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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Forces needed to move the Vehicle
Prepared for the Florida EAA
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.
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
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
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 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
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
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.
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.
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.
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.
There is still the transmission between the motor and the final drive
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
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 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.
More speed still requires more power.
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.
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.
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.
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.
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
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.
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.