Year: 2024

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Hand Speed Index for Pickleball Paddles

ATTENTION: The calculator originally contained an error. I had erroneously equated 90° and π/4 radians instead of π/2 radians. This understated the angular power by a factor of four and made it appear that adding weight to the butt-end of a paddle could reduce the power required to flip it. Correcting the error eliminated that phenomenon. The results of this calculation now seem much less interesting, but I’m leaving it up for now.

John of Johnkew Pickleball reached out a while back for feedback on an idea to combine the static weight, balance point, and swingweight of pickleball paddles into a Hand Speed Index (HSI). Being an engineer, I noticed that the units didn’t work out in his proposed formula. Since it wasn’t based on physics, I was worried that it would be possible to “game” the number, but I liked the idea. After some thought and trying some not-so-great concepts, I had the idea to calculate the power required to maneuver a paddle through a certain motion in a given time. Paddles that require more power will be slower to maneuver, and vice-versa.

Just want the numbers? Go to the Hand Speed Index Calculator. Interested in the details? Read on.

I picked the motion of moving the paddle through 180°, from backhand to forehand (or vice-versa), while moving the hand a defined distance. The power requirement is calculated as two components: angular and linear.

The angular component of power is exerted as a torque that accelerates and decelerates the paddle in rotation. In the formula for the angular component of power below, RW is recoilweight in kg·m² , Θ is the angle of rotation in radians (180° is π radians), and t is the duration in seconds. The angle and duration are both halved, as the first half of the motion is acceleration and the second half is deceleration, but the power required for each phase is identical.

P_{a}=\frac{2RW(\Theta/2)^{2}}{(t/2)^{3}}

RW is a moment of inertia value similar to swingweight (SW) but about an axis through the center of mass. The formula below calculates RW from SW, mass (m), and balance point (BP). The units are kg·m², kg, and meters. These are not customary units for these values, but they make the power formula work out to Watts. The 0.05 m (5 cm) is the distance from the end of the handle to the SW axis.

RW=SW-m(BP-(0.05\ m))^{2}

The linear portion of the power is exerted as force that accelerates and decelerates the paddle center of mass laterally. If there were no force and just a torque was applied, the paddle would rotate about its center of mass. Just to keep the center of the hand stationary during the motion requires force. You can see this yourself by trying to flip a paddle from the forehand to backhand side as quickly as possible. Your hand will move opposite of the direction that the tip of the paddle moves unless you try to keep it stationary.

In the formula for the linear component of power below, m is the mass in kg, d is the distance that the hand moves in meters, BP is the balance point in meters, and t is the duration in seconds.

P_{l}=\frac{2m(d/2+BP-(0.05\ m))^{2}}{(t/2)^{3}}

I originally thought we’d pick a single hand move distance for the HSI, but the results are quite different at different distances. For short distances, adding mass at the bottom of the handle actually reduces the power requirement. The effect of the lowered balance point is greater than the effects of adding mass and swingweight. As the distance increases, mass becomes a larger factor, so the same mass at the bottom of the handle increases the power requirement. We settled on three distances to start, and the durations were selected to make the power requirements similar across the distances for a typical paddle. The actual power values aren’t necessarily meaningful, but the relative differences should be. I’m interested to hear if the relative differences match your subjective feel.

Per Harry’s comment below, I’m adding the derivation of the angular power formula. I’ve used basic equations from the AP Physics 1 Equation Sheet.

Here’s an equation for angular position (Θ):

\Theta=\Theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^{2}

The first two terms are zero. Solving for α, we get:

\alpha = \frac{2 \Theta}{t^{2}}

Then, an equation for angular velocity (ω) is:

\omega = \omega_{0} + \alpha t

The first term is zero, so substituting α from above, we get:

\omega = \frac{2 \Theta}{t}

An equation for kinetic energy (K) is:

K = \frac{1}{2} I \omega^{2}

Substituting ω from above, we get:

K = \frac{1}{2} I (\frac{2 \Theta}{t})^{2} = \frac{2 I \Theta^{2}}{t^{2}}

An equation for average power (Pavg) is:

P_{avg} = \frac{\Delta E}{\Delta t} = \frac{K}{t}

Substituting K from above, it becomes:

P_{avg} = \frac{2 I \Theta^{2}}{t^{3}}

This is the equation for angular power in the post above, where I is RW and both Θ and t are halved, as I explained there. Actually, I had originally omitted the “2” in the equation (fixed now). It was correct in the calculator, but I made an error when creating the equation for this post. The linear power equation can be derived similarly with the linear motion equations.

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Can Twistweight be Too High?

I was watching Episode 42 of the Pickleball Effect podcast while assembling SW1s today, and Braydon asked if it was possible for twistweight to be too high. He went on to mention that he felt that a high twistweight slowed down the acceleration of his backhand flick. I’m a bit surprised that he was able to notice it, but that is an expected effect of higher twistweight.

Consider mass added at the 4 and 8 o’clock positions of a paddle and reference the sketch below. The effect of that mass (m) on swingweight is m·a². For example, if adding 6 g (3 g per side), and a is 18 cm, then the swingweight will increase by (0.006 kg)·(18 cm)² = 1.944 kg·cm². The effect of that mass on twistweight is m·b², so if b is 9.5 cm, the twistweight will increase by (0.006 kg)·(9.5 cm)² = 0.542 kg·cm².

There is a third axis about which moment of inertia is interesting. This has conventionally been called spinweight, and it’s measured about an axis 90 degrees from the swingweight axis. In the sketch above, it would be about an axis coming out of the screen through the vertex a-c. It’s also the axis you’d get by installing a paddle into a swingweight machine with the paddle face parallel to the ground (more on that after the break). The effect of the added mass on spinweight is m·c². Length c is √(a² + b²) = 20.35 cm, so the added spinweight is (0.006 kg)·(20.35 cm)² = 2.485 kg·cm².

Note that the sum of the added swingweight and twistweight is equal to the added spinweight: 1.944 + 0.542 = 2.486 kg·cm². This will always be the case, assuming a paddle is approximately planar, and is described by the perpendicular axis theorem.

Back to Braydon’s backhand flick, there’s a large component of acceleration about the spinweight axis for this shot. Given two paddles, with everything equal except that one has a twistweight of 6.0 kg·cm² and the other has a twistweight of 7.0 kg·cm², the second paddle will have a spinweight that is 1.0 kg·cm² higher. The difference is there, but it’s small.

I wanted to add a bit more about measuring swingweight and spinweight in the real world. Theoretically, you could measure both swingweight and spinweight with the SW1. Practically, for paddles, the results are a bit misleading, as the effect of air resistance is very significant in the swingweight orientation. For example, my current main paddle has a swingweight of 114.3 kg·cm² and a twistweight of 7.5 kg·cm², but the spinweight measures 117.7 kg·cm². That’s lower than the expected result of 114.3 + 7.5 = 121.8 kg·cm², but that’s because the swingweight measurement is inflated due to the effect of air resistance. To minimize the effect of air resistance on swingweight, you could measure spinweight and subtract twistweight. In my case, that gives a result of 110.2 kg·cm². Of course, that’s only useful if comparing to paddles measured similarly.

If you’re interested, I have a bit more about the relationship between swingweight, twistweight, and spinweight for tennis racquets in the post Racket Twistweight from Spinweight and Swingweight.

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New SW1 App Features in Open Beta

I recently improved the calibration process in the Briffidi SW1 app. I also added saved tare values to make switching between adapters easier. These features are available to try now in open beta versions of the apps. Install the appropriate app by following one of these links: iPhone Open Beta or Android Open Beta.

Calibration Process

The calibration process for the Briffidi SW1 is confusing to many users. It’s unintuitive and was heavily influenced by the required app development effort. In my defense, I didn’t know if I’d sell many SW1s, and there were many other things to do to release a product. It worked.

Recently, I spent some time figuring out a more intuitive calibration process. Instead of creating measurement groups and taking specific measurements in each group, the calibration measurements are taken directly from the Calibrate tab in the app. There are sections for each configuration of the calibration rod, and each section incudes a dedicated Measure button and a dedicated measurement group.

When there is at least one measurement in each calibration group (I recommend at least two measurements of each), the Calibrate button will become active. After the Calibrate button is tapped, the Calibration Results below will update, and a confirmation will be displayed. If the calibration results are outside of normal ranges, the confirmation will indicate that, the abnormal result will be highlighted in red, and possible solutions will be displayed below. For example, users commonly extend only three of the four internal sections of the extendable calibration rod. When this happens, the Spring Constant result will be abnormally high. The results and confirmation display as shown below.

Things to check are displayed below the results.

Saved Tare Values

As I used the SW1 more with the twistweight and pickleball adapters, I often found myself forgetting to tare out the adapter before mounting a racquet or paddle. To make this process easier, the three latest Tare values are available for recall. When switching back to an adapter you’ve previously tared-out, long-press the Tare button to select the appropriate tare value. Note that the saved tare values are cleared during calibration.

Additionally, it wasn’t always clear that the Tare function was active. Now, when active, in addition to the button being filled in blue, the button text will indicate the value being subtracted from the measurement result.

Feedback Requested

If you try out a beta app and have any problems or suggestions for further improvement, please let me know in an email to support@briffidi.com.

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How Does the Joola “Propulsion Core” Work?

How does the Joola “Propulsion Core” design, with a strip of soft foam around the top/sides, work to increase pop/power?

Since I find it interesting, I thought I’d write a bit about my hypothesis. I have a mechanical engineering degree, but I never took any vibration courses. I probably had some basic introduction, but mostly, I’ve just picked things up from reading papers about coefficient of restitution and the like. I’m far from an expert in this area, and I’d be grateful for feedback from those with more knowledge.

First, a paddle has a first bending mode (FBM) that is similar to that of a free-free beam. The animation below shows the FBM of a simulated, typical paddle. The brightly colored areas indicate the nodes, or places where displacement is minimal, of this vibration mode. You can find these nodes on a paddle by holding the paddle loosely between your thumb and forefinger at approximately the node on the handle. Let the paddle hang vertically and tap up and down the face. The paddle will vibrate more or less, depending on where you strike it. The vibration will be minimal at the node.

The FBM node is a “sweet spot” and, based on my reading about tennis racquets, the most important “sweet spot”. When a ball is struck at the node, no energy is converted into vibration of the FBM, so more energy is returned to the ball. The vibration of the FBM is easy to feel, so a player will naturally attempt to hit the node.

The FBM is too “slow” to return energy to the ball. If a ball is struck between the nodes, you might expect the paddle to act like a trampoline and provide extra energy to the ball as it vibrates back into the ball. It won’t happen, though, as the ball has already left the paddle before it cycles back.

There are vibrations modes that are sufficiently “fast” to return energy to the ball. There is a membrane mode of the paddle face that is approximately centered on the paddle face, as shown in the animation below. The frequency of this membrane mode is probably the most important value for the “trampoline effect” of a paddle. I don’t yet have any data to back this up, but I suspect that on thicker (16+ mm) paddles, the frequency is higher than optimal for energy return, and it becomes more optimal as the thickness, and thus frequency, decrease.

So, how does the foam around the top and sides of the Joola “Propulsion Core” create more power?

In the vibration animations above, notice that the location of the FBM node is higher up the paddle face than the center of the membrane mode. As a result, if you hit the FBM node, you avoid energy loss to FBM vibration, but you don’t get the full benefit of the membrane mode. If you hit the center of the membrane mode, you get the full effect of that mode, but you lose some energy to FBM vibration. What if the two were aligned? I think that’s what the soft foam of the propulsion core achieves. Because the rigid honeycomb portion of the core is shorter, the FBM node will shift down the paddle face and be better aligned with the center of the membrane mode. The new, better-aligned “sweet spot” will also be closer to the paddle center of gravity than a typical “sweet spot” would be, resulting in a higher effective mass for the collision. The result of all that is more energy returned to the ball.

Is that all good news? I’m not sure. When the FBM node and the center of the membrane mode are misaligned, it elongates the area where energy returned is approximately equal. That misalignment may be better for consistency, as if you play anything like I do, you don’t hit the same spot on the paddle every time.

What do you think? I’ve not touched one of these paddles yet. If you have one, does the FBM node seem lower on the paddle face than on a typical paddle? Does it feel like the power falls off more quickly than a typical paddle if you hit near the tip of the paddle? I’d expect it to.