I’m going to keep this simple:

Let’s pretend that the earth is flat (1); for our purposes
it clarifies nothing to introduce spherical trig.

Here you see part of a grid that marks
locations. The horizontal lines are latitude
lines. They measure distances north and
south. The vertical lines are longitude
lines. They measure distances east and
west.

Now let’s say that the locational
precision in our system is

**two**decimal places. That means that no position on our flat earth can be described with numbers more precise than one one hundredth of a degree. All lat/lon values must be in one of these forms:
0.nn

N.nn

NN.nn

NNN.nn

Our system forbids anything else.

So let’s say that the lines in our grid
representation are exactly two fractional decimal places apart – a line (lat or lon) every
one one hundredth (0.01) of a degree.

I’ve put in some suggested numbers for us to work with. The first thing that you must notice is that
unless a location is sitting on one of the vertices (that’s where the lines
actually cross) it is

**in our system.***NOT REPRESENTABLE*
Let’s look at an example.
Let us suppose that there is a place called, oh I don’t know, let’s make up some unusual name like ‘Prophitis Elias’.

Now PE is a disadvantaged site because it happens to be located exactly halfway between all the vertices; at position 34.345 N and 22.835
E. Our system only allows us two decimal
points of representation and so the position of PE is rendered as 34.35, 22.84. (2) That
shifts the position of PI onto the vertex at that position. This is a deliberately introduced mistake
(called ‘aliasing’) that tries to keep the town of PE in the system. But it’s still a falsification which arises out of the constraints of our system and so it’s up to users to determine how serious this aliasing is.

How serious is it? How far is PE from its true position under this kind of aliasing?

I’m going to introduce some simplifying assumptions. I
calculate that the circumference of the earth at 35.34 N is 20,291.484 miles [3] and so one one
hundredth of a degree longitude at that latitude is 0.56365 miles or

**2976.0851**feet.
One one hundredth of a degree in latitude is

**3652.14666**feet. Since our town of PE is sitting exactly equal distances from the vertices the error would be the hypotenuse (*) of the triangle shown here***A**
The aliasing error for the town of PE would be 1826.0733
feet in latitude (NS) and 1488.043 feet in longitude (EW). These numbers, 1826.1 and 1488.0, are

**half**of the distances mentioned just above because PE is**half-way**from all the vertices. The actual error (A) is merely the hypotenuse of the resulting right triangle. Solving for A by the Pythagorean theorem gives us the maximum aliasing error in this system which is**2355.592**feet or 717.984 m. So the maximum error in this system is almost ¾ of a kilometer. I emphasize that this is true only for this latitude; these errors would grow smaller towards the poles and larger towards the equator.
More important: what is the average error? How far off will we be in the usual case?

The average error has to be less than the maximum error but
how much less? Is it half?

To answer this question I created a simulation and ran 1,000,000
trials several times. It turns out that
the error is not quite normally distributed (a little skewed to the low end) and the average (the arithmetic median) size of the error converges
on ~

**1272**feet (**388**m.)Here's a bar chart of 5,000,000 runs. Each cell from L to R represents an additional 73 m. in error.

To make a long story short – In a positional system limited
to two decimal places in the fraction you can

*expect*an aliasing error of nearly 400 m.
Who would create such a limited cockamamie geographical positioning
system and expect it to be used for precise work such as describing the position of, oh I don't know, say something like Bronze Age find sites?

Who indeed?

**Notes**

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