J.Norad Makes Paper Ball Joints

This post is more for personal reference than of possible use to any of you. Today, I'll be covering some parameters for designing a ball joint. A ball and socket joint provides excellent movement capabilities but isn't an easy element to build with paper.

The key to building a robust ball joint with paper is to have the appropriate materials needed. A good ball joint should provide adequate joint stiffness, but it may vary on your application.

Designing the Ball and Socket
You'll need to know a few things first:

• Figure out how big your joint is going to be.
• Figure out how much mobility you want.

Here's some calculations to help assist you in figuring out the dimensions of your ball and socket joint.

Figure 1: The Ball Joint Schematic

There's four design parameters that dictate the performance of your ball joint:

1. Ball radius (r1 in Figure 1) (You'll hopefully know this value first)
   
2. Socket depth (Hl in Figure 1) (somewhat adjustable, has a lower limit)
   
3. Support rod diameter (2P in Figure 1) (adjustable, has a lower limit)
   
4. Range of motion (phi in Figure 1) (defined by values 1-3)
   

For a given set of construction variables, r1, Hl and 2p, you can figure out how much motion your ball joint will provide. We'll use some trigonometry to solve for the angle phi, and use this to determine the total angle your joint will provide.

First, we must acknowledge that the support rod diameter 2P and the lower socket height Hl limit our angle. This is because the rod hits the edge of the socket, defined by how deep (Hl) the socket is. Second, we'll define the angle the joint provides as the angle that the center of the support rod makes with the line parallel to the bottom of the socket. This angle will be phi.

From Figure 1:

1. The red angle Beta formed from the point of contact with the socket edge (Hl) and the support rod walls is less than our desired angle, phi. Since it's a right triangle, we know from the pythagorean theorem that the length of from the center to the tip of Hl is the square root of r1^2 + Hl^2.
2. The angle formed from the line OP to the support rod centerline is the difference between angles phi and Beta. We know the thickness is 2P, and that OP forms the hypotenuse, and half the rod thickness P forms the opposite wall. Therefore, the angle Phi- Beta = arcsin(P/sqrt(r1^2 +Hl^2)).
3. From trigoneometry, the angle Beta =arctan(Hl/r1)

Solving for the angle phi, we get Phi = arctan(Hl/r1) + arcsin(p/sqrt(r1^2 +Hl^2)).

The angle we actually want is the complimentary angle to Phi, since that determines the angle relative to the null position. So we take twice the compliment to Phi (two directions) to find the angle of our joint.
Total angle range for the ball and socket joint = 2(90° - Phi)

Where Phi = arctan(Hl/r1) + arcsin(p/sqrt(r1^2 +Hl^2))

This equation tells us some obvious relations, which help support the validity of the result:

1. If Hl is longer than r1, the angle decreases
   
2. if p increases, the angle decreases
3. increasing r1 increases the angle.
   

Now for the actual construction!

The Socket
The socket should ideally consist of a durable material. 110lb cardstock will not work, as it wears out fast and easily over a few cycles. The smooth varnished surface of a Magic: the Gathering card is an excellent material. It will withstand more cycles and is fairly strong. Since paper (and Magic cards) does not have "negligible" thickness anymore (you're now working with a system that will requires a few thousandths of an inch in terms of tolerance to work well), you need to account for the overlap of paper. Magic cards have a thickness of 0.30988 mm (experimentally measured), which translates to 0.0122 in. This is enough to make your cylinder a slight oval if there's overlap. (If you consider that the accuracy of hand building has a tolerance in the range of half a millimeter anyways, it might not matter in the long run. And you can always correct for it later...)

Figure 2: The socket.

By acknowledging paper overlap, we form our socket by cutting out a strip of Magic card of a length equal to our projected ball diameter so that when we curl it up, both ends sit flush with each other, thereby eliminating the overlap. Once you have your cylinder, you can freely complete the cylinder with additional layer of Magic card without worrying as much about overlap. Removing overlap helps reduce wear of the ball and socket over time, since the raised edge is most likely to wear first.

Remember: paper is not incompressible. You'll lose a few thousandths over time. Adjust accordingly.

The Ball
The ball part is perhaps the most difficult part to build. Not also do you need to have the dimensions as close as possible for a tight fit, it needs to be built well and uniform.

First, you'll need a decent support rod to use. 1/8" (3.175 mm) diameter bamboo sticks are a good choice. They're usually $2 for a pack of 100. Pick one with low eccentricity if possible, and look for non-slivering/splintering ones. Those will snap first over time.

Next, you'll need to use the excel sheet for making cylinders. I suggest using 110lb cardstock for the ball, since it's easier to work with and tears less than printer paper. However, printer paper glued together with superglue will provide a nice solid sphere. I like to sand the ball after it's made, so I use 110lb.

Figure 3: Tapering the strip. Note the 5cm allowance before tapering.

To make the spherical shape, you need to taper the strip to a triangular shape. If using the 1/8" rod method, start the taper from 5cm from the starting edge and taper it linearly to 3mm to the other side.

Roll the paper around the rod as tight as possible. Any gaps or loosely bonded sections will result in failure in the rod axial direction, meaning it will start to deform and separate as you push it in the socket.


After you've finished, your sphere will be somewhat octagonal in cross section. Break out those calipers and sand that sphere down to as best as you can to a uniform diameter throughout. Irregularities will result in uneven performance, where certain positions are looser than others. You ideally want the ball to be a few thousandths (0.003-0.010 in) larger than the socket for a nice snug fit.

Adjusting the Fit
Your ball and socket joint may be loose or come loose over time due to thermal expansion, humidity, wear or other factors. You can easily adjust the joint to regain stiffness. Options include:

1. Adding some additional material to pad out the socket to reduce the inner diameter. I suggest using a small section of paper (printer or 110lb works, depending on the looseness) inserted into the joint
2. Thickening the ball with superglue. Make sure the ball is dried completely before re-inserting.

Miscellaneous
If there are any other adjustments or updates, I'll add them as necessary to this page.

Making Simple Spheres and Tubes Overly Complicated

Personal math reference time! Today's post deals with making revolved solids out of paper. This post will assist you in building your own tubes, cylinders and spheres for use in making ball joints, miniguns, and staves.

If you have a good understanding of math or don't care about the proof, skip to the end. If you're curious to the method behind the result, keep reading.

The basis of this method lies in the formula for a circle based on the diameter:

EQ. 1: Circumference = π* diameter

What we're doing when we roll paper into a tube is stacking a lot of thin walled tubes together to form a solid one. The goal is to make a formula to add up all the lengths of paper that consist of each individual tube together, so we can simply measure a length, roll it up around an object, and make a tube of the approximate dimensions.

I've been making my ball joints out of 3.175 mm (1/8") diameter bamboo sticks. For this example, I'll be using that as a baseline inner diameter for my measurements. You can choose to use other inner diameters later on. My goal for this example is to generate an 8mm diameter cylinder/sphere. First, I need to calculate how many times I'll need to wrap my strip of paper around this stick. Unfortunately, I need a vital measurement that most of you will be hard pressed to find: the thickness of paper. Fortunately, I've done that work for you.

Thickness of 110lb cardstock is roughly 0.252095 mm (or 0.009925"), calculated by a series of measurements, averaged to compensate for the lack of a digital caliper. (Dial calipers do have their drawbacks, but I love watching the spinning dial when I use mine.) Regular printer paper is roughly 0.004" to 0.005" for reference. You may notice, I kept all the decimals as far out as possible. I did this since the later part is going to vary a lot based on how accurate my values are.

Now. To the formulas! Time to pay attention to the equations, page skimmer! The number of revolutions your strip will need to make is:

You'll get a non-whole number. That's fine. Your revolved solid may be a bit thicker on one side than the other, but if you're aiming to make precise paper Rolexes, you shouldn't be learning from a guy who inhales Magic: the Gathering card dust on a daily basis.

Armed with this number, we're going to mess with arithmetic series. As we add a layer of paper, we're incrementally adding a length of paper proportional to the thickness and a constant. The thickness of paper is the change in diameter we're adding, and based off our formula earlier for the circumference, we're changing the circumference with each increment of the diameter. That gives us:

Where C is the circumference, N is the number of revolutions you need to make. Each revolution generates a length of paper that we'll have to add up to form our overall length to cut. So, if N=10, we'd add up C1, C2, C3... all the way to C10. Great for small values, terrible if you discover N=100. Arithmetic sums to the rescue!

The sum of an arithmetic series is simply:

With this, you just need to input your thickness, inner and outer diameters into the formulas above, and get the sum. That is the length of the strip of paper you need to cut out and roll around to get your tube.

Armed with this length, you can now control the size of the following revolved objects!


For small ball joints, the reference labeled "triangle" suffices. For large ones, you should probably just buy wooden beads or modify the "convex" one. I'd recommend using sandpaper to smooth out the spheres to be less octagonal.

For the socket part of a ball joint, I'd suggest using a sphere with a diameter larger than the inner diameter of the tube by 0.254mm (0.01"). That will give you a snug fit. You can always enlarge the sphere by adding a thin coat of superglue over it, and sanding it to fit when dry. I highly discourage adding material to the inside of a tube. It's too messy.


EDIT: Here's a microsoft Excel file for the formula for the extra-apathetic.

Just go to the proper tab and fiddle with the settings. And if you delete a formula, just download it again!

Miscellaneous reference posting, April 12, 2008

Given two vectors A and B with coords (x1,y1,z1) and (x2,y2,z2), the angle between these vectors from the origin is arcos((x1*x2 + y1*y2+ z1*z2)/||A||*||B||)

||A|| is the magnitude of A

Original formula is: Dot product of A and B = ||A||*||B||*cos(theta)
where theta is the angle between the vectors A and B

I'm sick of forgetting how to calculate the angle of a triangle inscribed inside a square prism. Not forgetting anymore.

Making Cone Frustums (Revised)

Today's lesson is on how to make approximations for spherical objects out of paper. Why? It's easier and lighter to make these to fill up the dimensions for a round part of the model (in my case, to round out the joints to make them more spherical) than by making them by layer.

Figure 1: Making something round into a polygonal mess! The item on the far right is the side view of this "frustum" I'll be talking about.

I've taken the essence of the idea from a paper modeler forum.

(Edit: URL doesn't exist anymore. No point making you click on a link that doesn't work. I might as well explain how it works here.
-J.Norad)

The principle is that you draw two concentric circles, cut out a sector, and when you glue that sector end to end, you get a cone frustum. How does it work?

Figure 2: All you need to know about the next paragraph.

First, you need to know three things: how tall your cone frustum is, and the smaller and larger diameters that compose the top and bottom of it. From Figure 2, you can see the arc segment for the smaller and larger circle shares the same arc angle. As a refresher, arc segments are:

Eq 1.) arc segment length = pi * radius* 2 * (arc angle)/(360 degrees)

We're going to use this to set up two equations: one being the arc segment length of the smaller circle and the arc segment length of the larger circle, then setting the angles equal to each other. This gets us

Eq 2.) (Arc length 1)/ (π*radius1* 2) = (Arc length 2)/(π* radius2 *2)

or simplified,

Eq 3.) (Arc length 1)/ (radius1) = (Arc length 2)/(radius2)

We know arc lengths 1 and 2, since we've chosen what the cone frustum's top and bottom diameters are. These diameters will correspond to the arc lengths, since when we cut and assemble the sector, the edges will form the diameters of our frustum. Confused? You'll understand when you try it out soon.

As a recap, the circumference of a circle = pi * diameter.

However, before we can proceed, we need to know the difference between the two radii of the concentric circles we're going to draw. This is where the height of the frustum comes into play. We'll be using the Pythagorean theorem to do this.

As a refresher, the hypotenuse of a right triangle is equal to the square root of the sum of the squares of the two legs. (a^2+ b^2 = c^2)

Figure 3: Frustum geometry and how it relates to our arc sector.

We want the length of the sloped side of the cone frustum. We have the height and the bottom leg, which is HALF the difference between the two diameters since we have two sides of the trapezoid. Substituting these into the Pythagorean Theorem and referencing Figure 3, we get the following:

Eq 4.) Radius2-Radius1 = square root of (height^2 + ((larger diameter- smaller diameter)/2) ^2)

Hooray. Now we combine Equation 4 with Equation 3 and we can solve for one of the radii. Once we know what one radius is, we can determine the other using Equation 3, since we know the arc lengths= circumference of our cone frustum tops and bottoms. Once we know the radius and arc length of the inner or outer circle, we can solve for the arc angle using Equation 1.

Still confused? Can't be bothered solving equations? I have simplified the process down for you, with less potential error involved by turning it into a Microsoft Excel spreadsheet!

Edit: Added a download link to a much more updated cone calculator. This one's a bit more silly, as it includes my attempt to calculate the shape of offset cones.

Figure 4: Excel spreadsheet. Copy as directed.

What you'll need to do to use this is enter the smaller frustum diameter and the larger frustum diameter along with how tall you want the section to be. It will spit out two circle diameters which you should then draw concentrically, along with an angle. Use the angle to draw a sector on those circles. The resulting rainbow shaped arc will be the thing you need to cut out and glue together. Just remember to add your own tabs.

Here's the layout of the spreadsheet. Fill in the text fields as they are arranged, as the commands reference the numbers in the second column exactly.

Now paste in
B5: =B2/2*SQRT(1+4*B3^2/(B2-B1)^2)
B6: =B5-SQRT(((B2-B1)/2)^2 + B3^2)
B7: =B2*360/(2*B5)

E5: =B5*2
E6: =B6*2
E7 was a reference number for me, since some of the angles produced were over 180 degrees and I only needed to deal with the remainder. This value is =b7-180 in case you want to reference them in terms of increments of 180, since protractors tend to deal only with 180 increments.

Figure 5: Cone frustum side view on left; sector ready to cut and glue on the right

This image best shows the result of the exercise. Stolen from the site I linked from.

Once you try it out, you'll understand what's all behind this gibberish and can start making cones out of paper.

(-Revised by J.Norad, June 16/2009)