Temporary Work
sin(30) * r = ½ *r 30 degrees and 30 degrees cross at center of square but the length of 30*r is less than r How to tell when 2 lines cross? Use Linear Algebra??? _______ r in the form of a circle not a line must convert to a line for linear equation form. But shouldn’t there be a matrices form of polynomial??? _______ Completing the square of a quadratic formula. That is exactly what the circle formed by the chord is. It is just another form of the quadratic equation. Have to test this. f(x) = ax^2 + bx + c = f(x) = a(x-h)^2 + k Since many equations have been found that describe a circle without using the radius and one of them uses the quadratic equation, maybe there is another simpler alternative to the quadratic equation. We know that there is in this instance. _______ Even though the parabola of a polynomial equation may lie in negative or imaginary numbers, the parabola itself may still be measured using relative coordinates. Picture of parabola-key equation (handwritten for now) _______ So essentially the x distance from the vertex (the x distance is the length of the chord, which is smaller than the radius) will yield a “y” value on the parabola. This y-value is the radius. Picture showing the parabola. _______ Three Hundred Forty Fourth Post: Math Application Ok so we have the “Arched Doorway Problem” and the “Parabola Key” but other than finding a chord’s encompassing circle and hence the radius is there any application? Is there an application that the contest on Constructor’s Corner is based possible? And the answer is very simply yes. As the people work on entries to the contest, I continue to work on my own to create more interest. In with alternating electric current there is a current and a voltage wave. It is described by a sine wave and measured in angles. But what if you only knew one value on both the current and voltage graphs (both occurring at the same time)? Well with the “Parabola Key” (if it works) and the other methods for finding the radius of a chord, you would be able to find the radius that encompasses it. (The chord on a power graph will always be less than the radius.) Once you know this one radius the other parts of the sine graph can be determined because you know the time of when the chord occurred versus the difference of length from the radius. Find the radius of both the voltage and current graph and determining the phase angle by the positioning of that radius, you know have the knowledge to determine the power. If you were to find the area of both the voltage and current by the formula 2*pi*r^2 (two pi r squared), you would have a complete description of the power so that voltage area + current area = new circle = total power. And the perimeter (2*pi*r) around this power circle would describe the power at various instances. Much the same way as the hysteresis loop describes magnetic power. So with 2 single chords we have determined the radius, phase angle, and power. We have also added the voltage and current graphs. This was done since the area under the sine curve is equal to the area of a circle with the same values. I’m am still working on this but this is the main idea behind my work. It will be posted on both the Blog and message board. So with new and interesting ideas... May the Creative Force be with You |