[math-fun] Surveying a dig (or graphing on blank paper)
When we teach people how to use cartesian coordinates, we tell them to move x units along the x axis, y units up the y axis, and then take the intersection of the perpendiculars. The other day it occurred to me to wonder what would happen if instead, we drew a line from (x,0) to (0,y) and called the midpoint our destination. So if [a,b] are coordinates in Midpointesian coordinates (corresponding to the cartesian endpoints of the line (a,0) and (0,y)), and the point (x,y) is the corresponding point in Cartesian space, we could have a function f([a,b]) = (x,y). That function turns out quite trivial, but was nonetheless fun to try to visualize before I realized how straightforward it actually is, so I won't spoil it by giving it explicitly :) Then I realized it would actually be a rather useful thing to do. Say you're sketching out the graph of a function, but you don't have any graph paper. Use a compass and ruler to create X and Y axes, and mark at regular intervals. To plot a point (x,y) s.t. f^-1((x,y))=[a,b], draw a line from (a,0) to (0,y) and then use the compass to find the midpoint. Or say you're mapping out things you find at an archaeological dig. Create an X and Y axis at your site, and when you want to find the placement of an item, have a friend walk along each axis holding ends of a tape measure, with you holding it in the middle. When you read half the value seen by the friend holding the other end, they record their positions on the axes. (Or, in fact you could also record [a,b,r] for any line passing through the item, (a,0), (0,y) and showing a distance of r on the tape).
When we teach people how to use cartesian coordinates, we tell them to move x units along the x axis, y units up the y axis, and then take the intersection of the perpendiculars. The other day it occurred to me to wonder what would happen if instead, we drew a line from (x,0) to (0,y) and called the midpoint our destination.
I certainly miss something important here... The midpoint is (x/2, y/2), isn't it? So an alternative to the intersection of the perpendiculars would be the midpoint of (2x,0) and (0,2y). So now flame me...
Yup, it's really, really simple. But my brain still came up with some cool surfaces before I figured that out :) Mostly I think it's because there are an infinite number of lines through the midpoint, and until I saw the solution I kept thinking it would get weird, say, very near the +x axis for large x values. -J
On 1/19/07, Jason Holt <jason@lunkwill.org> wrote:
Or say you're mapping out things you find at an archaeological dig. Create an X and Y axis at your site, and when you want to find the placement of an item, have a friend walk along each axis holding ends of a tape measure, with you holding it in the middle. When you read half the value seen by the friend holding the other end, they record their positions on the axes. (Or, in fact you could also record [a,b,r] for any line passing through the item, (a,0), (0,y) and showing a distance of r on the tape).
I don't think I'm deliberately trying to hijack this discussion, to go banging on again about one of my hobby-horses ... But surely a better solution to this engineering problem would be to use distance geometry. Establish base points B_i for i = 1,2,3,4 at known locations on your site --- not concyclic and no three collinear --- for example, at the corners of a square enclosing it. To assign coordinates to a point X, measure the distances d(X, B_i) to the base points. The vector of squared distances [d(X, B_i)^2] will be a linear combination of the four vectors [[d(B_k, B_i)^2], for k = 1,2,3,4. The coefficients of this combination, obtained by multiplication by the inverse of this last matrix, give you coordinates for the point. These coordinates are a special case of classical (general) plane tetracyclic coordinates: a second linear transformation recovers the special tetracyclic form more usually employed, whose first two components are (ta-DAH) the Cartesian coordinates of the point. Under it, coordinates may be assigned to circles and lines as well as points; the linear transformations include the conformal Moebius group. However, there is now only a single inversive element at infinity; projective transformations other than similarities are no longer available. Finally, adding a fifth component --- the radius --- yields the still more general Lie system, the linear transformations of which include "equilong" Laguerre transformations as well as Moebius and Euclidean. Because of the redundancy, these systems are more robust than the standard Cartesian, special projective homogeneous, or barycentric options: for instance, small errors in pacing out the distances to your base points would result instead in a circle, whose radius estimates the uncertainty. These coordinate systems and their associated symmetry groups deserve to be better known. For instance, a spectacular member of the 3-D Laguerre group is the "offset" transformation computing the path of a milling machine cutter required to form a given contour. Fred Lunnon
On 1/20/07, Fred lunnon <fred.lunnon@gmail.com> wrote:
But surely a better solution to this engineering problem would be to use distance geometry. Establish base points B_i for i = 1,2,3,4 at known locations on your site --- not concyclic and no three collinear --- for example, at the corners of a square enclosing it.
Noooh! NOT the corners of a square (which are concyclic, and so linearly dependent ...) WFL
participants (3)
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Fred lunnon -
Jason Holt -
Pacher Christoph