What’s the force of a baseball bat swing by Mike Trout?

by

The sheer force of a baseball bat swing striking a ball by Mike Trout, one of the MLB’s elite players, is tremendous.

Join me as we dive into calculating the average force generated when Mike Trout smashes a line drive in the big leagues. It’s truly astounding to see the immense force a human can produce by simply striking a ball with a bat.

Calculation variables for determining average force on a bat during a collision

First and foremost, we’ll be concentrating on the crème de la crème of MLB players. So, let’s dive into some batting stats from MLB superstars.

According to Insider Baseball, the average exit velocity for a baseball in the MLB is 89 mph. In contrast, Mike Trout, a top-tier MLB player, has recorded a ball with an exit velocity of 118 mph on a line drive!

Moving on, the MLB states that an average fastball pitch clocks in at 92.3 mph, which we’ll use for our calculation. But that’s not all! We need to consider some crucial variables, such as:

  • A baseball weighs: 5.125 ounces (0.145 kg)
  • The contact time between the baseball and the bat is a mere 0.7 milliseconds (0.0007 seconds). It’s astonishing how brief the collision time between a baseball and a bat is.

Now, let’s crunch some numbers to calculate the average force in a bat and ball collision.

Important Note: To perform the calculation, I’ll make several assumptions since I do not have all the necessary data for Mike Trout’s remarkable hit. Additionally, there is also missing data on the pitcher that we need to consider.

Calculating the average force on the baseball bat during the collision

mike trout hitting baseball in game
Mike Trout swinging (Photo Credit: Erik Drost)

Our first step is to calculate the change in momentum of the ball. Once we’ve got that, we can figure out the average force exerted on both the bat and the ball during their collision.

Important Note: During a collision between a ball and a bat, momentum is exchanged. The momentum of a pitched ball is calculated by multiplying its mass by velocity. Similarly, a batter converts potential energy into kinetic energy by swinging a bat, giving the bat its own momentum. When the bat makes contact with the ball, the two exchange momentum while the total momentum remains constant, in accordance with the Law of Conservation of Momentum.

The change in the ball’s momentum

The momentum equation, p = mv, defines momentum as a product of mass (m) and velocity (v). When a baseball moves along a straight line, it reacts to a continuous force F = F(t), which is a function of time. The change in momentum over the time interval (t_{0}, t_{1}) is equal to the integral of F from t_{0} to t_{1}, as shown below:

p(t_{1}) - p(t_{0}) = \int_{t_{0}}^{t_{1}} F(t)dt, where the integral represents the impulse of the force over the time interval.

To calculate the change in the ball’s momentum, we use the formula F = ma = m\dfrac{dv}{dt}, where (a) is acceleration and (v) is velocity. Substituting this into the integral, we get:

\int_{t_{0}}^{t_{1}} F(t)dt = \int_{t_{0}}^{t_{1}} m(\dfrac{dv}{dt})dt = m\int_{v_{0}}^{v_{1}}dv

= (mv)|_{v_{0}}^{v_{1}} = mv_{1} - mv_{0} = p(t_{1}) - p(t_{0})

Next, we calculate the change in the ball’s momentum:

v_{1} = 118 \: mph = \dfrac{118 \: mile}{1 \: hour} \times \dfrac{5280 \: feet}{1 \: mile} \times \dfrac{1 \: hour}{3600 \: seconds} = 173.1 \: feet/second

v_{0} = -92.3 \: mph = \dfrac{-92.3 \: mile}{1 \: hour} \times \dfrac{5280 \: feet}{1 \: mile} \times \dfrac{1 \: hour}{3600 \: seconds} = -135.4 \: feet/second

Note the positive and negative signs of the velocities. The positive sign shows the baseball is moving away from the batter after the hit, and the negative sign shows the baseball is moving toward the batter after the pitch.

Finally, we convert the mass of the baseball to slugs, where 1 oz = 0.00194 slugs:

\dfrac{5.125 \: ounces}{1} \times \dfrac{0.00194}{1 \: ounce} = 0.0099 slugs

Using this mass and the velocity values, we calculate the change in momentum to be:

mv_{1} - mv_{0} = m(v_{1} - v_{0}) = p(t_{1}) - p(t_{0}) = 0.0099 \times [173.1 - (-135.4)] = 3.05 slug-feet

Calculating the average force on the bat

Let’s now calculate the estimated average force that acted on the bat during Mike Trout’s legendary line drive hit.

Using our integral from earlier and the change in the ball’s momentum we calculated, we get:

\int_{t_{0}}^{t_{1}} F(t)dt = \int_{0}^{0.0007} F(t)dt = p(0.0007) - p(0) = 3.05

This means that the average force over the 0 to 0.0007 second interval is:

\dfrac{1}{0.0007} \int_{0}^{0.0007} F(t)dt = \dfrac{1}{0.0007} \times (3.05) = 4,357.1 pounds of force.

What happens if Mike Trout hits the fastest recorded pitch speed?

mike trout hitting ball
Mike Trout hitting ball (Photo Credit: Erik Drost [image cropped])
On September 24th, 2010, Aroldis Chapman made history by throwing the fastest pitch ever recorded in MLB at a whopping 105.1 mph.

Now, let’s imagine the sheer force unleashed on the bat as Mike Trout takes a swing at this blazing fastball.

Dr. Alan Nathan suggests that for every 1 mph increase in pitch speed, the exit velocity jumps by 0.2 mph. Granted, this may be an oversimplification, but it’s the best information available. Keep in mind, as the pitch speed increases, it becomes increasingly challenging for a batter to make solid contact with the ball.

That being said, with lightning-fast pitches, the ball’s exit velocity can sometimes even decrease. However, we’ll assume that Mike Trout maintains the same contact with the ball as in our initial calculation.

We’ll also assume that the ball contact time remains at 0.0007 milliseconds.

Given the number of assumptions already made, it seems reasonable to rely on Dr. Alan Nathan’s exit velocity formula, as it’s the best data at our disposal.

So, with that settled, our pitch speed has surged from 92.3 mph to 105.1 mph, an increase of 12.8 mph.

According to Dr. Alan Nathan, this would result in an exit velocity boost of 2.56 mph. By adding 2.56 mph to the initial 118 mph, we arrive at 120.56 mph. Rounding up, we get an exhilarating exit velocity of 120.6 mph!

Calculating the average force on the bat

Let’s now plug in our new value to calculate the average force on the bat.

v_{1} = 120.56 \: mph = \dfrac{120.56 \: mile}{1 \: hour} \times \dfrac{5280 \: feet}{1 \: mile} \times \dfrac{1 \: hour}{3600 \: seconds} = 176.8 \: feet/second

v_{0} = -105.1 \: mph = \dfrac{-105.1 \: mile}{1 \: hour} \times \dfrac{5280 \: feet}{1 \: mile} \times \dfrac{1 \: hour}{3600 \: seconds} = -154.1 \: feet/second

Again, we need the mass of the baseball measured in slugs, where 1 oz = 0.00194 slugs.

\dfrac{5.125 \: ounces}{1} \times \dfrac{0.00194}{1 \: ounce} = 0.0099 slugs

mv_{1} - mv_{0} = m(v_{1} - v_{0}) = p(t_{1}) - p(t_{0}) = 0.0099 \times [176.8 - (-154.1)] = 3.28 slug-feet

\dfrac{1}{0.0007} \int_{0}^{0.0007} F(t)dt = \dfrac{1}{0.0007} \times (3.28) = 4,685.7 pound-force.

So, this increase in pitch speed results in an added 328.6 pound-force.

If we convert 4,685.7 pound-force to Newtons, we get 20,843 Newtons.

For comparison, I’ve calculated the jumping force of Zion Williamson’s vertical jump. It amounts to 5,134.25 Newtons.

Clearly, a bat and ball collision generates an insane amount of force!

Conclusion

There’s a reason why Mike Trout is one of the top players in the MLB, and I’d even venture to call him one of the most outstanding athletes of our generation!

The sheer force he generates with his hits is nothing short of astounding. Add to that the widely-held belief that hitting a baseball is the most challenging skill to master in any sport, and it’s enough to leave you in awe.

In a nutshell, Mike Trout has struck the perfect balance between exceptional hitting mechanics and raw power.

Moreover, our mathematical exercise demonstrates how pitch speed significantly influences the force generated by a hit. And let’s not forget another crucial variable: bat swing velocity. Although it wasn’t a controlled variable in our calculation, it’s undeniably vital in the grand scheme of things.

Ultimately, baseball is brimming with incredible physics, which is one reason why I think it’s such an amazing sport!

What are your thoughts on the force of a baseball bat swing? Do you think the force of a baseball bat swing will increase moving into the future?

Featured Image Photo Credit: Erik Drost (image cropped)

SUBSCRIBE TO ENGINEER CALCS NEWSLETTER

Get daily articles and news delivered to your email inbox

Leave a Comment