Thaksha Kulahan, Bob Tian, Kai Yang Page 1 of 16 Mathematical Model: Finding the Sweet Spot of a Baseball Bat Thaksha Kulahan Bob Tian Kai Yang University of Toronto Our approach to the problem is the conservation of energy. The sweet spot should be the location where maximum energy is transferred to the ball so that the ball attains best performance. Although hitting at the end gives you the greatest torque that gives you greatest force and greatest acceleration but high torque also means high rotational force on the handle which causes the handle to shake. Since hitters are human, their arms cannot be perfectly rigid when the ball hits at the end of the bat, so there will be some energy loss due to the movement of the bat because the ball does not hit on the center of percussion of the bat. Also the end of the bat is an anti-node of the bat which means hitting on the end of bat causes the greatest vibration, which is another place you lose energy. Given a math model, our goal is to find the center of percussion where the movement of the bat is minimized and a point near the node of fundamental mode of the bat where the vibration of the bat is minimized. And hence, the sweet spot should be in between the center of percussion and the point to minimize vibration so that the total energy loss it minimized. We will find the sweet spot first in terms of different models and then we will discuss what corking would do to our model and the impact of different material of the bat. I) The location of the sweet spot 1) A simple model Thaksha Kulahan, Bob Tian, Kai Yang Page 2 of 16 We would start with an easier model to illustrate our idea. After a close inspection on how batter hits the ball, we observe that the pivot of the bat is the wrist of the batter. Since the part of the bat below the pivot is negligible, we would just consider the part of the baseball bat above the pivot. Let us take our bat as a uniform cylinder with length L and radius R and density ρ, centered at the origin which is the pivot. When the ball strikes the bat, it will cause two motions in the bat depending on the point it is struck. The rotational velocity rω, and the translational velocity u. The two velocities are against each other, and so the difference of the two will be the final velocity of v the bat due to the ball. We can account for the energies lost due to this velocity v. Here, r is the distance of the centre of gravity to the pivot. v = u – rω Differentiate it with respect to time t: ① d is the distance from the pivot the point of interaction. F is the force between the baseball and the bat. Torque dF is moment of inertia times angular acceleration . Hence we have: ② According to Newton’s second law, where m is the mass of the bat. ③ Sub ② ③ into ①, we have Integrate on both sides, we have is the impulse experienced by the baseball. Thaksha Kulahan, Bob Tian, Kai Yang Page 3 of 16 When the difference between the rotational and transitional velocity is 0, this portion of kinetic energy is not dissipated. And d* that make v 0 is the distance from the pivot to the center of percussion. To derive the formula for center of percussion, we let the left hand side to be 0. Hence, Calculation of moment of inertia: where is the distance from the pivot to a point in the cylinder. We evaluate the triple integral in cylindrical coordinate, For a uniform rod the center of gravity would be just at L/2. The energy lost to shaking the bat Ec by hitting the ball at position x would be: The node of fundamental frequency of a uniform cylinder would be the node of the second harmonic would be , since the first two harmonics capture the majority of the energy of the vibration we simply ignore energy of harmonics of higher orders. The mid-point of the node of fundamental mode and second harmonic is , but the point to minimize vibration should be closer to the node of fundamental mode. Hence the point where the vibration of the bat caused by collision is minimized would be in the interval . Since the boundaries are really close, we might take the mean of the interval as the point to minimize vibration, but Thaksha Kulahan, Bob Tian, Kai Yang Page 4 of 16 where exactly the point to minimize vibration of the bat is left to be examined. Ideally we would be able to figure out mathematically how much energy will be lost if we do not hit the ball at but at point x. Suppose we can derive this function f(x), then We differentiate to find the critical point where energy lost is minimized. This critical point is the sweet spot according our mathematical model. Since f(x) is intractable for now this work cannot be done. But what we are certain according to our model, the sweet spot lies in the interval of . 2) A more complex model To simulate a baseball bat with a cylinder seems to be too inaccurate because a baseball bat has a much bigger end than the handle. We tried various models, but the only one that is still simple enough to present analytically is a combination of two cylinders. (actually this model is already very complicated) The radius of the bottom cylinder is r1 and the radius of the upper cylinder is r2. The length of the bottom cylinder is a1 and the length of upper cylinder is (a2-a1). The entire object has uniform density of ρ, centered at the origin which is the pivot. Using the same logic, we want to figure out the center of percussion and the mode of fundamental frequency, and try to minimize the energy dissipation, but the calculation is a lot more tedious. The formula used to calculate center of percussion does not change, to keep consistent we still use d* to denote center of percussion, r is the distance of the centre of gravity to the pivot, m as the mass of the object. Thaksha Kulahan, Bob Tian, Kai Yang Page 5 of 16 Calculation of moment of inertia: where is the distance from the pivot to a point in the cylinder, and V1 is the bottom(RHS) cylinder while V2 is the upper cylinder. We evaluate the triple integral in cylindrical coordinate, For this object, the center of gravity is much harder to calculate. where is the distance from the pivot to a point in the cylinder Where v1 is the bottom cylinder and v2 is the upper cylinder. With the help of computer, we are still able to calculate the integral, Thaksha Kulahan, Bob Tian, Kai Yang Page 6 of 16 m is easy to calculate: Although tedious we are still able to get the center of percussion which better simulates a baseball bat: As long as we hit the baseball at center of percussion, we could minimize the energy loss due to the translation and rotation of the bat. But unfortunately, we were not able to derive formulas to find out the nodes of the fundamental mode of this object. But what we are certain is that the sweet spot should still be somewhere even closer to the center of percussion because now the mass of the bat is more concentrated on the top and the top is much shorter than the thin handle. 3) A much more precise numerical model Thaksha Kulahan, Bob Tian, Kai Yang Page 7 of 16 We construct a much more precise numerical model from a 30-inch Little League wood bat. The barrel is a cylinder with radius of 1 inch and the handle is of radius 0.5 inch. The bat consists of three parts, the barrel which is 8 inches long, the difference of two cones which is 8 inches long and a handle which is 14 inches long. First let us consider the point where vibration is minimized. Thanks to Dr.Russell1, We have mode shapes for this 30-inch Little League wood bat. Fundamental bending mode (215 Hz) Second bending mode (670 Hz) From his experiment, it is pretty obvious that the node of fundamental bending is at approximately 7 and that of second bending mode is at 5. We are certain that the point to minimize vibration is in the interval of [6,7]. Take the mean of this interval, vibration-minimizing point should be approximately 6.5. Now we want to calculate the center of percussion of this particular bat. We have seen that the location of center of percussion does not depend on the material of the bat. Without losing generality, we assume the density of the bat to be 1 unit. 1 Cited from http://paws.kettering.edu/~drussell/bats-new/bend-sweet.html Thaksha Kulahan, Bob Tian, Kai Yang Page 8 of 16 The formula for center of percussion still applies: Calculation of moment of inertia: where is the distance from the pivot to a point in the cylinder, and V1 is the barrel while V2 is the conical frustum and V3 is the handle part. To solve the second triple integral we fix z first do the double integral of the circle and then integrate with respect to z, Using MATLAB, it is very easy to evaluate these numerical integrations.