For you math wizards

I need to draw an arc for a piece of trim. The end points of the arc are 5 feet apart and the depth of the arc at the center point is 2 inches. What is the radius of the circle?

There ought to be an equation for this that would be far superior to trial and error, oui?

Larry

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

error, oui?

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

error, oui?

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

error, oui?

R=226" according to Sketchup.

According to those cool calculators, 18'10".

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

error, oui?

Bend a piece of thin cutoff around the points?

-------------------------------------------------------------- Find a copy of Fred Bingham's book, "Practical Yacht Joinery" at the library.

A very easy graphical solution is shown.

I laid out all the deck cambers for my boat using it.

Lew

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

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

and error, oui?

!8' string with a pencil died around one end. Nail in the ground on the other end.

Or use sketchup to print out a template.

Trigonometry concurs. :)

Thanks to the O.P. for a pleasant lunchtime puzzle.

My sense of the physics involved tells me that that method will not produce an arc of a circle. Will it matter? Might the resultant curve be subtly nicer than a circular arc? That's up to the designer.

It was more fun to try to work it out for myself, but I just looked it up and there is (of course) a more direct way to find it.

Radius = H/2 + W^2/8H

Where H = the height of the arc (2") and W= the width of the base (60")

In our example that's:

"Gramp's shop" wrote in news:384921f5-292d-40cc-9e3e- snipped-for-privacy@googlegroups.com:

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

error, oui?

There is.

radius squared = (radius minus height) squared + (half the distance between endpoints) squared

In this case:

r^2 = (r - 2)^2 + 30^2 r^2 = r^2 -4r +4 + 900

news:384921f5-292d-40cc-9e3e-

feet apart and the

and error, oui?

Now we know how long a piece of string is!

My plan exactly, Larry. I'm going to start with drilling screws into the waste side of the end points and the top of the arc, attach a thin strip of one-by with spring clamps and then add a couple of screws/clamps along the arc.

The other Larry

news:384921f5-292d-40cc-9e3e-

feet apart and the

and error, oui?

I just drew that out. Clever. It's much more elegant solution than the one I posted. Good work.

Greg, We both drew the same picture. How much math background do you have (if you don't mind me asking)?

Bill

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

error, oui?

Pythagorean Theorem: 30^2 + (r-2)^2 = r^2. Solution is 226" exactly, which is 18' 10", as hasalready been disclosed,I believe. Didn't even need trig. (which surprised me).

Nothing too advanced. Algebra, geometry and trig in high school, a little calculus in college. That would all have been in the '70s.