Oh boy, another math question!

[jpierce]

I've turned to you guys for help with my math deficiency in the past, and I'm back again.

I have done more round and half circle trim in the past 8 months than in the last 20 years.
A very common trim around here is 2"X6" for exterior with 1/2 log or cedar bevel siding. Interior it is common to just have 1"X4" trim with no profile.

I have been mitering the materials and pocket screwing the miter joints then routing the radius with a router and trammel.

To figure out my miter cuts I draw the arch in a cad program. Overlay an octagon or however many sides I need. Then measure my pieces in the cad program.

I found the formulas online to figure out the length of my pieces, but.....I just don't understand them. (pay attention in school kids!)

My method works, but it's time consuming, and I can't do it on site.

Does anyone have any suggestions for a better method, or a good place online that explains it better. I've searched the net, but I just can't seem to get the formulas pounded into my thick skull.

Do any of the calculated industries calcs simplify this process. I've had the project calc for years, but have been thinking in light of my math skill a better version may be in order.

Thanks for your help.

[Joe Bartok]

To figure out my miter cuts I draw the arch in a cad program. Overlay an octagon or however many sides I need. Then measure my pieces in the cad program.

I'm not understanding your question. Are you trying to fit a polygon to an arc?

Regular Polygon Calculator ... but you can't carry this in the field and fitting the polygon to a given arc is hit-and-miss. Another method of fitting polygons and arcs employs the calcs developed in the Bowed Bay Window Framing thread in the Rough Framing forum. (Bow Window Calculator; this fits a polygon to an arc but is even less suitable for field use).

Do you have a rough sketch of an example layout?

Edit ... or, are you trying to fit an arc within a given width of material? Circular Arch Layout

[Joe Carola]

To figure out my miter cuts I draw the arch in a cad program. Overlay an octagon or however many sides I need. Then measure my pieces in the cad program.

I found the formulas online to figure out the length of my pieces, but.....I just don't understand them. (pay attention in school kids!)

My method works, but it's time consuming, and I can't do it on site.

Does anyone have any suggestions for a better method, or a good place online that explains it better. I've searched the net, but I just can't seem to get the formulas pounded into my thick skull.

Do any of the calculated industries calcs simplify this process. I've had the project calc for years, but have been thinking in light of my math skill a better version may be in order.

Thanks for your help.

Are you looking for figuring out the 8 side lengths for an Octagon?

[Joe Bartok]

I found the formulas online to figure out the length of my pieces, but.....I just don't understand them.

Can you point to where the formulas are? This might lend some insight as to what you are after and perhaps someone can explain the formulas.

[jpierce]

I figured you both would chime in on this one!

Sorry I was not more clear.

Basically I am trying to find the 8 side length, for example. I have not always used 8 segments. The number of sides varies.

Yes I am also trying to fit the arc in a specific width of material also.

A better way to explain might be;
Take a round window for example. I know what I need for an inside and outside radius of the finished trim.
I'm trying to calculate the size of the polygon I need to create. Say an octagon for this example.
I need to calculate the size of each side of the octagon.
Then after the octagon is fabricated I would rout the inside and outside radius.

I don't know if this description is any clearer than my last attempt.

I will see if I bookmarked any of the sites I found formulas on.

I will look at the links provided also.

Thanks for jumping in and trying to help guys.

[Joe Bartok]

There are many ways this can go, depending on the constraints such as the depths and widths of available stock, must all the stock be of equal size, etc. So let’s start with an example … the solution of a Circular Arch. Most of the calcs in the link devolved about the piece with the pendant and can be disregarded. We’re looking only at the section at the top of the arch (or the adoining section). The givens were:

Depth of timber = 13
Depth of Arch = 9
Maximum Radius = 70.25

What’s the length of the timber?

The arch height is 13 – 9 = 4
The radius to this point is 70.25 – 4 = 66.25

Looking at the attached drawing: from Euclidean geometry we know that an angle inscribed in a circle is half the central angle. Therefore the angle inscribed in a semi-circle is a right angle and triangle ABC is a right triangle. Furthermore, if we drop a perpendicular of length h to point P …

∆ ABC is proportional to ∆ BPC is proportional to ∆ APB

Therefore we can write the formula in the diagram. (The drawing is not to scale). With a bit of algebra we can rearrange the terms and find h = 23.36664, doubling the result is the length of the timber required to achieve a fit for the arch.

Edit ... another solution, using the Pythagorean Theorem:
2 × √(70.25² – 66.25²) = 46.73329

The saw miter angle at the outer corner of the board is:
arcsin (23.36664 ÷ 70.25) or
arctan (23.36664 ÷ 66.25) = 19.42795° ... start the cut at 4 down from the top of the board.


Checking the answers ... enter the 46.73329 arch width and 4 arch height in the Circular Arch Calculator, or use the formula. The script will return the original radius of 70.25.

An image of the arch is attached to Post #15 ... Algebraic solutions of Framing Problems.

[Joe Bartok]

Basically I am trying to find the 8 side length, for example. I have not always used 8 segments. The number of sides varies.

Oh yeah, you've got that right!

Check out the Bowed Baywindow Roof Framing thread ... the "polygons" are apt to be Star Polygons, with a lot more vertices than those illustrated in the "Mathworld" link. The real question is: can size the curve to the board? It's not worth getting hung up on trying to fit a specific shape around the arch.

The math for fitting a polygon to a curve is ill suited for field use. I have a programmable graphing calculator that will easily handle either the iterative calculations or the graphical solution using Newton's Method. But it's time consuming to pound out the numbers on a standard scientific calculator. And - the graphics only show a line for the arch. A real arch has a depth and that's going to introduce more complications.

See what you think of the solution in Post #6. If you have situation that isn't covered post an example and we'll work through it.

Edit: More food for thought and digital tools to check your calculations. Regular Polygons ... there's a link to a nifty calculator. And if you know the chord and arc length, this Sagitta Calculator will find the arch height (the sagitta or Radius × versine θ) and other circle dimensions.

[Joe Carola]

Basically I am trying to find the 8 side length, for example. I have not always used 8 segments.

Yes I am also trying to fit the arc in a specific width of material also.

A better way to explain might be;
Take a round window for example. I know what I need for an inside and outside radius of the finished trim.
I'm trying to calculate the size of the polygon I need to create. Say an octagon for this example.
I need to calculate the size of each side of the octagon.
Then after the octagon is fabricated I would rout the inside and outside radius.

There are many ways to figure out the sides for an Octagon that has 8 sides for example. It all depends on the math you're used to.

360/8 = 45/2 = 22.5 (90-22.5) = 67.5° (Working Angle)

67.5° is the angle the hip runs at in plan view.

The hip run is the diagonal measurement in plan view and is also the radius for a circle/ inside of your casing radius because the radius is to the outside of the hip corner, not the sides of the Octagon.

Let's say that you have a piece of round 2-1/4" casing and to the inside of that casing the radius is 2'. That means that your diameter is 4'. If you draw a circle in Sketchup with a 2' radius and then draw an Octagon with a 2' radius, the Octagon will fit right inside that circle with all 8 corners touching it.

Couple ways to figure the sides when you know the Radius:

24 x 2 = 48 [Diag] 67.5 [Pitch] [Run] = 18.3688" or 1' 6-3/8" (Side Lengths)

[Rise] = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)


Or you can do this:

24 x 2= 48 x 67.5 (Cos) = 18.3688" or 1' 6-3/8" (Side Lengths)

48 x 67.5 (Sine) = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)


I made a drawing with a 2' radius to the inside of a 2-1/4" round casing. You will see the triangle inside the Octagon with the side length, the rise from outside to outside of the sides and the Hypotenuse where you will see the 2' radius and 4' diameter.

What I'm getting at is that it's very easy to figure the sides when it's for an Octagon room or window.

If the plans call for an Octagon shaped room with the outside measurements from side to side of 12' for example, you can do this:

12' [Rise] 67.5 [Pitch] [Run] = 59.64675" or 4' 11-5/8" (Side Measurements)

Or

144 x 22.5 (Tan) = 59.64675" or 4' 11-5/8" (Side Measurements)


If you had an Octagon window to trim and the measurement from the inside of the jamb was 4' from side to side (not the outside corners) and you wanted an 1/8" reveal for the casing. Now you have 48-1/4" from side to side.

48-1/4" [Rise] 67.5 [Pitch] [Run] = 20" (Short Point side measurement for casing)

Or:

48-1/4 x 22.5 (Tan) = 20" (Short Point side measurement for casing)

[Joe Bartok]

Another example calculation, the same arch as yesterday but divided into five sections.

I prefer to work directly from the circular arc lengths so the angles will be expressed in radians. *For those who are more comfortable with degrees multiply the radian values by 180/pi.

Arch Width = 119.518146
Arch Height = 33.318771
Arch Radius = 70.25
Arc Length = 2 × 70.25 × arcsin ((.5 × 119.518146)/70.25) = 142.922934

Check: Circular Arch Calculator … enter arch width and height. The correct radius and arc length are returned.

The arc length and angle are divided into five equal parts:

Arc Length = 142.922934/5 = 28.584587
Working Angle = arcsin ((.5 × 119.518146)/70.25))/5 = .203449016 … save to calculator memory
*Saw Miter Angle = .203449016 × 180/pi = 11.656770° … starts at the 1.448869 sagitta depth
Timber Width = Chord = 2 × 70.25 × sin .203449016 = 28.387801
Sagitta = 70.25 × (1 – cos .203449016) = 1.448869

Check: Sagitta Calculator … enter arc length and chord. The correct sagitta and radius are returned.

Timber Depth = 9 + 1.448869 = 10.448869
Height from Circle Center = 70.25 – 10.448869 = 59.801131
Inside Radius of Arch = 70.25 – 9 = 61.25
Height of Inside Radius from Circle Center = 61.25 × cos .203449016 = 59.986752
Height of Inside Radius from Bottom of Timber = 59.986752 – 59.801131 = .185621 … the arch falls within the bounds of the stock. Technically this dimension can be subtracted from the stock depth but it’s best to leave some room for tolerances, kerf, etc. Ditto for the other timber dimensions ... leave a bit of room to spare to stay the safe side.
Horizontal Dimension of Inside Radius from Arch Center = 61.25 × sin .203449016 = 12.375465

Check: Bow Window Calculator … enter:
Width or Span = 119.518146
Projection = 33.318771 – 1.448869 = 31.869902
Number of Bays (the number of arch sections) = 5

Sides = 28.387801
Radius = 70.25
The Central Angle is 2 × Working Angle

The attached diagram was plotted using the WZ Function Grapher. Moving the cursor over the intersections confirms the calculations above. The formulas entered in the graph plotter:

sqrt(70.25^2-x^2);
sqrt(61.25^2-x^2);
70.25;
68.801131;
59.801131;
4.847232x

[Joe Bartok]

Here are the formulas arranged to find the radius if the stock width and depth are given. They are fine for checking but I have to work on these some more because they also assume that the angle is known. Ideally we should be able to begin with only the timber width and timber depth and find the arch radius to fit the stock.

θ = Working Angle = arcsin (Stock Width/(2 × Radius))

Radius = Stock Width/(2 × sin θ)

Radius = (Stock Depth – Arch Depth × cos θ)/(1 – cos θ)

[Joe Bartok]

This is an example of a graphical solution where the arch depth and stock dimensions are known. What's the radius that will fit?

Does anyone have any suggestions for a better method, or a good place online that explains it better. I've searched the net, but I just can't seem to get the formulas pounded into my thick skull.

I don't think there is an easy answer to this fit-the-radius-to-rectangular-stock problem. It may be possible to simplify the formula so it's easier to use in the field; maybe not. A graphing calculator will accept the equation as written.

[markhoni]

Sorry to bust all the super math brainiacs bubbles, but sometimes our heads get too big for our necks to hold up. So how about eliminating the math altogether, and saving yourself a lot of time and thinking. Make a jig with two pieces of scrap wood that is at 135 degrees, with legs a and b (see atachment below). Set up your trammel and draw the desired arch and trim to actual width. then lay your jig on the outside edge of the arch and draw the lines of an octagon section against the arch. Then rotate the jig so leg a is on the line first drawn to leg b, and draw another line on the new position that leg b is now on. That would give you the two corners and the length of your octagon pieces. Then place the jig so it rests against the inside of the arch and parallell to the lines drawn on the out side, measure the distance between the jig and the outside line and you have your minimum width of board to use.

[Joe Carola]

how about eliminating the math altogether, and saving yourself a lot of time and thinking. Make a jig with two pieces of scrap wood that is at 135 degrees, with legs a and b (see attachment below). Set up your trammel and draw the desired arch and trim to actual width. then lay your jig on the edge on the outside edge of the arch and draw the lines of an octagon section against the arch. Then rotate the jig so leg a is on the line first drawn to leg b, and draw another line on the new position that leg b is now on. That would give you the two corners and the length of your octagon pieces. Then place the jig so it rests against the inside of the arch and parallel to the lines drawn on the out side, measure the distance between the jig and the outside line and you have your minimum width of board to use.

Mark,

The formula is right below here and takes about 20 seconds to do. It gives you the side lengths of all 8 sides and gives you the total width of the Octagon. Compare that to the time it would take you to do what you've just described.

Couple ways to figure the sides when you know the Radius:

24 x 2 = 48 [Diag] 67.5 [Pitch] [Run] = 18.3688" or 1' 6-3/8" (Side Lengths)

[Rise] = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)


Or you can do this:

24 x 2= 48 x 67.5 (Cos) = 18.3688" or 1' 6-3/8" (Side Lengths)

48 x 67.5 (Sine) = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)

[markhoni]

Mark,

The formula is right below here and takes about 20 seconds to do. It gives you the side lengths of all 8 sides and gives you the total width of the Octagon. Compare that to the time it would take you to do what you've just described.

Couple ways to figure the sides when you know the Radius:

24 x 2 = 48 [Diag] 67.5 [Pitch] [Run] = 18.3688" or 1' 6-3/8" (Side Lengths)

[Rise] = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)


Or you can do this:

24 x 2= 48 x 67.5 (Cos) = 18.3688" or 1' 6-3/8" (Side Lengths)

48 x 67.5 (Sine) = 44.34622" or 3' 8-3/8" (Side to Side outside measurement)

Joe, how many guys are going to remember that formula in their head? So now we're talking about having it written down somewhere, and how many guys are going to forget where they wrote it down, or where they put the formula they wrote down since the last time they used it. Or where they put their calculator down when they got interrupted with a 5 minute phone call from the school telling them their kid didn't show up that day, and that they wanted you to come in and talk to them about his/her behavior problems?
The OP was looking for a practical easy way to figure the sizing of his boards in an on-site at work situation. I'm just saying that math isn't always the best solution to the problem, and sometimes an alternative method is easier to digest and apply. I believe the method I showed fits that category with little room for error.

[Joe Bartok]

Joe, how many guys are going to remember that formula in their head?

The post is addressed to the other Joe but I'll put in my $.02 worth. The math is never about remembering formulas. There is no such thing as a "stand alone" formula. All formulas are based on geometry. The idea is to begin with the geometry, then we can either stay with the geometry if that seems easiest (and often is) or build on it and solve whatever formulas we need.

The OP was looking for a practical easy way to figure the sizing of his boards in an on-site at work situation.

Some of the solutions I've posted aren't the best suited for field use. Some problems are like that ... sorry. I can understand why the O.P. was having difficulty with the math and had to use AutoCAD.

But let's not forget that not all layouts are necessarily based on an octagon or other known polygons. We typically cut our own timber from logs and have to be prepared to work with whatever lengths and widths are on hand. In such situations it's a bonus to be familiar with a variety of solutions and be able to test a proposed arch for "best fit" to the stock available. Math is faster and easier than trial-and-error.

Here are the diagrams and formulas posted in this thread thus far; a couple of sketches of the assembled arches drawings have been added. The math can be done on any scientific calculator (programmable graphing type preferred ... these calculators will annihilate "complex" problems).

Circular Arch Sections sized to Rectangular Boundaries