Ok, I CAN do this full scale, but lets have an exercise in math for fun.
I need to make a trellis, it requires a 4×4 half circle , out to out of 68″ (two actually, held by cross 2×4″ and lattice, on preset posts)
I’m gonna miter layers of 1.2″ cedar and with three of them achieve the desired thickness..I will segment the arch into say 6 at 60 degree joints..or 30 on the saw..how long do the rough segements need to be assuming I am using a 1.2 x 5.5″ stock?
I believe there is function on the CMII that can deduce the segments..or there is a rule of thumb formula that will get right close.
In the past I have just swung a trammel on a scrap and plotted the angle..but here I am also wiishing to maybe stagger the joints in the build up thickness..or more precisely, skip the middle lamination all together, and use spacers at the miters as both ventilation and reinforcement.
So my numbers are 34” RAD. and full circle would be 12 pcs. and I an using a 5.5″ wide stock..how LONG must the segments be?
I bet some of you can wing it off the top of your head, I have an idea..and wanna see how you would come up with an answer.
Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
You gonna play that thing?
http://www.youtube.com/watch?v=32Ln-SpJsy0
Replies
The length of chord C of a circular segment of angle A with radius R is
c = 2 x R x sin (A / 2).
For 6 segments in the entire circle (3 in the half circle) the length of the segment would be 34"; this is long point to long point of each section.
If you have 12 segments for the full circle, the length of each segment would be 17.60 inches.
BTW thats not off the top of my head - googled 'length of a chord' to find the formula.
In case I misunderstood part of the problem - if the 34" radius refers to the inside of the arc created by the 5.5" pieces, then its a bit trickier. 34" is correct for Short to short. Probably an inelegant way to determine long to long is to take the tangent of the saw angle (e.g. 30) multiply by the width of the stock 5.5" (result is 3.18") and add 2x this to the short point length to get a result of 40.35".
Wow..Ok, lets try this
View Image
I need 12-0-clock to 1-0 clock long point to long point (including waste) to describe a circle of 68" with a resulting 3.5" INNER arc..from a 1x6 (nominal).
Good googleing on yer part, I hadn't thought of that and flunked Algebra in HS, and became a Naval Pilot..LOL There ya go.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
You gonna play that thing?
http://www.youtube.com/watch?v=32Ln-SpJsy0
Not sure I understand exactly.
The drawing shows 16 segments making up the circle - is this what you want?
Also - how to interpret the parameter '3.5" inner arc'
Ok, I need a half circle 3.5" thick that bends to 68" diameter. Or a rad. of 34",I want to cuta few segments and join them to make the arch.
Ignore the supposed angles in the paint up I did, that was a visual clue only.
How many and what length pcs. are needed to cut that half circle?
BTW..your first guess or google was ....................Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
You gonna play that thing?
http://www.youtube.com/watch?v=32Ln-SpJsy0
LOL...I think it has been answered. As I told my students, draw pictures!
Geez Duane, you woke me up for this? You need somebody who teaches or something. Calling Carole...
Below is as far as I go. A buck and a quarter didn't cover your problem. And I never liked geometry, thank you Mr. Grove.PAHS works. Bury it.
Here's how I would do it, using a full-sized pattern:
Set trammel points to 34" and draw the semi-circle, off to one side of a 68"+ line. I'd probably use some 36" wide builder's paper for this.
Keeping the trammels set, pace off around the perimeter. This will give you a hexagon and half the points.
Reduce the length of the trammel anywhere between 34" and 18".
Put one of the trammel points on each of two adjacent points and draw intersecting arcs. Draw a straight line between the two points of intersection. This will go though the midpoint between the points.
Set your trammel to the distance between one of the original points and your newly found mid-point (on the perimeter of the semi-circle). Using this distance, mark off the intermediate points.
You now have the center of the original circle, half the circle, and half a regular Dodecahedron (12-sided figure).
From this, you should be able to lay out your arches, the rays into the corners of a dodecahedron, and the length of the segments.
I think you need to go by inside radius. 6 pieces 15 degree mitre= 1/2 circle
25" inside radius? 25x.5176=12.94" sp to sp
I am pretty sure a math guy will correct me.