Replies

1. 
   Danford_C_ | Sep 15, 2002 06:20pm | #1

   Are we to assume that the segment is based on a circle?  If so, what is it's diameter?  The radius will be half that.  The segment will be measured in degrees.  A chord could also be constructed from the points where two radii are tangent to the segment.  FWIW.

   Dano

   "Form and Function are One" - Frank L. Wright

   
   

   ---

   Edited 9/15/2002 11:25:58 AM ET by Danford C. Jennings

1. 
   RW | Sep 15, 2002 09:21pm | #3

   Well, maybe I didn't explain well enough. DJ restates the initial issue. It is a segment of an arc. I could draw an arc on a piece of paper, and then a rectangle around it which touched the arc at both ends and in the very middle. I can then give the length and depth of the rectangle. The issue is finding the radius of the circle which the arc comes from. This arc would fit a rectangle of 7 ft long by 3 inches in depth, so I suspect that translates into a fairly large circle. Not something which you're going to use a router and a trammel to recreate. At any rate - so how do I find the radius?

1. 
   riverr1 | Sep 15, 2002 09:39pm | #4

   I don't have any reference books handy so I can't give you the exact formula, but you need both algebra and geometry. Take any of the equations that are cord related that use radius in the equation. Rearrange the equation to isolate for the radius with the knowns that you have. You have the chord length and the distance from the chord to the perimeter and that should be all you need.

   Don

   found this:

   r = ((1/2 w) squared + h squared) 2h

   w is the length of the chord and h is the height

   I come up with 24.58' as the radius

   Edited 9/15/2002 2:47:05 PM ET by Don C.

   Edited 9/15/2002 2:51:31 PM ET by Don C.

1. 
   TomMGTC | Sep 16, 2002 12:40am | #5

   That formula is correct. I needed it a couple of years ago to do an arched molding for my kitchen and it took a while to figure out. I keep it handy in the shop.

2. 
   UncleDunc | Sep 16, 2002 01:36am | #6

   The first time I saw this formula I stared at it for a long time, trying to figure out why I didn't have to take the square root of something. I also knew I'd never use it enough to remember it, so if it was going to do me any good at all, I needed to figure out how to derive it. In case anyone is as puzzled as I was the first time, here's the derivation, in excruciating detail.

   I've used the '^' character for the exponentiation operator instead of writing out 'squared'. 'w' is the width of the arch and the length of a chord of the circle between the two endpoints of the arch. 'h' is the height and 'r' is the radius. A line from the center of the circle to the midpoint of the chord, a line from the center of the circle to an endpoint of the arch, and a line from the midpoint of the chord to that same endpoint form a right triangle, with the radius as the hypotenuse, so the square of the radius is equal to the sum of the squares of the other two sides.

   r^2 = (r - h)^2 + (w / 2)^2

   First expand the (r - h)^2 term.

   r^2 = (r^2 - 2rh + h^2) + (w / 2)^2

   Then subtract r^2 from both sides of the equation. This is why no square root is necessary.

   0 = -2rh + h^2 + (w / 2)^2

   Add 2rh to both sides.

   2rh = h^2 + (w / 2)^2

   Divide both sides by 2h, and voila.

   r = (h^2 + (w / 2)^2) / 2h

2. 
   HeidiDad | Sep 16, 2002 08:18pm | #12

   It doesn't sound like you are describing a circle, but rather an ellipse.  You might consider the thin tapered stick method and make a quarter pattern templet.  OR, if you were to get a basic drafting book, it will demonstrate exactly how to lay out an ellipse of any size. In any case, you are going to use a template, not a radius  Hope this helps.

3. 
   TKanzler | Sep 16, 2002 08:55pm | #13

   Since I spend a fair amount of time on AutoCAD anyway, and it's running most of the time I'm on the computer, I just draw things like this and ask the machine what I have.  For your problem, I drew a line 84 units long, offset it 3, drew an arc starting one end of original line, second point at midpoint of the offset, and ended at the far end of the original line, used "list" to ask what I had and it tells me the radius is 295.5 units.  Assuming each unit is one inch, that converts to 24.625 ft or 24 ft 7 1/2 inches.  The whole process took less than 30 seconds, but I didn't have to think about how to do it since I do this sort of thing all the time.   

   Be seeing you...

1. 
   DennisS | Sep 16, 2002 10:34pm | #14

   Totally off topic, perhaps, but what kind of work do you do? I've got my face plastered to the AutoCAD screen all day myself. And yes, my first inclination is to just re-create the geometry from the givens and the rest is a piece o' cake. Dennis in Bellevue WA

   [email protected]

1. 
   TKanzler | Sep 18, 2002 05:36pm | #19

   Totally off topic, perhaps, but what kind of work do you do?

   Since this thread is about dead anyway, I'll tell you.  I design and build custom crane and derrick machinery, and sometimes structures, primarily for high steel construction (skyscrapers, for instance), and occasionally the odd specialty machine.  Like an airport pavement test machine, which went into service for the FAA a couple of years ago.  Since I design mechanical components, weldments, hydraulics, electrical systems, and controls, I prefer to use AutoCAD straight-up, as a precision drafting tool, rather than getting involved with add-ons or specialty programs.  And since everything is pretty much one-off, there's little need to interface with fabricators or machine shops (except via large format prints and specs), so plain vanilla is best for me.  And that's why recreating the geometry and listing or measuring the feature of interest is my first inclination.  None of which, I might add, has anything to do with Fine Woodworking.  I've been slowly getting back into woodworking, which was my passion since I got a Mattel PowerShop as a kid, but dropped due to unconducive personal situations, and I mostly lurk around here sponging off some of the amazing depth and breadth of knowledge and experience.  I contribute only where I think I can add something of value, which isn't often.

    

   Be seeing you...

   Edited 9/18/2002 11:08:14 AM ET by TDKPE

2. 
   DennisS | Sep 18, 2002 06:39pm | #20

   I'm quite new to this forum myself - I'm pleasantly surprised at how many people are using CAD these days .... AutoCAD, TurboCAD,... you-name-it CAD. Which is good in a sense. But like figuring out the rise per foot of run of an arc, - I mention this to some of the younger folk with whom I work and their eyes glaze over in disbelief that it can be done with paper and pencil.

   None the less, given a problem of geometry, I reach for my mouse before the sketch pad myself.

   Dennis in Bellevue WA

   [email protected]

2. 
   riverr1 | Sep 16, 2002 11:41pm | #16

   I think I would have just put in the line, put in another line 3 units long @ 180 deg. at the mid point and then done a 3 point circle and gotten the radius. Those darn things sure are versitile.

   Don

1. 
   RW | Sep 17, 2002 03:37am | #17

   Wow. So that's what it's like to get hit with a firehose.

   I think I got it, and yeah, spreadsheet is good.

2. 
   observer | Sep 15, 2002 07:04pm | #2

   I find Audels' Carpenters and Builders Guide, Vol. 2 (copyright 1923) very useful for layout problems of that nature. Virtually every used bookstore has a copy.

3. 
   DonaldCBrown | Sep 16, 2002 03:24am | #7

   Maybe the attachment will help

   
4. 
   Joe_Fusco | Sep 16, 2002 04:33am | #8

   Here's some "stuff":

   

   View Image

1. 
   clampman2 | Sep 16, 2002 07:07am | #9

   Here's another alternative for those of you who use calculators, and the one which I can figure out without having to remember anything but my calculator.

   Using Joe's diagram side A A' is known (3") and side A' P is known (84"/2= 42". Therefore, (the hypotanuse) A P= 42.107. And A M = 1/2 that or 21.0535.

   The angle whose tangent = 3/42 is 4.085616 degrees. therefore the other angle = 90 degrees-4.085616 degrees or 85.91438, therefore smallest angle in the right triangle AMC is 4.085616.

   the Sin of an angle = opposite / hypotanuse. so the sin of 4.0856=

   21.0535 / radius. Or 0.071247036 = 21.0535 / radius = 21.0535/0.071247036 = 295.4986683 inches or /12 equals 24.6248feet or 24' 71/2 inches.

   The reason I like sin cos tan is that I know the tan is rise over run or opposite over adjacent. I can pick 2 degrees, say, stick it in my calculator and get a sin, cos and tan of 2 degrees. I know the smallest value will be opposite over hypotenuse, the next will be opposite over adjacent. This way I don't have to remember all the #### formulas.

   Clampman

5. 
   DennisS | Sep 16, 2002 08:49am | #10

   You can determine the radius from the formula:

   r=(4b^2 + c^2)/8b

   where

   b=rise at midpoint of the arc

   c=chord length.

   You can approximate the curve by calculating the rise per unit length of run along the chord using:

   y = b - r + sqrt(r^2 - x^2)

   where

   y=rise

   b=total arc rise at midpoint of arc

   r=radius

   x= distance from arc endpoint measured along chord

   Plot the points out on a large sheet of paper then use a thin rip of some straight even grained wood or any flexible material to loft between the points. We use spray glue to glue the paper to something like masonite first, plot the points, rough out with band or saber saw then finish by hand to the line for templates, etc.

   Hope this makes sense and the server doesn't hose the formatting.

   Dennis in Bellevue WA

   [email protected]

1. 
   BSover21 | Sep 16, 2002 04:04pm | #11

   If you have ExCel try the attached.

   
1. 
   floman47 | Sep 16, 2002 10:39pm | #15

   Hi Bob

   Thank you for that Excel program I can yse it alot.

   Jeff in so cal

6. 
   trialnut | Sep 17, 2002 07:44pm | #18

   RW

   I see you have been given much good advice already. I find the easy way for me is this formula: 1/2 the chord squared plus the rise squared divided by 2 times the rise = radius.  No fancy sins, tangents, square roots etc. just plain multiplication and division. Can be done even with a graphite filled calculator.

   Have fun with all your new "private" math

   Rich

   The Professional Termite, aka Woodbutcher Extrodinaire