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Basic Motion calculations. Forces needed to move the Vehicle Prepared for the Florida EAA October 2008 David Kerzel. Basic Information about the Vehicle. Weight of the car Maximum load it can carry Tire diameter Final drive ratio Gear ratio for each transmission speed. Linear Inertia.

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basic motion calculations

Basic Motion calculations.

Forces needed to move the Vehicle

Prepared for the Florida EAA

October 2008

David Kerzel

basic information about the vehicle
Basic Information about the Vehicle
  • Weight of the car
  • Maximum load it can carry
  • Tire diameter
  • Final drive ratio
  • Gear ratio for each transmission speed.
linear inertia
Linear Inertia

Inertia needs to be over come to accelerate

It all begins with Newton’s Second Law


F in Lbf, M in Lb, a in ft/sec2

Most people think acceleration in mph/sec

To convert MPH/sec

divide MPH/sec by 21.95 => ft/sec2

M Mass is the weight of the vehicle in Lb after conversion with batteries and occupants.

linear inertia4
Linear Inertia

For a good starting point use the curb weight and the maximum load weight, mine is 3,800 Lb


My driving is at 35-45 MPH and it takes about 10 seconds for me to get to 40 MPH.

That leads to an acceleration of 4 MPH/sec.

Acceleration needs to be ft/sec2 so divide 4 MPH/sec by 21.95 to get 0.1822 ft/sec2.

Fal=Ma = 3800 X .1822 = 692Lbf

rotating inertia
Rotating Inertia

All rotating parts also need to be accelerated. The wheels and ties are obvious, but the axels and shafts. Motor armature and fly wheel all must be considered. These calculations are impossible without weights and dimensions of the items. This rotational inertia is typical 5 to 20% of the linear inertia for production vehicles.

Whether you are moving the car with the motor or the car is causing the motor to rotate this rotational inertia is the same so we can look at this like a force and make an assumption it will be 10%.

Far = 0.10 X Fal = 69Lbf

Far = 0.10 X 692 = 69Lbf

aerodynamic drag
Aerodynamic Drag

As the car moves through the air there is more force needed.

Fdrag = (Cd X A X v2)/391

Cd is the drag coefficient A is frontal area of the car in ft2v is speed in MPH 391 fixes the units

For my donor car it is Cd = 0.34 and A is 20 ft2

wind drag
Wind Drag

Wind causes more aerodynamic drag. Basically a head wind is like moving faster. So driving at 40 MPH with a 10 MPH head wind give the aerodynamic drag of 50 MPH. similarly a tail wind reduces aerodynamic drag. I am going to ignore it because it is a small factor in all this.


Even here in South Florida there are some minor inclines and a few steeper ones. Some Rail road crossings are steep, entrance ramps for I-95, where I work has a 25% grade from street to parking lot.

Fclimb = WX sin (θ)

θ = ArcTan (rise/run)

For 10% we get a angle of 8˚32’

The sin of 8˚32’ is 0.0996

3800 X 0.096 = 378Lbf

I would estimate 5% would be reasonable


A 5% incline needs more force than the aerodynamic drag at 80 MPH

rolling drag
Rolling Drag

Tires have rolling resistance that ends up being a form of drag. Different tire designs have higher or lower rolling resistance. Higher air pressure reduces rolling resistance, there are tradeoffs.

Frolling = CrXWX cos (θ)

The angle in the formula reduces the rolling resistance as the road incline angle increases.

Low rolling resistance tires have Cr of about 0.01 and normal tires are 0.02.

On a flat surface:

Frolling = 0.02 X 3800 X 1 = 76Lbf

total acceleration force
Total Acceleration Force

All of the above forces need to be added up for the total force.

These are the 4 MPH/sec acceleration forces. As long as force can be provided the acceleration will continue. No power source has unlimited force and at some point the required force cannot be provided.

constant velocity force
Constant Velocity Force

Once cruising speed is reached the acceleration forces no longer apply so the totals reduce.

I have also eliminated the climbing factor.

These cruising forces show how traveling at 70 MPH requires about two times the force at 20 MPH.

It shows how acceleration requires 5 to 10 times the force that cursing at constant speed does.

torque at the wheels
Torque at the wheels

Regardless of the number of driven wheels the same amount of torque needs to be delivered to the wheels.

Torque is measured in Ft-Lb which is a weight of I Lb at the end of a 1 foot lever.

In a vehicle on wheels the length of the lever arm is the radius, half the diameter of the tire, and the weight is the required force.

My tires have a radius or 12.45 inches or 1.0375 ft.

If I combine that with the 1027Lbf from the acceleration calculations I get

1027 X 1.0375 = 1065 ft-Lb of torque for starting and accelerating at 4MPH/sec.

torque at the motor acceleration
Torque at the motor Acceleration

The gears in the final drive and transmission trade torque for shaft speeds. As the speed is decreased the torque increases at the same ratio.

I need 1065 Ff-Lb at the wheels for 4 MPH/sec acceleration. My final drive is 3.87:1 ratio so the input torque is 1065 / 3.87 = 275 Ft-Lb to the final drive.

There is still the transmission between the motor and the final drive.

The final drive and transmission are not 100% efficient but this will be ignored for simplicity.

torque at the motor acceleration15
Torque at the motor Acceleration

There is still the transmission between the motor and the final drive

torque at the motor acceleration16
Torque at the motor Acceleration

A 9 inch Advance DC motor has 240 Ft-Lb peak torque at low speed. It would give the initial acceleration well in 3nd gear. In 2nd gear more than target acceleration could be reached and maintained as speed increases.

2nd gear will actually provide 6.8 MPH/Sec Acceleration to 10 MPH

torque at the motor acceleration18
Torque at the motor Acceleration

The AC motor and controller I plan to use at 150% continuous torque has 90 Ft-Lb output so I will need to start in first gear. The AC motor does not have the incredibly high low speed torque that a series wound DC motor has. My motor is capable of higher torque but getting a controller is the issue today. The AC advantage is the torque stays constant over the full rated speed range. A Series wound DC motor starts with very high torque but that torque is reduced as RPM increases.

power acceleration
Power & Acceleration

Power is Torque X RPM / 5252

Power in in HP, Torque is in Ft-Lb

If the acceleration was reduced to 2MPH/sec the Fal and Far inertia forces would be half of the original levels. Everything else would be the same.

It would take twice as long to get to speed.

With a electric vehicle it takes the same amount of power to get to a speed. There is no penalty for aggressive acceleration in electric vehicles.

Actually rapid acceleration may be slightly more efficient since less time is spent in acceleration.

power at constant speed
Power at Constant Speed

More speed still requires more power.

vehicle speed
Vehicle Speed

Vehicle speed was in a few calculations but has not been looked at.

For this calculation we start at the motor, change that speed with the transmission and final drive. The vehicle speed / motor speed is locked by these ratios.

vehicle speed22
Vehicle Speed

If we look at the 9 inch Advance DC motor running in second gear we see a reasonable speed range for a commuter car with no shifting. This motor can reach 6,000 RPM so this is a good fit.

The 9 inch Advance DC motor torque drops to about 50 Ft-Lb at 3,000 RPM. Torque continues to drop as speed increases. That reducing torque at increasing speed reduces acceleration as speed increases. At 30 MPH the 6.8 MPH/sec acceleration we started with has dropped to 2 MPH/sec acceleration . At some point the speed and torque will be in balance and there will be no more acceleration.

vehicle speed23
Vehicle Speed

For the AC drive and 90 Ft-Lb to 4,000 RPM, I need to change gears. In first gear I will get my expected 4 MPH/sec acceleration, but I won’t have it in second gear. In second gear I will get about 2.5 MPH/sec acceleration and in 3rd gear it will be 1.8 MPH/sec acceleration. I doubt I will ever need the higher gears.

torque at the motor constant velocity
Torque at the motor Constant Velocity

From the constant velocity note 123 Ft-Lb is needed for 45 MPH at the wheels.

My final drive is 3.87:1 ratio so the input torque is 123 / 3.87 = 32 Ft-Lb.

torque at the motor constant velocity25
Torque at the motor Constant Velocity

A 9 inch Advance DC motor has 240 Ft-Lb peak torque at low speed. As the RPM increase the torque available decreases.

These input torques are well within the torque produced by this motor at higher speeds for driving at 45 MPH

Motor torque must exceed the Input Torque

torque at the motor constant velocity26
Torque at the motor Constant Velocity

If the vehicle speed increase to 60 MPH more torque and more motor speed is needed.

The required torque needed to maintain these speeds is not available from the motor in 2nd or 3rd gears.

Even the DC motor powered vehicle could benefit from at transmission at higher speeds.

torque at the motor constant velocity27
Torque at the motor Constant Velocity

The AC motor has a continuous output of 60 Ft-Lb to 4,000 RPM and drops to 30 Ft-Lb at 8,000 RPM.

Based on the previous torques the AC motor could power the vehicle at 60 MPH in 2nd, 3rd or 4th gear.

your project
Your Project
  • The information presented here is basic physics and uses information from my conversion project collected from the internet. It may not exactly apply to your project.
  • You need to gather the data for you project and work through it all.
  • Do calculations at different speeds.
  • Do not ignore any of the data.
  • These calculation should help avoid a conversion that has poor acceleration, can’t go up a hill, or can’t keep up with traffic.
  • There are still a multitude of things that can rob performance but this will take care of the elementary mechanics.
  • You are in charge and responsible for what you design and build. Please do it responsibly.