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

The force of a baseball bat swing striking a ball is tremendous. The force is even greater when you consider Mike Trout hits the ball.

We’re going to calculate the average force when Mike Trout blasts a line drive in the big leagues.

I find it amazing how much force a human can generate through hitting a ball by swinging a bat.

Calculation variables for calculating average force on the bat during the collision

Let’s go over the variables in our calculation.

First, we’re going to focus on the best of the best in the MLB. So, let’s look at batting stats from MLB stars.

According to Insider Baseball, the average MLB baseball exit velocity is 89 mph.

While Mike Trout, one of the top MLB players, had a ball clocked with an exit velocity of 118 mph on a line drive.

We’re talking about a baseball traveling 118 miles per hour after its hit with a bat. That’s insane!

Now, according to the MLB, an average fastball pitch is 92.3 mph. We’ll assume this pitch speed in our calculation.

Next, several other important variables we need to consider:

  • A baseball weighs: 5.125 ounces (0.145 kg)
  • The baseball will have a contact time of 0.7 milliseconds (0.0007 seconds) with the bat. Quite surprising how small the collision time between a baseball and bat actually is.

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

Important Note: I’m going to make several assumptions in my calculation. Because frankly, there’s a lot of data I simply don’t have on Mike Trout’s amazing hit. Plus, I’m missing data on the pitcher. 

Calculating the average force on the baseball bat during the collision 

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

To start, we first calculate the change in the ball’s momentum. This will then allow us to calculate the average force on the bat in the ball and bat collision.

Important Note: there’s an exchange in momentum when a ball and bat collide. 

A ball thrown by a pitcher has a momentum calculated by mass x velocity. At the same time, a batter converts potential energy into kinetic energy by swinging a bat. The bat now has a velocity, and thus momentum. 

Finally, when the bat strikes the ball, the ball and bat exchange momentum. The total momentum remains equal though. This is because of the Law of Conservation of Momentum. 

The change in the ball’s momentum

The momentum equation is p = mv, where the variables are:

p = momentum
m = mass
v = velocity

Now, a baseball moving along a straight line reacts to a force F = F(t). The force is a continuous function of time.

We’ll first show the change in momentum over the following time interval: (t_{0}, t_{1}) equals the integral of F from t_{0} to t_{1}. In other words, we want to show the below mathematical statement is true.

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

This integral is the impulse of the force over our time interval.

Let’s do some math!

F = ma = m\dfrac{dv}{dt}

Using the substitution rule, we can substitute into our integral.

\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})

Now, let’s 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

Notice the positive and negative signs with our values. The positive sign shows the baseball moving away from the batter after the hit. The negative sign shows the baseball moving towards the batter after the pitch.

Now, the mass of the baseball we need to convert to slugs.

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 [173.1 - (-135.4)] = 3.05 slug-feet

Calculating the average force on the bat

We now finally calculate the estimated average force that acts on the bat in Mike Trout’s line drive hit.

We’ll use our integral from earlier and the change in the ball’s momentum we calculated.

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

Thus, 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 recorded the fastest pitch in MLB history. His pitch clocked in at 105.1 mph.

Let’s now see how this would impact the force on the bat when Mike Trout swings.

According to Dr. Alan Nathan, every 1 mph pitch leads to a 0.2 mph increase in exit velocity.

I know this is an overgeneralization, but it’s the best data I have. You need to keep in mind, the faster a pitch is, the harder it is for a batter to square up too.

In other words, it simply becomes too difficult for the batter to hit the ball.

Thus, with lightning pitches, ball exit velocity sometimes even goes down. But, I’m assuming Mike Trout makes the same contact with the ball, as he did in our first calculation.

Also, I’m also assuming the ball contact time will remain to be 0.0007 milliseconds.

I’ve made a lot of assumptions as it is. Thus, this makes it’s more acceptable to use Dr. Alan Nathan’s exit velocity math. Because it’s simply the best data I have.

With that out of the way, our pitch speed increased from 92.3 mph to 105.1 mph. That’s an increase of 12.8 mph.

Thus, per Dr. Alan Nathan, our exit velocity increases by 2.56 mph. So, adding 2.56 mph to 118 mph gives us 120.56 mph. We’ll round up to 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, the mass of the baseball we need measured in slugs.

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.

As a comparison, I’ve calculated the jumping force of Zion Williamson’s vertical jump. It calculated 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. I’d go as far as labeling him as one of the top athletes of our generation.

The amount of force he generates in his hits is tremendous. Then throw in the belief that hitting a baseball is the hardest skill to pull off in any sport.

It all blows your mind!

In short, Mike Trout has the perfect balance of great hitting mechanics and sheer power.

What’s more, our math exercise shows how the pitch speed greatly affects the force generated by a hit.

I can’t forget to mention, the other great variable is the bat swing velocity. This wasn’t a controlled variable in our calculation though.

In the end, baseball is a whole lot of amazing physics. It’s one reason I think it’s such a great 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)

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