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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Mon Jul 03, 2006 5:11 pm Post subject:
Re: Distribution of slope in a terrain sample



Doug Goncz (I) wrote:
Quote:  To find an eikonal between points A and B on for a bicycle on arbitrary
terrain, I have been considering, in news:sci.physics.research, how to
best establish that there is, first and foremost, some advantage to be
gained by deviating from the (as seen from above) straight line
connecting A and B.
I've got a simple routine in Mathcad that takes a reasonable time to
compute which assigns
( rnd( 2 )  1 ) / sqrt( i * j )
amplitude to the real and imaginary components of the Fourier
decomposition of synthetically generated terrain, with i,j < 9 and
generates the terrain with the icfft transform. The routine then
selects ponts A and B on the terrain, and rightturning and
leftturning paths from A to B parallel to the edges of the patch. In
addition to simulating the bicycle travel on the path, I intend to
compute the distribution of slope on the path and take a short cut to
travel time, comparing the two methods.
I want to generate paths with smooth turns, not right angle turns,
initially with A and B at the ends of the elllipse which has foci X and
Y
I figure an ellipse with eccentricity 0.5 will do. Since for any point
C on such an ellipse,
(1) C,X + C,Y = k
and at the intersection of the minor axis with the ellipse,
(2) C,X = C,Y and
(3) sqrt( C,X^2  ( X,Y / 2 )^2 ) / X,Y = e = 0.5
and at point A (assigning A to be near X rather than Y)
A,B = X,Y + 2 * X,A
A,B / X,Y = 2k
that I can solve (perhaps with Mathcad or by scribbling) for X,Y as
some function of A,B and a parmetric function of angle theta around
the midpoint of AB generating the elliptical path within the patch.
I will consider that solution and post here for a check later.

OK, I have an answer. For an ellipse with eccentricity 0.5, the
constant of proportionality between the major axis and the length is
sqrt(5)1.
Here's how I got that:
Since the proportions do not change, we may assume the major axis is 4
without loss of generality. That makes the minor axis 2. So the
distance from a focus to a midpoint along the side is sqrt(5) =
sqrt(2^2 + 1^2), because half of the major axis is 2 and half of the
minor axis is 1.
The sum of distances from foci to points on this ellipse is 2sqrt(5).
So the distance between the ends of the ellipse is, subtracting the
major axis, which is counted twice, from twice the constant distance
4sqrt(5)  4
and since the major axis is 4, the ratio is sqrt(5)1. In other words,
with major axis XY end X near focus A and Y near B, (AY + BX  AB) /
XY = sqrt(5)1
Doug 

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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Sun Jul 02, 2006 9:45 am Post subject:
Distribution of slope in a terrain sample



To find an eikonal between points A and B on for a bicycle on arbitrary
terrain, I have been considering, in news:sci.physics.research, how to
best establish that there is, first and foremost, some advantage to be
gained by deviating from the (as seen from above) straight line
connecting A and B.
I've got a simple routine in Mathcad that takes a reasonable time to
compute which assigns
( rnd( 2 )  1 ) / sqrt( i * j )
amplitude to the real and imaginary components of the Fourier
decomposition of synthetically generated terrain, with i,j < 9 and
generates the terrain with the icfft transform. The routine then
selects ponts A and B on the terrain, and rightturning and
leftturning paths from A to B parallel to the edges of the patch. In
addition to simulating the bicycle travel on the path, I intend to
compute the distribution of slope on the path and take a short cut to
travel time, comparing the two methods.
I want to generate paths with smooth turns, not right angle turns,
initially with A and B at the ends of the elllipse which has foci X and
Y
I figure an ellipse with eccentricity 0.5 will do. Since for any point
C on such an ellipse,
(1) C,X + C,Y = k
and at the intersection of the minor axis with the ellipse,
(2) C,X = C,Y and
(3) sqrt( C,X^2  ( X,Y / 2 )^2 ) / X,Y = e = 0.5
and at point A (assigning A to be near X rather than Y)
A,B = X,Y + 2 * X,A
A,B / X,Y = 2k
that I can solve (perhaps with Mathcad or by scribbling) for X,Y as
some function of A,B and a parmetric function of angle theta around
the midpoint of AB generating the elliptical path within the patch.
I will consider that solution and post here for a check later.
It's like the old problem of slamming two nails into a piece of 4x8
plywood, taking a bit of string, and jigsawing out an elliptical sign
that gets painted nicely with the name of the construction project you
are working on. You want the sign to have maximal area to maximize
visibility, and you want it to fit on the plywood. Where do you put the
nails?
I think the first step is to consider whether the nails have to be
along the longer central axis. That in itself could be a calculus
problem. I'd think (visualizing) they'd be on the diagonal.
In this case, having maximal area isn't a consideration. A and B are
picked randomly within a square patch of synthetic terrain and they are
the endpoints. not the foci.
Anyway, I think I made the above conditions right. Does anyone here see
a setup error, before I proceed?
Doug Goncz
Replikon Research
Seven Corners, VA 220440394 

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