Sorry if this is too elementary, if $R$ is the radius how do I visualize $1/R$? Thanks.

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    $\begingroup$ What are you trying to use this for? Some more specifics would help give a good answer. $\endgroup$ – Potato Dec 27 '11 at 1:09
  • $\begingroup$ The reciprocal of the radius is often called the curvature of the circle. If we have a more general curve $\mathcal{C}$, the reciprocal of the radius of the circle which hugs $\mathcal{C}$ most closely at $P$ is called the curvature of $\mathcal{C}$ at $P$. $\endgroup$ – André Nicolas Dec 27 '11 at 1:17
  • $\begingroup$ @Potato I am looking at the orbit rule written by Newton (currently known as Kepler's Third law) in its proportional form and I am trying to write all its permutations, one is, 1/R=RR/TT with radius R and period T, and I would like to draw a line that represents 1/R. Is this possible? $\endgroup$ – Zeynel Dec 27 '11 at 1:28
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    $\begingroup$ You should have mentioned this in your question to begin with. Most of us here don't have mind-reading abilities. $\endgroup$ – J. M. is a poor mathematician Dec 27 '11 at 2:05
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    $\begingroup$ @Pat: I don't have those abilities, but I also don't want to categorically state that none of us do. :) $\endgroup$ – J. M. is a poor mathematician Dec 27 '11 at 3:25

$1/R$ can be seen in many different ways.

One way to think of $1/R$ is that it is the curvature of the circle (seen as a curve in the plane).

If you wish to "draw" the length $1/R$ by using straightedge-compass constructions, it is possible to "look" at $1/R$, too. I'll shoot a picture of this here :

enter image description here

You start with a circle of radius $R$, and then draw a tangent line segment to the circle that has length 1 in both directions from the tangent point. This gives you the triangle formed next, and you can use compass & straightedge construction to draw the lines perpendicular to the triangle sides that goes through the point lying on the middle of the sides. These three lines intersect at the center of a circle that goes through all three vertices of this triangle, and there is a theorem in Euclidean geometry that says that if two straight lines go through a circle, we have $$ ac = bd $$ (in my drawing, we could replace the straightlines by any straightlines, and the roles of $R,1/R,1,1$ could be replaced by $a,c,b,d$, respectively, and the intersection needs not to be orthogonal). Therefore, the length that I pointed in the drawing to be $1/R$, call it $x$, satisfies $$ Rx = 1 \cdot 1 = 1 $$ thus $x = 1/R$.

Hope that helps!

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    $\begingroup$ Thanks so much. Exactly what I was looking for. $\endgroup$ – Zeynel Dec 27 '11 at 2:04
  • $\begingroup$ This is my blackboard by the way. This is the second time I'm using it to post answers on MSE... last time the picture was kind of blurry. Is it readable to you? I'm looking for feedback. =) $\endgroup$ – Patrick Da Silva Dec 27 '11 at 2:54
  • $\begingroup$ Yes, I think it is a great idea. It looks very nice on my computer. $\endgroup$ – Zeynel Dec 27 '11 at 3:13
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    $\begingroup$ I like your use of the blackboard for depicting answers. Classy. :) $\endgroup$ – J. M. is a poor mathematician Dec 27 '11 at 3:24
  • $\begingroup$ Thanks for the comments =) a blackboard at home helps a lot though. =P $\endgroup$ – Patrick Da Silva Dec 27 '11 at 3:31

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