You don’t have to be a scientist to understand that it’s easier to flip a tire than lift it completely. But how much easier? While some will shrug off the question, inquiring minds want to know, so we had Brittney Saline link up with fitness experts and physicists to take a closer look at tire flipping, turnarounds in run courses, sled physics and fat bars. This is not the high science of a physics lab but rather a rough link between that lab and the guy who can’t figure out why he’s struggling to push a sled on a really hot day. Call it “gym science.” This is Part 2 of 4.

The tire flip: It’s classic backyard CrossFit.

Big, awkward and dirty, tires are all about one thing: moving heavy shit. And as impressive—or intimidating—as a tire three times your body weight might be, you’ve got physics on your side.

The hardest part is getting started: It’s you against inertia. You place your feet in a squat stance, about a foot between your toes and the tire. Dropping your hips, you wedge your fingers beneath the tire. But don’t think of the tire as weighing 550 lb.—with the floor supporting the tire as well, you need only lift half the weight.

ALT TEXTPhysics will confirm what you’ve felt: It requires more force to move the tire when it’s flat on the ground.

Think of the tire as a lever. It’s a class 2 lever for all you nerds, where the end on the floor is the fulcrum, the weight being lifted—or the center of gravity—is smack in the middle of the tire, and the lifting force—you—is on the other end. While gravity pulls down on the tire from its center, you get its entire diameter as a lever.

“So the easy way to look at it (is) since you are pulling from twice the lever arm, you only have to lift half as much,” said Steve Schuh, a physics instructor at Ball State University and the Indiana Academy for Science, Mathematics and Humanities.

The amount of weight you actually lift decreases the higher you lift the tire, or as the angle between the tire and its upright position decreases (assuming you consider the prone tire’s starting point to be at a 90-degree angle in relation to true vertical). So how much weight are you actually lifting? In other words, how much force do you need to apply at any given point in the lift?

It’s all about torque. Torque is an effort to rotationally accelerate something, Schuh said. In this case, you’re trying to rotate the tire around the pivot point—the end still on the floor. You need to apply torque to lift the tire, while the weight of the tire itself needs torque to fight against you. Here’s how we calculate the force you need to lift it, assuming the tire weighs 550 lb. and is 1.5 meters (about 60 inches) in diameter. (Because the force you are applying is perpendicular to the weight, the angle in consideration is 90 degrees.)

  • Torque = lifting force x lever arm x sin(θ)
  • θ is the angle between the lifting force and lever arm.
  • The tire’s torque is equal to its weight x 0.75 meters x sin(90°)
  • Lifting force x 1.5 meters sin(θ) = 550 lb. x 0.75 meters x sin(90°)
  • The sine terms cancel out because they both equal.
  • Therefore, lifting force = 550 lb. x 0.75 meters/1.5 meters, which equals 275 lb. in this case, or 50 percent of the tire’s total weight.

ALT TEXTAt 45 degrees, much less force is needed to move the tire.

As long as you keep applying perpendicular force as you lift the tire—think more offensive linemanthan deadlift—the angle of your applied force will remain 90 degrees. However, the law of gravity dictates that the weight of the tire continues to push downward, giving us a new formula as the tire rises (let’s assume the angle between the tire and true vertical is now 45 degrees).

  • Lifting force x 1.5 meters x sin(90°) = 550 lb. x 0.75 meters x sin(45°)
  • Therefore, lifting force = 194.4 lb., or about 35 percent of the tire’s total weight.

By the time the tire reaches true vertical, the tire’s torque is zero, and “you can practically spit on this thing and knock it over,” Schuh said.

ALT TEXTAs the tire approaches vertical, almost all the work is done.

Still find tires daunting? Let’s take a look at how much energy is actually required to flip that baby over (energy, measured in joules, is the capacity to do work, which is expressed as force multiplied by distance).

The force you apply moves the tire’s center of gravity from the ground—in reality, from a point in the very center of the tire, where the midpoint of the tire’s diameter meets the midpoint of the tire’s width. When you lift the tire, you transfer energy from yourself to the tire, and to calculate how much energy that is, we must find the tire’s new potential energy. Here’s the equation, courtesy of Ed Bacher, former physics instructor at Trinity-Pawling School in Pawling, New York, and The Lawrenceville School in Lawrenceville, New Jersey:

  • Change in potential energy = mg (0.5 tire diameter – 0.5 tire width)
  • m = the mass of the tire and g = the acceleration of gravity (9.8 meters/second²)

Our 550-lb. tire is about 249.5 kg, and let’s say it’s 0.6 meters (about 22 inches.) in width. Plug all that in and you get 1,100 joules, or about 0.26 calories. For most people, that’s less than a single pull on the ergometer.

Reference: “Tire Technique”

ALT TEXTThe trigonometry of moving heavy stuff.

You can do all the math in the world, but whether you can flip that tire over depends largely upon your strength and technique. But if you’re a dedicated CrossFit athlete with a good coach, it’s likely you’ve done it before and will do it again even if you don’t appreciate the equations behind your labor.

“Humans instinctively understand physics,” Schuh maintained. “We just don’t instinctively understand the math.”

About the Author: Brittney Saline is a freelance writer contributing to the CrossFit Journal and the CrossFit Games website.

Credit: www.journal.crossfit.com