Hello All,
I have the need to cut a very large radius (like 80 ft.) with a router, And I'm in a shop that doesn't have the room to use the obvious 80 ft. piece of rope, so... is there some mechanical contraption that would allow me to use some levers to build a jig to accomplish my task. I know some of you math guy's (or gal's) have been itchin' to share this with somebody. Seriously, if someone knows of a "math" site or formula to do this, I would be very grateful.
TIA, Muggs
You could layout some plot points, say 6" apart, and then connect the dots with a jig saw to create a jig for the router. You could use this calculator at
Roland
Assuming the piece you're cutting *does* fit in the shop, why not make a template to guide the router on a similar sized piece of plywood by laying out the curve with the 80 feet of rope while *outside* the shop. Cut the curve on a bandsaw and clamp the template to the workpiece.
Art W.
You could also plot the curve after creating it in a CAD package. I don't know how long the cut is, and therefore don't know if 36" wide plotter paper is enough, but an 80' curve is going to be fairly flat over a short distance.
WT
I know that some cabinet shops have CNC routers they can be 6' x 12' tables I am sure you could pay a shop to cut that for you. A friend of the family in Detroit, MI has one in the shop he works at. They do Corian Counter tops.
Corey Scheich
"Muggs" wrote in news: snipped-for-privacy@enews3.newsguy.com:
Find a shop with a CNC router table, it's a pretty inexpensive service for what your after.
Zander
There are several ways to do this, but it depends on the following: What is the size and shape of the area to be cut? Is the radius convex or concave? What is the material and tolerance?
Sorry about rushed description, I'm supposed to be elsewhere
If you weld or glue together two straightedges, end to end, almost but not quite collinear, lay them on a sheet of ply, butt them against two pegs protruding through the ply, and hold a pencil in the intersection of the straight lines, it will describe a VERY LARGE radius circular arc as you slide the edges along in contact with the pegs. Make sure the router rotation and feed direction provide a force against rather than away from the pegs and you should be sweet.
There is a simple relationship between distance between pegs and angle between lines which I can't remember right now, but a simple parametric study model in SldWks will establish that. (I think you draw a normal to each straightedge through the points where they contact the pegs and where the normals intersect is the center of the big circle)
Good Luck!
Sorry about rushed description, I'm supposed to be elsewhere
If you weld or glue together two straightedges, end to end, almost but not quite collinear, lay them on a sheet of ply, butt them against two pegs protruding through the ply, and hold a pencil in the intersection of the straight lines, it will draw a VERY LARGE radius circular arc, as you slide the edges along in contact with the pegs. Make sure the router rotation and feed direction provide a force against rather than away from the pegs and you should be sweet.
There is a simple relationship between distance between pegs and angle between lines which I can't remember right now, but a simple parametric study model in SldWks will establish that. (In your study model, position the "bevel gauge" thingy midway between the pegs, and draw a 3 point arc through the contact points and the intersection of the straight lines. The radius of that arc will be the radius of the curve you will rout, and you can use this - by sliding the angled lines from side to side - to prove to your satisfaction that their intersection follows the arc exactly)
Good Luck!
Thank You, Andrew, and to all who commented.
All the other solutions were very good but already thought of. I should have made it clearer that what I'm was looking for was a formula for determining a large radius, so that I can make something variable depending on the radius, which will always be different.
I'll try to find the relationship between the angle and the two pins, but if someone else has another suggestion, or know the formula for the two pins method, please enlighten me.
Again, sorry for the off topic, but this group is a terrific resource for very intelligent peeps.
Thank You All, Muggs
If you have SW which I assume you do then make a 100% scale print of the required radius. Set the paper size as appropriate and send it to a service bureau like CopyMax or Kinkos that have roll fed plotters and can work from pdf files. They can print the radius on a long sheet and you can then glue it to a backing to get your radius.
The other source for a "math" soluti> Hello All,
If only you knew the subtended chord length and the arc length ...... .
A fairly stiff strip of anything the length of the arc bent to fit the chord length *might* come close enough (but would probably actually be a parabola).
Hmmm. If the beam (the material you bent to form the guide) was in a buckling mode pinned at both ends, wouldn't you have a circular arc? I'm thinking that stress throughout the beam would be uniform, so curvature would have to be uniform. (assuming ideal pinned joints of course).
just takes a little bit of trigonometry.
first, assign some names: R is the radius you want A is the amount of arc you want (in degrees) D is the distance between the pins (which you're trying to find) L is the length of the bar that you're going to bend (from center to center on the pins)
You need to decide how much of the arc you want to use (e.g.--10 degrees of arc). This sets a minimum length for your bar--
Lminimum= R* A * pi / 180 anything longer than that is OK
given a known length for L, the distance between the pins is:
D= 2 * L * sin (A / 2) / (A * pi / 180)
note1: the pi / 180 just converts degrees to radians note2: for this to work out, you need to use the same units for R, D & L--you can't use feet for one and inches for the other
oops--correction.... ignore the previous
first, assign some names: R is the radius you want A is the amount of arc you want (in degrees) D is the distance between the pins (which you're trying to find) L is the length of the bar that you're going to bend (from center to center on the pins)
You need to decide how much of the arc you want to use (e.g.--10 degrees of arc). This sets a minimum length for your bar--
Lminimum= R* A * pi / 180 anything longer than that is OK
If you use the minimum length for L, the distance between the pins is:
D= 2 * Lminimum * sin (A / 2) / (A * pi / 180)
you can use any length of bar (L) that you want, as long as it's more than the mimimum. If you use anything other than the minimum, however, you need to calculate an additional variable--let's call it B (the actual amount of arc that you'll get with your bar that's longer than the minimum)
B= ( Lactual / R ) * (180 / pi )
then,
D= 2 * Lactual * sin( B/2) / (B * pi / 180 )
note1: the pi / 180 just converts degrees to radians note2: for this to work out, you need to use the same units for R, D & L--you can't use feet for one and inches for the other
I doubt you wish it to buckle.
A Catenary Curve should be the result. Lofting of ship and aircraft surfaces in the past using splines ...
In this case, a parabola I think ... is the tolerance close enough for what's needed in this case? We don't know the chord or tolerance AFAIK.
Dale,
You're thinking of a four-point bending load. If you apply opposing loads with equal offsets on either end of a beam, then the moment is constant, as will be the curvature of a uniform beam, throughout the section between the two inner loads.
Jerry Steiger Tripod Data Systems
SNIP....
Muggs
I've now had a moment to do a bit of trig, and come up with the following:
NB: Read this in conjunction with the SECOND of my two messages above. I think the first was a red herring, it came off the top of my head.
IF the diameter of the pins is small in relation to the radius of the circle you wish to draw, the following approximation will hold:
Included angle between straightedges will be (180- asin(pin cr.dist/Diameter of circle to be drawn)) where asin is the inverse sin, or "angle whose sin is....". The "included angle" will be the lesser of the two possible angles (given that you can measure in either of two directions between two almost collinear lines).
To turn it from an approximation into an exact solution, you need to input the distance between the contact points between straightedges and pins, rather than simply the center distance of the pins. This contact distance can easily be checked from a sketch in SolidWks, using 2 circles and 2 lines and tangency relations, showing the straightedges in mid-position at the calculated included angle. Sketch a point coincident with each circle and coincident with the associated line, and put a driven dimension between the points.
Because this contact distance varies when you correct the included angle between straightedges, this correction requires iteration (going several times around the mulberry bush), but for an 80ft radius and small pins, the initial approximation would be very close indeed, and the correction minuscule.
Your included angle will be close to 180 degrees: the easiest way would probably be to NC machine the thing out of one piece, including a register for the router.
If the arc is very long, you may need to leapfrog the pins around a PCD so the "wingspan" doesn't have to be too excessive.
HTH
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