For you math wizards

Trial and error isn't needed, neither is knowing the radius of the arc's circle.

Put two nails 5' apart in a piece of ply. Put another nail 2" above the line formed by the first two. Take a batten, bend it between the nails and draw the arc.

The potential flaws I see in that method are... ...you might get a peak/angle in the curve at the center nail ...you can't always count on getting equal bending at every point along the length of a piece of wood.

I know it's a standard method to use a long, flexible piece and something to mark out a curve so I'm certain it works. I would just want to double check and practice a few times to make sure the bendable thing was bending equally.

Perhaps the curve could be checked by flipping the layout stick end for end?

Did he say wanted the full circle or the 5' arc?

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"dadiOH" wrote:

--------------------------------------------------------------- Nice try but no cigar.

The end result needed is a cambered beam shape which requires more than the three points you suggest.

Bingham outlines the method that works in his book.

Have used the method to define the deck cambers which varied from

BTW, a batten is needed, I used a 3/4" x 3/4" x 1/16" x 96" aluminum angle which provides a knife edge for fairing out the profile with a fairing board.

Lew

He said he wanted the 5' Chord!

Good advice.

feet apart and the depth of the arc at the center point is 2 inches. What is the radius of the circle?

error, oui?

No trig or anything real advanced needed here. Merely that the square of the hypotenuse of a right triangle = the sum of the squares of the other two sides

Picture your arc, and a line between the ends of the arc. Now draw another line from the center of your circle to the midpoint of the arc, and a third line from the circle center to one of the endpoints. Now if you have a circle of radius R, you have just drawn a right triangle with hypotenuse R, one side of R-2, and the other side of 30 (converting the 5-ft width to inches and dividing by 2.

900 + R^2 -4R + 4 = R^2

R=226"

Actually, he said he wanted the radius so he could draw the arc. Arc, not chord. If he wanted to wind up with a 5' chord all he would need is a 5' straight edge.

"dadiOH" wrote in news:kkj81j$5dj$ snipped-for-privacy@dont-email.me:

Bill's point is that the 5' dimension is the measurement of the chord, not the arc.

He really needs to make about 20 of them. To save time, he will array the stock in a regular twenty-sided polygon, presumably on the basketball court of the local high school. This will allow him to mark all of the pieces in one step with a string and pencil, or if the ceiling height is sufficient, a compass. :)

Actually LOL, I think Bill was filling me in with an accurate answer to my question that I posed to Richard. I was not really asking for an answer so to speak. I mentioned a template printed from Sketchup and Richard questioned an 18' radius template. I believe he was thinking about printing an 18' foot long template, maybe not. LOL

The desk I just completed I useed the printing template technique for an

A few strips of hardboard screwed together, and that 18 foot template should come together in seconds! The OP may wish to include a micro-adjuster at one end. : )

Bill

radius :-) I have a couple of ideas and will post pix of the process.

Two lengths of PVC pipe. Drill a hole in one end, attach a pencil to the other at 18'. Draw the arc on the ground, spiking the pivot end in the ground and putting the wood at the pencil end.

...

Note picture UL 2nd page... :)

========================================== "Just W> Two lengths of PVC pipe. Drill a hole in one end, attach a pencil

========================================= You've obviously never done this.

Lew

Ah, OK, got it. Mea culpa.

I couple of "trammel points" on a piece of steel pipe, along with two humans, may work. What is your solution?

Bill (who just happens to have 40 feet of Schedule 40 black pipe)