how do you make a net with a radius of 4 in and height of 10 in?

Answer 1

As I write this, the question has been posted for 3 hours without an answer, so I'll give it a try. When you suspend a slack (not taut) string between two points, the string, having weight, will naturally form a parabolic shape between the points.

Earlier I did the math, figuring out how long the string would be if strung across the 8-inch diameter and falling parabolically to a depth of 10 inches in the center. This is a "length of curve" problem that involves some very messy calculus.

That's not the only problem. Having done the math for the longest string -- the one that crosses the diameter -- I realized that the same problem would have to be repeated for each of the chords that will have shorter distances between the endpoints and shorter depths.

As a result, after reading your question again, I decided to take a new tack, focusing on the "how do you make" part of your question. There's still some math, but not nearly as much as there would've been otherwise.

To begin, think of a circular tennis racket with loose strings sinking in the center. That's the kind of net we're going to design. For orientation, think of the 8-inch circular rim as the face of a clock. with 12:00, 1:00, ... 11:00 positions.

For the general shape, long strings will connect the 12:00 and 6:00 positions, falling parabolically to a central depth of 10 inches, and also connect the 3:00 and 9:00 positions, falling to the same depth.

There can be shorter strings connecting 1:00 and 5:00, 11:00 and 7:00, 10:00 and 2:00, and 8:00 and 4:00, and falling to a shallower depth (less than 10 inches). These four strings will all be the same length. Another set of four still-shorter strings will connect 2 and 4, 10 and 8, 11 and 1, and 7 and 5, again all of equal length and falling to a still-shorter depth.

All strings will be parabolic. The longer and deeper strings will be in the center, and the shorter and shallower strings (the chords) will be toward the sides. There will be an odd number of both vertical (e.g., 12-6) and horizontal (e.g., 9-3) strings because there's one central string in each direction and an even number of strings on the sides due to symmetry.

To form the net, the strings will all be woven as they interlace with each other. I imagine that your net may be constructed with fishing line.

Having described the general structure of the net, we now have to do some math to figure out the spacing of the strings. Going back to the tennis racquet idea, let's suppose we want the strings to be equally spaced

For an example, suppose the strings are a half-inch apart. Let the 12-6 line be the y-axis and the 9-3 line be the x-axis. We want vertical strings at x-values of 0, 0.5, 1.0, 1.5, ... 3.5 (plus and minus) for a total of 15 strings. There'll also be 15 horizontal strings for a total of 30.

We've already said that the central strings will attach across diameters (plus and minus 4) and descend to a depth of 10. In three dimensions, suppose the bottom of the net is at (0,0,0), and the top circle is in the plane z = 10. The equation of the parabola for the center strings is z = ax^2, passing through (4,10), so 10 = 16a, or a = 5/8, and the final equation in the x-z plane is

z = (5/8) x^2 (Equation 1)

Let's look at the placement of the string at x=3. In the x-y plane, the circle has radius 4, and by the Pythagorean Theorem, the y-value is sqrt(4^2 - 3^2) = sqrt 7. This string will be attached at (3, sqrt 7, 10) in 3 dimensions, with the other end at negative sqrt 7 in the y-dimension. The distance between the attachment points is 2 sqrt 7.

How far down will this string go? To get that, we evaluate Equation 1 at x=3:

z = (5/8)(3^2) = 45/8 = 5 5/8 = 5.625

This means the at x=3 is z= 5.625 inches. But the top of the netting is at z=10, so the is 10 - 5.625 = 4.375 inches (from the top).

This particular string (and there are four just like it) attaches at points 2 sqrt 7 = 5.29 inches apart, and drops 4.38 inches.

By repeating this procedure seven times -- each repetition gives you four strings -- you'll get the lengths of 28 of the 30 strings. (The other two are the long ones in the center.)

That solves your problem according to the net description we gave. However, there's one other thing to consider. The way we set it up, the strings are equally spaced This does mean the small squares in the net are all the same size. That's because the squares are all tilted at different angles. The squares at the bottom of the net are smaller than the ones at the top.

It is possible to make the squares all the same size. (And if I were making a net, that's what I'd want.) But to do that, we'd need to do some more trigonometry in three dimensions. I can do that, but only if that's what you're looking for. If that's what you want, you need to say so.