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.
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.
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.
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):
The first two terms are zero. Solving for α, we get:
Then, an equation for angular velocity (ω) is:
The first term is zero, so substituting α from above, we get:
An equation for kinetic energy (K) is:
Substituting ω from above, we get:
An equation for average power (Pavg) is:
Substituting K from above, it becomes:
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.
