“A pound of rotating weight is equal to ten pounds of static weight”. Or seven pounds. Or three pounds depending on who you ask. And they’d all be correct and incorrect. As with so many things in life, “it depends.” We are going to look into the concept of mass factor and figure out where you get the most bang for your rotating mass reduction buck. Isaac Newton and physics tells us force equals mass * acceleration (F=ma). This equation is basically the definition of inertia. The more mass there is, the greater the force is required to accelerate that mass. Therefore, more mass = more inertia.
When it comes to cars, a straight application of F=ma does not work because there are many round bits that spin such as wheels, tires, flywheels, clutches, pulleys, etc. The equation for making round things spin faster about an axis is torque = moment of inertia * angular acceleration. With vehicles, it gets complicated because we have to apply force to get it to accelerate in a direction plus apply force to get the round things to spin. There is a concept called mass factor which calculates an effective mass that takes into account the inertia of all the rotating parts. For example, a car may have a mass of 3000lbs, but when you take into account the inertia of all the rotating bits, it’s like trying to accelerate a car that weighs 4000lbs with no rotating parts. For those of us who like to accelerate faster, we’re going to figure out the best places to reduce mass of the spinning bits to gain the most acceleration and learn about the concept of mass factor.
Visualize a tire. If you tried to apply just F=ma in calculating how fast you could get the tire to accelerate along the ground, you’d end up with the wrong answer because you’d leave out the force required to also make the tire spin. We can use a yo-yo as a great example. If you just dropped the yo-yo, it would accelerate due to the force of gravity, f=mg where gravity (g) has a value of 9.81m/s^2. Now take that same yo-yo and let it fall down while unwinding from the string. The yo-yo will fall down slower because some of the force is going towards making the yo-yo spin. So, it’s the same with all the parts that spin on a car. We’re going to start off with a little bit of math and then crunch the numbers to see how much the rotating bits of a car contribute to the mass factor.
I’m borrowing this slide from my old professor Dr. Longoria at the U. of Texas at Austin from the course ME 379M/397 Cyber Vehicle Systems. It’s part of a presentation on vehicle performance modeling that you can download. It shows the math equations for the 2d motion of a torque driven wheel. The first equation is F=ma for the Force causing translation in the x direction. The dot over the ‘v’ basically means the first derivative; so, the derivative of a velocity profile/equation is acceleration. The second equation says that Force Z is equal and opposite of the force of gravity, so the wheel isn’t going up or down. So, no displacement in Z direction. The third equation says that Torque_d minus the torque created by Ftx * radius of the wheel (remember, torque = force * distance) is equal to the rotational inertia times the derivative of the angular velocity (= angular acceleration).
This slide shows a few of the rotating bits to be taken into consideration along with the equation for mass factor. Of importance, note that all the rotating parts that are spun up by the transmission (flywheel, clutch, engine) are multiplied by the Gear Ratio (GR) squared! Remember, gears are torque multipliers. As the equation is written in the slide, it calculates the effective mass which is equal to the vehicle mass (mv) plus the effective mass of all the rotating parts.
I made approximate (lots of assumptions…) calculations for the big rotating parts of Project S2000 and assumed a vehicle mass for 1350kg with driver based on my corner weight session.
A paper titled, “Problems of Rotational Mass in Passenger Vehicles” that you can download calculates a mass coefficient which just divides all the terms in the previous mass factor equation by the vehicle mass. Notice the flywheel has the most inertia of the engine connected parts because it’s fairly heavy and has a big radius. Inertia is a function of radius^2, so having a lot of mass out at a far radius exponentially increases the inertia. Comparing the tires to the wheels, the tires only weigh about a third more but have around three times the inertia because the mass is all hung out at a larger radius. The crank and rods account for much less inertia compared the flywheel.