Measure across the two ends of the arc. Let's say that is 48". This is the cord of the arc ( C ). Go to the middle of this (24" from each end) and measure up from this cord line to the top of the arc. Let's call this value H, for height. Let's say this value is 2"

R = [(C/2)^2 + H^2]/(2H)

In this example:

C = 48"
H = 2"

R = [(48/2)^2 + 2^2](2×2) = {24^2 + 4]/4 = [576 - 4]/4 = 580/4 = 145"

Just make sure you always use the same units, inches, feet, mm, cm whatever for both C and H. This of course is for a simple circular arc (constant radius).

This is an EXACT calculation, not an approximation, so it works no mater how much angle of the circle makes up the arc. Well at least up to a full 180 deg of arc.

I did this calculation quickly, but I do believe it is correct. But once you get an answer, get out a piece of string or similar, and make sure the radius you calculated works. Just swing this radius from each end of the arc. Where these two cross is the center of the arc. Then go from this center and check that the string follows the arc.

 

If it is a circular arc, and you know the width and height of the arc, you can use this formula:

R= (x^2 + y^2) / 2x

(x^2 means x squared--I don't know how to type exponents)

where x= height of the arc, and y = half of the width:

 

wildbill001,

After quick read of clin's post the method I describe is that same as that described by clin; and I agree with his ancillary comments. He beat me to the punch but I will post mine with a sketch.

If my geometry and algebra is correct, a couple of measurements and a calculation will result in the radius of the circle from which the arc is derived.

Draw a line connecting the ends of the arc. Measure the distance from the midpoint of this line to one end of the arc. This measurement is represented by y in the sketch.

Draw a line perpendicular to the line that connects the ends of the arc. This perpendicular line starts at the midpoint and ends at the arc. This is represented by x in the sketch.

Plug these values into the formula shown in the sketch. I think the resulting value will be the radius of the circle from which the arc is derived.

 

Yeah.

What they said. And sterff.

 

A mathless method is that you select two random chords and swing an arc both ways to find the perpendicular bisector. Where the two bisectors cross is the center. All with string, no unit of measurement, no math. Egyptian technology.

I can also find true north with a stick, a string, & a sunny day …

M

 

wildbill001

SAY WHAT,,,!

It looks to me that to simplify; Y & Z are the same length which makes X plus Z equal to the radius. ?!?

 

It looks to me that to simplify; Y & Z are the same length which makes X plus Z equal to the radius. ?!?

- TheTurtleCarpenter

Nope. For example, if the height of the arc (X) is 6, and half the length (Y) is 12, then the radius ( R) would be 15, not 18. X^2=36, Y^2=144. 36+144=180. 180/2X = 180/12=15.

Here is a good web page that lets you visualize it fairly well:
http://www.mathopenref.com/arcradius.html

Cheers,
Brad

 

TheTurtleCarpenter,

It looks to me that to simplify; Y & Z are the same length which makes X plus Z equal to the radius. ?!?

I see the confusion, the lengths do appear to about the same length in the sketch. I am not smart enough to know whether there could be a special case where Y = Z; if so it would be a special case and therefore not generally applicable. For example, move the cord (Y) either north or south on the sketch. If the cord is moved very close to the center, Y grows longer while Z grows shorter. The method and formula I and others propose works whenever the cord is north of the center of the derived circle.

The derivation of the formula does rely on the observation you made, that Z + X = Radius, which remains true no matter where the chord is placed north of the radius.

MadMark described an alternative method which also works for finding the center of an arc. But it has a broader application, that of the locating the center of a circle whose center is otherwise unknown. One additional note is that using more than 2 chords can improve the method's reliability. All the chord bisectors should intersect at the center.

 

for a circular arc

lay a framing square (inside corner) up against the arc

whatever it reads on both legs
is the radius

 

I KNEW I should have been paying attention in Geometry all those years ago!

A heart-felt thanks to everyone. This is awesome. Now, where did I put my tape-measure…..

 

I see what your saying Jbrow.

 

If you measure the chord (I thought that's how it is spelled) and get the height of the segment, there is no end of math websites to plug those numbers in and get your answer.

 

Fred Hargis,

Yep, you are correct! My spelling is awful, one of my many downfalls. Anyway, here is what the Grammarist say on the subject (cord vs chord):

A cord is (1) a string or rope, (2) an electrical cable, (3) a measure of wood equal to 128 cubic feet, (4) a ribbed fabric (short for corduroy) or pants made from the fabric, and (5) one of several types of cords found within the bodies of animals (e.g., the spinal cord and the umbilical cord)

Chord is usually a musical term (though it is sometimes used metaphorically) denoting any combination of three or more pitches played at the same time, and it also has a few rare uses in geometry and science.

 

This may be saying the same thing as many of the aforementioned, not sure.

If your arc is a circular arc, you can draw a circle through any three points not in a straight line. Just pick any three points on the arc; then draw a straight line between point A and point B and between B and C. With a compass, bisect those two lines; the intersection will be the center of the circle.

Hope this helps.

 

Here is a method I use to draw an arc
http://lumberjocks.com/projects/49135
If you are copying and arc simply lay a straight edge across the bottom of the arc and measure up from the centre to the top of the arc and you have the three points which will allow you to duplicate the existing arc .
Hope this helps .

Klaus

 

The classic tools for layout & such are straightedge, compass (string), plumb bob. No math, just proper technique.

Look at all the things that were built without fancy measuring tools and are still geometrically perfect.

M

 

Here is a method I use to draw an arc
http://lumberjocks.com/projects/49135
If you are copying and arc simply lay a straight edge across the bottom of the arc and measure up from the centre to the top of the arc and you have the three points which will allow you to duplicate the existing arc .
Hope this helps .

Klaus

- kiefer

That's an interesting method and looks to be useful. I had to develop a geometric proof to prove to myself it was correct. It is. but it isn't intuitive. I like that there is no need to find the center.

I actually have a need to draw an arc on my current project. This may be a perfect way to do that.

Thanks

 

If it is a circular arc, and you know the width and height of the arc, you can use this formula:

R= (x^2 + y^2) / 2x

(x^2 means x squared--I don t know how to type exponents)

where x= height of the arc, and y = half of the width:

- jerryminer

To type an exponent, enter "ALT" + 0178 for square (²) or 0179 for cube (³).

 

To type an exponent, enter "ALT" + 0178 for square (²) or 0179 for cube (³).

- MrRon

Thanks MrRon, but it doesn't seem to work for me. I type "x (ALT)0178" and get "x"

 

All seriousness aside, I'm really glad MadMark brought up the perpendicular bisector. As an engineer, graphical solutions are poetically beautiful. I can still remember generating cognate linkages…