I have a question for all of the math guys here.
Thanks in advance for your help.
A few months ago we were trimming a house with custom moldings. We
had several arched top windows that needed trim to match. These
windows had a top section that was arched and the arch was really just
a part of a true circle but less than a half circle. In other words
if you knew the radius you could make the trim piece to match.
Now, I've done a lot of these and I normally mark a plumb line in the
center of the window and use a stick long enough to move up and down
the line and change the length until my stick follows the window.
This gives me the radius and I'm good to go.
In this case, since the moldings were to match, the millwork sent out
the sales guy and he got out his tape and measured the length across
the arch (where the arch hit the vertical sides) then measured the
distance from that line up plumb in the center of the window which
would be the highest point of the arch. He wrote down the
measurements and left. My partner and I both had this look on our
faces that said "I'll believe it when I see it!". A few days later
(very much to our surprise) the trim pieces arrived and were correct.
My question is how the hell did he do that?
Does anyone know what formula might be used to find the radius when
you only have the two measurements mentioned above?
I have two arched windows to trim in the house we're currently
Window A arch length 57 1/2" rise 10 7/8"
Window B arch length 53 1/2" rise 9 1/2"
My preliminary stick method on window A is 43 3/8" radius
and surprisingly window B is about the same.
If anyone could post a formula for me I would appreciate it very much.
If not, I'll just keep poking a stick at it.:-)
r = (cē / 8h) + (h / 2)
Where c = the chord (your "Length" above) and h= height (your "rise" above)
My brain is too tired to attempt to prove it at the moment ... try it and
see if it comes close to your expectations. if it is wrong, rest assured it
will be well pointed out, ad infinitum. :(
Swingman (in firstname.lastname@example.org) said:
| r = (c2 / 8h) + (h / 2)
| Where c = the chord (your "Length" above) and h= height (your
| "rise" above)
| My brain is too tired to attempt to prove it at the moment ... try
| it and see if it comes close to your expectations. if it is wrong,
| rest assured it will be well pointed out, ad infinitum. :(
Check the link below. Works for both coves and arches.
DeSoto, Iowa USA
It's right. I used "c" for semi-chord; you use it for the entire
chord length. I use one fraction, you separate it into two.
Otherwise it's the same end result.
Radius = "R", semi-chord = "c". By difference the other side of a
right triangle from chord to center = "R - h".
By Pythagoras... R^2 = (R-h)^2 + c^2
Expand and simplify for "R" in terms of "c" and "h".
It worked great! My stick was only off by 1/16!
I also went to DJ's site and plugged in the numbers and everything
Thank you very much and thanks to everyone else who answered!
BTW, now I just feel like a dumb ass! ;-)
I have to agree with Leon. He probably used a cad program, though it
certainly doesn't have to be done that way. Most anyplace that build
anything these days uses cad. Work like this generally go through the
planner, or programmer if done on a CNC, and that would be the cad man. If
he had it, no reason he wouldn't use it.
I used to be able to do all the geometry and trig stuff in my head.
Since using CAD I can't do any of it anymore but I can solve even more
complex problems without even thinking about it. I recall when I was
just a pup and I solved some nasty hip and valley bracing intersections
in about an hour using CAD. It took the seasoned checker 2 days, he
built a cardboard model of some parts and still couldn't get every
dimension. However, he signed off on the few he couldn't prove because
he couldn't find even one error in the ones he could prove.
This was for fabricated steel that was fab'd in California and shipped
to Hawaii for installation on the roof of a high rise. It all fit.
As usual, anything works well in the right hands. My older brother
[rest his soul] became wealthy as a consultant correcting structural
steel drawings done by young engineers using CAD. I don't know if he
could turn on a computer. His background was the old trade
school/steel works apprencticeship /drafting office; a good solid on
the ground foundation then the theory. Also, he wasn't afraid of the
fudge factor, instinctively knowing the limitations from his
When he got really stuck, and needed more advanced math or
verification, he used me. Remember, I said *he* was the one who got
rich. His last job involved deciding whether a bent piece at the NY
site was supposed to be that shape. He said I saved him hours and
hours of work when I showed it was from his dimensions. I taught
math, and can still teach *all* high school math without a computer.
They're useful, and I sue one all the time, but base knowledge, not
the tool, is the key. That's why I'm here. There are some people
with some excellent down-to-earth useful advice willing to share, and
I like to listen in.
Sorry, after posting I regretted what was probably to much of a
W/ so many automated millwork shops these days and the plethora of CAD
designers, it is quite probable a custom mill shop <did> do it that
Yes, it can be done that way:
1. Draw the chord.
2. Draw the height from chord center.
3. Join the height top point to the chord ends.
4. Draw the perpendicular bisectors of those two smaller chords.
Where they meet will be the circle center. Dimension the radius from
there to a chord end point.
With AutoCAD the radius dimension tool will show the radius.
Simply draw a line the width of the window. From the midpoint draw a
perpendicular line the height of the arc. Then draw an arc through all 3
points and use the radius dimension tool to show the radius of the arc.
I believe the formula is as follows:
Let a = half the arch length
Let b = rise
then radius = (a*a + b*b)/(2*b)
where * means multiply.
In your two examples above I get 43.44" and 42.21" respectively.
It took Pythgoras and a bit of manipulation.
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