to find the direction and distance between any two points or the point at a given direction and distance from another. Before any calculations, set the distance units to nm and the Earth model to "Spherical (1'=1nm)". This means that all calculations made by the calculator will follow great-circle paths on a spherical model of the Earth.

Put the pole of the small circle at $$1^\circ$$ N, $$0^\circ$$ E. Enter $$1{:}00.00$$ N for Lat$$1$$ and $$0{:}00.00$$ E for Lon$$1$$ under "Compute true course and distance between points." The north pole is at $$90^\circ$$ N, $$0^\circ$$ E. Enter $$90{:}00.00$$ N for Lat$$2$$ and $$0{:}00.00$$ E for Lon$$2$$. Click "Compute". This gives us the great-circle distance $$5340$$ nm ($$89$$ degrees) from the small circle's pole to the north pole. So the small circle consists of all points whose great-circle distance from the point at $$1^\circ$$ N, $$0^\circ$$ E is $$5340$$ nm.

Now find the point $$5340$$ nm due east from the pole of the small circle; that is, $$5340$$ nm along the great-circle path that is heading due east as it passes through the small circle's pole. Enter $$1{:}00.00$$ N for Lat$$1$$ and $$0{:}00.00$$ E for Lon$$1$$ under "Compute lat/lon given radial and distance from a known point." Enter $$90$$ for Course$$1$$-$$2$$ and $$5340$$ for Distance$$1$$-$$2$$. Click "Compute". In the result, Lat$$2$$ is $$0{:}1.0471$$ N and Lon$$2$$ is $$89{:}0.0091$$ E, that is, the point is at $$0^\circ 1.0471'$$ N, $$89^\circ 0.0091'$$ E. So this is a point on the small circle that is almost on the equator. (Note, we started at $$1^\circ$$ latitude but ended up at about $$0.017^\circ$$ latitude because, as you know, the great-circle path cannot follow the line of latitude.)

Now compute the back course using "Compute true course and distance between points." Enter $$0{:}1.0471$$ N for Lat$$1$$, $$89{:}0.0091$$ E for Lon$$1$$, $$1{:}00.00$$ N for Lat$$2$$, and $$0{:}00.00$$ E for Lon$$2$$. (Make sure the longitudes are E, because the default is W and this makes a big difference for Lon$$1$$.) The resulting Course$$1$$-$$2$$ is $$270.99984772352923.$$ That's about $$89$$ degrees west of true north, but as you know, the direction of travel along the small circle is not exactly true north. Your intuition that it's roughly true north when you are near the equator is correct, however; it's only about one degree east of north.

Next, try the same thing for a point halfway from the first point we tried to the north pole. That is, try the point $$5340$$ nm northeast of the small circle's pole; put $$45$$ instead of $$90$$ for Course$$1$$-$$2$$ under "Compute lat/lon given radial and distance from a known point" and follow the rest of the steps. The result is that the point on the small circle is at $$45^\circ 0.4337'$$ N, $$89^\circ 35.1489'$$ E, and the back course from that point to the pole of the small circle is $$270.41418555715956^\circ,$$ about $$89.6$$ degrees west of true north, as shown in the screenshot below.

This tells us that to go in a great-circle arc (the shortest path) from a certain point at a little more than $$45$$ degrees north latitude to a certain other point at $$1$$ degree north latitude, we want to start out in a direction that is not southwest (as you might expect) but is actually slightly north of due west. The results for Course$$2$$-$$1$$ and for Distance confirm that the great-circle distance is indeed $$5340$$ nm (plus a roundoff error) and that when you finally arrive at $$1$$ degree latitude your direction of travel has changed; the back course is $$45$$ degrees, which means that coming into that point the direction is $$225$$ degrees, which is exactly in the southwest direction as you might expect.

So, about halfway from the equator to the north pole, a $$90$$-degree difference between the direction of travel and the direction to the small circle's pole is plausible. The direction to the small circle's pole isn't exactly due west, but the tangent direction along the small circle isn't exactly due north either. They're merely close to due west and due north, respectively.

Let's try a point much closer to the pole. Put $$1$$ for Course$$1$$-$$2$$ under "Compute lat/lon given radial and distance from a known point" and compute the point on the small circle and the back course. The point on the small circle is $$89^\circ 0.0091'$$ N, $$89^\circ 59.4764'$$ E and the course from there to the pole of the small circle is $$270.00872619391873^\circ,$$ almost exactly due west. Meanwhile the direction of travel along the small circle is not exactly due north/south (it never is, except in the limit at the pole itself, where the small circle is tangent to a line of longitude), but it's almost exactly due north.

Here's where intuition tends to steer us wrong in spherical coordinates: we expect the direction from point $$A$$ to point $$B$$ to be $$180$$ degrees different from the direction from point $$B$$ to point $$A,$$ but in general it is not. The intuition is accurate if both points are at the same longitude or if both are on the equator; it is nearly accurate if both points are near the equator or not too far from each other. In many other cases, this intuition is horribly inaccurate. Do not let yourself be fooled by it.

• If we pass a great circle through the tangent point (when we are at anything other than some single unknown point that is some unknown degree north, between 0 and 1 degrees), it will NOT be orientated in the same direction as due north. Definitely WRONG. – steveupson Jan 12 at 16:00
• Actually, the unknown point will be slightly greater than 1 degrees north. You must allow for spherical excess. – steveupson Jan 12 at 16:06
• I think that the problem with your analysis is that in a correct representation, the southernmost point on the circle will have a tangent that is perpendicular to the longitude, while the north pole will be the only point that has a tangent that is parallel with the longitude. What I said in the two previous responses in not correct. In any event, when I am at 45 degrees north, or south for that matter, the equator (or a point one degree from it) is not to my west. – steveupson Jan 12 at 16:21
• Indeed when you are far south of the equator on that small circle, the tangent will not be oriented anywhere near due north or south. At the southernmost point it is indeed due east or west, and the center of the small circle is due north--$90$ degrees difference in direction. As you pass the equator the direction is slightly more than one degree different from true north. What you have not yet identified is any point on the small circle where the direction of travel and the direction to the center can be shown not to be $90$ degrees apart. – David K Jan 12 at 16:26
• Yes, I've done them. Thank you. I appreciate your explanation and I do think that the original answer, solving the spherical triangle is the straightforward method. That's much more of the answer that I was looking for, I just didn't think about it correctly. Do you know of a function that relates these variables to one another? – steveupson Jan 12 at 22:40