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Kd0afk
09-25-2012, 12:31 PM
Has anyone used this method to make a radius cutter for the lathe?
http://gicl.cs.drexel.edu/people/sevy/luthierie/compass/Long_compass.html

http://www.youtube.com/watch?v=tvGemvizrz4

winchman
09-25-2012, 03:30 PM
I wonder what he does with a piece of MDF with a 15' radius on one side and a 25' radius on the other. I see some sort of musical instrument being assembled in the background, but it doesn't seem to fit there. It might be useful for astronomy, though.

I'm not sure how you'd go about adapting that method to a metal working lathe. It'll be interesting to see what people come up with.

dp
09-25-2012, 03:58 PM
I wonder what he does with a piece of MDF with a 15' radius on one side and a 25' radius on the other. I see some sort of musical instrument being assembled in the background, but it doesn't seem to fit there. It might be useful for astronomy, though.

I'm not sure how you'd go about adapting that method to a metal working lathe. It'll be interesting to see what people come up with.

He's a hobbyist luthier and I think this is what he uses as a form for making curved guitar tops and backs.

KIMFAB
09-25-2012, 04:52 PM
He's a hobbyist luthier and I think this is what he uses as a form for making curved guitar tops and backs.

In his narrative he mentions that

Lew Hartswick
09-25-2012, 05:31 PM
Might be a way to make a solar concentrator, line the "dish" with the shiny side
aluminum foil.
...Lew...

CCWKen
09-25-2012, 05:35 PM
I guess I'm missing something that may be obvious to everyone else but what is he talking about? I keep hearing 15 foot, 25 foot and 60 foot radii. I see how his jig works but I don't see where he's getting a radius in feet. Did I miss a class in new math or science? Quantim physics perhaps?

michigan doug
09-25-2012, 06:10 PM
One method is to take a piece of wire, at whatever length/radius you want, and stretch it tight and use that to scribe your radius on your jig boards. It depends on what precision you need. Primitive, but effective.

HTH,

doug

cuslog
09-25-2012, 06:29 PM
I've never used that method on the lathe but in response to some others questions about how its used; (I built my first guitar last Winter)
Its may be used as a form, holding pieces (guitar top or back and their respective braces) in an arch while the glue sets.
I believe some are also lined with sandpaper so a builder can sand an accurate arch onto the top and bottom of the body to get a tight fitting joint when joining top / back and sides of a guitar.
The only way I'd know to lay it out is with a string or wire on the actual radiius or do the layout in CAD and do a full-size print if you've got a big enough printer !

dp
09-25-2012, 07:04 PM
I guess I'm missing something that may be obvious to everyone else but what is he talking about? I keep hearing 15 foot, 25 foot and 60 foot radii. I see how his jig works but I don't see where he's getting a radius in feet. Did I miss a class in new math or science? Quantim physics perhaps?

He is worrying a 15' radius concave face into the MDF by making incrementally deeper cuts as the router works toward the center. He's facing the MDF, in other words.

Tony Ennis
09-25-2012, 08:25 PM
He needs a drill with a rubber wheel on the end to spin that large disk.

darryl
09-25-2012, 09:16 PM
Interesting method. He's climb cutting, so with just the right feed the disc will turn by itself :)

Seems to be a compound action going on as the sled moves along. I don't know if there's any math that can describe the actual curve being generated, although I'm sure it's good enough for the application.

Jaakko Fagerlund
09-25-2012, 11:53 PM
From what I understood the sled has two angles on the bottom that meet up under the cutter, and the sled is supported from two points that let the sled change its position. Now, it is morning and my head is not working, so I'm not capable of thinking if the line drawn by this is a true radius or not, but I think it is.

dp
09-26-2012, 12:14 AM
Imagine a 30' cylinder sitting between the uprights. Now rotate the cylinder such that it remains in constant contact with both uprights. Any point on the exterior surface of the cylinder will move in a perfect circle about the center of the cylinder. Same as a wheel on an axle.

Now remove all the parts of the cylinder that you don't need and slide the remaining segment between the uprights as before. The same point still carves a circle segment, same diameter.

Edit: Removed an erroneous interpretation - the irregular under surface of the rails was necessary to provide clearance. That part does not come in contact with the uprights.

Paul Alciatore
09-26-2012, 02:30 AM
His method does generate a true circular arc. If you are interested, here is the math and the derivation of his skid.

http://img.photobucket.com/albums/v55/EPAIII/aLongCompassProof1.jpg

The drawing shows a circular arc with three points on it, L, M, and N. These three points form two lines that are joined at an angle at point M. Warning, math ahead.

Assuming the arc is circular and that it passes through these three points (three points do define a unique circular arc).
Angle A is the overall angle between points L and N. It is divided into two arbitrary angles B and C, both of which are smaller than A and B + C = A.

By drawing chords LM and MN, we form two isosceles triangles. Each of the angles B and C is bisected by a line that is perpendicular to one of the two chords.

These bisectors form right triangles with angles b and c such that b = B/2 and c = C/2.

These smaller right triangles provide the following about angles d and e:
d = 90-b
e = 90-c

And d + e = F. F is the angle between the two chords formed by LM and MN. We are now going to show that wherever point M is on the arc between L and N, the angle F is the same value. It is a constant. It is the same angle weather M is next to L or next to N or at any and EVERY other point in between them.

b = B/2
c = C/2

but since C = A-B

c = (A-B)/2

Now since we created right triangles with the two bisectors,

d = 90-b
and
e = 90-c

F = d+e
substituting the values of d and e from above,
F = (90-b)+(90-c)
F = 180-b-c

again substituting for b and c

F = 180-B/2-(A/2-B/2)
F = 180-B/2-A/2+B/2

and finally

F = 180-A/2

Thus, the angle F, between the two chords, is only dependent on the original, overall angle A between L and N. It does not change no matter how we divide this angle A up in to angles B and C.

What this means is once we pick two outer points L and N we can fix an angle F between two straight lines that go through these two outer points and the intersection of these two straight lines will always lie on the same circular arc. So two straight edges that are fixed at a constant angle can be supported by two points and when this angle is moved across the points of support, the point where they intersect will form a circular arc.

I have extracted the two chordal lines and the three points in the middle section of the drawing. The angle between them is still the same value, F.

The device in the video has two straight edges on the boards that support the router. One of them is apparently the original side of the board and the other is at a slight angle to the first. This forms his angle F. He has the router's cutter at the intersection of these two straight edges.

The bottom section of the drawing shows these two chordal lines and three points incorporated into his boards. I have used a smaller radius to make the ideas here more obvious so my drawing shows the two straight chordal lines at a greater angle than is apparent in the video. But the principle is the same. We did not assume any particular radius above so the proof is good for any radius. The arc I added between these two chordal lines is to illustrate the irregular cutout he used to provide clearance so that the boards do not hit the workpiece. The larger arc is the cut line and is exactly the same radius as shown in the top section of the drawing.

If you want to design a mechanism like this, you could proceed like this.

First determine the radius you want to cut.
Then make a good guess at the distance between the two outer, support points you will use. This would need to be as large as possible for better accuracy, but would be constrained by the dimensions of the machine you are using it on. So make it as large as possible. Draw a line between them.

http://img.photobucket.com/albums/v55/EPAIII/aLongCompassLayOut1.jpg

Since the middle point, where the cutting tool would be located, can be at any point on the arc between the two outer points, you can choose the point half way between them. This simplifies the drawing and the trig.

So on the line between the outer points that you drew above, draw a perpendicular bisector to it. This perpendicular line does not need to be too long if the radius you want to cut is large.

Now a bit of math. Take 1/2 of the distance between your outer two points (X) and divide it by the desired radius of the cut (R). This will be the tangent of the angle (H) between the mid point (M) and each of the outer points (L and N). Find this angle from trig tables or a calculator.

Now, from each of the outer points, draw a second line at that angle from the original line between them, towards the perpendicular line drawn above. These two lines should intersect at that perpendicular line. These two lines form the angle you need between two straight surfaces to generate the desired radius.

In any practical mechanism for cutting an arc with this method, you will need to allow some clearance around the tool's cutting tip and the two straight lines used to guide it.

darryl
09-26-2012, 03:17 AM
Thanks Paul. I actually understood much of what you laid out there. I think I've ended up with a little higher regard for the method. It's one to put in the 'how to do' book.

Barrington
09-26-2012, 09:49 AM
Paul, a shortcut:-

http://i564.photobucket.com/albums/ss82/MrBarrington/radiussled.png

Given the two large triangles are isosceles (so angles 'd' and 'e' are mirrored) and that the three angles of a triangle add up to 180 :-
B + 2d = 180
C + 2e = 180

Adding both sides of the above:-
(B+C)+ 2(d+e) = 360
i.e.
A + 2F = 360
F = 180 - A/2



To calculate the required dimensions:-

Where 'S' is the distance between supports, and 'D' is the circle diameter.

A vertical dropped from the centre of the circle divides both angle A and distance S in half, giving:-
sin(A/2) = (S/2)/(D/2) = S/D

Now, from earlier, A/2 = 180 - F, so:-
sin(180-F) = S/D

(- and remembering sin(180-X) = sin(X))


1: Given support distance and required diameter:-

'Sled angle', F = 180 - arcsin(S/D)


2: Given 'sled angle' and required diameter:-

Support distance, S = D * sin(F)


3: Given support distance and 'sled angle'

Resulting diameter, D = S / sin(F)


Cheers

.

Paul Alciatore
09-28-2012, 06:00 PM
Paul, a shortcut:-

.....
.

Always more than one way to prove a theorem in math. I once saw a whole book full of different proofs of the Pythagorean theorem (a^2 + b^2 = c^2) for a right triangle. Thanks for the help.