Layout problem

Re: Layout problem 

Re: Layout problem

So is that 60" diameter

Re: Layout problem

I get a 64" diameter.

OOPS! 64 looks right

Re: Layout problem

Yes, that's the correct approach. But how big is it?

Re: Layout problem

"I get a 64" diameter."

How? With CAD, I can draw the solution just a hair larger than that. (64.4 will work, but 66 won't.)

But it seems that this problem should yield a mathematical solution. So far I have failed in that quest.

Re: Layout problem

64 was a mathematical solution. I don't have time right now, but will post the method later. It uses the angle bisector theorum and the Pythagorian theorum.

Re: Layout problem

Thanks. I'll look forward to seeing that. Now I am worried about my CAD program (or its operator!) since I get an answer somewhere between 64 3/4 and 65.

Re: Layout problem

I did it in Adobe Illustrator (graphics CAD) and the closest I could get was 64.919 with 0.004 between the halves. If I had more time ... maybe 64.920 or 64.921. :)

Re: Layout problem

Ed, we are in agreement on the answer derived graphically, and I finally have a mathematical answer, but it is UGLY! The radius in feet is given by r^4 - 24 r^3 + 144 r^2 - 352 r + 320 = 0. (Don't ask, I probably can't recreate the trail!) Using a spreadsheet to solve for a root of this equation, I get an answer that translates out to a diameter of 64.93" And this works very close to perfectly on my CAD program.

Re: Layout problem

OK, if we take the triangle ABC with B being at the right angle and C at the smaller of the other two. Call the center of the semicircle point O. Draw 2 lines from point O to the points of tangency of the semicircle. They are the same length because they are both radii of the same circle. They are also right angles because the line from the center of a circle is perpendicular to a line tangent to the circle.if both of those are right angles, and B is a right angle because ABC is a right triangle, then they end up forming a square because the angles are all right angles and two adjacent sides are the same length. (both radii) This means that a diagonal from angle B to the center bisects angle B. The bisector theorum states that the bisector of an angle of a triangle divides the opposite side in the same ration as the other two sides, hence AO/BO=48/96. You can use the Pythagorean theorum to find AC and from that, OC (2/3 of AC) If you call the point of tangency with the 96" long side Y, then triangle ABC is similar (simple to prove) to triangle OYC and hence OC/OY =AC/AB. If you plug in the numbers, you end up with side OY = 32" or a diameter of 64". I left out some of the really simple proofs and a bit of the algebra, but can lay them all out if necessary for some of the assumptions. I realize this is a bit disjointed, but if you cannot see it, maybe I can write it up with all of the math and scan it in. I plugged all of the numbers into a spreadsheet for the calculations, so it is possible that there might be some very small amount of roundoff error in the spreadsheet calculations. I guess I could do the whole thing in algebraic form and make sure it is absolutely exact. I didn't feel like doing complete solutions to all of the equations ;-) The answers that you got were close enough to what I got that it is possible there is a bit of error in the spreadsheet, but at least, this is a math method for calculating it.

Clear as mud?

Re: Layout problem

Went back and checked the numbers and I get 32.0000" for the radius out to 4 places.

Re: Layout problem

If I follow this, you have determined the size of the largest semicircle that can be cut from a half sheet, where the half sheet is divided along the diagonal. I agree that is exactly 64".

But you can get slightly larger semicircles than that. Go back to Ed's post, second in this thread and look at his picture. The slanted line goes from long-side to long-side, near but not at the corners. This is the way to squeeze a slightly larger circle out of the MDF sheet.

hmmm

OK, will have to take a look and see if I can figure out a way to solve that one.

Two approaches

I have tried:

1) Note that as the "attachment point" of the slanted line moves away from the corner, the available radius grows then decreases. So if I could express the radius as a function of the distance from the corner, I could differentiate to get a max. This line of attack has not be fruitful for me. (An equivalent approach that I have not tried is to express radius as a function of the angle of the slanted line.)

2) In the attached picture, consider that the height of the rectangle is 4 (I worked in feet) and the upper left square and the hypotenuse of the two triangles on the diameter of the semicircle are all r. With help from Pythagoras you can get the dimensions of those two triangles, the slightly larger third one, and the distance from the corners. Adding the horizontal distances and setting the total to 8 gives a very messy equation, one root of which is the solution. When I manipulated it to get rid of the sqrt(8r-16) term that kept showing up, I ended up with an equation with r to the 0th-4th powers. Yuck!

Now you did it ;-)

Unless I can figure out a way to forget about this, it is going to drive me nuts until I can figure out a way to solve it ;-)

Misery loves company ;-)

Re: Now you did it ;-)

My math skills suffer from disuse, but I'll guess the center point of the diagonal (board) has to be factored into the equation.

Re: Now you did it ;-)

I think that you are correct. I also think that the angle is an important part. I have a couple of ideas that I want to try that probably will just get rid of what hair I have left, but at least it will be fun ;-)

A solution

I don't see how the center point can help. Hopefully it will.

I tried the angle when I was trying the calculus approach (maximum of a function describing R in terms of the slanted line.) But I now see WHY the calculus approach won't work (any such formula is probably continuous but not differentiable at the point in question, since what affects the radius is different on each side of that point.)

Here is the diagram I posted earlier, with some dimensions filled in.

The horizontal side of the triangles on the diameter can be determined by x^2 +y^2 = z^2.

The distance from the slanted line to the corner (at each end) is r- the length of the horizontal side above.

The third triangle is similar to the other two, so its horizontal side can determined using the ratio of the vertical sides.

Adding all those horizontal measures up and setting the sum equal to the length of the sheet of MDF gives you an equation that can be solved (numerically) for r.

So I have a solution. I'd like to have a more elegant one, but maybe that doesn't exist.

BTW, the picture shows measurements in inches. In actually working with the formulas, I used feet, because inches get kinda unwieldy in the formulas. In fact, I probably should have worked in "quadrafeet". If my sheet were 1x2, I probably would have had fewer mulligans due to arithmetic errors as I worked the formulas.

Re: A solution

I think you are right to just use a numeric solution. The main way that I saw the center point as useful is that the diagonal has to rotate about that point because the two sides have to be symmetrical. Did you ever look at the path the center of the circle takes as the diagonal is rotated from the corner to vertical? Interesting problem.

Path of the center

"Did you ever look at the path the center of the circle takes as the diagonal is rotated from the corner to vertical?"

No, I didn't. But I see the problem with that being the same as my differentiation idea: what constrains the movement of the center or the radius is different on either side of the optimum point. When the slanted line is more vertical than optimal, the radius is 48/(1+cos(theta)). When less vertical, the radius is given by finding the point on the line of the form (r, 4-r) (where the lower left corner of the rectangle is the origin). Maybe there's hope in setting those two equal?

Re: Path of the center

And expressing one formula in inches and the other in feet just adds more challenges!