Replies

1. 
   AlanWS | Sep 23, 2004 07:37pm | #1

   A straightedge is awfully close to a good template. For an 8' length, with a radius of 150' (150' = 1800") you will deviate from a straight line by 0.05332", less than 1/16". If you take a jointed narrow board, clamp it in the middle and bend it to the right position at the ends and maybe a few other spots you measure, and use that as a template to rout along, you'll get a smooth curve. What kind of tolerances do you need for this? Hand sanding to a line would be pretty fast and easy with a long fairing board.

1. 
   JohnWW | Sep 23, 2004 08:10pm | #4

   Alan,

   I think your math is wrong, though it's possible mine is, but I calculate that 150 ft radius arc will rise almost three inches over 8 feet.

   John W.

   Post script, 9/28/04:  Turns out my math was wrong, the actual rise at the center of an 8 foot long section of the arc is only .64224 inches.  Calculated this with trigonometry and checked it a couple of times, think I've got it right now.

   Edited 9/28/2004 8:43 pm ET by JohnW

1. 
   AlanWS | Sep 23, 2004 09:49pm | #5

   I did make a mistake, since I didn't convert back to inches the 0.0533'. This is 0.640" or about 5/8". I meant the center of an 8' span is off by this much. I suspect that the reason this doesn't match your number is that I meant the center of the span, so you could consider this deviation to be over 4' rather than 8'. The deviation of the center of a 16' span would be about 2.6", matching what you said.

   Thanks, by the way for this, and other answers to my questions.

   math, for those who care:

   Considering an 8' chord of a 150' diameter circle, imagine a right triangle consisting of the center of the circle, the center of the chord, and one end of the chord. The deviation of the center of the chord from the circle is the difference between the radius and the long leg of the triangle: 150 - sqrt(150^2 - 4^2) = 0.0533 feet or 0.640 inches. If I were plotting points to draw an arc, I would use trigonometry rather than the pythagoran theorem.

2. 
   Pondfish | Sep 23, 2004 07:43pm | #2

   I did something like this recently.  It involves marking out the curve and shaping it with a combination of tools.  You need a sharp pencil, a ruler, and the equation for a circle.

   To mark out the curve, I used excel to set up a calculation of the offsets of a circular arc from a tangent, given the distance along the tangent. If that makes no sense to you, imagine your circle nudged up along a long straightedge, with "0" on the straightedge placed next to the curve. The offset at "0" is, obviously, zero, since the circle touches the straighedge.  But go down to where it says 16 inches on your straightedge. The distance from the straighedge to the circle, measured 90 degrees from the straightedge, is the offset.  This distance depends on the the circle's radius.

   I set this up where I had measurements every 1cm along the line (I tend to do things like this in metric units).  Close to the zero point I could mark out the offset every 5cm or so (since it would be only 1mm).  Further out I would mark it every 2-3cm on a piece of 1/2" MDF.

   Once done, I used a ruler to connect the dots.  Short arcs of large circles are very close to linear, at least within woodworking tolerances.  Once done, I cut the curve on a bandsaw, and smoothed it (to the line) using a block plane and spoke shave.  A couple coats of 2lb shellac to seal it, and it was ready for use.

   The template I made can be seen in use at http://home.comcast.net/~paulchapko/sofaTable.html (the 4th picture down).

   If you want to try this but don't know how to set up the excel spreadsheet, I could send you mine.  You only need to enter in the circle's radius.

   Hope this helps more than confuses!

   Paul

1. 
   wblohm | Oct 25, 2018 07:53pm | #7

   Spreadsheet

   Paul,

   Would like to see your Excel spreadsheet.  Please send to [email protected].  Thank you.

   Walter

3. 
   JohnWW | Sep 23, 2004 08:03pm | #3

   There is a way to do this without needing a football field to lay it out, but it's a bit complicated to describe without a fair amount of text and a few drawings, more than is practical on a bulletin board.  The actual jig for guiding the router is surprisingly simple and the arc will be a true radius not an approximation. 

   I could make you a template for a specific radius in about an hour, less time than it would take for me to write up the technique.   

   I could make you a template and ship it for not much money or I'd be glad to teach you how to do it in my shop, which is in Connecticut.

   Just curious, why do you need a curve with a 150 foot radius and what kind of accuracy is needed?

   John W.

   
   

   ---

   Edited 9/23/2004 1:12 pm ET by JohnW

4. 
   utternovice | Sep 24, 2004 12:04am | #6

   You can lay out the curve quickly as follows -

    

   A second degree curve, such as a circular arc, requires 5 pieces of information about the curve.  In most cases 5 points along the curve are easiest, let’s call the points A, B, C, D, E.  Assuming you know the radius (R) and the length of the chord (L) of the arc it is relatively easy to determine the x (horizontal),y(vertical) positions of the A, B, C, D, E. The coordinates of A and E are the endpoints of the chord so no calculations are required (y=0). 

    

   The equation for a circle is: (x-h)^2+(y-k)^2=R^2.  For our case h = 0 (assumes the circle center lies on the y-axis and the y-axis extends through the midpoint of the chord).  The center of the circle is located a distance d below the chord.  Point A, the midpoint of the chord and the center of the circle for a right triangle with a hypotenuse R, and the short leg having a dimension of L/2.   The distance d is therefore given by:

    

   d = sqrt(R^2 - (L/2)^2).

    

   The y coordinate for Point C is located at the midpoint of the chord (x=L/2) and it distance (d) from the chord is given by

    

   y = R - d

    

   Points B and D are located at x = L/4 and 3L/4 respectively.  The y coordinate for points B and D is given by

    

   y = sqrt(R^2-(L/4)^2)-d

    

   To draw the arc, lay out these 5 points and use a flexible spline to connect them.

   Using the numbers you provided I get:

    

   L	96
   R	1800
   d	1799.359886
   	X	Y
   A	-48
   B	-24	0.480106707
   C		0.640113818
   D	24	0.480106707
   E	48

    

   Please check the math before cutting anything.

   Hope this helps.

    

   Offsets for making jigs can be computed similarly.