Sunday, December 1, 2013







ELLIPTICAL TRIANGLE


This is a figure I created upon desiring to take the cusps out of the Reuleaux Triangle. The radius of curvature constantly changes unlike for the Reuleaux Triangle where the radius of curvature is fixed (excepting at the cusps). This is based on the ellipse as opposed to the circle. Since names get attributed so freely, we could call this the Blankenhorn Triangle, lol, and even as per below, apply a similar name to elliptical convex polygons.


It is accomplished by attaching to each side of an equilateral triangle, a portion of an ellipse so that the curve is continuous with no cusp.

There are an infinite number of possibilities of fitting with portions of elliptical arcs, therefore, I seek to characterize a "standard" form. This would not preclude there being any application where a different set of elliptical curves would be desired.

In seeking a unique solution in defining a "standard form" we could to have the center of the ellipse at a vertex of the equilateral triangle and the elliptical arc defined by the ellipse spanning the other two vertices. In this way I think it may provide for such a "standard" form as it would also relate more directly with the Reuleaux triangle in how an arc is constructed. In both cases however, I think for both using circle arcs and elliptical arcs, even numbered sides of regular polygons could be extended and not just odd numbered sides as the midpoint of an opposing side of two vertices could serve as the center for defining the arc placement for the two vertices.

And so if we set a vertex of an equilateral triangle having side length=1 at the origin and the rest of the triangle to the right such that the triangle would be symmetric with respect to the y-axis, the two other vertices would be at (sqrt(3)/2, 1/2), and (sqrt(3)/2, -1/2). The ellipse related to these points making the derivative of point (sqrt(3)/2, 1/2) = -sqrt(3)/3 yields a=sqrt(6)/2, and b=sqrt(2)/2.

y = sqrt(1/2-x^2/3)

In polar coordinates: r=1/sqrt(2/3+4/3 (sin(theta))^2)

And so the other two portions of the Blankenhorn Triangle would come from the portion of the ellipse defined above from x=sqrt(3)/2 to sqrt(6)/4 and rotated both pi/3 and 2pi/3. The ellipses in polar form would be:

r=1/sqrt(2/3+4/3 (sin(theta-2pi/3))^2), and r=1/sqrt(2/3+4/3 (sin(theta-4pi/3))^2)

After copying and pasting into a paint program then adjusting the scale for all three ellipses, losing some detail along the way then connecting them up in a photo editing program, I obtain this figure:





The above figure looks skewed as I encountered difficulty in the work-up. I will try to derive a parametric equation to trace the entire elliptical triangle on a single screen to confirm the shape.


The portions of the ellipses to piece together in analytical terms requires a bit of work. As for the first rotation by 2pi/3:

( dr)/( dtheta) = -(sqrt(3/2) cos(2 theta+pi/6))/(2 cos^2(1/6 (pi-6 theta))+1)^(3/2)

We desire the value of theta for which dr/dtheta = infinite and also for when dr/dtheta=-sqrt(3)/3. Of course there would be two values for each of these (not counting redundant rotations), but we would choose those that would in the second and possibly first quadrant. Now I made a bad assumption as we would actually require dy/dx to be the values infinite and -sqrt(3)/3. And so we would find the Cartesian equivalent and maybe even make the appropriate translation to connect to the first portion of the three total ellipse sections. Before going directly to the Cartesian equivalent, we could make use of the following relationship:


dy/dx= (dr/dθ sin(θ)+rcos(θ))/(dr/dθ cos(θ)−rsin(θ))

When dy/dx=infinite, 0 = dr/dθ cos(θ)−rsin(θ).

Well, this is getting involved. I think perhaps the better way of going about this is to simply make the translation to the appropriate location in the plane and then the range for theta would have to be from pi/3 to pi for this second portion of three total of ellipse sections.

Using the point rotation formula:

x'=cos(alpha)*x+sin(alpha)*y
y'=-sin(alpha)*x+cos(alpha)*y

We can map the point (sqrt(3)/2,1/2) to the new coordinate through rotating 2pi/3 (alpha=-2pi/3):

x'=(-1/2)(sqrt(3)/2)+(-sqrt(3)/2)(1/2)
y'=(sqrt(3)/2)(sqrt(3)/2)+(-1/2)(1/2)

(x',y')=(-sqrt(3)/2,1/2)

This (x',y') point needs to map to (0,0) for proper alignment to the equilateral triangle where the triangle as discussed above has one of its vertices at (0,0).

This then requires adding (sqrt(3)/2,1/2) to the equation: r=1/sqrt(2/3+4/3 (sin(theta-2pi/3))^2) to obtain the second portion of a total of three of the elliptical arcs to form the Blankenhorn Triangle. The third portion would be simply a flip around the x-axis of the translated ellipse forming the second portion of the curve.

This may still lead into more complications then what it's worth.

So in simply converting the transformed polar equation for the first rotation of the primary ellipse I get:

1=y^2+(5/3)x^2+2sqrt(3)xy/3.

From here I would solve for y and make the simple transformation and then convert back into polar coordinates. Being all values for y would be positive for the location I am interested in, I wouldn't be concerned with negative values for y.

I thus obtain:

y=(sqrt(3-4x^2)-x)/sqrt(3)


The translation of (sqrt(3)/2,-1/2) would thus be:


y=-1/2+(sqrt(3-4(x-sqrt(3)/2)^2)-(x-sqrt(3)/2))/sqrt(3), 0<=x<=sqrt(3)/2 And then directly following, the third portion of the elliptical arc would be: y=1/2-(sqrt(3-4(x-sqrt(3)/2)^2)-(x-sqrt(3)/2))/sqrt(3), 0<=x<=sqrt(3)/2 I'm not certain if there would be any gain in converting the above two equations into polar coordinates but I shall try. One step before I proceed would at least combine the above two equations: y^2=(1/2-(sqrt(3-4(x-sqrt(3)/2)^2)-(x-sqrt(3)/2))/sqrt(3))^2, 0<=x<=sqrt(3)/2 With the equation being reduced to: y^2=4xsqrt(3)/3-x^2-(2/3)x*sqrt(3-4(x-sqrt(3)/2)^2) This above graphs beautifully on wolfram alpha:



Graph of 2 of 3 portions of elliptical curves forming the Blankenhorn Triangle, noting 0<=x<=sqrt(3)/2)







After a very long time trying to figure out Wolfram Alpha's mysterious input format, this should work:
<br /> Plot [{{y^2=(1/2-(sqrt(3-4(x-sqrt(3)/2)^2)-(x-sqrt(3)/2))/sqrt(3))^2, 0<x<sqrt(3)/2},{y^2=1/2-x^2/3, sqrt(3)/2<x<sqrt(6)/2}}] from 0 to sqrt(6)/2
However, the scale is skewed. In trying to counter for the skewness, I expand the last number for the overall plot:
<br /> Plot [{{y^2=(1/2-(sqrt(3-4(x-sqrt(3)/2)^2)-(x-sqrt(3)/2))/sqrt(3))^2, 0<x<sqrt(3)/2},{y^2=1/2-x^2/3, sqrt(3)/2<x<sqrt(6)/2}}] from 0 to 3
And thus you see the scale got adjusted and can be viewed at:

THE COMPLETE BLANKENHORN TRIANGLE or ELLIPTICAL TRIANGLE

The image, noting the extraneous lines beyond the curve are merely extensions of the functions. Note also how smoothly the curves meet at x=sqrt(3)/2~0.866:







Showing the set of three elliptical curves circumscribing the equilateral triangle:






Well, at least this much got completed! Still more to go but this should serve as a good starting point for other researchers interested in this special curve.

In modeling this new geometric figure, one could circumscribe an equilateral triangle in such a way as to have the tighter radius of curvature at the vertices of the triangle. It doesn't produce anything different except perhaps a scaling factor of I would guess around sqrt(3)/2 or so. However, this may be the more desirable approach if wanting to apply elliptical curves to non-equilateral triangles.



Using the aforementioned method as the standard, we would have an overall area of the Blankenhorn Triangle with side length=1 to be:

sqrt(3)/4+6*(3*pi-6)/(16*sqrt(3))=SQRT(3)*(3*pi-4)/8~1.174498881

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Now, one could also go about constructing such an elliptical triangle in a converse fashion (using an equilateral triangle as the basis) by taking any ellipse with semi major axis of length 'a' and minor axis of length 'b'. The elliptical arc to the right (in normal number line configuration) when in standard position (or make the appropriate adjustments if not) of the value:

x=a^2/sqrt(3b^2+a^2)

would form one of the three sets of elliptical arcs in which to make the construction and then the other two elliptical arcs would be the same arc but rotated 2pi/3 and 4pi/3 and making the appropriate translations for continuous connections.


The side length of the 'inscribed" equilateral triangle would thus be:

s=2b^2*sqrt(3/(3b^2+a^2))


Note that id a=b, we do not obtain circular arcs creating a Reuleaux Triangle but simply the basic circle circumscribing a triangle as the constraints set forth was to have a smooth curve circumscribing the triangle, with no cusps.


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I was hoping it would have a consistent diameter but from rough calculations it failed - I would still like to re-check.


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This elliptical type of geometric curves could be extended to regular polygons and further to convex polygons:

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See also:

BLANKENHORN STEP POLYNOMIALS - A new type of polynomial with non-integral coefficients - instrumental in modeling processes where regular polynomials and other special equations prove to be inadequate

Football Quarterback Rating Formula with mathematical basis - compare with popular formula

Arithmetic-Geometric Mean Calculator

Advances in Ellipse Circumference Formula - Blankenhorn Ellipse Circumference formula arguably better than the popularized Ramanujan II formula


And many more!