I have lost my notes on how to determine the stretch length of an arc using a compass. Does anyone remember how to get the length of the portion of the circumference using a compass & straight edge?
Attached is a sketchup drawing which I hope will make things a little more clear.
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Replies
Try this:
Length = number of degrees, divided by 360, times pi times diameter.
Thanks guys,I know how to find the circumference of a portion a circle. What I'm hoping to remember is how to do it with just a compass.
You need the angle of the arc for that. Easy. If you don't have the angle, though ... I googled a few sites on the topic.
http://www.themathpage.com/atrig/arc-length.htm
http://www.sparknotes.com/testprep/books/sat2/math2c/chapter6section4.rhtml
http://www.mrexcel.com/forum/showthread.php?t=10482
1] I don't have sketchup [too long a download - need disk] so I can't open your picture.
2] If your straight edge is long, flexible and is graduated, then you can bend it to the arc and measure it.
3] Other than than, you resort to geometry. Draw the chord. Bisect the chord. This should run from the centre of the circle to the circle.
a] If you have the center of the circle, then just treat the radiuses and the length of the chord as a triangle and sic trigometry on it.
b] If you don't have the center, then draw another chord, bisect it and the two bisectors had better intersect at the center.
c] Just for kicks, I drew a chord from the beginng of the arc to the intersection point of 3] above and bisected it. This gives you a bunch of right triangles. Use the trigonmetry to find the angle from the beginning of the arc to the center to the end point of the arc.
d] Or just use a protractor.
4] Now that you have the central angle of the arc, you divide that by 360* to get the percentage.
5] The radius times two = diameter. The diameter times pi = circumference.
6] Circumference times percent = length of arc.
~Peter
Here is a pdf that can hopefully be opened.TFB (Bill)
bump for you
I did some googling and came up with noting. Have you seen it done before with a compass? I haven't but would like to see/learn how.
Is this the what you're after?
The right column is an explanation of the left column. I think fig, 2 on page 10 is what you are after
Hope this helps
Ott
Edited 1/19/2009 8:51 am ET by ottcarpentry
Here's the previous page for the first part of the explanation.
Ott
Sorry for all the posts, I just realized page 11 is even clearer. Here's all three.
Thanks for the posts. I have Collins book and that may be where I first saw that approach. There is an easier way, I'll have to look when I get to the shop today.
Kochanski's Approximation ... estimates the value of pi to four decimal places.
The line AF can be scaled to any length, hence multiplying pi by the same factor. Conversely, a line can be subdivided into equal parts. So we can produce lengths that are fractions of multiples of pi.
Having said that I can't see a really easy intuitive way to solve your problem for any random length.
How to unroll a Circle with a Straightedge and Compass
Ohhh Dude..marry me.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
Sorry, I'm already spoken for. :)Joe Bartok
Well it was worth a shot..LOL'
Hey so.. I need to make a semi circular vent that intersects an 8/12 roof ..I'd like to make a full size pattern of that ring segment..I used to "Loft" the curve by drawing the semi..extending lines to the proposed slope angle and joining the dots.
You got a smoother way? I'm all ears..well, not quite..but I'm listening..
BTW. my project is 3' wide.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
Sphere, there's a new thread at JLC where I explained the lofting of a cylinder intersection with a slope (incorrect response to the intent of the OP, as it turned out! All he wanted was the ellipse).
The only "smoother" way I can think of to plot the sinusoid is to use a programmable graphing calculator or the downloadable version of this WZ Function Grapher. These are the grapher formulations posted in the JLC thread ...
Select “Function” on the “Graph” drop down menu.For the “Displayed Range” enter …x min = 0x max = Radius × pi = Semiperimeter of cylindery min = 0y max = Roof Pitch × Diameter of CylinderThe formula to be graphed is …Roof Pitch × Radius × cos (x / Radius + pi) + Roof Pitch × RadiusFollow the graph with the cursor to find the horizontal (x) and vertical (y) dimensions; the default precision is two decimal places.8 inch pipe and 12/12 roof:Inchesx min = 0x max = 4 × pi = 12.56637y min = 0y max = 12/12 × 8 = 8Formula = 4*cos(x/4+pi)+4Centimetersx min = 0x max = 4 × 2.54 × pi = 31.91858y min = 0y max = 12/12 × 8 × 2.54 = 20.32Formula = 10.16*cos(x/10.16+pi)+10.169 inch Cylinder intersects a 7/12 Slope:Inchesx min = 0x max = 4.5 × pi = 14.13717y min = 0y max = 7/12 × 9 = 5.25Formula = 2.625*cos(x/4.5+pi)+2.625Centimetersx min = 0x max = 4.5 × 2.54 × pi = 35.9084y min = 0y max = 7/12 × 9 × 2.54 = 13.335Formula = 6.6675*cos(x/11.43+pi)+6.6675
I think the portion of the semicircle I'm showing may be wrong for your purposes; you likely have to plot the entire sine wave and slab off the top. The attachments show the formulas I used in my scientific calculator to plot the development of the 8 inch cylinder and 12/12 roof intersection. It has a "Replay" or "Edit" function so I can keep going back to the original formula and edit only the number of the "Station" ... makes plotting curves a breeze.
Joe Bartok
Edited 1/20/2009 9:21 am ET by JoeBartok
Edited 1/23/2009 4:18 pm ET by JoeBartok
Thanks, good stuff to chew on there..I'll be busy for awhile.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
The easiest means of finding the arc length is to set your calculator in "Radian" mode. The angle is treated as a ratio (which is what it actually is) ... the arc length/radius. This eliminates the extra steps involving degrees.
Either one of the following formulas will solve for the circular arc length:
Length of Arc = 2 × Radius × arcsin(½ Chord/Radius)Length of Arc = Diameter × arcsin(Chord/Diameter)
Joe Bartok
Edited 1/20/2009 2:46 pm ET by JoeBartok
Edited 1/20/2009 2:47 pm ET by JoeBartok
A link to an excellent Geometric Constructions site.Joe Bartok
Joe,I can't agree that the equations you put up are the "easiest". If all the information is given, your equation requires 5 steps not including switching the calculators mode.The one I gave requires 4 steps.AL = ((r x pi x 2)/ 360)x A.http://www.josephfusco.org
http://www.constructionforumsonline.com
"Joe,
I can't agree that the equations you put up are the "easiest". If all the information is given, your equation requires 5 steps not including switching the calculators mode.
The one I gave requires 4 steps.
AL = ((r x pi x 2)/ 360)x A."
Joe, the only information needed for the formula I posted is the chord length and diameter.
Length of Arc = Diameter × arcsin(Chord/Diameter)
"Switching the calculator mode" doesn't count as a step since angles are correctly expressed in radians. All calculators and computers use algorithms which solve internally angles in radians and a calculator should never be set in degrees mode to begin with. Calculating angles in degrees throws an extra conversion factor of 180°/pi into the trig formulas.
The sexagismal system used for degrees-minutes-seconds in timepieces and angular measurements dates back to ancient Sumeria and Babylon. The math served well in its time but it's now outdated, awkward to use and inefficient. No doubt this is why many folks have trouble learning trigonometry. The simplest way to express an angle is the ratio of the arc length/radius and leave the 180°/pi (or pi/180°, as the case may be) out of the equation.
These are not my opinions ... anyone can Google this information on the Internet or find it in a decent textbook.View Image
Joe Bartok
Joe,It doesn't matter what information is supplied, your calculation still needs either 5 with a radius or 4 steps with the diameter.Mine will either need 4 or 3 steps respectively.""Switching the calculator mode" doesn't count as a step since angles are correctly expressed in radians. All calculators and computers use algorithms which solve internally angles in radians and a calculator should never be set in degrees mode to begin with. Calculating angles in degrees throws an extra conversion factor of 180°/pi into the trig formulas.""If it requires an action, it's a step.http://www.josephfusco.org
http://www.constructionforumsonline.com
Joe, “If it requires an action, it's a step” … setting your calculator in Degrees mode also requires a step. No saving in effort there.
<!----><!----> <!---->
I don't see how your formula is any shorter. This is a breakdown by keystrokes:
<!----><!----><!---->AL<!----><!----> = ((r x pi x 2)/ 360)x A
(1) =
(2) (
(3) (
(4) r
(5) x
(6) pi
(7) x
(8) 2
(9) )
(10) /
(11) 3
(12) 6
(13) 0
(14) )
(15) x
(16) A
<!----> <!---->
That’s sixteen keystrokes. Here are the steps my formula:
<!----> <!---->
Length of Arc = Diameter × arcsin(Chord/Diameter)
(1) =
(2) Diameter<!----><!---->
(3) ×
(4) arcsin
(5) (
(6) Chord<!----><!---->
(7) /
(8) Diameter<!----><!---->
(9) )
<!----> <!---->Nine keystrokes.
Joe Bartok
Edited 1/23/2009 4:56 pm ET by JoeBartok
Joe,Since you're concerned with keys strokes and not steps, this only takes 8. . .Just entering the info in order removes the need for the parentheses. Sorry to say you will need yours. . .
AL = d x pi / 360 x A
(1)= There really is no need for this as well.
(2)d
(3)x
(4)pi
(5)/
(6)360
(7)x
(8)ahttp://www.josephfusco.org
http://www.constructionforumsonline.comEdited 1/23/2009 7:16 pm ET by Joe
Edited 1/23/2009 7:17 pm ET by Joe
Joe, each operator, parenthesis and number entered requires a keystroke or a step.
When working with angles in degrees there is always that conversion factor of pi/180 or 2*pi/360 (or the reciprocals of these factors). In addition, using these conversion factors means inserting parentheses in the formulation. Expressing a formula in degrees automatically means the formula will be lengthier than if we write the formula in radians.
EDIT ... yes, both of our formulations can be shortened when entering them in a calculator by arranging the terms to our advantage. My calculator (and I presume yours as well) accepts juxtaposition so the multiplication sign can be dropped. And I can leave out closing parentheses. So my formula looks like this in the calculator:
= Diameterarcsin(Chord/Diameter
You can probably shorten your formula as well ... but you can't get away from extra keystrokes (steps) needed to enter those degree conversion factors!
Joe Bartok
Edited 1/23/2009 7:30 pm ET by JoeBartok
Edited 1/23/2009 7:32 pm ET by JoeBartok
Joe,I'm sure if you entered that formula into a calculator you'd get an error.I'm sorry to say that all things being equal my formula is one step faster then yours no matter how you wish to slice and dice it. But, if you want to think otherwise, go right ahead.http://www.josephfusco.org
http://www.constructionforumsonline.com
Any formula that employs degrees to find an arc length needs that 180/pi conversion factor ... that's five extra keystrokes or steps.
(1) 1
(2) 8
(3) 0
(4) /
(5) pi
The formula I posted; the actual keystroke sequence is Diameterarcsin(Chord/Diameter=, works directly from the given diameter and chord, no conversion factors necessary.
Joe Bartok
Edited 1/24/2009 9:25 am ET by JoeBartok
Edited 1/24/2009 9:41 am ET by JoeBartok
Here are a few more examples of math involving arc lengths:
Radius of a Circle given the Arc Length and the Chord, Apothem or Sagitta
Circular Arch sized to a Rectangular Bounday
Solution of Bow Window Central Angle
Solving the Arc Length of an Ellipse
Note that the angles are given in radians (arc length/radius) as this is the fundamental ratio. Were the angles expressed in degrees there would be multiplication(s) or division(s) by pi/180, making these problems more difficult or impossible to solve.Joe Bartok
Joe,In your drawing of the Trig function of a unit circle along the x-axis, from edge of the right angle to the point of the arc length is what I believe is called the versed sine. Are you familiar with it and if so how is it determined.Thanks
Edited 1/24/2009 8:05 pm ET by mathewson
It's also called the sagitta ... as in arrow, with the arc being the bow and the chord the string.
Wikipedia Article ... Versine
How is it determined? Depends on the givens.
Sagitta = Radius(1 – cos theta)
I gave the iterative solution of the radius using Newton's Method if we know the arc length and sagitta in this link to my online notes:
Circle Radius given the Chord, Apothem or Sagitta
Here is a link to my Sagitta Calculator given the arc length and chord. Newton's Method requires an understanding of the graph to implement reliably (human judgement ... something sadly lacking in a computer) so the Javascript uses a brute force method of crunching the numbers:
Sagitta Calculator
Joe Bartok
Edited 1/26/2009 12:11 pm ET by JoeBartok
Edited 1/26/2009 12:13 pm ET by JoeBartok
Edited 1/27/2009 9:42 am ET by JoeBartok
Thanks Joe,I'm going to have to spend some time going over the links tonight. Might take more than just tonight....
Keith,You might also want to look at this page first and then follow the links from there.http://mathworld.wolfram.com/Chord.htmlhttp://www.josephfusco.org
http://www.constructionforumsonline.com
Joe & Joe,Thanks for the links guys. In all modesty & fairness, in college calculus was the weeder class for my major. It took me 2 tries in order to get a passing grade to graduate. That was almost 30 years ago and I haven't used it since, so some of this will most likely pass me by.
Call me stupid, but if you have the arc and want to determine the length and you have a compass ..or dividers.
Could you not walk the dividers at a pre set dim. say..1/2" and count the steps till you either run short or over and reset for the last leg? It could be say 4 walks at a 1/2" plus 2 walks at 3/16" so the length would be 2 and 3/8THs ..???Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
Sphere,I may not have stated my question correctly or misunderstood the answer. What I was getting at comes from an example in the Audel's books where they were showing a round deck and how to size the boards given the radius and a given shoulder width. It said that with a right angle and versed sine one could know the required board width. I couldn't find how to determine the versed sine.
I see now. I was the one confused. carry on, I'm learning a lot.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
"I may not have stated my question correctly or misunderstood the answer. What I was getting at comes from an example in the Audel's books where they were showing a round deck and how to size the boards given the radius and a given shoulder width. It said that with a right angle and versed sine one could know the required board width. I couldn't find how to determine the versed sine."
I misspoke in my versine post. Technically the versed sine formula is:
1 - cos theta
The value of the versine is between zero and one. Old tables of trig functions used to give the versine but it's usage (along with tables of trig functions, logarithms, etc) has fallen out of fashion in recent years with the advent of inexpensive calculators. In my formulas and the calculators linked to I was multiplying through by the radius. Which is O.K., because that's what we have to do in a real would application such as your board width problem.
The working angle, theta, may be found by the formula:
arcsin (Shoulder Width / Diameter)
Board Width = Radius × versine theta
The drawing will help visualise the geometry.
Another way to think of this: the angle formed by the sagitta and half the chord width, or 4/h, is the half of the angle formed by h/Radius, where arcsin (h/Radius) = theta, the working angle. (This diagram doesn't show the radius in the correct position for this solution ... from point B to the center ... but you can easily draw one yourself).
View Image
Suppose our arch has a thickness, that is, we need a double radius. Now the plot thickens. For the amusement and edification of one and all here is a link to my Circular Arch sized to fit a Rectangular Boundary Page. There are links to a couple of calculators at the bottom of the page.
Joe Bartok
Edited 1/27/2009 10:12 am ET by JoeBartok
Edited 1/27/2009 10:16 am ET by JoeBartok
Edited 1/27/2009 10:27 am ET by JoeBartok
More about the Sagitta of an ArcJoe Bartok
Thanks Joe,I can see it better now. Will have to do some drawing before I get to the point where I can use it on a job.Keith
Joe, you don't know who much I wish math would come easy to me. All of that stuff is Greek to me. Oh well.
Time is money and knowing the formulas and how to apply them to worksheets, Javascript and calculators sure speeds things up. But the so called "higher math" has its basis in geometry. Just about anything a person can do with a calculator can be done with compass and straightedge. For centuries, geometry was the only mathematical tool the average craftsman had to work with.
Here is how to develop the sinusoid representing the intersection of a cylinder with the slope of a roof ... compass and straightedge only. Kochanski's Approximation ... if we make line AF the length of the radius, then line FE will represent the arc length of the semicircle. Subdivide line FE into as many possible divisions as convenient, and number the stations or points on the line.
Construct a circle equal to the cylinder diameter. Subdivide the semicircle into the same number of angles as the stations or points on line FE. Draw the line(s) of the run(s) as shown in my diagram below; find the rise(s) for the run(s) by drawing the pitch triangle. Transfer the length(s) of the rise(s) to the appropriate station(s) on line FE. Connect the points with a fair curve.
The completed development will be as precise as the developments I constructed using a calculator.View Image
Joe Bartok
Edited 1/20/2009 10:28 am ET by JoeBartok
Only wish I knew how Joe. I guess somethings come natural to some people.
If you want to get the arc length by calculation, a simple method is:((r x pi x 2) / 360) Then take that number and multiply it by the number of degrees in the arc.http://www.josephfusco.org
http://www.constructionforumsonline.com
Hope attachment this works.
Just A Guy With A Hammer
Thanks.
I'm not sure that I fully understand the problem, but having done a lot of arc work, always by trial and error, I was thrilled to find the following formula. The formula was in the Roof Framer's Bible. It suits the situation I run into a lot, where I want to cut an arc in a horizontal board ie. for a fence top, I know the length I want and the height, but not the radius. The formula where C=the length from left to right, and H=the height of the arc. (C2 / 8H) + (H / 2) = RadiusThe C2 means c to the power of two.
I have this formula listed in my notes on my phone as well since I won't have the excell chart to play with.
Just A Guy With A Hammer
I'm unsure of your quest, but I get the feeling that it can be done by walking off with the compass, trail and error wise till you get an even amount of steps..I know, it's a drag..but damm..Joe's dissertation has me reeling in the splendor of Fibbinacci and Aristotle..if ya need an arc segment or chord length and don't have a CM2 calculater..I used a wheeled "traveller" and a compass to get the radial length.
A simple screen tool or a pizza cutter, make a mark, aline it w/the handle, walk it around the project and count the times it passes the handle..a 3" wheel will give you just over 9" per rotation..then usin the compass take the remainder of the mark to handle, and measure that distance. Add that to the revolutions and wallhalla you have a gross number.
I think then the compass dim. is an arc segment equal to a NINTH of the radius..so usig a 3" wheel and a 3" circle, it traverses one and a bit of revolution, that bit ought to be a factor of pii to the 3rd..divided by the number of segments you aim for.
Of course, I could be all wet..it's been a long time and I have slept since then.
I see you lofting the rise and I get a headache, I just eyeball most of it, you show real tenacity in your approach to garner the correct rise and run by the book..hat's off to you.
Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
I puzzled it out today. Sorry the pic is so bad
All radiuses are the same.
Points are struck on the horizontal line 1/2 the dia of the semi-circle
Arc's are struck on either side of the semi-circle drawn
A line is drawn from the intersections of the arc's & edge of semi-circle on both sides
A line drawn tangent to the apex of the semi circle and terminating at these lines equal the circumference of the portion of the circleI have a few stairbuilding books from the 1800's and in it they show beautiful work for the 1600's and made me stop to think how they did things with out a CM. So I'm teaching myself how to do the layouts without a calculator as well...
Edited 1/19/2009 10:21 pm ET by mathewson
Isn't it great as we get older we want to get back to basics. . . I've got some nice drawings as well from when I did some handrail work back in another life. The technique you show here is good, but I think it only works for semi circles.http://www.josephfusco.org
http://www.constructionforumsonline.com
Edited 1/20/2009 8:54 am ET by Joe
It does help to keep the interest level up.I remember reading about it somewhere being used for laying out handrail curves.
When I drew it out yesterday I briefly looked at a smaller section and believe it works. Will look at it closer today
Joe,It only works when laid out for a semi-circle. By using the centerline one can find the stretch out for a 90° turn, but it still has to be laid out in 180° first.
Hey, Guy,
This thread has gone on for a while, and I’m getting a bit scared now that Joe and Joe Bartok have taken the guards off their calculators and thrown them away --
Someone’s about to lose a finger!
But I’m curious, though. Looking at ToolFreakBlue’s drawing, are you trying to construct arc DBE, given points D, B, and E?
If so, you don’t need a compass. The old carpenter’s trick for this was to drive nails at all three points, set a 1x against nails D and B, set another, overlapping 1x against B and E, and tack the two pieces of wood together.
Remove nail B, hold a pencil into the inside corner at B, and slide the sticks over in both directions until you have drawn arc DBE.
Usually, you don’t really need to know the radius, or the center of your circle, you just need the arc, and this is a simple way to construct it.
Also, since it’s a physical, real-world solution, you can often use it to actually build the finished product. For instance, plasterers used to use these sticks as a physical guide to help set the bead on arches to a perfect arc.
AitchKay
Yes, the guys are getting a little intense, but they both have a good depth of knowledge.What I was trying to do is determine the circumference of a portion a circle without using a calculator. I.E. how did they lay out stairs a couple of hundred of years ago.Good tip with the sticks approach.
Gotcha. Then my tip wouldn't help you (well, maybe you could use it to guide a map measurer!).AitchKay
Sorry guys. All I was trying to establish is that an arc length calculation is best carried out using the briefest possible formula and a minimum of information.
This is not to say that Joe Fusco's formulation is wrong. But an arc length equation involving degrees is guaranteed to be lengthier than the same formula using radians. His method still gives perfectly good answers. (EDIT: We should just agree to disagree. I'm using a relationship that converts the chord and diameter ... linear measurements ... to an arc length. Joe Fusco's formula uses angles ... arc lengths ... converted to degrees. So we are trying to compare apples and oranges).
And yes, the original point of this thread was about compass and straightedge construction, not how to calculate the arc length ... apologies for derailing the thread.
Joe Bartok
Edited 1/24/2009 12:38 pm ET by JoeBartok
Edited 1/24/2009 12:40 pm ET by JoeBartok
Edited 1/24/2009 3:06 pm ET by JoeBartok
Edited 1/24/2009 3:09 pm ET by JoeBartok
Joe,This is a simple matter of addition, nothing more. You're free to use whatever you like, but if you use mine it will be quicker ;-). I'm done now.http://www.josephfusco.org
http://www.constructionforumsonline.com
"This is a simple matter of addition, nothing more. You're free to use whatever you like, but if you use mine it will be quicker ;-).I'm done now."Joe ... you were "done" when you first posted your formula. Forget about which formula is shorter or longer. The two formulas don't ask for the same input variables and it makes no sense whatsoever to try to compare them. There's no point!Joe Bartok