# Homework Help: Kinetic energy of a vehicle's constituent parts

1. Dec 30, 2011

### dusty22

1. The problem statement, all variables and given/known data

I have set myself the problem of modeling a KER's technology applied to road cars.

I am looking to establish the kinetic energy of a vehicle corresponding to a specific drive cycle for instance the NEDC or similar.

I have distance speed and time data for the drive cycle and also general data about the vehicle such as mass rolling radius gear ratios etc.

Obviously the kinetic energy into the cycle is 0.5*mv^2 + energy required to overcome rolling resistance and aerodrag.

The energy I can extract from the system is 0.5*mv^2 - rolling resistance - drag.

Obviously there is drive train efficiencies in both cases.

My question relates to the 0.5*mv^2

rotating parts of the car will have a rotational kinetic energy 0.5*IW^2

If i was to count the wheels in both the 0.5*mv^2 and the 0.5*IW^2 would I be counting them twice or in fact would i just be accounting for the I guess lateral and rotational energies.

The wheels are really my only concern because this is energy that is harvestable other rotating components will be disengaged from the driveling during energy recovery.

just 0.5*mv^2 is probably sufficiently accurate for my purposes but if I can make it more accurate I would like too.

Cheers for any help.

2. Dec 31, 2011

### AlephZero

The KE of each particle of the body depends on the velocity of that particular particle. When the wheel is rotating, different points on the wheel have different velocities. For eaxmple the point touching the road has zero velocity and the point at the top of the wheel has velocity 2v.

But you don't need to worry about all that detail, because (as you correctly said) the total KE adds up to the translational KE of the center of mass of the wheel, plus the rotational KE about the center of mass.

You will probably find the rotational KE of the wheels is small compared with the total KE of the car, though.

3. Dec 31, 2011

### Tea Jay

The weight of the wheel and tire can be significant in some cases though.

Essentially, the larger the tire and/or rim diameter, the more impact the mass at the outer circumference will have. Using a heavier rim and/or tire will then magnify this force.

The force closer to the axle/center of the hub is less, and it increases proportionally as you go to the outside of the circle.

There have been some experimental studies that have indicated that for a given diameter, adding a pound of rim/wheel weight is the moral equivalent of adding 3-4 lb of cargo (Per tire/rim), as far as performance is concerned.

The smaller the diameter, and the lighter the combination, the less of an impact.

As tire's are further out from the hub, their mass has proportionally more impact than the rim, and so forth.

The rotational drag of the drivetrain itself (Transmission, differentials, etc) can be significant as well, depending on the type.

4. Dec 31, 2011

### dusty22

Thanks for the responses.

I guess its time to weigh some wheels and tires to calculate approximate I values.

5. Dec 31, 2011

### LawrenceC

Don't forget the brake rotors. They are made of cast iron on most vehicles. Some are hollow for added heat dissipation while others are solid.

You know you could put the car on a known constant slope and let it start rolliing down the hill. Time it and note speed over known distance. For low speed, the air drag is of no consequence. With a standard tranny with clutch depressed, much of the drive train is being turned (rear wheel drive). Comparing this to what it should do with no friction, you could come up with some definitive numbers.

6. Dec 31, 2011

### LawrenceC

Or you could drive it on flat ground at a low speed and check how long (distance) it takes for it to coast to a stop on its own accord. That might also provide some needed data.

7. Dec 31, 2011

### Tea Jay

I suppose that would work after it was built.

If it DOES get built, this would be what I'd like to see as a check against the calculations for QA/QC purposes. That would tell you if the calculations used were likely to be applicable to other applications, or if there was something missing/over compensated for.

8. Dec 31, 2011

### dusty22

LawrenceC

Great idea although I don't own the vehicle. I am doing a mock exercise based on the LT-TX4 London Taxi Cab.

The thought is the stop start nature of a typical Taxi drive cycle combined with extended idle periods could lend itself nicely to a KERS system ideally mechanical based with some start stop engine technology.