Simulating Inertia of a Vehicle

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SUMMARY

This discussion focuses on designing a brake inertia dynamometer to simulate the braking system of a vehicle. The required inertia value is approximately 2000 kgm/s, which corresponds to a quarter car model weighing 250 kg moving at a velocity of 8 m/s. The kinetic energy needed is calculated to be 8000 joules, and a tire diameter of 0.584 meters is assumed for calculations. The discussion outlines the process of determining the flywheel's size and mass through trial and error, using the relationship between kinetic energy and moment of inertia for a cylinder.

PREREQUISITES
  • Understanding of moment of inertia for cylinders
  • Basic principles of kinetic energy and momentum
  • Familiarity with vehicle dynamics and braking systems
  • Knowledge of rotational motion and angular velocity
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  • Calculate the moment of inertia for a cylinder using the formula I = (1/2) * m * r^2
  • Explore the relationship between kinetic energy and angular velocity in rotating systems
  • Research different methods for driving a flywheel in dynamometer setups
  • Investigate the effects of gear ratios on the inertia of rotating systems
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Engineers, automotive designers, and researchers involved in vehicle dynamics, particularly those focused on braking system testing and simulation.

Oli166
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Hi all

I am in the process of designing a brake inertia dynamometer to test one complete corner of a car's braking system.

I'm having some trouble with what size the spinning disc of metal, that will simulate the energy of the moving car, needs to be. I have the inertia value that I need to simulate (roughly 2000 kgm/s)

I also have a formula to calculate the moment of inertia for a cylinder but i have a feeling that these are two different things.

So any help with how I can calculate the size and mass of the disc and also how I would calculate the speed it needs to be rotated at.

Cheers

Oli
 
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A brake dynameter needs to simulate both momentum and kinetic energy. The braking forces are determined by the momentum and the energy absorbed is determined by the kinetic energy. There is not enough information given to solve this, so we need to make some assumptions.

Assume a small car that weighs only 1000 kg, so the quarter car model is 250 kg. If the momentum (not inertia) of the quarter car model is 2000 kg-m/s, then the velocity is 8 m/s, or just under 18 MPH. The kinetic energy is 8000 joules.

Next, we need to assume a tire diameter, so randomly pick a 155/80R13, which is 0.584 meters diameter (dimension from tirerack.com). At 8 m/sec, the tire is rotating 4.36 revolutions per second, or 27.4 rad/sec.

Next, simplify by assuming direct drive from flywheel to brake (no gear reducer). We need a flywheel that has 8000 joules kinetic energy at 4.36 rev/sec (262 RPM, 27.4 rad/sec). Since the kinetic energy of the flywheel is ##I\omega^2##, and you have both the kinetic energy and speed (27.4 rad/sec), you can solve for I. Then use the equation for inertia of a cylinder to pick the diameter and length of your flywheel. This is trial and error because the length determines the diameter (or the diameter determines the length), so you choose a combination of length and diameter that work for your system.

There are several ways to drive the flywheel. You can mount it directly on the output shaft of a gear reducer, you can run a V-belt around the OD of the flywheel, and there are many other ways. Note that the inertia of the motor driving the flywheel is part of the inertia of the flywheel, and that the motor inertia is reflected to the flywheel by the square of the gear (or drive) ratio.
 
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