cone problem

[Jon]

Distance on a cone

You have a 45-degree cone with base diameter 2 inches and height 1 inch. The
cone is sitting on a table with the apex pointed up. What is the minimum
distance of travel across the top curved surface between two points on the
bottom edge 180 degrees apart?

the half angle of the cone is pi/4

suppose the vertex of the cone is at (0,1,0)

cos(pi/4) =
(0,1,0)*(x-0,y-1,z-0)/sqrt[(0^2+1^2+0^2)((x-0)^2+(y-1)^2+(z-0)^2)] * dot
product

(sqrt 2)/2 = (y-1)/sqrt(x^2 + (y-1)^2 + z^2) equation of cone.

Suppose a plane section passes through the points (-1,0,0) and (1,0,0)
having normal (a,b,c).

(a,b,c)*(x,y,z)=0 (a,b,c)*(-1,0,0)=0 (a,b,c)*(1,0,0)=0 => a=0

(0,b,c)*(x,y,z)=0 by + cz = 0 => equation of plane

(b/c)y + z = 0 if (b/c) =D, Dy + z = 0 or z = – Dy substituting
in the equation of the cone,

(sqrt 2)/2 = (y-1)/sqrt(x^2 + (y-1)^2 + (-Dy)^2) => equation of curve
resulting from the intersection of the plane with the cone.

Solving for y,

(1/2)(x^2 + (y-1)^2 + (-Dy)^2) = (y-1)^2

x^2 – (y-1)^2 + (Dy)^2 = 0

(Dy)^2 - y^2 + 2y + x^2 – 1 = 0

(D^2 – 1)y^2 + 2y + (x^2 – 1) = 0

y = (1/(2(D^2 – 1)))(-2 +/- {2^2 – 4(D^2 – 1)(x^2 – 1)}^(1/2) ) by the
Quadradic Formula.

y = (1/(D^2 – 1))(-1 +/-{1 – (D^2 – 1)(x^2 –1)}^(1/2))

y = (1/(D^2 – 1))(-1 +/-{1 – [(Dx)^2 – D^2 - x^2 + 1] }^(1/2))

y = (-1 +/-{ x^2 – (Dx)^2 + D^2}^(1/2))/(D^2 – 1)

From Calculus, Arclength A is,

A = the integral of sqrt( 1 + (dy/dx)^2) dx from x= –1 to x=1

dy/dx = [{ x^2 – (Dx)^2 + D^2}^(-1/2)][x – (D^2)x]/(D^2 – 1)

(dy/dx)^2 = (x^2)/(x^2 – (Dx)^2 + D^2)

A = the integral of { 1 + (x^2)/(x^2 – (Dx)^2 + D^2) }^(1/2) dx from x=-1 to
x=1

A = the integral of { (2x^2 – (Dx)^2 + D^2)/(x^2 – (Dx)^2 + D^2) }^(1/2) dx
from x=-1 to x=1

A = the integral of { ((2 – D^2)x^2 + D^2)/((1 – D^2)x^2 + D^2) }^(1/2) dx
from x=-1 to x=1

A = the integral of { ((D^2 - 2)x^2 - D^2)/((D^2 - 1)x^2 - D^2) }^(1/2) dx
from x=-1 to x=1

[(D^2 – 2)^.5 x + D][(D^2 – 2)^.5 x – D]
A = the integral of
{ --------------------------------------------------- }^.5 dx from x=-1 to
x=1
[(D^2 – 1)^.5 x + D][(D^2 – 1)^.5 x – D]

u = (D^2 – 1)^.5 x x = (D^2 – 1)^-.5 u dx = (D^2 – 1)^-.5 du

D^2 – 2 D^2 – 2
[ {-----------}^.5 u + D ][
{-------------}^.5 u – D ]
D^2 – 1 D^2 - 1
A = the integral of
{-----------------------------------------------------------}^.5 (D^2 –
1)^-.5 du
(u + D)(u – D)
D^2 - 2
B = {----------}^.5
D^2 – 1

(Bu + D)(Bu – D)
A = (D^2 – 1)^-.5 * the integral of { ------------------------}^.5 du
(u + D)(u – D)
B^2 u^2 – D^2
A_u = {---------------------}^.5
u^2 – D^2
A_0 = (D^2 – 1)^-.5

u^2 – D^2
2u 2B^2 u
A’_u = .5{---------------------}^.5 { (B^2 u^2 – D^2)(-1)------------------
+ ------------------- }
B^2 u^2 – D^2
(u^2 – D^2)^2 u^2 – D^2

u^2 – D^2 -2B^2 u^3 + 2D^2 u
2B^2 u
A’_u = .5{---------------------}^.5 {-----------------------------
+ ----------------}
B^2 u^2 – D^2 (u^2 – D^2)^2 u^2 –
D^2
A’_0 = 0

B^2 u^2 – D^2 (u^2 –D^2)(2B^2 u)
2u -2B^2 u^3 + 2D^2 u 2B^2 u
A’’_u = .5*.5{--------------------}^.5 { ----------------------------
+ --------------------}{--------------------------- + --------------} +
u^2 – D^2 (B^2 u^2 – D^2)^2
B^2 u^2 – D^2 (u^2 – d^2)^2 u^2 – D^2

u^2 – D^2 2u(-2B^2 u^3 + 2D^2 u) -2*3 B^2
u^2 + 2D^2 2u(2B^2 u) 2B^2
.5{---------------------}^.5 { - 2----------------------------------
+ ----------------------------- + ------------------- + -----------------}
B^2 u^2 – D^2 (u^2 – D^2)^3
(u^2 – D^2)^2 (u^2 – D^2)^2 u^2 – D^2

2 2B^2 1 – B^2
A’’_0 = .5{ ------- - -------- } = ------------
D^2 D^2 D^2

A = A_0 + A’_0 u + A’’_0 u^2

1
– B^2
A = {D^2 – 1}^-.5 * the integral of [1 + 0*u + ------------u^2 ] du from u
= - (D^2 – 1)^.5 to u = (D^2 – 1)^.5

D^2
1 1 - B^2
A = {D^2 – 1}^-.5 [ u + ----- * ------------u^3 ] from u = - (D^2 – 1)^.5
to u = (D^2 – 1)^.5
3 D^2

A = {D^2 – 1)^-.5 [ 2(D^2 – 1)^.5 + (2/3)(1 – B^2)[{D^2 – 1}^(3/2)]/D^2 ]

A = 2 + (2/3)(1-B^2)(D^2-1)/D^2

D^2 - 2
B = {----------}^.5 B^2 = (D^2 – 2)/(D^2 –1)
D^2 – 1

A = 2 + (2/3)( 1 – (D^2 – 2)/(D^2 – 1) )(D^2 – 1)/D^2

A = 2 + (2/3)( ( (D^2 – 1) – (D^2 – 2) )/D^2

A = 2 + (2/3)/D^2

dA/dD = (2/3)(-2)/D^3 = 0 to minimize A.

D approaches infinity for this condition to be met. That is, D = b/c = oo
or c = 0, the z-component of the normal. This means the shortest path is a
straight line from (1,0,0) to (0,1,0) and from (0,1,0) to (0,0,-1).

check:

2sqrt 2 < pi : true

[Vurgil]

Post by Jon
Distance on a cone
You have a 45-degree cone with base diameter 2 inches and height 1 inch. The
cone is sitting on a table with the apex pointed up. What is the minimum
distance of travel across the top curved surface between two points on the
bottom edge 180 degrees apart?

Unroll it to a plane surface, a segment of a circle of radius sqrt(5)
and arc length 2*pi, and find the straight line distance from the end of
the arc to its midpoint.

[Peter Webb]

As others have (implicitly) assumed, the surface of a cone has no curvature.
The shortest line between two points on a cone is exactly the same as for a
cone which has had a slit cut down one side and then laid down flat as a
segment of a circle.

A very complicated 3D problem becomes a trivial question about the length of
a chord in a circle.

This trick is quite common in puzzles.
http://www.jimloy.com/puzz/spider0.htm shows a common version. Its easier to
see why this holds true for the walls of a room than the surface of a cone,
but the underlying principle is the same.

[Jon]

This is the formula for the length of a plane section on a cone. The plane
passes through two points on the base pi radians apart. The normal to the
plane is at angle theta with the base.
http://jons-math.bravehost.com/coneproblem.html I didn't solve the original
problem.

"Peter Webb" wrote in message news:jfgdnh$7cv$***@news.albasani.net...

As others have (implicitly) assumed, the surface of a cone has no curvature.
The shortest line between two points on a cone is exactly the same as for a
cone which has had a slit cut down one side and then laid down flat as a
segment of a circle.

A very complicated 3D problem becomes a trivial question about the length of
a chord in a circle.

This trick is quite common in puzzles.
http://www.jimloy.com/puzz/spider0.htm shows a common version. Its easier to
see why this holds true for the walls of a room than the surface of a cone,
but the underlying principle is the same.

[Tim Little]

Post by Jon
This is the formula for the length of a plane section on a cone.
The plane passes through two points on the base pi radians apart.

Do you have some reason for believing that plane sections of cones
have any relevance to shortst paths?

[Tim Little]

Post by Jon
You have a 45-degree cone with base diameter 2 inches and height 1 inch. The
cone is sitting on a table with the apex pointed up. What is the minimum
distance of travel across the top curved surface between two points on the
bottom edge 180 degrees apart?

[ Snip 162 lines of calculation ]

Cones are surfaces of zero curvature. Their geodesics (shortest
paths) can be found by cutting along a suitable line and flattening
into a plane.

In this case, the planar shape is a sector of a circle with radius
sqrt(2) inches. Opposite points on the bottom edge map to points on
the perimeter having a subtended angle of pi / sqrt(2), which yields a
shortest distance of 2 sqrt(2) sin(pi / (2 sqrt(2))) inches.