Measuring a Radius

The Radius R = b + [(a² - b²)/2*b]

What is this and how did I get it????

Measure across the chord of the circular segment i.e. across the screen [as you would measure the diameter].

Let this dimension be 2*a so the midpoint is half 2*a = a

At the midpoint [a], measure the perpendicular distance [using a straight edge & measuring tape]from the line to the inside of the curvature, Let this distance = b.

Let the centre of the arbitrary circle be O [naturally] and the distance from O to the inside of the screen = R = b + c [where c is the missing [unknown] bit]

If you draw the segment with a & b, you can follow the following equation:

R = b + c

Then R² + [b+c]² = b² + 2*b*c + c²

But in right angled triangle, R² = c² + a²

Then b² + 2*b*c + c² = c² + a²

c = [a² - b²]/ 2*b

So R = b + c, as above.

Of course as a final check, measure the radius on a tape and check the curvature.

Same as above

This site has pictures

http://www.woodweb.com/knowledge_base/Radius_of_Convex_Wall.html

Thanks for both of your replies. They are both a great help.

This is similar to a problem that the traditional platelayer came across every day, in calculating the radius of a curved railway line, or in setting out a desired curve when access to the centre of the curve was not possible.

Let A, B, C be three points on a curve of constant radius. Let B be in the middle of the curve.

Strike a chord across A and C with a piece of taut, light string. Let the string length be c.

At the centre of the chord measure the offset between the centre of the string and the centre of the curve. Let's call this V (for 'versine').

The radius of the curve R = c²/8V. Now this equation is approximate. Although it is derived from Pythagoras'_theorem, it ignores a term in V², as V² is tiny in comparison with R² and the approximation is justifiable in terms of the accuracy needed in railway engineering.

Platelayers used to remember the formula and the technique because it was so simple. Of course, a rigorous application of Pythagoras and trigonometry would yield a more accurate result. It's time to dig those maths textbooks out for some refreshment!

I use the formula R=CxC + (4 xHxH) / 8xH where C is the lenght of the chord and H is the distance from centre of chord to centre of radius. That is the formula from "Machinery Handbook"

If the same symbols are followed as in comment #1 it simplifies as follows :

R = b + (a²-b²) /2b or 2Rb = (2b²+a²-b²) = (a² + b²)

R = (a² + b²⁾ / 2b .

This is also derivable from the fact that if two chords AB and CD within a circle intersect at a point 'O' then, AO X OB = CO X OD which by using the symbols as #1 in your case gives

a*a = (2R-b)*b

or a² = 2Rb-b²

R = (a2+b2) / 2b

Here is another approach. Create yourself a measuring template by drawing various arcs in CAD and printing out on a plotter or engineering printer set to full scale. Use a transparent or translucent material like mylar or vellum paper that will allow you to observe the curve you are trying to measure.

Find the arc which most closely fits your curve. Most manufactured curves in the US probably are based on integer,fractional (as in denominators based on powers of two, ie, halves, quarters, eighths, sixteenths, 32nds, or 64ths), or compound fraction (integer plus fraction) dimensions of the inch or foot. If it is an odd-ball fraction you can narrow it down to being between two standard fractions, then plot decimal-based curves between the two, which should get you, as we used to say, "close enough for government work!".

Of course this may not work very well if your radius is HUGE, in which case your arc would approximate a straight line anyway, but you did not say what approximate probable range for the radius. Most engineering departs will have at least a printer capable of 11 X 17 paper, so this could accommodate arcs nearly the length of the paper. Just be sure that all of your arc lines nest and you clearly identify their radii and leave sufficient space between them so you can discern between adjacent arcs. Encoding your lines by color, thickness, or line-type could help if they are closely spaced.

Provide that the manufactured part has a true circular arc, and not parabolic, elliptical or some other, shape (in which case the other methods won't work either!) you should be able to visually fit the printed curve to the manufactured one by placing the two ends of your printed arc onto the part. If the part bulges out more than your printed arc, then go to a smaller radius. If it bulges less at the center and in fact remains entirely under the printed arc (inside its concavity), then you need to move to a larger radius. In this way you should be able to quickly determine the correct arc by this high/low method.

From my experience, it may not be a perfect arc. If you really want to get as close as possible, measure as many points on the curve as possible and fit them with multiple arcs.

Put a straight edge/line across the ends of the curve. Mark maybe every 6" along the line then measure curve distance to the line at each point. You can then put the points in your CAD software and join them with arcs. The more points you measure, the more accurate it got.

Pineapple

"You can then put the points in your CAD software and join them with arcs."

Don't you mean "join them with line segments"? Two points do NOT determine an arc, basic geometry. After all, the radius is unknown at this point and the problem would be compounded if it is not "a perfect arc".

I applaud your approach, since it does consider the case if the radius is not constant. However, 6 inch sampling along the 96" length may be far less accurate than finding the closest true arc which best fits most of the curve.

If, in fact, the radius varies so much that the arc-fitting visual method does not work, then you will need a lot more than 16 data points. Even doubling that number by finding the approximate center point of each of the 16 arcs would still give a rather bumpy series of arcs, especially if the variation was due to a manufacturing flaw or wear and tear and not a design variation!

After all, calculus, which allows mathematicians and engineers to determine the formula of complex curves in a similar fashion, is based on infinitesimally small increments (dy/dx).

I'm confused with the point of all these formulas. Sounds like he has an actual screen. Wouldn't it be much more straight forward to say... use a string, swing an arc on the radius he is trying to determine, once the 2 arcs match, measure the length of string and he will have determined the radius he is looking for.

"I'm confused with the point of all these formulas. Sounds like he has an actual screen. Wouldn't it be much more straight forward to say... use a string, swing an arc on the radius he is trying to determine, once the 2 arcs match, measure the length of string and he will have determined the radius he is looking for."

Sounds like you remember a bit of something from a geometry or drafting class you once took. You can determine a line that bisects an arc from this method, even if you do not know the radius of the arc. Swing an arc from each end of the first arc. Connect a line between the two points where the new arcs cross each other and the point where this line crosses the arc is the center point of the arc.

Yes, the end of your radius will fall somewhere on this line, but where? This cannot be determined by the method you attempted to describe, unless I missed something in your description. Please clarify if you can.

Hi all,

Im am not a science student, just a newbie designer, who wants to give a measurement for a carpenter of a curve in my drawing. The curve is not a perfect arc, so cannot just use a simple radius of it.

How can I dimension this curve to illustrate to him?

please can this sbe explained in a non-formula way...

Cheers

Dan