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 (P_{avg}) 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.

Let me see if I can convince you that the power equation is in error.

1. The power equation uses as its basis the recoil weight RW. The recoil weight is also known as the inertia at the center of mass Icom. We know that Icom or RW remains the same regardless of where the paddle is gripped. Whether a user chokes up on the paddle or grips it near the butt Icom and RW remain the same and therefore the power calculation remains the same. But we know from experience that choking up on the handle allows faster rotation and/or lower power. Thus, something seems amiss with the power calculation.

2. In a previous post I mentioned sword design. The pickleball paddle can be considered a sword and the same equations of motion apply. There is an instructive paper on sword design here:

https://armor.typepad.com/bastardsword/sword_dynamics.pdf

I’m looking at the diagram on page 24 where the linear and rotational forces are defined as:

F=mrα

Mc=(Icom-mrx)α

Let’s move the desired point of rotation to handle meaning r=-x. The two equations distill down to:

F=-mxα

Mc=(Icom+mx^2)α or boiling it down further

Mc=SW*α where SW is the swing weight (which changes with grip location).

There are two forces in play. One force is a torque that would normally rotate the paddle around the center of mass. The second linear force is necessary to move the point of rotation from the center of mass to the handle. If I’m interpreting the equations correctly the forefinger and the pinky provide a couple for the rotation and both share in providing the linear force.

In response to #1, the center of the hand is assumed to be 5 cm from the end of the handle (location of the SW axis). If you choke up on the handle, you can increase the “0.05 m” term in the linear power formula, which will reduce the power. The angular component won’t be affected. Of note, you wouldn’t change the “0.05 m” in the angular power formula, as that’s based on the measurement axis.

For #2, I didn’t read all of the context, but I believe that diagram is for rotating the sword around a point that differs from the CoM. That’s different, because in my calculations, I don’t care about rotating about a certain point. The player can achieve the angular rotation and linear translation is whatever way is most efficient. The CoM is allowed to translate fore and aft during the motion.

Can you point to an article that derives the equations? Is the equation taken from sword design?

I have no knowledge of sword design. I added a derivation of the angular power formula.

I guess replies can only go four levels deep.

I started by using SW but realized that was wrong. If you apply only a torque to an object, no matter where it’s applied, the object will rotate about its center of mass (CoM). The SW1 applies a torque to the paddle PLUS forces required to keep the rotation axis location fixed. It won’t work if there isn’t enough friction between its feet and the surface below.

By breaking the motion into angular and linear components, I’m handling the torque and the forces separately. If using SW for the angular component, the mass at the CoM is double-counted.

Of note, the linear component only considers lateral forces. The fore/aft forces are ignored, as it would be acceptable for your hand to move fore/aft during the motion.

I’m having problems with John’s assertion that adding a Slyce Speed Cap (~1 oz) to a paddle will decrease the power needed to swing the paddle and increase hand speed. We know that adding weight to a paddle at any place away from the pivot point will increase the inertia and therefore increase the power and time needed to do a 180 degree rotation. Do you agree?

I agree that it will increase the moment of inertia about the pivot point, but I don’t agree about the power requirement. I was surprised that it worked out the way it does. For short distances, the power requirement goes down. As the distance increases, that effect goes away. If there’s a flaw in my reasoning above, I’m happy to discuss.

Love to discuss! My thinking is that the MOI goes up so any rotational motion takes more power. Also the static weight increases so any translational motion requires more power.

What’s your thinking? Do you have equations or measurements that show something different?

The rotation does take a bit more power. However, the balance point moves down, so the center of mass does not have to translate as far. The mass goes up, but the reduced distance more than makes up for it. Equations? They’re right above in the post.

Why use the inertia around the balance point (recoil weight, RW) to determine the required power? The equation above is valid if trying to determine the power required to rotate the paddle around the balance point. But the racquet is not rotating around the balance point. It’s rotating around the pivot point and the inertia and required power is defined, I believe, by the swing weight not the recoil weight. Let me think on this a bit and derive the equation for power for myself.