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

]]>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.

]]>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.

]]>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.

]]>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?

]]>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.

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