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hemispheric mountain monorail
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Mark Spahn
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Joined: 07 Jul 2005
Posts: 62

PostPosted: Tue Jun 13, 2006 6:54 am    Post subject: hemispheric mountain monorail Reply with quote

From John Derbyshire "May Diary" column...
=== QUOTE == For this month, I commend to you the puzzles of Bob Pease, who does the “Pease Porridge” column in the magazine Electronic Design. A pal in Ohio introduced me to Bob’s puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob’s puzzles here. For solutions, see here.
=== UNQUOTE==
In trying to solve this problem, I thought that "a 4% grade" means
an angle of 4% of straight up, that is the angle designated by 4 GRAD
on a scientific calculator. This is wrong. A "4% grade" refers to
a slope with a right triangle whose "vertical height ("rise") is 4% of
its horizontal distance ("run"). That it, it is a slope whose angle from the
horizontal is arctan(.04) = arctan(1/25) = 2 degrees 17 minutes 26.20 seconds.

If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.



SPOILER SPACE







SPOILER SPACE









SPOILER SPACE






Solution:
The top part of the hemispherical mountain is nearly flat, with
a grade of less than 4%. At a certain distance from the top,
the surface of the hemisphere has a grade of 4%. By playing
with similar triangles, you will find that this distance along the
surface of the hemisphere is the arc of a circle of radius R 1 mile = 5280 feet that subtends a central angle of arctan(.04),
namely R*arctan(.04) = .039978687 mi = 211.087468 ft.
The height of a point on the hemisphere where the grade to
the top is 4% is R*cos(arctan(.04)) = .999200958 mi.
A right triangle of this height whose grade is 4% has a
hypotenuse of length h = R/tan(arctan(.04)) = R/.04 = 25R = 25 mi.
By embedding this triangle inside the mountain, wrapping it
around, we get a track on the surface of a hemisphere of
radius R. The total length of this track is 25R + .039978687R
= 25.040 R.

Question: When seen from above (a plan view), what shape
does this track make? From the top of the mountain to a point
where the surface of the hemisphere has a 4% grade to the top,
the track makes a straight line, and from there down it spirals either
clockwise or counterclockwise, maintaining a 4% grade,
until it reaches the bottom of the mountain. Thus, there are
two mirror-image solutions for the shape of this track on a map.
But how many times does the track wrap around the mountain?

The surface of the hemisphere obeys the relationship
x^2 + y^2 + z^2 = R^2, from which z(x,y) = sqrt(R^2 - x^2 - y^2).
The slope this surface makes in direction theta is
lim [z(x+t cos theta,y+t sin theta) - z(x,y)]/t as t approaches 0,
which, by L'Hopital's Rule, works out to (-1/z)*(x cos theta + y sin theta).
To find the direction in which the slope of the surface of the
hemisphere above point (x,y) equals a given slope s (= .04), we set
(-1/z)*(x cos theta + y sin theta) = s, from which
x cos theta + y sin theta = -sz = -s*sqrt(R^2 - x^2 - y^2).
Solving for theta gives the direction in which the track must point.
But I don't see how to proceed further.
(Would polar coordinates simplify the problem?
Is this a differential-equations problem?)

-- Mark Spahn
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matt271829-news@yahoo.co.
science forum Guru


Joined: 11 Sep 2005
Posts: 846

PostPosted: Tue Jun 13, 2006 9:36 pm    Post subject: Re: hemispheric mountain monorail Reply with quote

Mark Spahn wrote:
Quote:
From John Derbyshire "May Diary" column...
=== QUOTE ===
For this month, I commend to you the puzzles of Bob Pease, who does the "Pease Porridge" column in the magazine Electronic Design. A pal in Ohio introduced me to Bob's puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob's puzzles here. For solutions, see here.
=== UNQUOTE===

In trying to solve this problem, I thought that "a 4% grade" means
an angle of 4% of straight up, that is the angle designated by 4 GRAD
on a scientific calculator. This is wrong. A "4% grade" refers to
a slope with a right triangle whose "vertical height ("rise") is 4% of
its horizontal distance ("run"). That it, it is a slope whose angle from the
horizontal is arctan(.04) = arctan(1/25) = 2 degrees 17 minutes 26.20 seconds.

If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.



SPOILER SPACE







SPOILER SPACE









SPOILER SPACE






Solution:
The top part of the hemispherical mountain is nearly flat, with
a grade of less than 4%. At a certain distance from the top,
the surface of the hemisphere has a grade of 4%. By playing
with similar triangles, you will find that this distance along the
surface of the hemisphere is the arc of a circle of radius R =
1 mile = 5280 feet that subtends a central angle of arctan(.04),
namely R*arctan(.04) = .039978687 mi = 211.087468 ft.
The height of a point on the hemisphere where the grade to
the top is 4% is R*cos(arctan(.04)) = .999200958 mi.
A right triangle of this height whose grade is 4% has a
hypotenuse of length h = R/tan(arctan(.04)) = R/.04 = 25R = 25 mi.
By embedding this triangle inside the mountain, wrapping it
around, we get a track on the surface of a hemisphere of
radius R. The total length of this track is 25R + .039978687R
= 25.040 R.

Question: When seen from above (a plan view), what shape
does this track make? From the top of the mountain to a point
where the surface of the hemisphere has a 4% grade to the top,
the track makes a straight line, and from there down it spirals either
clockwise or counterclockwise, maintaining a 4% grade,
until it reaches the bottom of the mountain. Thus, there are
two mirror-image solutions for the shape of this track on a map.
But how many times does the track wrap around the mountain?

The surface of the hemisphere obeys the relationship
x^2 + y^2 + z^2 = R^2, from which z(x,y) = sqrt(R^2 - x^2 - y^2).
The slope this surface makes in direction theta is
lim [z(x+t cos theta,y+t sin theta) - z(x,y)]/t as t approaches 0,
which, by L'Hopital's Rule, works out to (-1/z)*(x cos theta + y sin theta).
To find the direction in which the slope of the surface of the
hemisphere above point (x,y) equals a given slope s (= .04), we set
(-1/z)*(x cos theta + y sin theta) = s, from which
x cos theta + y sin theta = -sz = -s*sqrt(R^2 - x^2 - y^2).
Solving for theta gives the direction in which the track must point.
But I don't see how to proceed further.
(Would polar coordinates simplify the problem?
Is this a differential-equations problem?)

-- Mark Spahn

Doing it in polar coordinates I get the plane view to be given by

d theta/dr = -1/(r*g) * Sqr((r^2*(1 + g^2) - g^2*m^2) / (m^2 - r^2))

where (r, theta) are the usual polar coordinates, m is the (adjusted)
radius of the mountain (= 5280 ft), and g is the gradient (= 0.04). The
sign of d theta/dr is arbitrary, depending on whether you want a
clockwise or anticlockwise spiral. I just happened to choose it
negative.

This has solution

theta = Sqr(g^2 + 1) / g * Atn(a) - Atn(b)

where

a = Sqr(g^2 + 1) * Sqr(m^2 - r^2) / Sqr(r^2*(g^2 + 1) - m^2*g^2)
b = g*Sqr(m^2 - r^2) / Sqr(r^2*(g^2 + 1) - m^2*g^2)

I make the total angle of rotation to be 2*pi*(Sqr(g^2 + 1)/(4*g) -
1/4), or about 6.005 complete revolutions.

I made a graph at
http://img155.imageshack.us/img155/7108/monorail9me.gif.
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Mike1161
science forum addict


Joined: 14 Oct 2005
Posts: 50

PostPosted: Tue Jun 13, 2006 9:54 pm    Post subject: Re: hemispheric mountain monorail Reply with quote

In article <1150234606.238836.302720@f6g2000cwb.googlegroups.com>, matt271829-news@yahoo.co.uk says...
Quote:
Mark Spahn wrote:
From John Derbyshire "May Diary" column...
=== QUOTE ===
For this month, I commend to you the puzzles of Bob Pease, who does the "Pease Porridge" column in the magazine Electronic Design. A pal in Ohio introduced me to Bob's puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the
mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob's puzzles here. For solutions, see here.
=== UNQUOTE===

....
If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.

....
I make the total angle of rotation to be 2*pi*(Sqr(g^2 + 1)/(4*g) -
1/4), or about 6.005 complete revolutions.

I made a graph at
http://img155.imageshack.us/img155/7108/monorail9me.gif.

From the wording of the question it appears possible that the monorail could zig-zag up the hill (a train that climbs

out of Cusco in Peru actually does this in real-world-land), so there is no need for the track to circle the hill even
once.

Mike
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matt271829-news@yahoo.co.
science forum Guru


Joined: 11 Sep 2005
Posts: 846

PostPosted: Tue Jun 13, 2006 10:08 pm    Post subject: Re: hemispheric mountain monorail Reply with quote

Mike wrote:
Quote:
In article <1150234606.238836.302720@f6g2000cwb.googlegroups.com>, matt271829-news@yahoo.co.uk says...
Mark Spahn wrote:
From John Derbyshire "May Diary" column...
=== QUOTE ===
For this month, I commend to you the puzzles of Bob Pease, who does the "Pease Porridge" column in the magazine Electronic Design. A pal in Ohio introduced me to Bob's puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the
mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob's puzzles here. For solutions, see here.
=== UNQUOTE===

...
If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.

...
I make the total angle of rotation to be 2*pi*(Sqr(g^2 + 1)/(4*g) -
1/4), or about 6.005 complete revolutions.

I made a graph at
http://img155.imageshack.us/img155/7108/monorail9me.gif.

From the wording of the question it appears possible that the monorail could zig-zag up the hill (a train that climbs
out of Cusco in Peru actually does this in real-world-land), so there is no need for the track to circle the hill even
once.

Mike

In theory the spiral could reverse direction at any point, any number
of times, but the corners of the zig-zag would have to be sharp points
(not curves), which wouldn't be very practical for a monorail. I'm
fairly confident the question expects a smooth spiral as the answer.
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Chip Eastham
science forum Guru


Joined: 01 May 2005
Posts: 412

PostPosted: Tue Jun 13, 2006 10:39 pm    Post subject: Re: hemispheric mountain monorail Reply with quote

Mike wrote:
Quote:
In article <1150234606.238836.302720@f6g2000cwb.googlegroups.com>, matt271829-news@yahoo.co.uk says...
Mark Spahn wrote:
From John Derbyshire "May Diary" column...
=== QUOTE ===
For this month, I commend to you the puzzles of Bob Pease, who does the "Pease Porridge" column in the magazine Electronic Design. A pal in Ohio introduced me to Bob's puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the
mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob's puzzles here. For solutions, see here.
=== UNQUOTE===

...
If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.

...
I make the total angle of rotation to be 2*pi*(Sqr(g^2 + 1)/(4*g) -
1/4), or about 6.005 complete revolutions.

I made a graph at
http://img155.imageshack.us/img155/7108/monorail9me.gif.

Mike wrote:

Quote:
From the wording of the question it appears possible that the
monorail could zig-zag up the hill (a train that climbs out of
Cusco in Peru actually does this in real-world-land), so there
is no need for the track to circle the hill even once.

Would it make any difference to the length of track required?
It seems to me the zig-zag arrangements will use the same
length of track as the "spiral" solution.

regards, chip
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matt271829-news@yahoo.co.
science forum Guru


Joined: 11 Sep 2005
Posts: 846

PostPosted: Wed Jun 14, 2006 12:29 am    Post subject: Re: hemispheric mountain monorail Reply with quote

Chip Eastham wrote:
Quote:
Mike wrote:
In article <1150234606.238836.302720@f6g2000cwb.googlegroups.com>, matt271829-news@yahoo.co.uk says...
Mark Spahn wrote:
From John Derbyshire "May Diary" column...
=== QUOTE ===
For this month, I commend to you the puzzles of Bob Pease, who does the "Pease Porridge" column in the magazine Electronic Design. A pal in Ohio introduced me to Bob's puzzles. Sample:

Little Egbert bought a mountain. Its shape was a perfect hemisphere with a 5270-ft radius, set on a flat plain. He decided to build a railroad to transport him to the top. It was a monorail, which made it very easy to plan, with minimum width, and only one rail. The rail was offset 10 ft away from the surface of the mountain, to avoid digging, so the radius was exactly 5280 ft. The train could only ascend a 4% grade. The base station was at the very bottom of the
mountain, so the train could not get a running start up the hill. How much track did Little Egbert have to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

There are more of Bob's puzzles here. For solutions, see here.
=== UNQUOTE===

...
If you want to solve this on your own, stop here. I will leave some spoiler
space before outlining my solution, along with a further question
about what a plan view of the track (seen from the top) looks like.

...
I make the total angle of rotation to be 2*pi*(Sqr(g^2 + 1)/(4*g) -
1/4), or about 6.005 complete revolutions.

I made a graph at
http://img155.imageshack.us/img155/7108/monorail9me.gif.

Mike wrote:

From the wording of the question it appears possible that the
monorail could zig-zag up the hill (a train that climbs out of
Cusco in Peru actually does this in real-world-land), so there
is no need for the track to circle the hill even once.

Would it make any difference to the length of track required?
It seems to me the zig-zag arrangements will use the same
length of track as the "spiral" solution.

regards, chip

Yes, I think the length of track would be the same no matter how many
times you zig-zagged.
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William Elliot
science forum Guru


Joined: 24 Mar 2005
Posts: 1906

PostPosted: Wed Jun 14, 2006 2:47 am    Post subject: Re: hemispheric mountain monorail Reply with quote

On Tue, 13 Jun 2006, Mark Spahn wrote:

Quote:
From John Derbyshire "May Diary" column...

Little Egbert bought a mountain. Its shape was a perfect hemisphere with
a 5270-ft radius, set on a flat plain. He decided to build a railroad to
transport him to the top. It was a monorail, which made it very easy to
plan, with minimum width, and only one rail. The rail was offset 10 ft
away from the surface of the mountain, to avoid digging, so the radius
was exactly 5280 ft. The train could only ascend a 4% grade. The base
station was at the very bottom of the mountain, so the train could not
get a running start up the hill. How much track did Little Egbert have
to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

At the base of the mountain the slope will be 90 degrees.

Thus we'll have to edge track at 4% grade around the hill.
Now however is problem because monorails sit on top of the rail.

However the rail will be horizontal from the immediately vertical
surface which mean the cab will be not resting on monorail but
be rolled over as if laying on the ground and supported by just
one edge of whatever support monorails have when functing in
proper upright position.
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matt271829-news@yahoo.co.
science forum Guru


Joined: 11 Sep 2005
Posts: 846

PostPosted: Wed Jun 14, 2006 9:23 am    Post subject: Re: hemispheric mountain monorail Reply with quote

William Elliot wrote:
Quote:
On Tue, 13 Jun 2006, Mark Spahn wrote:

From John Derbyshire "May Diary" column...

Little Egbert bought a mountain. Its shape was a perfect hemisphere with
a 5270-ft radius, set on a flat plain. He decided to build a railroad to
transport him to the top. It was a monorail, which made it very easy to
plan, with minimum width, and only one rail. The rail was offset 10 ft
away from the surface of the mountain, to avoid digging, so the radius
was exactly 5280 ft. The train could only ascend a 4% grade. The base
station was at the very bottom of the mountain, so the train could not
get a running start up the hill. How much track did Little Egbert have
to buy? (a) 25.00 miles (b) 25.020 (c) 25.040 (d) 25.40

At the base of the mountain the slope will be 90 degrees.
Thus we'll have to edge track at 4% grade around the hill.
Now however is problem because monorails sit on top of the rail.

Not necessarily. Some are suspended.

Quote:

However the rail will be horizontal from the immediately vertical
surface which mean the cab will be not resting on monorail but
be rolled over as if laying on the ground and supported by just
one edge of whatever support monorails have when functing in
proper upright position.

I don't see why it would have to be "rolled over". It could be either
sitting on top of, or (with limited room) suspended from, a rail
cantilevered out from the mountain.
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BookWight
science forum beginner


Joined: 28 Feb 2006
Posts: 3

PostPosted: Wed Jun 14, 2006 6:13 pm    Post subject: Re: hemispheric mountain monorail Reply with quote

Wandering along the edges of alt.math.recreational, I found the
following bit of electronic flotsam written by "Chip Eastham"
<hardmath@gmail.com> in
news:1150238387.140300.279230@f6g2000cwb.googlegroups.com:

Quote:
Mike wrote:

From the wording of the question it appears possible that the
monorail could zig-zag up the hill (a train that climbs out of
Cusco in Peru actually does this in real-world-land), so there
is no need for the track to circle the hill even once.

Would it make any difference to the length of track required?
It seems to me the zig-zag arrangements will use the same
length of track as the "spiral" solution.

regards, chip


I respectfully disagree. In the process of zig-zagging, there needs to be
a length of monorail track equal to the length of the monorail train at
each zig-zag point, unless you expect the train to make a hairpin turn on a
curve of exceddingly minute radius. The length of track needed for the
train to switch directions is a constant, therefore the more zigzags, the
greater the excessive length of the track over & above that length needed
for the spiral solutions.
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