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Treebeard's Stumper Answer
16 April 1999

Orchard Rows

However you feel about the land use, it's impressive how many vineyards there are in our area. I like driving past the new plantings watching the clear rows between the stakes that seem to move along with me. This effect is definitely best with new grapes. Now look closer. There are the real plowed rows, but you can also see clear diagonal rows to the sides, and even more rows between them, though they become less distinct. How many different rows are visible through the orchard stakes? What math is at work here?

Orchard Rows Picture

The hypnotic rows of stakes we see when driving past new vineyards correspond to simple fractions. Look at the picture below and sight down the rows. The numbers give the mathematical slope of each row. For example, the 2/1 line goes up two trees for every one tree over. The rows straight up and over are the widest. Next is the 1/1 row, and then the 2/1 and 1/2 rows. The rows get narrower and more crowded as the numbers get larger. With thin enough poles and a limitless orchard, there would be rows beyond counting!

Orchard Rows Graph
Figure 1


I took my computer to school on Friday as part of my project for the Dunn Middle School Science Fair, so I couldn't work at home this weekend. And this turned out to be a tough stumper! The real problem is to figure the widths of the many rational rows through the lattice. Given a row with slope m=y/x, is there a general formula for the apparent width? This is easy for slopes of the form m=y/1 and m=1/x, but the general case is tricky. I'll add to these notes as I find time (and inspiration). Consider this a stumper-in-progress.

(Update, 9 May 1999)

The complex pattern of orchard rows disappears quickly when the grapes sprout and obscure the virtual rows. Otherwise I'd be out with a transit!

There's actually two kinds of lines through the infinite orchard that don't hit any (infinitesimal) stakes. Lines with irrational slopes miss them all. But so do lines with rational slopes that run parallel to the rational diagonals close enough to avoid neighboring stakes. These lines may miss the mathematical origin, but they're close enough to perceive as a row. At first I thought the visible rows would correspond to the irrational lines that don't hit any stakes. But now I'm sure that the visible rows follow the rows where the stakes line up. You can see this in the photo if you carefully count rows and columns of stakes. We need the visual cue of aligned stakes to get the impression of rows.

That's assuming we're elevated a bit so we can see the aligned stakes towards the back of the orchard. At ground level in the orchard, there are no visible rows since the first stake of each row obscures all those behind it. The first stake corresponds to the fraction y/x, where x and y are coprime. Hidden behind it are all the other stakes with fractions that simplify to that value. But there are still clear paths corresponding to irrational slopes. Cantor's Diagonal Proof shows that there are infinitely more irrationals than the already infinite rationals, c > w, and so there are infinitely more irrational lines through the orchard than rational rows. But that's not what we see driving by.

The real problem is to figure the widths of the many rational rows through the lattice. Given a row with slope m=y/x, is there a general formula for the apparent width of the row? At first I thought this would be easy because I thought only of lines of the form n/1 and 1/n.

I'm placing the observer at (or near?) the origin. Driving by an orchard is different, but it doesn't effect the math if you assume the first few rows are invisible. I'm assuming a square grid, so the row spacing is the same as the spacing along the rows. I'm also ignoring the height of the stakes, though this may be really important for what we see. In the photo, the main row going straight back on the left actually includes several rows of stakes that seem to converge by perspective. It seems that if you can see over one orchard row to the grass at the base of the next row over, we perceive them as being part of the same apparent row. This is really complicated!

Consider a line of orchard stakes with slope 3/1, three up for one over:

Figure 2

We sight along the line BA. The apparent width of the row is limited by the red stakes at E and on the other side, so we see the row in yellow. The width of the row is given by the altitude h of the upper triangle AED.

The geometry of this case is not hard to figure out by using similar triangles. Consider a right triangle ACB:

Figure 3

CD is the altitude from right angle C to the hypotenuse AB, so CD is perpendicular to AB. Right triangles ACB, ADC, and CDB are all similar triangles because of shared angles.

The altitude CD is the mean proportional to the segments of the hypotenuse since

       AD   CD            m   h
  1)   -- = --     or     - = -        so:  h^2 = mn
       CD   DB            h   n
In fact, useful proportions abound in these similar triangles:
       AD   AC            m   b                 b^2
  2)   -- = --     or     - = -        so:  m = ---
       AC   AB            b   c                  c

       BD   CB            n   a                 a^2
  3)   -- = --     or     - = -        so:  n = ---
       CB   AB            a   c                  c

       CD   CB            h   a                 ab
  4)   -- = --     or     - = -        so:  h = --
       AC   AB            b   c                 c
These proportions provide an easy proof of the Pythagorean Theorem:
       n   a
       - = -       so: a^2 = cn      by (3)
       a   c

       m   b
       - = -       so: b^2 = cm      by (4)
       b   c

       a^2 + b^2 = cn + cm           adding equals
                 = c(n+m)            distributive law
                 = c(c)              because n+m=c by definition

  5)   a^2 + b^2 = c^2               QED
Combining these proportions with a bit of algebra gives a useful formula for the altitude of a right triangle:
           ab    ab         ab            abc
       h = --  = ---  =  ---------  =  ---------
           c     n+m     a^2 + b^2     a^2 + b^2
                         ---   ---     
                          c     c
By the Pythagorean Theorem (5), c^2 = a^2 + b^2 and c = sqrt(a^2 + b^2), so:
  6)   h = ---------------
           sqrt(a^2 + b^2)
This is a useful equation for our problem.

Consider figure 2 again. Since angle A is common and angles E and C are both right angles, triangles ACB and AED are similar. Therefore

       n   x               x
       - = -      so:  n = -     (this is the inverse of the slope y/x)
       1   y               y
In equation (6) for the altitude, a is n and b is 1, so we get
                 ab                (n)(1)               x/y
  7)   h = ---------------  =  ---------------  = ----------------
           sqrt(a^2 + b^2)     sqrt(n^2 + 1^2)    sqrt{(x/y)^2 + 1}
This is of the form:
For a>1, this approaches 1, but for a<1, it gets small. As the slope y/x gets large (steep lines), the value x/y gets small, and the altitude h from (7) gets very small. This is as expected. The 1/1 main diagonal is the widest, and the rows get narrower to the sides, something like a fan.

But this is only part of the story. This analysis doesn't generalize for more complex slopes of the form y/x when both x and y are not equal to 1. The problem is figuring which stakes along the hypotenuse will be closest to the visible line, and I don't see any easy pattern for this. For example, consider the rows with slope 5/n for n = 1 to 5:

Figure 4

The question is which stakes along the hypotenuse between the end points will be closest to the line and so limit the apparent width. They're shown here in red. Is there a pattern? Slopes are real numbers, but lattice points are integers, so this becomes a problem in discrete mathematics which is beyond me. Or am I missing somethng? The pattern is so easy to see, it shouldn't be so hard to figure!

The pattern I actually see in the fields is more complicated than just "widest in the middle getting smaller to the sides." The photo at the top of this page shows that this fan-pattern is actually repeated between each of the diagonals as fans within fans. This is what makes the visual pattern of orchard rows so attractive when driving by. I think there's a fractal here with a fine structure.

A brute force computer program could probably sort this out. My approach would be to consider every rational line with slope y/x for x and y up to some maximum value. For each line, I would then find the altitude for every lattice point on each side of the hypotenuse to find the smallest visible width. Maybe this program could also deal with the height of the stakes. Plotting the many rows in different colors corresponding to width should make a revealing graph. Maybe I'll find time this summer.

I didn't find anything on this stumper on the Web, which surprised me. I did find a few sources on related problems in my library.

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