How much force

[RIstar]

How much force does a pitcher need to create to throw a 80 and 90 mph fastball?

I've alway's wanted to know this does anyone know?

[fastbal95]

Assuming that you apply force for .2 seconds, or 2 tenths of a sec, it would take 6.6 lbs of force to release the ball at 132 ft/sec, which equals 90 mph. Now you apply force for less time, then it would take more force; apply force for a longer time period, then it would take less force.

velocity=force time the time period you appy that force. I went into detail about this in the thread called, Best Pitching Coach Ever.

[son who is sidearm]

i like this explanation better. a little easier to understand.

fasbal i admire your understanding of all that u know and have researched but sometimes speaking a little plainer is easier for the us nontechnical/intellicual type to absorb.

i would really like to say ty to everyone from a person who real just knew what i grew up on and watching in mlb games. i have learned alot and tried to apply what i have learned for my son and so far so good

[Chris O'Leary]

Assuming that you apply force for .2 seconds, or 2 tenths of a sec, it would take 6.6 lbs of force to release the ball at 132 ft/sec, which equals 90 mph. Now you apply force for less time, then it would take more force; apply force for a longer time period, then it would take less force.

velocity=force time the time period you appy that force.

Someone check my physics, but I don't think this math is quite right.

While there is some relevance to the idea of applying force over time (that is related to injury prevention), what really matters is the amount of force that is being applied at the release point. It's not as if the ball is able to accumulate the force that has been applied over time. Rather, the benefit of applying force over a greater period of time is that it is less stressful at any moment in time.

Think about the load on an engine if your goal is to get a car going 100 MPH. In case A, you have 10 seconds to get there. In case B, you have 20 seconds to get there.

You could argue that case B will place less stress on the engine.

[fastbal95]

Here ya go Chris,


In formula form, the law of acceleration is: a = F/m

Where:
(a) stands for acceleration which equals the velocity of baseballs at release (vr) divided by the time (t) pitchers apply toward-home-plate force.
(F) stands for straight-line toward-home-plate force.
(m) stands for mass which equals weight (wt) divided by gravity.

Baseballs weigh five and one-quarter ounces or 0.328 pounds and gravity approximates 32 ft/sec2. Therefore, the mass of baseballs equals 5.25/16/32 or 0.0102539 ft. lbs./sec2.

Substituting these two factors into the law of acceleration formula creates the release velocity formula for baseball pitchers.

1. Formula: a = F/m
2. Multiple both sides by mass (m): (m)(a) = (F)(m)/(m)
3. Because (m)/(m) = 1: (m)(a) = F
4. Substitute (0.01) for mass: (0.01)(a) = F
5. Substitute (vr) / (t) for a: (0.01) (vr)/(t) = F
6. Multiply both sides by time (t): (0.01)(vr)(t)/(t) = F(t)
7. Because (t)/(t) = 1: (0.01)(vr) = F(t)
8. Divide both sides by (0.01): (0.01)(vr)/(0.01) = F(t) / (0.01)
9. Because (0.01)/(0.01) = 1: (vr) = F(t)/(0.01)

The release velocity of baseball pitches equals the amount of straight-line toward-home-plate force that pitchers apply times the time period over which pitchers apply their forces divided by the mass of the baseballs or 0.01. Therefore, the variables that determine what release velocities pitchers achieve are their straight-line toward-home-plate force applications and the time period or distance over which they apply their forces.

My uniform acceleration study determined that pitchers apply force from leverage through release for approximately 0.2 seconds. If we assume that pitchers want release velocities of at least 90 miles per hour, then we can determine how much straight-line toward-home-plate force they must apply for two-tenths of second. To determine the feet per second of ninety miles per hour, we multiply 1.467 times 90 and learn that 90 mph equals 132 ft/sec. Therefore, for vr, we substitute 132 ft/sec.

1. Release Velocity Formula: (vr) = (F)(t)/(0.01)
2. Substitute known quantities: (132) = (F)(0.2)/(0.01)
3. Divide both sides by (0.2): (132)/(0.2) = (F)(0.2)/(0.2) / (0.01)
4. Because (0.2)/(0.2) = 1: 660 = (F)/(0.01)
5. Multiply both sides by (0.01): (660)(0.01) = (F)(0.01)/(0.01)
6. Because (0.01)/(0.01) = 1: 6.6 = F
7. Change sides: F = 6.6 lbs.

The Release Velocity Formula shows that when pitchers apply 6.6 pounds of straight-line toward-home-plate force for two-tenths of a second, they achieve release velocities of ninety miles per hour. To better understand the interrelationship between release velocities and force and application time, let us assume some other numbers.

When pitchers decrease their application time forces by five-hundreds of a second, the force required for pitchers to achieve ninety miles per hour release velocities increases to 8.8 lbs. Therefore, the application time directly influences the amount of force that pitchers have to apply to achieve their desired release velocities. For example, when pitchers uniformly apply 6.6 lbs. of force for 0.22 seconds, then they achieve release velocities of 145.2 ft/sec or 98.98 mph.