Difference between revisions of "Covering Circle Radius versus Extreme Spread"

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So if a shot happens to fall in one of the green regions, then the covering circle radius will have to be more than 1/2 of the ES.  Of course if one shot was in the upper green zone and one shot in the lower green zone, then those two shots would be further apart than AB.
 
So if a shot happens to fall in one of the green regions, then the covering circle radius will have to be more than 1/2 of the ES.  Of course if one shot was in the upper green zone and one shot in the lower green zone, then those two shots would be further apart than AB.
 
<br clear=both>
 
<br clear=both>
Consider the radius of the circle circumscribing the top green region and points A and B. It has an diameter of:<br />  
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Consider the radius of the circle circumscribing the top green region and points A and B. Let C be the point where circles A and B intersect. Thus ABC form an equilateral triangle. The circumscribing circle has an diameter of:<br />  
  
 
&nbsp;&nbsp;&nbsp;&nbsp;Diameter = <math>\frac{2 \sqrt{3}}{3} AB \approx 1.155 AB</math>
 
&nbsp;&nbsp;&nbsp;&nbsp;Diameter = <math>\frac{2 \sqrt{3}}{3} AB \approx 1.155 AB</math>

Revision as of 21:15, 8 June 2015

The figure below illustrates the difference between the Covering Circle Radius and the Extreme Spread measurements. Given that AB is the extreme spread measure, a back circle which has AB as a diameter has been drawn. At points A and B, red circles have been drawn with a radius of AB.

Covering circle.jpg

Thus points in the green region are:

  • Outside circle that has AB as a diameter
  • Closer to A than B
  • Closer to B than A

So if a shot happens to fall in one of the green regions, then the covering circle radius will have to be more than 1/2 of the ES. Of course if one shot was in the upper green zone and one shot in the lower green zone, then those two shots would be further apart than AB.
Consider the radius of the circle circumscribing the top green region and points A and B. Let C be the point where circles A and B intersect. Thus ABC form an equilateral triangle. The circumscribing circle has an diameter of:

    Diameter = \(\frac{2 \sqrt{3}}{3} AB \approx 1.155 AB\)

Thus:
    \( 0.577 ES \geq CCR \geq 0.5 ES\)