Sorry if this is too elementary, if $R$ is the radius how do I visualize $1/R$? Thanks.
What is the geometrical representation of $1/R$?
-
2What are you trying to use this for? Some more specifics would help give a good answer. – 2011-12-27
-
0The 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 $P$. – 2011-12-27
-
0@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? – 2011-12-27
-
1You should have mentioned this in your question to begin with. Most of us here don't have mind-reading abilities. – 2011-12-27
-
0@J.M. : Are you implying that some of us do? Sometimes great faith makes me believe that Arturo Madigin does. XD – 2011-12-27
-
1@Pat: I don't have those abilities, but I also don't want to categorically state that none of us do. :) – 2011-12-27
1 Answers
$1/R$ 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 :
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!
-
1Thanks so much. Exactly what I was looking for. – 2011-12-27
-
0This 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. =) – 2011-12-27
-
0Yes, I think it is a great idea. It looks very nice on my computer. – 2011-12-27
-
3I like your use of the blackboard for depicting answers. Classy. :) – 2011-12-27
-
0Thanks for the comments =) a blackboard at home helps a lot though. =P – 2011-12-27
-
0Hi, I asked a similar question for a geometrical visualization of the inverse of volume. Do you have an answer for that too? Thanks! – 2011-12-30
-
0@Patrick, I envy you (blackboard at home?) My +1 – 2012-05-14
-
0@Artin : Yes, blackboard in my bedroom! – 2012-05-15
-
0@PatrickDaSilva: is it a slate blackboard? That's the best. – 2015-01-04
-
0@robjohn : Since then, things changed! I am now in Berlin with mathematician + physicist roommates and a 3m x 1,2m blackboard in the living room. I only know the name of the surface in German.. but trust me it's high quality. ;) – 2015-01-05
-
0@PatrickDaSilva: I think a slate blackboard in German is Schiefertafel. – 2015-01-06
-
0Oh, so you do know some german? Yes, that would be the translation, I guess. – 2015-01-06