# Thread: formula for A Shape

1. We know, fix a nail put a loop around it, we can draw a circle. Two nails give an ellipse.
Consider a rectangle of size AB, fix 4 nails in a rectangle shape, put a loop around it then draw a curve. How can I find the optimum size of rectangle CD for the nails which is inside AB? There can be a number of rectangles CD. Give me one with a smooth curve, for a table top.

2.

3. Originally Posted by sak
We know, fix a nail put a loop around it, we can draw a circle. Two nails give an ellipse.
Consider a rectangle of size AB, fix 4 nails in a rectangle shape, put a loop around it then draw a curve. How can I find the optimum size of rectangle CD for the nails which is inside AB? There can be a number of rectangles CD. Give me one with a smooth curve, for a table top.
You need to define what you mean by optimal. If you want the maximum area within your given rectangle, put one nail on each corner, and stretch the string tight. Then you get the whole thing.

4. Let me see if I understand what you want to do. If you put 4 nails at ABCD then stretch a string loosely around the 4 nails, then follow that outline with a pencil, you will trace a shape. When your pencil is midway between A and B, you are tracing an ellipse with foci A, B. A little further on, the string lifts up off of nail B and you are tracing an ellipse with foci A,C. Later on the string touches B and then the foci are B, C etc. Is that correct?

5. I think I see what he means though. There should be one free variable in this system, assuming the bounding rectangle is fixed.

Let's say the piece of wood is a rectangle from (0,0) to (A,B). Inside that piece of wood, you place 4 nails, also arranged in an axially-aligned rectangle. Assuming you want the string to touch the center of the 4 sides of the outer rectangle, the placement of the nails and the length of the string should all depend on a single free variable (sort-of a roundness factor, correct me if I'm wrong though).

Obviously, if this is 0, the nails go in the corners, and the string fits snugly all around, but as this value increases, the nails move inward, but not in such a simple way.

Let's say the nail closest to the origin is at (C,D). When the string stretches to the bottom center, it'll form a pentagon with side lengths . Similarly, when it's on the left side, it'd have sides . Since the string is the same length in either case, these two cases must sum to the same length. Since A and B are fixed, this gives us one equation in two variables (or if you add that the length of the string is equal to either of these, two equations in three variables). Solving these equations should give you all the information you need, technically, but you may have to apply some substitutions to get a more meaningful free variable.

Well, for now I've got to go, but it's pretty likely someone else here can poke holes in my equations and help you figure the rest of this out.

6. Originally Posted by MagiMaster
I think I see what he means though. There should be one free variable in this system, assuming the bounding rectangle is fixed.

Let's say the piece of wood is a rectangle from (0,0) to (A,B). Inside that piece of wood, you place 4 nails, also arranged in an axially-aligned rectangle. Assuming you want the string to touch the center of the 4 sides of the outer rectangle, the placement of the nails and the length of the string should all depend on a single free variable (sort-of a roundness factor, correct me if I'm wrong though).
In that case, the co-ordinates of the corners of the inner rectangle should be , , , (the first pair being the co-ordinates of the nail closest to the origin).

The length of the string is

7. I'm a little confused. Did you mean , etc. or something more like , etc. or am I missing something?

8. Originally Posted by sak
We know, fix a nail put a loop around it, we can draw a circle. Two nails give an ellipse.
Consider a rectangle of size AB, fix 4 nails in a rectangle shape, put a loop around it then draw a curve. How can I find the optimum size of rectangle CD for the nails which is inside AB? There can be a number of rectangles CD. Give me one with a smooth curve, for a table top.
I am still not sure what you mean by optimal. But if you simply want some sort of a nice smooth curve that you can cut from a rectangular piece of wood for a table table top, why not an ellipse with major and minor axes conforming to the long and short lengths of the piece of wood ? That would simply be incribing an ellipse inside of a rectangle. You can do that with one calculation and two nails. To do that all that you need to do is follow the recipe in this Wiki article: find the center of your board and lay out a straight line throuh the center parallel to the long side of the board. Then calculate c following the article. Put two nails along the line that you have drawn a distanc c from the center. Then size a piece of string to loop around the nails that reaches the edge of the board (or a little less to give yourself a wee bit of wiggle room) along the line. Then draw the ellipse. It will automatically also have a minor axis that is the same as the short edge of your board. Here is the Wiki article with the simple formula for c (you will need a calculator to take the square roots): http://en.wikipedia.org/wiki/Ellipse

9. Harold14370 is correct

10. Originally Posted by MagiMaster
I'm a little confused. Did you mean
Yes, I did.

11. Originally Posted by MagiMaster
I think I see what he means though. There should be one free variable in this system, assuming the bounding rectangle is fixed.

Let's say the piece of wood is a rectangle from (0,0) to (A,B). Inside that piece of wood, you place 4 nails, also arranged in an axially-aligned rectangle. Assuming you want the string to touch the center of the 4 sides of the outer rectangle, the placement of the nails and the length of the string should all depend on a single free variable (sort-of a roundness factor, correct me if I'm wrong though).

Obviously, if this is 0, the nails go in the corners, and the string fits snugly all around, but as this value increases, the nails move inward, but not in such a simple way.

Let's say the nail closest to the origin is at (C,D). When the string stretches to the bottom center, it'll form a pentagon with side lengths . Similarly, when it's on the left side, it'd have sides . Since the string is the same length in either case, these two cases must sum to the same length. Since A and B are fixed, this gives us one equation in two variables (or if you add that the length of the string is equal to either of these, two equations in three variables). Solving these equations should give you all the information you need, technically, but you may have to apply some substitutions to get a more meaningful free variable.

Well, for now I've got to go, but it's pretty likely someone else here can poke holes in my equations and help you figure the rest of this out.
Maybe I'm the only one here who doesn't understand what you are trying to do. But I don't.

If you are trying to somehow get a smooth (and by smooth I mean at least differentiable, so no cusps or corners) then you will probably have to abandon the 4 nails or even 3. At any point where the tension of the string changes from being supported by one set of nails to another you are likely to get an point of non-differentiability. As I noted earlier you can use 2 nails and inscribe a smooth ellipse in your rectangle and have it touch the center point of each edge. You can use 4 nails at the corners and a tight string and recover the complete rectangle. Or you can do anything in between.

If what you want is to lay out 4 nails that define a rectangle that is centered in your original rectangle so that you can lay a string around them, put a pencil in it and pull it tight and have the pencil at the edge of the larger rectangle precisely in the center of each edge of the big rectangle then you can do that as follows. Consider the big rectangle laid out so that the lowerleft-hand corner represents (0,0). Then proceeding conter clockwise the corners are (A,0), (A,B), and (0,B). To center a smaller rectangle in that larger rectangle it is sufficient to find the coordinates of the lower left-hand corner. Call that corner (x,y). Then there are many solutions but they all satisfy the following equation:

This equation is obtained by finding (x,y) so that the distance from (xy) to the mid-points of the long and short sides of the large rectangle nearest to (x,y) are equal.

Is that what you are looking for or is there some other objective ?

12. Setting those two lengths to be equal won't give the right answer, since the rest of the string will be different lengths at those two points. (The part around the rectangle will be 2U+V at one point and U+2V at the other, for some U and V.) Anyway, JaneBennet's definition of C and D works better than mine, so use those.

Also, I think smooth in this case just means , or maybe . I think the 4 nails would give a sufficiently smooth curve, though I can't prove it.

13. Originally Posted by MagiMaster
Setting those two lengths to be equal won't give the right answer, since the rest of the string will be different lengths at those two points. (The part around the rectangle will be 2U+V at one point and U+2V at the other, for some U and V.) Anyway, JaneBennet's definition of C and D works better than mine, so use those.

Also, I think smooth in this case just means , or maybe . I think the 4 nails would give a sufficiently smooth curve, though I can't prove it.
You are correct. I screwed it up. That would only work for a square.

14. Originally Posted by sak
We know, fix a nail put a loop around it, we can draw a circle. Two nails give an ellipse.
Consider a rectangle of size AB, fix 4 nails in a rectangle shape, put a loop around it then draw a curve. How can I find the optimum size of rectangle CD for the nails which is inside AB? There can be a number of rectangles CD. Give me one with a smooth curve, for a table top.

You can lay out your ellipse with a grid rather easily.

http://www.Rockwelder.com/Flash/Ellipse/Ellipse.htm

It is funny but with only three digits of precision. There was actually a slight deviation in the ellipse made with a grid and one done with just one calculation. Or I messed up slightly on the math. Ha-ha. But usually it is compound error.

For most applications no one would notice and no one does by actual experience. However if you are going for the gold you have to use many digits of accuracy.

That is usually the problem in the real world. Compounding error through a few digits of accuracy.

I also have some other information about laying out ellipses. I will try to post that tomorrow. It is pretty cool. It shows the focal point with a triangle. I am not fully understanding it yet. As to how it could help me lay them out. But it may be of interest.

I believe they might be showing a simultaneous double axis angle shift.

Sincerely,

William McCormick

15. Originally Posted by sak
We know, fix a nail put a loop around it, we can draw a circle. Two nails give an ellipse.
Consider a rectangle of size AB, fix 4 nails in a rectangle shape, put a loop around it then draw a curve. How can I find the optimum size of rectangle CD for the nails which is inside AB? There can be a number of rectangles CD. Give me one with a smooth curve, for a table top.
You could probably also work out a pattern using the method to create a volute from a geometric figure. I believe you can use anything from a triangle to polygon to create a volute.

I believe you could incorporate that in principle, to create an ellipse from some simple shape drawn in the center of your ellipse. But maybe not. Ha-ha.

Sincerely,

William McCormick

16. No one is making ellipses or volutes William. Stick to the topic if you feel you must post something.

On topic:
I just tried punching JaneBennet's formulas into Matlab and the answer it gave for D in terms of A, B and C is about a mile long. I haven't tried solving it by hand yet, but surely a formula like this shouldn't have that complicated of an answer, right?

17. Originally Posted by MagiMaster
No one is making ellipses or volutes William. Stick to the topic if you feel you must post something.

On topic:
I just tried punching JaneBennet's formulas into Matlab and the answer it gave for D in terms of A, B and C is about a mile long. I haven't tried solving it by hand yet, but surely a formula like this shouldn't have that complicated of an answer, right?
I haven't played with this since you indicated that you were ready to roll with Jane's equations, but from what I did up to that time I think I can say that yes, it could be that complicated. It is not clear to me what conditions permit a solution. I have about convinced myself (a step or two remain) that if you want the inner rectangle to be similar to the big rectangle that there are no non-trivial solutions.

18. Sorry. I guess I gave the wrong impression. Anyway, I suspect there might be some substitutions or acceptable simplifications that could make this problem a little easier, but if there are, they aren't obvious.

19. Rectangle CD is inside AB. If the side C is A - X and side D is B - Y, then length of loop is given by a formula

2D + C + 2 * ( 0.5C + 0.5Y)**0.5 = D + 2C + 2 * ( 0.5D + 0.5X )**0.5

Is that clear?

20. Yes, that's basically what JaneBennet said earlier, but trying to go from there to a solution to C (or D) in terms of A, B and D (or C) isn't straightforward.

21. Originally Posted by sak
Rectangle CD is inside AB. If the side C is A - X and side D is B - Y, then length of loop is given by a formula

2D + C + 2 * ( 0.5C + 0.5Y)**0.5 = D + 2C + 2 * ( 0.5D + 0.5X )**0.5

Is that clear?

I have been assuming that C and D are the dimensions of the inner rectangle, and that the inner rectangle is symmetrically placed inside the bigger one. Is that the case?

22. Originally Posted by JaneBennet
Originally Posted by sak
Rectangle CD is inside AB. If the side C is A - X and side D is B - Y, then length of loop is given by a formula

2D + C + 2 * ( 0.5C + 0.5Y)**0.5 = D + 2C + 2 * ( 0.5D + 0.5X )**0.5

Is that clear?

I have been assuming that C and D are the dimensions of the inner rectangle, and that the inner rectangle is symmetrically placed inside the bigger one. Is that the case?
As long as the inner rectangle has an axis parallel to the axis of the outer rectangle you must have symmetry in order to get a solution. The equations that you have are a necessary condition for a solution. But it is not obvious that a solution exists for an arbitrary choice of A,B,C,D. By a scaling argument you may assume that A=1 but that doesn't help much.

23. Originally Posted by MagiMaster
No one is making ellipses or volutes William. Stick to the topic if you feel you must post something.

On topic:
I just tried punching JaneBennet's formulas into Matlab and the answer it gave for D in terms of A, B and C is about a mile long. I haven't tried solving it by hand yet, but surely a formula like this shouldn't have that complicated of an answer, right?
The first post was talking about making an ellipse with four nails. I thought I was rather on topic.

I don't think you can do it.

There are some rather easy ways to make an ellipse from a circumscribed rectangle. A few compass splitting tricks. Draw a couple rays and you have yourself the placement of points along the ellipse.

I have some other ones I will try to post them tomorrow.

Sincerely,

William McCormick

24. JaneBennet is correct.

William McCormick, you did a good job.

25. May be, William, but with 4 nails, it's not an ellipse anymore. It's something more like a rounded rectangle, although it won't have any straight sides.

26. Originally Posted by MagiMaster
May be, William, but with 4 nails, it's not an ellipse anymore. It's something more like a rounded rectangle, although it won't have any straight sides.

Did you see the rectangle method. You draw up a rectangle basically, dissect the small sides of the rectangle. And the major axis. Into equal parts.

Then draw rays though those intersections from each end of the minor axis. That is also a pretty easy way to actually lay one out.

Sincerely,

William McCormick

27. That is a fairly good way of laying out ellipses, but what if you wanted to make something else?

28. Originally Posted by MagiMaster
That is a fairly good way of laying out ellipses, but what if you wanted to make something else?
Like what else?

Sincerely,

William McCormick

29. Originally Posted by William McCormick
Like what else?
Like the shape we've been discussing with the string and four nails.

30. Originally Posted by Harold14370
Originally Posted by William McCormick
Like what else?
Like the shape we've been discussing with the string and four nails.

In most cases it seems to make an ellipse. However to make something nice I doubt you would want to use a string. It stretches and offers in about twenty variables, from my experience with actually working with string. You might do it with a small metal ribbon. But then there is the bending of the ribbon to take into consideration at the corners.

You might be able to use a cloth tape, like they used in the old 50 foot tape measures with the real. But I would use something else myself.

If you are going to lay out a table top, and you do not want an exact ellipse.

I would pencil in a shape onto the table. Then take thin metal and pick key points that make the metal, with a slight spring to it, like 26-24 gauge galvanized sheet metal, two inches wide, adhere to your drawn in arc, between the key points.

You can place nails or nail small pieces of slightly arced wood at the key points.

Then just mark the table and cut it. Do the same for the other side. Or use the cut off from the one side to mark the other. Even that would probably be more accurate then the string.

I thought you guys were just discussing the idea of using a string. Not really using a string. I know it is not a very good tool for exacting measurements.

If I had to make twenty of them, I would use a router and a smaller offset guide figure, to cut the shape.

It would be hard to cut out an ellipse on a band saw, with any kind of automation or template. You wold need a rather complex cam system to make it happen. Because of the angles compared to the blade.

Sincerely,

William McCormick

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