Figuring a Radius on Odd Piece

Usng the versine formula, for finding the radii of railway lines with a tape measure and a bit of string, is one way. The radius is the chord squared divided by eight times the versine. The versine is the distance perpendicularly from the chord to the apex of the arc.

It derives from applying Pythagoras' Theorem to a triangle, one point of which is at the centre of the curve, and ignoring terms in versine-squared, which is usually tiny in comparison with radius-squared. It's close enough for all practical purposes.

That might work for railway lines, but not accurate enough here. It's the property of intersecting chords, and reduces to R = chord²/h/8 when h << R. But if the sketch is anything like to scale, for the blue circle, taking chord = 1, height h (or versine) ~ 0.5, R ~ 1. But using R = chord²/h/8 gives R = 0.25. Full formula needed in a case like this.

let L be the length of the chord (dotted line) and H be the perpendicular distance from the midpoint of the chord to the radius. Then (R-H)^2 +(L/2)^2 = R^2. Solving for R (the R^2 terms cancel) R= (H^2 + (L/2)^2)/2H

Yep. That's the same sort of thing. Ignoring terms in H² just simplifies it a bit where H is very small w.r.t. R.

OK.

i will look at this tonight and get back to you all when i try the formulas.

thank you both.

One question though, what does the ^ mean ?

Thank you all and i will be back in touch tonight ( tomorrow morning)

nm

squared

How does one type a superscript 2

Use the ^{superscript function} in the reply box thus: ².

"..raised to the power of.."

Pythagoras shows the way.

^ means raised to the power for an example 5^3=5*5*5=125

If the drawing is to scale, you can work the radii out as a multiple of each chord as other posters have said. Then when you have your measurements, you're there. But if not, you can't do it, there aren't enough constraints. You need chord length and height from chord to top of curve.

Your compass and straight edge analysis to construct the perpendicular is elegantly simple. GA

All of the approaches though assume that the curves are symmetrical and circular. Most of the time this will be an acceptable assumption but it should be remembered to be an assumption.

A useful added step would be to also constructs the perpendicular to the line BE. This perpendicular should intersect at O if the points A, E, & B are all on the same circle.

IS THIS CORRECT?

nm

Using the following formula (from gringogreg in post #2) I come up with 6.5

R= (H^2 + (L/2)^2)/2H

R=(4^2 + (12/2)^2)/2*4

R=(16+36)/8

R=6.5

The formula should read [(b_xb) + (c_xc)]/2c

6x6+4x4=52, 2x4=8 and 52/8 = 6.5

I agree with the others 6.5

I don't know if the drawing is meant to be to scale, but it looks like the height or bulge is much nearer to 2 than to 4 (compared with the 6 to either side). ??

(If so, then R = 10)

If the drawing is on scale the radius is 8.5 square sides (4x2 on the horizontal and 1 on the center) b²= 4²+1²= 17 and c=1 thus R=17/(2x1)=8.5

It looks to me that you have correctly shown a tangent; the angle on both sides is the same as the the circle continues on both sides of the end point at each end of the arc. What is incorrect is the length shown on the truncated line at A & B. They must be extended until they intersect. I will try to scale the arc and calculate the units of the radius which will be a proportion to the partial circumference of the arc.

I am a great fan of Ted Williams - He is quite right about 3/10 = 30%

Quite true. You point out that as netmaker actually posed the question, knowing just the vertices as these curves intersect will never provide sufficient information to solve this problem. This is because knowing only two points on a graph one can only define the straight line that intersects them. If the function through these two points is anything but a straight line then if one knows the curve is specifically a circle or another specific geometrically defined curve (hyperbola, catenary, exponential, etc.) then a third point will uniquely define the known curve. When one does not know the curve, then as many data points as possible should be acquired.

GA for the first right answer - how come everyone else gets to miss the main point, and there is a high GA for a fairly obvious answer (to anyone who knows anything about geometry).

All those curves could be anything and still connect to their end points unless additional information (i.e. constraint) is provided.

If netmaker's arcs already exist, then of course nick name's construction is simple and correct. On the other hand (as some of us have interpreted), if netmaker's arcs are only example possibilities, than further info is needed. Thanks to several for the review of chord length plus arc bulge formulas.

I disagree - any "two" radii" intersect at the center of a circle and are "Tangent" at the point of intersecting the circumference of the portion of the circle.

You're missing my point. Any two radii of a circle will intersect at only one point is true. However, we are assuming that the curve is a circle if only one pair of tangents are used.

I printed the graph on Power Point, and the radii at each end of the arc, and they are the same length and intersect at the same point as is the center of the circle.

Try an ellipse or a hyperbola.

You can check if the curve is a circle arc or not in a very simple way:

look at the sketch in my answer, it is enough to repeat the construction of the line on the other side between E and B. If the 3 straight lines cross at the same point it is a circle arc if not it is an other curve.

It can be an ellipse and the 3 lines will cross at same point if CD is one of the ellipse axis.

In this case the 3 segments between crossing and curve will not be equal only 2 will have same length, the 3rd will be shorter.

Infinite solutions, or indeterminated.

Assuming that the drawing is accurate, to scale & the arc is part of a perfect circle.

I would find the centre of the arc & dotted line as it shows on this web page, so you can measure the distance between the dotted line & the arc.

Then use the 2 distances you now have on this web page, by moving the orange dots to those measurements. It will then calculate the radius for you & it even shows the formula of how to calculate the results. Or further down the page enter the distances in the calculator.

The formula from the web page above is:

where: W is the length of the chord defining the base of the arc H is the height measured at the midpoint of the arc's base.

It works for arcs that are up to a semicircle, so the height you enter must be less than half the width.

If I have assumed wrong in my first line then this is useless good luck

Reply to all,

This quickly became something outside my level of compitence and confidence.

The drawing is NOT to scale.

I am not sure of the exact measurements as this is a piece that was broken and I had to guess at some of the curves.

I believe i can figure what i need , close enough, from the formulas provided.

thank you all.

nm

I have read several of the previous Reply, and would generally agree they would lead to the right answer. A carpenter might take the following approach: Draw a line "tangent" to the arc at each end. The graph should contribute to the accuracy of these lines. Using a "square", draw a line perpendicular to each of these two lines. Where they intercept is the center of the circle that has this arc. the length of each line has the same length and because they terminate on the circle, they are defining the "radius".

Mr. Larry