# The ant and honey problem on a spherical bowl - shortest paths?

1. Sep 2, 2014

### nixed

In considering the shortest paths between two points on a sphere I came across the following interesting problem:

An ant sits on the outside of a glass bowl of spherical curvature (radius R), at a distance d from the lip of the bowl. It sees a drop of honey on the inside of the bowl directly opposite it (i.e. also a distance d from the lip and 180° round the bowl from the ant. What is the shortest path for the ant to take to get the honey? Is it unique?

I know that the shortest distance between two points on the surface of a sphere is along a great circle which passes through the two points and is centered at the sphere origin O.

There are three distinct cases which depend on the depth of the bowl (i.e. the vertical distance from rim to base).

case 1. depth < R

case 2. depth = R

case 3. depth > R

What are the shortest paths for each of these cases and how do they differ from each other?

2. Sep 2, 2014

### HallsofIvy

Staff Emeritus
It is well known- and can be shown from, say, Calculus of Variations, that the shortest distance between two points on a sphere is along the great circle containing them. In this case that would be the intersection of a vertical plane with the bowl, the path passing through the bottom of the bowl. The only distance between the three cases is that in the case (3), depth> R, technically, the shortest path would be over the (non-existant) top of the bowl but, still, the shortest possible path would be path through the bottom of the bowl.

3. Sep 2, 2014

### Bacle2

But, if the curvature is constant, with value 1/R, can the depth then be greater than R? Maybe I misunderstood?

4. Sep 2, 2014

### Bacle2

Can the depth be different from R when the curvature is constant, with value 1/R; I would have thought the bowl would have to be a hemisphere (so that the vertical distance from rim to base would then be R) ; in this case the lower hemisphere. Am I missing something here?

5. Sep 3, 2014

### nixed

By depth of bowl ,D, I mean the vertical distance from the rim to the bottom of the bowl so that
0≤ D ≤ 2R cover all the possible depths.

The bowl is always a part of a sphere of radius R, just that case 1 is a sort of dish, case2 a hemisphere and case 3 a larger bowl that is more than a hemisphere.

6. Sep 3, 2014

### nixed

NOT True! The route on the great circle going straight up to the rim then tracking down through the bowl is not the shortest route between A and H! The shortest route over the rim and to H must leave the plane of AOH- the question is with the constraint of not being able to walk paths that go north of the rim how to choose a route?

Also the case 2 differs from case 1 eg think about the route that crosses the rim at 90deg to the original vertical plane through AOH.

I agree with you about case 1 shortest route is straight up to the rim and down through the bowl.

7. Sep 3, 2014

### MrAnchovy

For Case 1 the route down across the bottom of the bowl is always shortest.

Case 2 is geometrically eqivalent to travelling around the outside of a sphere (imagine the top half of the sphere reflected in the equator). The shortest route is therefore the great circle route to an imaginary point the same distance above the rim on the opposite side - in other words exactly opposite the ant. There are an infinite number of such routes of length πr, the simplest being straight up and straight down across the bottom of the bowl with the others heading up meeting the rim at an angle and heading down towards the honey at the same angle.

Case 3 has two subclasses: Case 3a where the ant and honey are above the widest part of the bowl when a route around the rim is obviously shorter than across the bottom, and any route that meets the rim at an angle is obviously shorter than one that goes around the rim: in this case the great circle route that meets the rim at an angle and the great circle route down to the honey at the same angle is the shortest.

Case 3b I will leave for others...

8. Sep 4, 2014

### nixed

For case 3a the Ant in the northern hemisphere I think I understand. The shortest route always passes through the P(∏/2) point on the rim. If we imagine the initial ant position as the pole of a sphere radiating 'longitude lines' (all the great circles from A) we are only interested in those that hit the rim since the ant must cross it somewhere between 0 ≤ ø ≤∏/2 . Of these possible routes the one with the longest path along the great circle to the rim will be optimum because it will minimize the walk along the rim. So we are looking for the great circle from A that just touches the circle of the rim tangentially. This will define the point of intersection of these two circles
C(ΘA) on the rim. This is where the ant leaves a great circle starts to walk along the constant latitude of the rim to P(∏/2). The symmetry of the problem then gives the length as just 2 times the A to C to P distance as the shortest A to H distance.

The position of C around the rim, øc , depends on the angle ΘA of the initial ant's position with respect to the vertical. If ΘA= Θrim the ant and honey starts on the rim and C is obviously at ø=0 and the shortest distance is just the rim ∏RSinΘrim. If the ant starts at ΘA= ∏/2 it is on the equator of the bowl and C has moved around to ø=∏/2 and the AH shortest distance is just ∏R.In general we need to find point C if we are to calculate the actual distance of the shortest route for a particular ant position ΘA. This is a tricky bit of coordinate geometry!

(I found that if the ant starts further than 10o from the rim the true minimum distance is less than 1% less than a distance obtained as follows: find the great circle curve which passes through the two points A(ax,ay,az) and P(∏/2) (px,py,pz) Find the point A' on this curve where it's z value is = pz. The approximate shortest distance from A to H is then 2x(the arc length AA' on the great circle + rim distance A'P)).

Case 3b. The ant starts in the southern hemisphere. The shortest route remains the path passing the rim at P(∏/2). this remains true for all ΘA<∏. Proof: The direct ø= 0 route has distance AH= L(o)/R= 2(∏-Θrim). We can always find a great circle from A direct to P(∏/2) so no walk along the rim is required. The distance AH = L(∏/2)/R = 2δ where δ is the angle between the OP vector and the OA vector so δ=arcos{(-1)*(cosΘA)(cosΘrim)}. Now for all ∏/2<ΘA<∏ and 0<Θrim<∏/2

δ < ∏-Θrim so L(∏/2) < L(0). Also small variation of point P along the rim away from this position increases the AH distance even though the rim is crossed at a single point so the whole journey is on two great circles.

Obviously the distance from A to H for case 3 depends on the distance of the ant from the rim but interestingly this is not so for cases 1 and 2

I agree that case 1 has a unique route (ø=0) crossing the rim directly above the A position. The distance from A to H is then equal to the arc length of the bowl rim to rim 180 deg apart. This distance is independent of how far the ant is initially below the rim.

Case 2 all routes which travel along great circles from A to point P(ø) on the rim (with 0≤ø≤∏ )and then on great circles from P(ø) to H are the same length L(ø)/R = ∏ and are shorter than any other route over the rim. So as you say there are infinitely many shortest routes (and infinitely many longer ones!). Again the distance does not depend on the initial distance of the ant from the rim.

Last edited: Sep 4, 2014
9. Sep 5, 2014

### nixed

What then is the shortest path between any two points A and B on a sphere, if the shorter arc of the great circle connecting them is interrupted by a restriction that we cannot go above a certain latitude?

The insight from Case 3a above may suggest the answer.

10. Sep 6, 2014

### nixed

The general solution for the shortest path from A to H case 3 is a path A-C-P-C'-H via point P(∏/2) on the rim as discussed above. The critical point C where the great circle from A just touches the rim tangentially occurs at an angle ψc around the great circle from A and at an angle øc around the rim from A determined by the latitude of the rim,(set by Θrim) and the latitude of the ant (set by ΘA both measured from the vertical). It is an exercise in vector analysis to show that the tangent point occurs at:

Cos ψc= Cos ΘA/ Cos Θrim

Cos øc = Tan Θrim / Tan ΘA

∴ For ant positions Θrim< ΘA <∏/2

The shortest AH distance is

L(∏/2)/R = 2{ ψc + (∏/2 - øc)*sinΘrim}

and for ∏/2≤ ΘA <∏ the rim is crossed at point P(∏/2) (no walk along the rim is required) so

L(∏/2)/R = 2 arcos((-1)*(Cos ΘA * Cos Θrim)).

11. Sep 7, 2014

### nixed

Consider the attached diagram

If point B is on the sphere surface outside the shaded spherical triangle, then the shortest route B to A is the shorter arc of the the great circle passing through A and B. (N.B. there is always one and only one such great circle for a given A and B unless they are polar opposites, collinear with the sphere center O, in which case there are an infinite number). For B outside the shaded region, the great curve connecting A and B never goes further north than the rim so the latitude constraint does not act.

If B is in the shaded triangle the shorter arc of the great circle B to A goes north of the rim latitude. What is the shortest route in these cases?

There is no other great circle route direct from A to B (uniqueness of the great circle). The best we can do is to travel on two different great circles (in planes BOI and AOI say) which intersect at some intermediate point I on the sphere surface. We travel along arc B to I and then I to A. To minimize the distance B to A we must orient the plane BOI with respect to AOI so the intersection point moves as far north as possible without leaving the sphere surface or going above the latitude of the rim. To see this, imagine a different great circle plane BOI’ such that I’ is further north than I. Point I’ can then be arrived at from B either by B to I on great circle BOI then I to I’ on great circle AOI or directly from B to I’ on a single great circle BOI’. The latter is always shorter. The best orientation of the great circle from B is that which just touches the rim latitude tangentially so defining a point of intersection with the rim, D, the same way C was obtained from point A).

Referring to the plan diagram, the arc BD along this great circle is shorter than B to t1 then t1 to D for all t1 on the rim latitude arc B’toD (uniqueness of great circle between two points again). You might wonder why choosing a great circle route B to t2 doesn’t reduce the distance to A still further. This is because any great circle route from B which hits the rim beyond D has already risen above the plane of the rim and has intersected it already at some point t1 between B’ and D so actually lengthening the path.

So the optimum route from B to A is the great circle arc BD then the latitude line DC then the great circle arc CA. We can write an expression for the length of this route in terms of the polar and azimuthal angles which define the A and B positions. See diagram.

This has obvious applications for optimum aircraft routes between cities if the planes cannot fly higher latitudes over the poles.

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12. Sep 7, 2014

### nixed

Given two points A(θAA) and B(ΘBB) and a latitude restriction Θrim how do we tell if the great circle between A and B will rise to higher latitudes than Θrim?

The parametric form of the great circle passing through two points was derived in the post "great circle problem". The z(ψ)/R coordinate is

Z(ψ)/R = Ca Cos ψ + α Sin ψ

where 0≤ψ≤ 2∏

α = { Sa2Cb - CΔCaSaSb} / Sinδ

δ = arcos(CΔSaSb + CaCb}

where Δ= øb - øa the longitude difference of A and B

Ca = CosΘA ; Sb = Sinθb etc

the maximum value of this occurs when d [z(ψ)/R]/dψ =0 ∴ tanψm = α/Ca

as Cosψ = 1/√(tan2ψ +1) and Sinψ = tanψ/√(tan2ψ +1)

Z(ψm)/R = √(α2 + C2a). If this is larger than cosθrim then the great circle tracks north of the latitude restriction and the formula for L/R under constraint must be used. If

√(α2 + C2a) < cosθrim

then the shortest distance between A and B is the great circle arc length

LAB /R = δ

13. Sep 8, 2014

### nixed

The analysis is shown in the attached sheet for those interested in following the details.

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14. Sep 9, 2014

### nixed

Here is another helpful picture of great circles tangent to a given latitude

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15. Sep 9, 2014

### nixed

I forgot to post the attached picture of all great circles from a point which cross a given latitude.

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