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parametrize circle
#1
i have found a way to parametrize a circle whose radius is unknown using only a part of the arc. it is helpful if you draw this as you read.

we place the arc so that it is concave directly down, then connect the two endpoints. we take a tangent line to the circle where it starts on the right endpoint. we call the angle it makes with the horizontal line joining the two endpoints: "alpha". we make a perpendicular bisector through the horizontal line, and measure the distance from the point of intersection of the two lines, to the intersection of the bisector and the curve. we call this distance "h"

now that we have defined h and alpha, we need to relate these two bits to the radius.

from geometry, we know that any tangent to a circle is perpendicular to the radius.

since we assumed the arc is part of a circle (a top portion), we can call the horizontal line "r sin(@)" where r is the unknown radius of the circle because it is that many units above the origin of the circle.

so now, h+r sin(@)=r --> h=r[1-sin(@)]

and since we measured (approximated) the height h of the arc, we can solve for r --> r=h/[1-sin(@)]

but we do not know the value for @.

so we now need a relationship between alpha and @.

let us introduce an x,y coordinate system, and have it centered at the origin of the circle. again, from geometry, since the x axis is parallel to r sin(@), and the radial line r, intersects these parallel lines, the angle between r sin(@) and the radial line will be the same as @.

now we will call this angle "gamma"

once more, from geometry, the right angle made from the radial line and the tangent is made of alpha and gamma. so these must add to pi/2

and since gamma=@, pi/2-alpha=@ --> alpha=pi/2-@

now we can plug in a value for alpha that is in direct correlation to @ which the radius depends on.

our formula for radius becomes: r=h/[1-sin(pi/2-@)]

and from trigonometry, sin(pi/2-**)=cos(**)

so this becomes: r=h/[1-cos(alpha)]

so we can now compute radius.

but we need a value for alpha. there are two ways that work as well as you want/need them to: ruler, or ruler and protractor. you can use a protractor to measure the angle, or you can draw a triangle with the ruler and use a ratio of>> 1/2 the horizontal line:hypotenuse.

plug in your knowns and let it rip.

once you are convinced you have a pretty good approximation of the radius, you can parametrize a circle and find anything you want about it.

r(t)=< cos(t) h/[1-cos(alpha)], sin(t) h/[1-cos(alpha)] >
or r(t)= h/[1-cos(alpha)] < cos(t), sin(t) >

and the curvature will be: 1/radius of circle --> [1-cos(alpha)]/h

there you have it. you can also use a part of the arc of a curve to approximate its curvature at a point using the same method.

play around with it and let me know if yo uguys find anything interesting.
-joe
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#2
Hello valohim,

valohim6 Wrote:i have found a way to parametrize a circle whose radius is unknown using only a part of the arc. it is helpful if you draw this as you read.
...

I just indeed drawed your description

.png   circleparam.png (Size: 10.36 KB / Downloads: 650)
so it may become a bit more accessible.

Quote:so this becomes: r=h/[1-cos(alpha)]

so we can now compute radius.

but we need a value for alpha. there are two ways that work as well as you want/need them to: ruler, or ruler and protractor. you can use a protractor to measure the angle, or you can draw a triangle with the ruler and use a ratio of>> 1/2 the horizontal line:hypotenuse.

plug in your knowns and let it rip.

once you are convinced you have a pretty good approximation of the radius, you can parametrize a circle and find anything you want about it.

r(t)=< cos(t) h/[1-cos(alpha)], sin(t) h/[1-cos(alpha)] >
or r(t)= h/[1-cos(alpha)] < cos(t), sin(t) >

and the curvature will be: 1/radius of circle --> [1-cos(alpha)]/h

there you have it. you can also use a part of the arc of a curve to approximate its curvature at a point using the same method.

play around with it and let me know if yo uguys find anything interesting.
-joe

Your computations look all correct to me.

In plane geometry there is the classical approach of constructing with compass and straightedge and the different analytic approach, where you compute lengths and angles by formulas derived from the geometric configuration.

You follow a somewhat mixed approach, given an arc on a piece of paper, how to construct-compute the radius. Can you give a purely compass and straightedge construction for the center M or for a line with the length of the radius?
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#3
where can i find a compass? i have looked everywhere but all i can find is a little plastic piece of crap that kids use to draw circles. they dont even know how to make use of one. i know how to bisect a line with only a straight edge and compass, yes. but im not sure what else you are asking.
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#4
valohim6 Wrote:where can i find a compass?

Perhaps ask your parents to get one at christmas? Wink

Quote:i know how to bisect a line with only a straight edge and compass, yes. but im not sure what else you are asking.

Given an arc on the paper, construct the center of the circle to which the arc belongs (with straight edge and compass).
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#5
valohim6 Wrote:where can i find a compass?

At most art supply stores you can find nice compasses.


bo198214 Wrote:Given an arc on the paper, construct the center of the circle to which the arc belongs (with straight edge and compass).

Draw two different secants in the arc.  Construct the perpendicular bisectors of the resulting chords.  These bisectors intersect at the center of the circle.

[Image: circle-center.png]
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#6
pat Wrote:Draw two different secants in the arc.  Construct the perpendicular bisectors of the resulting chords.  These bisectors intersect at the center of the circle.

Hey, that was a question for valohim! Wink
But it seems he didnt expect to solve exercises, so thanks for the solution and welcome on board!
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#7
bo198214 Wrote:Hey, that was a question for valohim! Wink

Sorry, I didn't realize it was an exercise. I thought you'd know how to do it, but I wasn't sure. On the tetraspace forum, it's been odd to me that the slices of retina thing doesn't seem more intuitive to you, while on every other topic, your math-sense blows me away. They don't teach compass/straightedge construction everywhere these days, so maybe you'd never done that one. Oops.
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