- posted on November 29, 2007, 5:06 pm

Chris

- posted on November 29, 2007, 5:50 pm

The only formulae I can find is to subtract the minor axis from the major axis and divide it by two. That give you the distance you need between the center sliding

This is because you need 2 centers to make an elipse. These will slide in the slots cut in the wood.

In your case you would subtract 4 from 8, leaving 4 or 48 inches divide tha by 2 is 24 inches. So your three points on your 48 inch trammel would be your center point, one at 24 inches and one at 48.

Back to the center board. it will need to be a bit over 24 inches square. (you don't want the guide points to fall out of it.) cut your guide tracks diagonally across the guide board so it's a big X.

Secure the guide board to the center of the board to be cut into an elipse. (using sticky tape or screwes if you are cutting from the back or bottom.) Insert the guide points then connect them to the trammel. Cut your elipse.

On Thu, 29 Nov 2007 17:06:38 +0000, Chris

- posted on November 29, 2007, 6:47 pm

On Thu, 29 Nov 2007 09:50:44 -0800, snipped-for-privacy@iinet.com wrote:

I'm sorry I did not make it clear. I have made a board 32 inches square with crossed dados ( not corner to corner ) in which bearings run. however the centre where the dados cross is only 16 inches from the edge of the square which would put the centre point of the trammel outside the square when the end point was in the cross track.

Clear as Mud ?

Chris

I'm sorry I did not make it clear. I have made a board 32 inches square with crossed dados ( not corner to corner ) in which bearings run. however the centre where the dados cross is only 16 inches from the edge of the square which would put the centre point of the trammel outside the square when the end point was in the cross track.

Clear as Mud ?

Chris

- posted on November 29, 2007, 10:36 pm

So by using the math I said before and reversing it. You can only
have a 32 inch difference between the long axis and the short axis.

I can't see a way to make the whole table with the undersized trammel. I can only suggest that you look at rebuilding it. Sadly I think the rebuilt trammel would be almost as big as the table top at about 50 inches square

R--------------------------------------------------P1----------------------P2

For the home built trammel to work the distance between P1 and P2 needs to be 24 inches. Where R is the router and P1 and P2 are the

On Thu, 29 Nov 2007 18:47:12 +0000, Chris

I can't see a way to make the whole table with the undersized trammel. I can only suggest that you look at rebuilding it. Sadly I think the rebuilt trammel would be almost as big as the table top at about 50 inches square

R--------------------------------------------------P1----------------------P2

For the home built trammel to work the distance between P1 and P2 needs to be 24 inches. Where R is the router and P1 and P2 are the

On Thu, 29 Nov 2007 18:47:12 +0000, Chris

- posted on December 1, 2007, 9:33 pm

Like the OP, I too prefer the trammel method. It's simple and direct.

I've built the elliptical mirror designed by David Marks on Woodworks. I didn't do the string method and cut to the line as he described. Instead, I did what the OP says he wants to do and built a trammel to use my router and a 1/2" spiral carbide end mill to cut the ellipse directly.

While I was figuring out how to do that, I used Autocad to draw my planned trammel. For any ellipse, there is a long axis & a short axis. Because an ellipse has two centers, the trammel you're building will move on two points.

R--------------------------------------------------P1----------------------P2

Formula:

The distance between those two points (P1 and P2) is (long axis - short axis)/2. This is the distance from P1 to P2.

Since you'll be cutting it directly from your material, you'll likely be useing a router with a bit of some diameter. To locate the center of the router R from P1, use the answer from the formula above and subtract it from 1/2 of the long axis of the ellipse. Add to this 1/2 of the cutter diameter. That gives you the distance from P1 to R.

Using those answers, I knew I would build an arm containing P1 & P2 and would attach that arm to a router base to get point R.

The base I built is an MDF square with two dados at 90 degrees to each other in the manner of a big X. Into those dadoes, I put aluminum T-track. The T-track crosses at the center of the X. To allow the two points to move past this, I mitered the T-track together at the center of the X. Each point, P1 & P2 are affixed to a T shaped slide. What I mean is two separate T shaped pieces. I chose to use a steel dowel pin in each of those sliders to locate my arm.

I know that if I were to size the trammel base to contain the point P2 so it remained inside the T-track, that when I would use the trammel to guide my router, that the cutter may strike said base at the narrow part of the ellipse. I got around this problem by sizing the base to allow point P2 to go outside the base. You can do this by making the sliding part long enough that you have several inches of it remaining inside the T-track. Remember, as you move the arm around, and the slides are sliding, all they have to do is miss at the center.

That's pretty much it. To get a better grasp of what I've attempted to describe with words, I suggest that you make full scale drawing of what I've tried to describe.

I have an Autocad drawing of all this from my own trammel design.

On Thu, 29 Nov 2007 14:36:30 -0800, snipped-for-privacy@iinet.com wrote:

I've built the elliptical mirror designed by David Marks on Woodworks. I didn't do the string method and cut to the line as he described. Instead, I did what the OP says he wants to do and built a trammel to use my router and a 1/2" spiral carbide end mill to cut the ellipse directly.

While I was figuring out how to do that, I used Autocad to draw my planned trammel. For any ellipse, there is a long axis & a short axis. Because an ellipse has two centers, the trammel you're building will move on two points.

R--------------------------------------------------P1----------------------P2

Formula:

The distance between those two points (P1 and P2) is (long axis - short axis)/2. This is the distance from P1 to P2.

Since you'll be cutting it directly from your material, you'll likely be useing a router with a bit of some diameter. To locate the center of the router R from P1, use the answer from the formula above and subtract it from 1/2 of the long axis of the ellipse. Add to this 1/2 of the cutter diameter. That gives you the distance from P1 to R.

Using those answers, I knew I would build an arm containing P1 & P2 and would attach that arm to a router base to get point R.

The base I built is an MDF square with two dados at 90 degrees to each other in the manner of a big X. Into those dadoes, I put aluminum T-track. The T-track crosses at the center of the X. To allow the two points to move past this, I mitered the T-track together at the center of the X. Each point, P1 & P2 are affixed to a T shaped slide. What I mean is two separate T shaped pieces. I chose to use a steel dowel pin in each of those sliders to locate my arm.

I know that if I were to size the trammel base to contain the point P2 so it remained inside the T-track, that when I would use the trammel to guide my router, that the cutter may strike said base at the narrow part of the ellipse. I got around this problem by sizing the base to allow point P2 to go outside the base. You can do this by making the sliding part long enough that you have several inches of it remaining inside the T-track. Remember, as you move the arm around, and the slides are sliding, all they have to do is miss at the center.

That's pretty much it. To get a better grasp of what I've attempted to describe with words, I suggest that you make full scale drawing of what I've tried to describe.

I have an Autocad drawing of all this from my own trammel design.

On Thu, 29 Nov 2007 14:36:30 -0800, snipped-for-privacy@iinet.com wrote:

- posted on November 30, 2007, 3:30 am

On Nov 29, 12:50 pm, snipped-for-privacy@iinet.com wrote:

I've used a framing square instead of a cross-slotted base board. Clamp it down, draw one quarter of the ellipse, flip it over, draw the next quarter, and so on. You can do the same with two lengths of 1 x 4 butt jointed together into a sufficiently large "L" if your framing square isn't large enough.

I've used a framing square instead of a cross-slotted base board. Clamp it down, draw one quarter of the ellipse, flip it over, draw the next quarter, and so on. You can do the same with two lengths of 1 x 4 butt jointed together into a sufficiently large "L" if your framing square isn't large enough.

- posted on November 29, 2007, 11:41 pm

: I've built a trammel (Less than 4ft square) to try to make an 8ft by
: 4ft oval table.

It might be easier to use the string and nails method rather than the tramel method. The hard part is finding string that will not stretch.

http://benchnotes.com/Laying%20out%20an%20oval/laving_out_an_oval.htm

--- Chip

It might be easier to use the string and nails method rather than the tramel method. The hard part is finding string that will not stretch.

http://benchnotes.com/Laying%20out%20an%20oval/laving_out_an_oval.htm

--- Chip

- posted on November 29, 2007, 11:57 pm

"Chip Buchholtz" wrote:

Forget the string, layout the nails with a tape measure and use a batten.

Trim proud, then clean up fith a fairing board.

Formula for an elipse: X^2/A^2 + Y^2/B^2 = 1

This will give you an elipse for 1/4 of the table.

Make a 1/4" hardboard template and use it to duplicate the other 3 sides.

Lew

Forget the string, layout the nails with a tape measure and use a batten.

Trim proud, then clean up fith a fairing board.

Formula for an elipse: X^2/A^2 + Y^2/B^2 = 1

This will give you an elipse for 1/4 of the table.

Make a 1/4" hardboard template and use it to duplicate the other 3 sides.

Lew

- posted on December 1, 2007, 9:43 pm

RE: Subject

As I remember, this is a project to construct a 48" x 96" eliptical table.

Using the formula given, divide the 96" dimension ("X" Axis) into 3" intervals, then calculate "Y" Axis values which locates the nail location for all four quadrants.

After that, it is batten and fairing board time for a full size, 1/4" hardboard template for use with a router and a pattern bit. ++++++++++++++++++++++++++++++++

Lew

As I remember, this is a project to construct a 48" x 96" eliptical table.

Using the formula given, divide the 96" dimension ("X" Axis) into 3" intervals, then calculate "Y" Axis values which locates the nail location for all four quadrants.

After that, it is batten and fairing board time for a full size, 1/4" hardboard template for use with a router and a pattern bit. ++++++++++++++++++++++++++++++++

Lew

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