How Do You Calculate the Length of an Arc for Hanging String Lights?

[cgeorgiadis89]

Need help finding the length of an arc...must be dependent on a specific diameter.

Basically, I want to hang outdoor string lights (I have a rectangular backyard). These lights will be attached to a flexible bar that will go from one fence to another. The height of the fence is 4 feet tall, but I want the center of the arc of the flexible bar to be 8 feet from the ground. The distance from one fence to another is 20 feet.

Please let me know what equation I would use to find the length of the flexible bar of which to hang these lights from, so I know how much to buy at the store.

Thank you

 

[zoobyshoe]

A rough and ready solution for practical purposes:

It's a simple proportion. The length of the arc is to the length of the circumference as the angle subtended by the arc is to 360 (the whole circle).

You start by geometrically finding the diameter of the circle of which the arc is a part of the circumference. Given any three points you can easily find that unique circle whose circumference contains those three points (and your three points are the two endpoints of the 20 foot span and the point 4 feet* perpendicular to the center of that span):

http://www.mathopenref.com/const3pointcircle.html

Do this carefully on paper drawn to scale. Once you have the circle, measure the diameter so you can find the circumference.

Then draw lines connecting the center of the circle to the ends of the 20 ft chord and simply measure that central angle with a protractor. The measure of that angle will be set in a ratio to 360. (If you're a reasonably careful draftsman, these measurements should be close enough for a practical back yard construction.)

That angle will be in a ratio to 360 degrees as the length of the arc is in a ratio to the circumference of the circle. angle/360=arc/circumference. Solve for the arc length.

*(I'm assuming the ends of the arc are going to be sitting on the fence, here, and not on the ground. If they will be on the ground then change the 4 foot dimension to 8 feet.)

This method is based on measurements of a diagram which, although you're making it to scale, can't be made or measured perfectly. If a tight algebraic solution is not forthcoming, this should still work well enough. (I don't get the feeling it's critical that the height be exactly 8 feet from the ground, just "about" 8 feet. So, buy long and cut back if it seems too high.)

 

[zoobyshoe]

edit: I just remembered there's a way to find the diameter algebraically with the given information:

Your 20 x 4 span and height can be broken into two right triangles, legs 10 and 4. By Pythagorus the square of the hypoteneuse of those triangles will be = 10² + 4²

By Galileo, the square of that hypoteneuse will also be equal to the length of the segment of the diameter (of the circle in which the triangles are inscribed) cut off by the chord (in this case: 4) times the full length of the diameter.

Given 10 x 4 legs the hypoteneuse will be: 10.770

4(x) =10.770² = 116/4 ft.

Now the only "iffy" measurement you have to make from the diagram will be the central angle.

 

[chingel]

You can calculate it using these:

http://en.wikipedia.org/wiki/Circular_segment#Formulas

 

[cgeorgiadis89]

got it, thanks. I'm going to calculate the angle first by setting it equal to 2arccros(height of fence/R).
R will be the hyptoneuse from the ground of the middle of the yard to the top of a fence.
When I find the angle, I can calculate the arc length = (angle)(R)

Thanks again

 

[chingel]

You don't know that the radius will be on the ground. For example, you could have a segment of a huge circle 1 mile in radius connecting the two fences and it would still connect the two fences, nevermind that its center is somewhere inside earth. The radius could be whatever, you are looking for the one which would make the height of the segment what you need. So you are looking for a circle, whose chord length is 20, c in the wiki diagram, and the segment height would be whatever you would want the arc to be above the fence at its max, h in the wiki diagram.

 

[zoobyshoe]

got it, thanks. I'm going to calculate the angle first by setting it equal to 2arccros(height of fence/R).
R will be the hyptoneuse from the ground of the middle of the yard to the top of a fence.
When I find the angle, I can calculate the arc length = (angle)(R)

Thanks again

I guess I didn't explain the Galilean method for finding the diameter (hence radius) well enough.

Here's a related, and less obscure, method with a diagram:

http://mindyourdecisions.com/blog/2012/07/06/5-neat-math-proofs-without-words/

Scroll down to #4 Geometric mean is less than or equal to arithmetic mean.

Your span, 20 feet, divides the diameter into two unequal parts. 1/2 the span, 10 feet (represented by the orange altitude in the diagram at the link), is the mean proportional (called "geometric mean" at the link) between those two parts. In the diagram at the site they know the lengths of the segments of the diameter and are looking for the orange altitude, but we already know it, 10 ft. and don't know one of the lengths. Knowing that 10 ft. mean proportional allows us to find the missing part of the diameter and add it to the known part to get the full diameter, because the known part is to the mean proportional as the mean proportional is to the unknown part.

The known part: If your arc is resting on the fence (which is 4 feet tall) and you want the highest point of the arc to be 8 ft. off the ground then the known segment of the diameter is 4 ft (represented by the green line segment in the diagram at the link). The unknown segment in your problem is the third proportional to 4 ft. and 10 ft, with 10 ft as the mean proportional. By a simple proportion we can find the unknown segment and add it to the known segment to get the diameter:

4/10 = 10/x → 100/4 = 25

25 + 4 = 29 ft. = diameter of the circle

(This is all the Right Triangle Altitude Theorem which you can google. And a right triangle can automatically be conceived of as inscribed in a semi-circle with the hypotenuse as the diameter. Galileo simply pointed out an alternate set of relationships for this circumstance.)

29/2 = 14.5 feet (called the "arithmetic mean" at the link). This is your radius. Plug that into the formula chingel found to get the angle. (and d can't be the height of the fence. d is the radius - h, and h = 4). Chingel is right: the center of the circle is underground.

If the ends of the arc will be resting on the ground and not on the 4 foot tall fence (you haven't specified which), use 8 ft. instead of 4 ft. as the known segment of the diameter. That's a whole different circle.

 

[bahamagreen]

Get a model of this (a couple of feet of thin wood dowel or thick wire...) and try bending it by holding just the ends and notice that to hold a circular bend you have to hold hard and apply a firm twist with each hand - clockwise for right hand, counterclockwise for the left; in addition to maintaining their relative positions by applying a lot of pressure as if to pull your hands away from each other... there is a huge lateral torque that needs to apply at each end to make a circular arc... each end needs two mount points to make this torque or loading.

From a practical standpoint the flexible length is not going to describe a constant radius arc, it is going to approximate a parabola... more curved near the center, straighter at the ends -this is a function of leverage which increases with distance from the center. The huge lateral torques would be needed to offset the effect on the bend from the variable leverage.

So unless you have some plan and mechanism to mount the ends quite firmly in which they will be held by a pair if vices or something to pull the ends outward from each other and at the same time applying torque to bend the two halves outward hard enough to match the curve of a circular arc, you won't get a circular arc.

Why not get somewhat more length than you need and trim off the excess when you make your desired height specification, then mount using a simple "static" binding (two simple forces pressing pretty much to the center) without having to include extra loading forces to make the curve? Would you be happy with just a good looking but not perfectly circular arc?